USAPhO Problems (2007-2014)

United States Physics Team Contest Papers with Solutions

(2007-2014)

United States Physics Team F= ma Contest 2007

2007 F=ma Contest 1. An object moves in two dimensions according to r (t ) = (4.0t 2 − 9.0)iˆ + (2.0t − 5.0) ˆj , where r is in meters and t in seconds. When does the object cross the x-axis?

(a) (b) (c) (d) (e)

0.0 s 0.4 s 0.6 s 1.5 s 2.5 s

2. The graph shows velocity as a function of time for a car. What was the acceleration at time = 90 seconds? (a) (b) (c) (d) (e)

0.22 m/s2 0.33 m/s2 1.0 m/s2 9.8 m/s2 30 m/s2

3. The coordinate of an object is given as a function of time by x = 8t - 3t2, where x is in meters and t is in seconds. Its average velocity over the interval from t = 1 to t = 2s is (a) (b) (c) (d) (e)

-2 m/s -1 m/s -0.5 m/s 0.5 m/s 1 m/s

4. An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next second of time? (a) (b) (c) (d) (e)

h 2h 3h 4h h2

1

5. A crate of toys remains at rest on a sleigh as the sleigh is pulled up a hill with an increasing speed. The crate is not fastened down to the sleigh. What force is responsible for the crate’s increase in speed up the hill? (a) (b) (c) (d) (e)

the force of static friction of the sleigh on the crate the contact force (normal force) of the ground on the sleigh the contact force (normal force) of the sleigh on the crate the gravitational force acting on the sleigh no force is needed

6. At time t = 0 a drag racer starts from rest at the origin and moves along a straight line with velocity given by v = 5t2, where v is in m/s and t in s. The expression for the displacement of the car from t = 0 to time t is (a) (b) (c) (d) (e)

5t3 5t3/3 10t 15t2 5t/2

7. The chemical potential energy stored in a battery is converted into kinetic energy in a toy car that increases its speed first from 0 mph to 2 mph and then from 2 mph up to 4 mph. Ignore the energy transferred to thermal energy due to friction and air resistance. Compared to the energy required to go from 0 to 2 mph, the energy required to go from 2 to 4 mph is (a) (b) (c) (d) (e)

half the amount. the same amount. twice the amount. three times the amount. four times the amount.

8. When two stars are very far apart their gravitational potential energy is zero; when they are separated by a distance d the gravitational potential energy of the system is U. If the stars are separated by a distance 2d the gravitational potential energy of the system is (a) (b) (c) (d) (e)

U/4 U/2 U 2U 4U

2

9. A large wedge rests on a horizontal frictionless surface, as shown. A block starts from rest and slides down the inclined surface of the wedge, which is rough. During the motion of the block, the center of mass of the block and wedge (a) (b) (c) (d) (e)

does not move moves horizontally with constant speed moves horizontally with increasing speed moves vertically with increasing speed moves both horizontally and vertically

10. Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations about their respective hubs, how large must F2 be? (a) (b) (c) (d) (e)

0.25 N 0.5 N 1N 2N 4N

11. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a fixed axis through its center. Assume the hoop is connected to the rotation axis by light spokes. With the objects starting from rest, identical forces are simultaneously applied to the rims, as shown. Rank the objects according to their kinetic energies after a given time t, from least to greatest. (a) (b) (c) (d) (e)

disk, hoop, sphere sphere, disk, hoop hoop, sphere, disk disk, sphere, hoop hoop, disk, sphere

3

12. A 2-kg rock is suspended by a massless string from one end of a uniform 1-meter measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 0.20-meter mark? (a) (b) (c) (d) (e)

0.20 kg 1.00 kg 1.33 kg 2.00 kg 3.00 kg

13. A particle moves along the x-axis. It collides elastically head-on with an identical particle initially at rest. Which of the following graphs could illustrate the momentum of each particle as a function of time?

14. When the speed of a rear-drive car is increasing on a horizontal road, the direction of the frictional force on the tires is (a) (b) (c) (d) (e)

backward on the front tires and forward on the rear tires. forward on the front tires and backward on the rear tires. forward on all tires. backward on all tires. zero.

4

15. A uniform disk (I =½ MR2) of mass 8.0 kg can rotate without friction on a fixed axis. A string is wrapped around its circumference and is attached to a 6.0 kg mass. The string does not slip. What is the tension in the cord while the mass is falling? (a) (b) (c) (d) (e)

20.0 N 24.0 N 34.3 N 60.0 N 80.0 N

16. A baseball is dropped on top of a basketball. The basketball hits the ground, rebounds with a speed of 4.0 m/s, and collides with the baseball as it is moving downward at 4.0 m/s. After the collision, the baseball moves upward as shown in the figure and the basketball is instantaneously at rest right after the collision. The mass of the baseball is 0.2 kg and the mass of the basketball is 0.5 kg. Ignore air resistance and ignore any changes in velocities due to gravity during the very short collision times. The speed of the baseball right after colliding with the upward moving basketball is (a) (b) (c) (d) (e)

4.0 m/s 6.0 m/s 8.0 m/s 12.0 m/s 16.0 m/s

17. A small point-like object is thrown horizontally off of a 50.0-m high building with an initial speed of 10.0 m/s. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance. (a) 2.00 s (b) 1.50 s (c) 1.00 s (d) 0.50 s (e) The building is not high enough for this to occur.

5

18. A small chunk of ice falls from rest down a frictionless parabolic ice sheet shown in the figure. At the point labeled A in the diagram, the ice sheet becomes a steady, rough incline of angle 30 0 with respect to the horizontal and friction coefficient µ k . 3 This incline is of length h and ends 2 at a cliff. The chunk of ice comes to rest precisely at the end of the incline. What is the coefficient of friction µ k ? (a) (b) (c) (d) (e)

300

0.866 0.770 0.667 0.385 0.333

19. A non-Hookian spring has force F = −kx 2 where k is the spring constant and x is the displacement from its unstretched position. For the system shown of a mass m connected to an unstretched spring initially at rest, how far does the spring extend before the system momentarily comes to rest? Assume that all surfaces are frictionless and that the pulley is frictionless as well. ⎛ 3mg ⎞ (a) ⎜ ⎟ ⎝ 2k ⎠ ⎛ mg ⎞ (b) ⎜ ⎟ ⎝ k ⎠

1

1

2

2

⎛ 2mg ⎞ (c) ⎜ ⎟ ⎝ k ⎠

1

⎛ 3mg ⎞ ⎟ (d) ⎜⎜ ⎟ k ⎝ ⎠

2

1

3

⎛ 3 3mg ⎞ ⎟ (e) ⎜⎜ ⎟ k 2 ⎝ ⎠

1

3

6

20. A point-like mass moves horizontally between two walls on a frictionless surface with initial kinetic energy E. With every collision with the walls, the mass loses ½ its kinetic energy to thermal energy. How many collisions with the walls are necessary before the speed of the mass is reduced by a factor of 8? (a) (b) (c) (d) (e)

3 4 6 8 16

21. If the rotational inertia of a sphere about an axis through the center of the sphere is I, what is the rotational inertia of another sphere that has the same density, but has twice the radius? (a) (b) (c) (d) (e)

2I 4I 8I 16I 32I

22. Two rockets are in space in a negligible gravitational field. All observations are made by an observer in a reference frame in which both rockets are initially at rest. The masses of the rockets are m and 9m. A constant force F acts on the rocket of mass m for a distance d. As a result, the rocket acquires a momentum p. If the same constant force F acts on the rocket of mass 9m for the same distance d, how much momentum does the rocket of mass 9m acquire? (a) p/9 (b) p/3 (c) p (d) 3p (e) 9p

7

23. If a planet of radius R spins with an angular velocity ω about an axis through the North Pole, what is the ratio of the normal force experienced by a person at the equator to that experienced by a person at the North Pole? Assume a constant gravitational field g and that both people are stationary relative to the planet and are at sea level. (a) g Rω 2 (b) Rω 2 g (c) 1 − Rω 2 g (d) 1 + g Rω 2 (e) 1 + Rω 2 g

24. A ball of mass m is launched into the air. Ignore air resistance, but assume that there is a wind that exerts a constant force Fo in the –x direction. In terms of F0 and the acceleration due to gravity g, at what angle above the positive x-axis must the ball be launched in order to come back to the point from which it was launched? (a) tan −1 ( F0 mg ) (b) tan −1 (mg / F0 ) (c) sin −1 ( F0 mg ) (d) the angle depends on the launch speed (e) no such angle is possible

25. Find the period of small oscillations of a water pogo, which is a stick of mass m in the shape of a box (a rectangular parallelepiped.) The stick has a length L, a width w and a height h and is bobbing up and down in water of density ρ . Assume that the water pogo is oriented such that the length L and width w are horizontal at all times. Hint: The buoyant force on an object is given by Fbuoy = ρVg , where V is the volume of the medium displaced by the object and ρ is the density of the medium. Assume that at equilibrium, the pogo is floating. L (a) 2π g (b) π

ρ w2 L2 g mh 2

(c) 2π

mh 2 ρ L2 w2 g

(d) 2π

m ρ wL g

(e) π

m ρ wL g

8

Questions 26 – 38: Be sure to show all of your work on the corresponding Free Response Answer Form as well as to record your answer on the optical mark answer sheet.

26. A sled loaded with children starts from rest and slides down a snowy 250 (with respect to the horizontal) incline traveling 85 meters in 17 seconds. Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope? (5 pts.) (a) (b) (c) (d) (e)

0.36 0.40 0.43 1.00 2.01

27. A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains N astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength g. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength g ′ . What is the ratio g ′ g in terms of N? (5 pts)

(a) 2 N /( N − 1) (b) N ( N − 1) (c)

( N − 1) N

(d) N ( N − 1) (e) none of the above

9


Questions 28-30

A simplified model of a bicycle of mass M has two tires that each comes into contact with the ground at a point. The wheelbase of this bicycle (the distance between the points of contact with the ground) is w, and the center of mass of the bicycle is located midway between the tires and a height h above the ground. The bicycle is moving to the right, but slowing down at a constant rate. The acceleration has a magnitude a. Air resistance may be ignored.

Case 1 (Questions 28 – 29): Assume that the coefficient of sliding friction between each tire and the ground is µ , and that both tires are skidding: sliding without rotating. Express your answers in terms of w, h, M, and g. 28. What is the maximum value of µ so that both tires remain in contact with the ground? (5 pts) (a) (b) (c) (d) (e)

w 2h h 2w 2h w w h none of the above

10


29. What is the maximum value of a so that both tires remain in contact with the ground? (5 pts) wg h wg (b) 2h hg (c) 2w h (d) 2wg (e) none of the above (a)

Case 2 (Question 30): Assume, instead, that the coefficient of sliding friction between each tire and the ground is different: µ1 for the front tire and µ 2 for the rear tire. Let µ1 = 2µ2

30. Assume that both tires are skidding: sliding without rotating. What is the maximum value of a so that both tires remain in contact with the ground? (5 pts) wg h wg (b) 3h 2wg (c) 3h hg (d) 2w (e) none of the above (a)

11


Questions 31 – 33

A thin, uniform rod has mass m and length L. Let the acceleration due to gravity be g. Let the rotational inertia of the rod about its center be md 2 .

31. Find the ratio L / d. (3 pts) (a) 3 2 (b) 3 (c) 12 (d) 2 3 (e) none of the above

The rod is suspended from a point a distance kd from the center, and undergoes small g oscillations with an angular frequency β . d

32. Find an expression for β in terms of k. (7 pts) (a) 1 + k 2 (b) 1 + k 2 k 1+ k

(c)

k2 1+ k (e) none of the above

(d)

33. Find the maximum value of β . (5 pts) (a) (b) (c) (d) (e)

1 2 1/ 2 β does not attain a maximum value none of the above

12


Questions 34 – 36

A point object of mass m is connected to a cylinder of radius R via a massless rope. At time t = 0 the object is moving with an initial velocity v0 perpendicular to the rope, the rope has a length L0, and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds Tmax. Express your answers in terms of Tmax, m, L0, R, and v0.

34. What is the angular momentum of the object with respect to the axis of the cylinder at the instant that the rope breaks? (6 pts) (a) mv0 R 3

m 2 v0 (b) Tmax (c) mv0 L0 Tmax R 2 v0 (e) none of the above

(d)

13


35.What is the kinetic energy of the object at the instant that the rope breaks? (4 pts) (a)

mv0 2

2

2

mv0 R (b) 2 L0 2

(c)

mv0 R 2 2L0 2

2 2

mv0 L0 (d) 2R2 (e) none of the above

36. What is the length (not yet wound) of the rope? (6 pts) (a) L0 − πR (b) L0 − 2πR (c) L0 − 18π R 2

mv0 Tmax (e) none of the above (d)

14


Questions 37 – 38

A massless elastic cord (that obeys Hooke’s Law) will break if the tension in the cord exceeds Tmax. One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m. If a second, smaller object of mass m moving at an initial speed v0 strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of vf. All motion occurs on a horizontal, frictionless surface.

37. Find vf/v0. (7 pts) (a) 1 12 (b) 1

2

(c) 1

6

(d) 1 3 (e) none of the above

38. Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial kinetic energy of the smaller mass prior to the collision. (7 pts) (a) (b) (c) (d) (e)

14 1/ 3 1/ 2 3/ 4 none of the above

15


Multiple Choice Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.

e b b c a b d b d d e c d a b b a b a c e d c b d b e a b e d e c b a d c d

16

2007 F=ma Contest SOLUTIONS Multiple Choice Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.

e b b c a b d b d d e c d a b b a b a c e d c b d b e a b e d e c b a d c d

Copyright © 2007, American Association of Physics Teachers

Page 1

Solutions to Free Response 26. Since the acceleration is constant and the sled starts from rest,

∆x =

1 2 at 2

(26-1)

so a = 0.588 m/s2. With the y-axis perpendicular to the incline, ay = 0, so the normal force is

N = mg cos θ

(26-2)

Applying Newton’s second law parallel to the incline with f = force of kinetic friction

mg sin θ − f = ma

(26-3)

Using f = µ N along with equations (26-2) and (26-3) we find that

µ = tan θ −

a g cos θ

0.588m / s 2 µ = tan 25 − = 0.40 (10m / s 2 ) cos 25

27. An accelerated reference frame is equivalent to a gravitational field. We will denote all quantities that change when the astronauts move with a primed superscript after the move. Due to circular motion and the fact that the radius does not change and that v = rω , we find that g ′ v ′ 2 ω ′2 = = g v2 ω 2

(27-1)

Angular momentum is conserved since there is no external torque acting on the system. Therefore, I ω = I ′ω ′

(27-2)

Since the corridors are long, we can consider the astronauts to be point masses. So, with r = the distance from the central hub to the living modules, m = the mass of one astronaut, and with two Copyright © 2007, American Association of Physics Teachers

Page 2

living modules each with N astronauts originally, we find that the rotational inertia before the astronauts move is

I = 2 Nmr 2

(27-3)

After the two astronauts climb into the central hub, I ′ = 2( N − 1)mr 2

(27-4)

When we substitute (27-3) and (27-4) into (27-2) we obtain 2 Nmr 2ω = 2( N − 1)mr 2ω ′

(27-5)

ω′ N = ω N −1

(27-6)

Finally, substituting (27-6) into (27-1), we find

g ′ ⎛ ω′ ⎞ ⎛ N ⎞ =⎜ ⎟ =⎜ ⎟ g ⎝ ω ⎠ ⎝ N −1 ⎠ 2

2

(27-7)

28. For static equilibrium in an accelerated reference frame, we need to calculate torques about the center of mass. Let N1 be the normal force on the front tire and N2 the normal force on the rear tire. Let f1 be the force of friction on the front tire and f2 the force of friction on the rear tire. If the front tire just barely remains in contact with the ground then N1 = f1 = 0. Then setting the counter-clockwise torque due to friction on the rear tire = the clockwise torque due to the normal force on the rear tire, we have f2h = N2

w 2

(28-1)

Substituting f 2 = µ N 2 into (28-1),

µ N2h = N2

w 2


(28-2)

Page 3

µ=

w 2h

(28-3)

29 – 30. Applying Newton’s Second Law to the horizontal direction, f1 + f 2 = Ma

(29-1)

Setting clockwise torque = counterclockwise torques: N 2 w N1w = + ( f1 + f 2 )h 2 2

(29-2)

Substituting (28-4) into (28-5) and solving for a, Mah =

a=

w ( N 2 − N1 ) 2

w ( N 2 − N1 ) Mh 2

(29-3)

(29-4)

The maximum acceleration will clearly occur when N1=0. In that case, N2 = Mg, and

a=

wg 2 h

(29-5)

(This is the answer to question 29. Note that this answer did not depend at all upon whether the coefficient of sliding friction for each tire and the ground is the same or different. Therefore, this is the answer to question 30 also.) 31. The rotational inertia of a thin, uniform rod about its center is I cm =

1 mL2 12

(31-1)

We are given that the rotational inertia of the rod about its center is md2. Setting this expression equal to (31-1), we obtain md 2 =

1 mL2 12


(31-2)

Page 4

Therefore, L2 = 12d 2

(31-3)

L = 12 = 2 3 d

(31-4)

and

32. The torque due to gravity is the same as if the entire mass were located at the center of mass. Therefore, the gravitational torque on the rod about an axis through the suspension point a distance kd from the center when the rod making an angle θ to the vertical is

τ p = −mgkd sin θ

(32-1)

where the subscript p denotes the pivot point. We now need the parallel axis theorem to find the rotational inertia about the pivot point. I p = I cm + mh 2 = md 2 + m(kd ) 2

(32-2)

I p = md 2 (1 + k 2 )

(32-3)

Now, apply the rotational analogue of Newton’s Second Law to the axis through the pivot point. Noting that the force of the pivot does not exert a torque about an axis through the pivot and using equations (32-1) and (32-3), we find

τ p = I pα − mgkd sin θ = md 2 (1 + k 2 )

(32-4) d 2θ dt 2

(32-5)

For small oscillations, sin θ ≈ θ . Therefore, d 2θ gk =− θ 2 dt d (1 + k 2 )


(32-6)

Page 5

Since an object oscillates with angular frequency ω when the object’s motion is governed by the differential equation d 2θ = −ω 2θ 2 dt

(32-7)

we find that gk k = 2 d (1 + k ) 1+ k 2

ω=

g d

(32-8)

g d

(32-9)

k 1+ k 2

(32-10)

ω=β where

β=

33. To find the value of k that gives the maximum value of β , square equation (32-10) and then differentiate both sides with respect to k. 2β

d β (1 + k 2 ) − k (2k ) 1− k 2 = = dk (1 + k 2 ) 2 (1 + k 2 ) 2

dβ = 0 when k = 1 dk

(33-1)

(33-2)

Substituting k = 1 into (32-10), we find that

β=

1 2

(33-3)

34 – 36. Since the velocity is perpendicular to the rope, the rope does not do any work on the object. Since the object is moving on a horizontal frictionless surface, the net work done on the object is zero and therefore the change in kinetic energy of the object is zero. Copyright © 2007, American Association of Physics Teachers

Page 6

Thus, the kinetic energy of the object at the instant that the rope breaks is the same as the initial kinetic energy of the object: K=

mv02 2

(34-1)

(This is the answer to #35.) Therefore, the speed of the object is always v0. The angular momentum of the object with respect to the axis of the cylinder is L = mv0 r

(34-2)

where r is the radius of the circular orbit (which is the length of the not yet wound rope.) At the time that the rope breaks, the tension is Tmax =

mv02 r

(34-3)

Solve equation (34-3) for r mv02 r = Tmax

(34-4)

(This is the answer to #36.) Substitute (34-4) into (34-2). L=

m 2 v03 Tmax

(34-4)

(This is the answer to #34.)


Page 7

37 – 38. The cord breaks when it has exceeded a certain tension, which happens when it exceeds a certain potential energy, U0. For the inelastic collision, when the cord is slack, we use conservation of momentum mv0 = 4mv′

(37-1)

v0 4

(37-2)

v′ =

The kinetic energy of the two masses immediately after the collision is K1 =

mv02 8

(37-3)

All of this kinetic energy gets transferred into potential energy so we know that the cord breaks when U 0 = K1 =

mv02 8

(37-4)

Now for the elastic collision: First, find the velocities of each mass immediately after the collision while the cord is slack. The easy way to do this is to find that the velocity of the center of mass is

vcm =

v0 4

(37-5)

Then in the center of mass reference frame, before the collision the velocity of m is

3v0 and the 4

−v0 . In a one-dimensional elastic collision, in the center of mass reference frame, 4 each block’s velocity after the collision is the same magnitude, but in the opposite direction, of its velocity before the collision. So the velocity of the 3m object right after the collision in the center of +v mass reference frame is 0 . Using (37-5) to transform back to the lab reference frame, we find that 4 v the velocity of the 3m object immediately after the collision is 0 and therefore its kinetic energy 2 immediately after the collision (while the cord is still slack) is

velocity of 3m is


Page 8

3mv02 K2 = 8

(37-6)

But since we know that the cord breaks, we know that U0 of the kinetic energy of the 3m block will be consumed by the cord. Therefore, the final kinetic energy of the 3m block, using conservation of energy along with equations (37-4) and (37-6) is K 3 = K 2 − K1 =

3mv02 mv02 mv02 − = 8 8 4

(37-7)

Now we can find that

K3 = vf v0

=

3mv 2f 2

=

mv02 4

1 6

(37-8)

(37-9)

The velocity of the object of mass m in the center of mass reference frame immediately after the −3v0 collision was . Transforming back to the lab reference frame, we find that the mass m has a 4 v velocity after the collision of 0 . Therefore, the kinetic energy of m after the elastic collision is 2 K4 =

mv02 8

(37-10)

The total kinetic energy of the system after the elastic collision and the cord is broken, using (37-7) and (37-10) is K3 + K 4 =

mv02 mv02 3mv02 + = 8 4 8

(37-11)

So, the ratio of the total kinetic energy of the system after the elastic collision and the cord is broken to the initial kinetic energy of the smaller mass prior to the collision is


Page 9

3mv02 8 =3 mv02 4 2

Copyright © 2007, American Association of Physics Teachers 10

(37-12)

Page

United States Physics Team Semi Final Contest 2007

2007 Semi-Final Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • • • • •

•

• • •

Work Part A first. You have 90 minutes to complete all four problems. After you have completed Part A, you may take a break. Then work Part B. You have 90 minutes to complete both problems. Show all your work. Partial credit will be given. Start each question on a new sheet of paper. Be sure to put your name in the upper righthand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie A1 – 1/3 A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s, or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. Questions with the same point value are not necessarily of the same difficulty. Do not discuss the contents of this exam with anyone until after March 27th. Good luck! Possibly Useful Information - (Use for both part A and for part B) Gravitational field at the Earth’s surface g = 9.8 N/kg Newton’s gravitational constant G = 6.67 x 10-11 N·m2/kg2 Coulomb’s constant k = 1/4πεο = 8.99 x 109 N·m2/C2 Biot-Savart constant km = µο/4π = 10-7 T·m/A Speed of light in a vacuum c = 3.00 x 108 m/s Boltzmann’s constant kB = 1.38 x 10-23 J/K Avogadro’s number NA = 6.02 x 1023 (mol)-1 Ideal gas constant R = NAkB = 8.31 J/(mol·K) Stefan-Boltzmann constant σ = 5.67 x 10-8 J/(s·m2·K4) Elementary charge e = 1.602 x 10-19 C 1 electron volt 1 eV = 1.602 x 10-19 J Planck’s constant h = 6.63 x 10-34 J·s = 4.14 x 10-15 eV·s Electron mass m = 9.109 x 10-31 kg = 0.511 MeV/c2 Binomial expansion (1 + x)n ≈ 1 + nx for |x| << 1 Small angle approximations sin θ ≈ θ cos θ ≈ 1 – 1/2 θ 2


Semi-Final Exam Part A A1. A group of 12 resistors is arranged along the edges of a cube as shown in the diagram below.

The vertices of the cube are labeled a-h.

d

c

b

a

h

g

e

f

a. (13 pts) The resistance between each pair of vertices is as follows: Rab = Rac = Rae = 3.0 Ω Rcg = Ref = Rbd = 8.0 Ω Rcd = Rbf = Reg = 12.0 Ω Rdh = Rfh = Rgh = 1.0 Ω What is the equivalent resistance between points a and h? b. (12 pts) The three 12.0 Ω resistors are replaced by identical capacitors. Ccd = Cbf = Ceg = 15.0 μF. A 12.0 V battery is attached across points a and h and the circuit is allowed to operate for a long period of time. What is the charge (Qcd, Qbf, Qeg) on each capacitor after this long period of time? Copyright © 2007, American Association of Physics Teachers

A2. A simple gun can be made from a uniform cylinder of length L0 and inside radius rc. One end of the cylinder is sealed with a moveable plunger and the other end is plugged with a cylindrical cork bullet. The bullet is held in place by friction with the walls of the cylinder. The pressure outside the cylinder is atmospheric pressure, P0 . The bullet will just start to slide out of the cylinder if the pressure inside the cylinder exceeds Pcr . a. There are two ways to launch the bullet: either by heating the gas inside the cylinder and keeping the plunger fixed, or by suddenly pushing the plunger into the cylinder. In either case, assume that an ideal monatomic gas is inside the cylinder, and that originally the gas is at temperature T0 , the pressure inside the cylinder is P0 , and the length of the cylinder is L0 . (8 pts)

i. Assume that we launch the bullet by heating the gas without moving the plunger. Find the minimum temperature of the gas necessary to launch the bullet. Express your answer in terms of any or all of the variables: rc , T0 , L0 , P0 , and Pcr .

(8 pts)

ii. Assume, instead that we launch the bullet by pushing in the plunger, and that we do so quickly enough so that no heat is transferred into or out of the gas. Find the length of the gas column inside the cylinder when the bullet just starts to move. Express your answer in terms of any or all of the variables: rc , T0 , L0 , P0 , and Pcr .

b. (9 pts) It is necessary to squeeze the bullet to get it into the cylinder in the first place. The bullet normally has a radius rb that is slightly larger than the inside radius of the cylinder; rb − rc = Δr , is small compared to rc . The bullet has a length h L0 . The walls of the cylinder apply a pressure to the cork bullet. When a pressure P is applied to the bullet along a given direction, the bullet’s dimensions in that direction change by Δx − P = x E for a constant E known as Young’s modulus. You may assume that compression along one direction does not cause expansion in any other direction. (This is true if the so-called Poisson ratio is close to zero, which is the case for cork.) If the coefficient of static friction between the cork and the cylinder is μ , find an expression for Pcr . Express your answer in terms of any or all of the variables: P0 , μ , h, E , Δr , and rc .


A3. A volume V f of fluid with uniform charge density ρ is sprayed into a room, forming spherical drops. As they float around the room, the drops may break apart into smaller drops or coalesce into larger ones. Suppose that all of the drops have radius R. Ignore inter-drop forces and assume that V f R 3 . (10 pts) a. Calculate the electrostatic potential energy of a single drop. (Hint: suppose the sphere has radius r. How much work is required to increase the radius by dr?). (4 pts) b. What is the total electrostatic energy of the drops? Your answer to (b) should indicate that the total energy increases with R. In the absence of surface tension, then, the fluid would break apart into infinitesimally small drops. Suppose, however, that the fluid has a surface tension γ . (This value is the potential energy per unit surface area, and is positive.) (4 pts) c.

What is the total energy of the drops due to surface tension?

(7 pts) d.

What is the equilibrium radius of the drops?


A4. A nonlinear circuit element can be made out of a parallel plate capacitor and small balls, each of mass m, that can move between the plates. The balls collide inelastically with the plates, dissipate all kinetic energy as thermal energy, and immediately release the charge they are carrying to the plate. Almost instantaneously, the balls then pick up a small charge of magnitude q from the plate; the balls are then repelled directly toward the other plate under electrostatic forces only. Another collision happens, kinetic energy is dissipated, the balls give up the charge, collect a new charge, and the cycle repeats. There are n0 balls per unit surface area of the plate. The capacitor has a capacitance C. The separation d between the plates is much larger than the radius r of the balls. A battery is connected to the plates in order to maintain a constant potential difference V. Neglect edge effects and assume that magnetic forces and gravitational forces may be ignored. (5)

a. Determine the time it takes for one ball to travel between the plates in terms of any or all of the following variables: m, q, d, and V.

(5)

b. Calculate the kinetic energy dissipated as thermal energy when one ball collides inelastically with a plate surface in terms of any or all of the following variables: m, q, d, and V.

(5)

c. Derive an expression for the current between the plates in terms of the permittivity of free space, ε 0 , and any or all of the following variables: m, q, n0, C, and V.

(5)

d. Derive an expression for the effective resistance of the device in terms of ε 0 , and any or all of the following variables: m, q, n0, C, and V.

(5)

e. Calculate the rate at which the kinetic energy of the balls is converted into thermal energy in terms of ε 0 , and any or all of the following variables: m, q, n0, C, and V.


Semi-Final Exam Part B B1. A certain mechanical oscillator can be modeled as an ideal massless spring connected to a moveable plate on an incline. The spring has spring constant k, the plate has mass m, and the incline makes an angle θ with the horizontal. When the system is operating correctly, the plate oscillates between points A and B in the figure, located a distance L apart. When the plate reaches point A it has zero kinetic energy, but then trips a small lever that instantaneously loads a block of mass M onto the plate. The block and plate then move down the incline to point B, where the force from the spring stops the plate. At this point, the block falls through a hole in the incline, allowing the plate to move back up under the force of the spring. Upon returning to point A it collects another block, and the cycle repeats. Both the plate and the block have a coefficient of friction μ with the incline for both kinetic and static friction. It is reasonable that the motion in either direction is simple harmonic in nature.

(10 pts) a. Let μc be the critical value of the coefficient of friction where the block will just start to slide under the force of gravity on an incline (without the spring acting on it). Then let μ =

μc

. Find μ in terms of g, the acceleration of free fall, and any 2 or all of the following variables: θ and M.


(14 pts) b. In order for this system to work correctly, it is necessary to have the correct ratio between the mass of the block and the mass of the plate. These masses are chosen so that the downward moving block and plate just stop at point B while the M upward moving plate just stops at point A. Find the ratio R = . m (13 pts) c. The system delivers blocks to point B with period T0 , until the blocks run out. T After that, the plate alone oscillates with a period T ′ . Find the ratio 0 . T′ (13 pts) d. The plate only oscillates a few times after delivering the last block. At what distance up the incline, measured from point B, does the plate come to a permanent stop?


B2. A model of the magnetic properties of materials is based upon small magnetic moments generated by each atom in the material. One source of this magnetic moment is the magnetic field generated by the electron in its orbit around the nucleus. For simplicity, we will assume that each atom consists of a single electron of charge –e and mass me , a single proton of charge +e and mass m p me , and that the electron orbits in a circular orbit of radius R about the proton. a. Magnetic Moments. Assume that the electron orbits in the x-y plane. (3 pts) i.

Calculate the net electrostatic force on the electron from the proton. Express your answer in terms of any or all of the following parameters: e, me , m p , R, and the

permittivity of free space, ε 0 , where 1 ε0 = . 4π k (k is the Coulomb’s Law constant). (5 pts) ii.

Determine the angular velocity ω0 of the electron around the proton in terms of any or all of the following parameters: e, me , R, and ε 0 .

(8 pts) iii. Derive an expression for the magnitude of the magnetic field Be due to the orbital motion of the electron at a distance z R from the x-y plane along the axis of orbital rotation of the electron. Express your answer in terms of any or all of the following parameters: e, me , R, ω0 , z , and the permeability of free space μ0 . (4 pts) iv. A small bar magnet has a magnetic field far from the magnet given by μ m B= 0 3, 2π z where z is the distance from the magnet on the axis connecting the north and south poles, m is the magnetic dipole moment, and μ0 is the permeability of free space. Assuming that an electron orbiting a proton acts like a small bar magnet, find the dipole moment m for an electron orbiting an atom in terms of any or all of the following parameters: e, me , R, and ω0 .


b. Diamagnetism. We model a diamagnetic substance to have all atoms oriented so that the electron orbits are in the x-y plane, exactly half of which are clockwise and half counterclockwise when viewed from the positive z axis looking toward the origin. Some substances are predominantly diamagnetic. (3 pts) i.

(6 pts) ii.

Calculate the total magnetic moment of a diamagnetic substance with N atoms. Write your answer in terms of any or all of the following parameters: e, me , R, N , and μ0 . G An external magnetic field B0 = B0 zˆ is applied to the substance. Assume that the introduction of the external field doesn’t change the fact that the electron moves in a circular orbit of radius R. Determine Δω , the change in angular velocity of the electron, for both the clockwise and counterclockwise orbits. Throughout this entire problem you can assume that Δω ω0 . Write your answer in terms of e, me , and B0 only.

(6 pts) iii. Assume that the external field is turned on at a constant rate in a time interval Δt . That is to say, when t = 0 the external field is zero and when t = Δt the external G field is B0 . Determine the induced emf E experienced by the electron. Write your answer in terms of any or all of the following parameters: e, me , R, N , B0 , ω0 , and μ0 . (6 pts) iv. Verify that the change in the kinetic energy of the electron satisfies ΔK = e E. This justifies our assumption in (ii) that R does not change. (6 pts) v.

Determine the change in the total magnetic moment Δm for the N atoms when the external field is applied, writing your answer in terms of e, me , R, N , μ0 and B0 .

(3 pts) vi. Suppose that the uniform magnetic field used in the previous parts of this problem is replaced with a bar magnet. Would the diamagnetic substance be attracted or repelled by the bar magnet? How does your answer show this?


2007 Semifinal Exam Solutions

AAPT AIP

1

UNITED STATES PHYSICS TEAM 2007

Solutions to Problems Part A Question 1 a. There is a high degree of symmetry present. Points b, c, and e are at the same potential; similarly, points d, f , and g are at the same potential. The circuit then reduces to a series connection of three parallel resistor clusters. The three parallel clusters have effective resistances of 1 Ω, 8/5 Ω, and 1/3 Ω. The effective resistance of the circuit is then 44/15 Ω. b. After a long time no current will flow through the branches of the circuits containing capacitors. The circuit then reduces to a parallel connection of three series resistor clusters. The effective resistance of the circuit is 4 Ω, the current through the circuit is then (12 V)/(4 Ω) = 3 A.

(A1-1)

The three branches are identical; each then carries 1 A. The potential drop across each capacitor is the same as the potential drop across the 8 Ω resistors, so VC = (1 A)(8 Ω) = 8 V. (A1-2) Finally, the charge on each capacitor is Q = (15 µF)(8 V) = 120 µC.

(A1-3)

Question 2 a. Two parts, solved individually. i. Isochoric compression: Pf Tf = , Pi Ti so T =

Pcr T0 . P0

c Copyright 2007 American Association of Physics Teachers

(A2-1)

(A2-2)


2

ii. Adiabatic compression: P V γ = const,

(A2-3)

where γ = CP /CV = (CV + 1)/CV = 5/3 for a monatomic gas. Consequently P0 Lγ0 = Pcr Lγ , and then

L = L0

P0 Pcr

(A2-4)

5/3

.

(A2-5)

b. The normal pressure on the bullet comes from δr E. rc

(A2-6)

δr E × (2πrc h), rc

(A2-7)

P = Therefore, the normal force on the bullet is FN =

and finally the force of friction is µFN . The force due to the pressure difference between the inside of the barrel and the outside must equal the normal force, so (πrc2 )(Pcr − P0 ) = 2πhµ δr E, and then Pcr = P0 +

2µEh δr. rc2

(A2-8)

(A2-9)

Question 3 a. If the sphere has radius r, it has charge 4 q = πρr3 3

(A3-1)

and thus its surface is at electrostatic potential V =

q ρr2 = 4π0 r 30

(A3-2)

To increase the radius by dr, an additional charge dq = 4πr2 dr must be brought in from infinity, requiring work 4πr4 ρ2 dU = V dq = dr (A3-3) 30 Thus to grow the sphere from r = 0 to r = R requires U=

Z R 4πr4 ρ2 0

30

dr =

4πR5 ρ2 150


(A3-4)


3

b. Each drop has volume Vd = 43 πR3 , so the number of drops is n=

Vf Vf = 4 3 Vd 3 πR

(A3-5)

Since we are ignoring inter-drop forces, the total energy of the drops is simply the sum of the energies of each individual drop: Ue,tot = nU =

Vf 4 3 3 πR

R 2 ρ2 4πR5 ρ2 = Vf 150 50

(A3-6)

c. Each drop has surface area 4πR2 and thus surface tension energy 4πR2 γ. As before, the total energy due to surface tension is just the sum of the energies of the individual drops: Vf 3γ Vf = 3 R πR 3

Us,tot = 4πR2 γn = 4πR2 γ 4

(A3-7)

d. The total potential energy from both sources is Utot =

R2 ρ2 3γ + 50 R

!

Vf

(A3-8)

Equilibrium is reached when the total energy is a minimum; since U → ∞ at both R → 0 and R → ∞, it must have an interior minimum. d Utot = dR

3γ 2Rρ2 − 2 50 R

!

Vf

(A3-9)

Setting this equal to zero, 2Rρ2 3γ = 2 50 R 15γ0 R3 = 2ρ2

R=

15γ0 2ρ2

(A3-10) (A3-11)

1

3

(A3-12)

Question 4 a. The electric field between the plates is given by E = V /d. The force on the charged ball is then F = Eq = V q/d. The acceleration of the ball is a = V q/md. Kinematics gives us d = at2 /2 for the time of flight. So t=

q

2d/a =

q

2md2 /qV .

(A4-1)

b. The kinetic energy collected by a ball will be K = qV as it moves between the plates. That’s what will be dissipated.



4

c. The current is given by I = ∆Q/∆t. The total number of balls is N = n0 A, where A is the surface area of a plate. The charge ∆Q is then ∆Q = n0 qA, so the current is I=

n0 qA ∆Q =p . ∆t 2md2 /qV

(A4-2)

We can’t stop here, since this is not in terms of the allowed variables. The problem is A and d, but since C = 0 A/d, we have I =

p

n0 qA , 2md2 /qV

A n0 q d

=

s

C n0 q 0

=

(A4-3)

qV , 2m

(A4-4)

qV . 2m

(A4-5)

s

d. R = V /I, so 0 V V = R= I Cn0 q

s

2m . qV

(A4-6)

We can simplify, slightly, with 0 R= Cn0 q

s

2mV . q

(A4-7)

e. P = V I, so P =V

C n0 q 0

s

qV = 2m

s

0 2 n0 2 C 2 q 3 V 3 . 2m

(A4-8)

Part B Question 1 a. To not slip, from a free-body diagram, we must have µmg cos θ ≥ mg sin θ

(B1-1)

µ ≥ tan θ.

(B1-2)

tan θ . 2

(B1-3)

so Therefore µc = tan θ and hence µ=

b. In one cycle the energy input into the system is M gL sin θ, the energy of the block dropping.


(B1-4)


5

The energy loss on the way up is Lµmg cos θ

(B1-5)

Lµ(m + M )g cos θ

(B1-6)

M gL sin θ = Lµmg cos θ + Lµ(m + M )g cos θ

(B1-7)

and the energy loss on the way down is

Thus and since 2µ cos θ = sin θ, M

m m+M + , 2 2 = 2m, =

R = M/m = 2.

(B1-8) (B1-9) (B1-10)

c. The period of a mass m oscillating on a spring of spring constant k is r

T = 2π

m . k

(B1-11)

In this case, the friction force is constant on both the up and down trips, and so each trip is simple harmonic (with different equilibrium points). Hence r

T0 T0 T0 /T 0

m +π = π k r m , = 2π k √ 1+ 3 . = 2

r

3m , k

(B1-12) (B1-13) (B1-14)

d. As mentioned in part (c), both the up and down trips are simple harmonic, this time with a mass of m both ways. The equilibrium points for the two trips are different, however. On the up trip, the equilibrium point is clearly at a distance L/2 from B, since the plate stops at both B and A and hence those are the endpoints of the oscillation and the equilibrium is halfway between. For the trip down, the equilibrium point will shift by a distance y such that ky = 2µmg cos θ = mg sin θ

(B1-15)

because 2µmg cos θ is the difference between the friction forces on the trip up and the trip down. The place where the plate finally comes to a stop is the first place that is at the end of an oscillation (either up or down) and where the total force being exerted by gravity and the spring is less than the maximal force of friction. For that to happen, the plate needs to not have gone past the other equilibrium point during that oscillation. So we start by determining where the endpoints of the oscillations are. For the first trip up these are B and A. For the following trip down, the plate stops at a distance of (2mg sin θ)/k from B (because the equilibrium shifts up by (mg sin θ)/k. For the following trip up, the c Copyright 2007 American Association of Physics Teachers


6

plate stops a distance L − (2mg sin θ)/k from B, since the equilibrium point is again in the middle of the incline. And so forth. Thus the stopping points are located at n(2mg sin θ)/k and L − n(2mg sin θ)/k

(B1-16)

for integer n. The plate will stop permanently once either n(2mgsin)/k > L/2

(B1-17)

L − n(2mg sin θ)/k < L/2 + (mg sin θ)/k,

(B1-18)

or whichever happens first. (The first condition corresponds to going down and ending up above the midpoint at the end of the down trip, the second condition corresponds to going up and stopping below the upper equilibrium.) The second condition can be rewritten as 1 (2mg sin θ)/k > L/2. 2

n+

(B1-19)

Question 2 a. Magnetic Moments i. From Coulomb’s Law, F =

e2 4π0 R2

(B2-1)

ii. For circular motion, me v 2 = me Rω02 , R The force is provided by the Coulomb force, so F =

e2 , 4π0 R2

me Rω02 =

s

ω0 =

e2 4π0 me R3

(B2-2)

(B2-3) (B2-4)

iii. From the law of Biot and Savart, ~e = B Be = ≈

µ0 i d~s × ~r , 4π r3 R µ0 i 2πR 2 , 4π (z + r2 )3/2 µ0 iR2 . 2z 3 I

For the current, i, we can write i= Then Be =

q eω0 = . t 2π µ0 eω0 R2 . 4πz 3


(B2-5) (B2-6) (B2-7)

(B2-8)

(B2-9)


7

iv. By substitution, m=

eω0 R . 2

(B2-10)

b. Diamagnetism i. If half go one way and half go the other, M = 0. ii. Additional force from magnetism, FB = qvB0 = eRωB0

(B2-11)

modifies previous central force problem to give me Rω 2 =

e2 ± eRω0 B0 , 4π0 R2

(B2-12)

where the positive sign corresponds to anticlockwise motion, the negative to clockwise motion. A little math, me R(ω 2 − ω02 ) = ±eRωB0 , me (ω − ω0 )(ω + ω0 ) = ±eωB0 , me (∆ω)(2ω0 ) = ±eω0 B0 ,

(B2-13) (B2-14) (B2-15)

where in the last line we have used the approximation ω ≈ ω0 . Then ∆ω = ±

eB0 . 2me

(B2-16)

iii. The emf is given by ∆n ∆Φ = Φ, (B2-17) ∆t ∆t but ∆n/∆t is a measure of the number of turns made by the electron in a time interval ∆t, so ω0 R ω0 ∆n = = . (B2-18) ∆t 2πR 2π Then 1 ω0 B0 πR2 = ω0 b0 R2 . (B2-19) E= 2π 2 E =n

iv. The change in kinetic energy is given by 1 me ω 2 R 2 , 2 = me R2 ω ∆ω,

∆K = ∆

2

≈ me R ω0 ∆ω, eB0 2 , = me ω 0 R ± 2me = eE.


(B2-20) (B2-21) (B2-22) (B2-23) (B2-24)


8

v. ∆M = N δm, where N is the number of atoms, and ∆m the change in magnetic moment in each. The change is eω0 R . ∆m = ∆ 2 eR ∆ω, = 2 e2 R2 B0 = , 4me

so ∆M = N

e2 R2 B0 . 4me

vi. Repelled, by Lenz’s law.


(B2-25) (B2-26) (B2-27)

(B2-28)

United States Physics Team F = ma Contest 2008

2008 F = ma Exam

AAPT AIP

1


2008 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Use g = 10 N/kg throughout this contest. • You may write in this booklet of questions. However, you will not receive any credit for anything written in this booklet. • Your answer to each question must be marked on the optical mark answer sheet. • Select the single answer that provides the best response to each question. Please be sure to use a No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer, the previous mark must be completely erased. • Correct answers will be awarded one point; incorrect answers will result in a deduction of point. There is no penalty for leaving an answer blank.

1 4

• A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • This test contains 25 multiple choice questions. Your answer to each question must be marked on the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1 through 25 are to be used on the answer sheet. • All questions are equally weighted, but are not necessarily the same level of difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after February 20, 2008. • The question booklet and answer sheet will be collected at the end of this exam. You may not use scratch paper.

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN


2008 F = ma Exam

2

1. A bird flying in a straight line, initially at 10 m/s, uniformly increases its speed to 18 m/s while covering a distance of 40 m. What is the magnitude of the acceleration of the bird? (a) 0.1 m/s2 (b) 0.2 m/s2 (c) 2.0 m/s2 (d) 2.8 m/s2 (e) 5.6 m/s2 2. A cockroach is crawling along the walls inside a cubical room that has an edge length of 3 m. If the cockroach starts from the back lower left hand corner of the cube and finishes at the front upper right hand corner, what is the magnitude of the displacement of the cockroach? √ (a) 3 2 m √ (b) 3 3 2 m √ (c) 3 3 m (d) 3 m (e) 9 m 3. The position vs. time graph for an object moving in a straight line is shown below. What is the instantaneous velocity at t = 2 s?

Position (m)

4 2 Time (s) 0 1

2

3

−2

(a) −2 m/s

(b) − 12 m/s (c) 0 m/s

(d) 2 m/s (e) 4 m/s


2008 F = ma Exam

3

The information below is for the next two problems Shown below is the velocity vs. time graph for a toy car moving along a straight line.

Velocity (m/s)

4 2 Time (s) 0 1

2

3

−2

4. What is the maximum displacement from start for the toy car? (a) 3 m (b) 5 m (c) 6.5 m (d) 7 m (e) 7.5 m

Time (s) 0 1

2

3

Acceleration

Acceleration

5. Which of the following acceleration vs. time graphs most closely represents the acceleration of the toy car?

Time (s) 0 1

Time (s) 0 1

2

3

(c) Acceleration

3

(b) Acceleration

Acceleration

(a)

2

Time (s) 0 1

2

3

(d)

Time (s) 0 1

2

3

(e)


2008 F = ma Exam

4

6. A cannon fires projectiles on a flat range at a fixed speed but with variable angle. The maximum range of the cannon is L. What is the range of the cannon when it fires at an angle π6 above the horizontal? Ignore air resistance. (a) (b) (c) (d) (e)

√

3 2 L √1 L 2 1 √ L 3 1 2L 1 3L

7. A toboggan sled is traveling at 2.0 m/s across the snow. The sled and its riders have a combined mass of 120 kg. Another child (mchild = 40 kg) headed in the opposite direction jumps on the sled from the front. She has a speed of 5.0 m/s immediately before she lands on the sled. What is the new speed of the sled? Neglect any effects of friction. (a) 0.25 m/s (b) 0.33 m/s (c) 2.75 m/s (d) 3.04 m/s (e) 3.67 m/s 8. Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall? (a) 0.0012 (b) 0.056 (c) 0.11 (d) 0.53 (e) 8.9


2008 F = ma Exam

5

9. A ball of mass m1 travels along the x-axis in the positive direction with an initial speed of v0 . It collides with a ball of mass m2 that is originally at rest. After the collision, the ball of mass m1 has velocity v1x x ˆ + v1y yˆ and the ball of mass m2 has velocity v2x x ˆ + v2y yˆ. Consider the following five statements: I) II) III) IV) V)

0 = m1 v1x + m1 v2x m1 v0 = m1 v1y + m2 v2y 0 = m1 v1y + m2 v2y m1 v0 = m1 v1x + m1 v1y m1 v0 = m1 v1x + m2 v2x

Of these five statements, the system must satisfy (a) I and II (b) III and V (c) II and V (d) III and IV (e) I and III

The following information applies to the next two problems An experiment consists of pulling a heavy wooden block across a level surface with a spring force meter. The constant force for each try is recorded, as is the acceleration of the block. The data are shown below. Force F in Newtons acceleration a in meters/second2

3.05 0.095

3.45 0.205

4.05 0.295

4.45 0.405

5.05 0.495

10. Which is the best value for the mass of the block? (a) 3 kg (b) 5 kg (c) 10 kg (d) 20 kg (e) 30 kg 11. Which is the best value for the coefficient of friction between the block and the surface? (a) 0.05 (b) 0.07 (c) 0.09 (d) 0.5 (e) 0.6


2008 F = ma Exam

6

12. A uniform disk rotates at a fixed angular velocity on an axis through its center normal to the plane of the disk, and has kinetic energy E. If the same disk rotates at the same angular velocity about an axis on the edge of the disk (still normal to the plane of the disk), what is its kinetic energy? (a) (b)

1 2E 3 2E

(c) 2E (d) 3E (e) 4E 13. A mass is attached to the wall by a spring of constant k. When the spring is at its natural length, the mass is given a certain initial velocity, resulting in oscillations of amplitude A. If the spring is replaced by a spring of constant 2k, and the mass is given the same initial velocity, what is the amplitude of the resulting oscillation? (a) (b) (c)

1 2A √1 A 2

√

2A

(d) 2A (e) 4A 14. A spaceborne energy storage device consists of two equal masses connected by a tether and rotating about their center of mass. Additional energy is stored by reeling in the tether; no external forces are applied. Initially the device has kinetic energy E and rotates at angular velocity ω. Energy is added until the device rotates at angular velocity 2ω. What is the new kinetic energy of the device? √ (a) 2E (b) 2E √ (c) 2 2E (d) 4E (e) 8E


2008 F = ma Exam

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15. A uniform round tabletop of diameter 4.0 m and mass 50.0 kg rests on massless, evenly spaced legs of length 1.0 m and spacing 3.0 m. A carpenter sits on the edge of the table. What is the maximum mass of the carpenter such that the table remains upright? Assume that the force exerted by the carpenter on the table is vertical and at the edge of the table.

1.0 m

3.0 m

4.0 m

(a) 67 kg (b) 75 kg (c) 81 kg (d) 150 kg (e) 350 kg 16. A massless spring with spring constant k is vertically mounted so that bottom end is firmly attached to the ground, and the top end free. A ball with mass m falls vertically down on the top end of the spring, becoming attached, so that the ball oscillates vertically on the spring. What equation describes the acceleration a of the ball when it is at a height y above the original position of the top end of the spring? Let down be negative, and neglect air resistance; g is the magnitude of the acceleration of free fall. (a) a = mv 2 /y + g (b) a = mv 2 /k − g

(c) a = (k/m)y − g

(d) a = −(k/m)y + g (e) a = −(k/m)y − g


2008 F = ma Exam

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17. A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant k inside a box as shown:

The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box? (a) (a/g)L (b) (g/a)L (c) m(g + a)/k (d) m(g − a)/k (e) ma/k

18. A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring at a distance much greater than R and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same (linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum speed of the particle? (a) (b)

1 2v √1 v 2

(c) v √ (d) 2v (e) 2v 19. A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and at t = t0 its acceleration is a0 . What is its acceleration at t = 2t0 ? Ignore energy loss due to friction. (a) (b)

1 2 a0 √1 a0 2

(c) a0 √ (d) 2a0 (e) 2a0 c Copyright 2008 American Association of Physics Teachers

2008 F = ma Exam

9

20. The Young’s modulus, E, of a material measures how stiff it is; the larger the value of E, the more stiff the material. Consider a solid, rectangular steel beam which is anchored horizontally to the wall at one end and allowed to deflect under its own weight. The beam has length L, vertical thickness h, width w, mass density ρ, and Young’s modulus E; the acceleration due to gravity is g. What is the distance through which the other end moves? (Hint: you are expected to solve this problem by eliminating implausible answers. All of the choices are dimensionally correct.) (a) h exp ρgL E 2

(b) 2 ρgh E √ (c) 2Lh (d) (e)

3 ρgL4 2 Eh2 √ EL 3 ρgh

21. Consider a particle at rest which may decay into two (daughter) particles or into three (daughter) particles. Which of the following is true in the two-body case but false in the three-body case? (There are no external forces.) (a) The velocity vectors of the daughter particles must lie in a single plane. (b) Given the total kinetic energy of the system and the mass of each daughter particle, it is possible to determine the speed of each daughter particle. (c) Given the speed(s) of all but one daughter particle, it is possible to determine the speed of the remaining particle. (d) The total momentum of the daughter particles is zero. (e) None of the above. 22. A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of length L with a horizontal velocity v0 . The collision is perfectly inelastic and the bullet sticks to the bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a circular vertical loop. √ (a) 2 Lg √ (b) 5Lg √ (c) (m1 + m2 )2 Lg/m1 √ (d) (m1 − m2 ) Lg/m2 √ (e) (m1 + m2 ) 5Lg/m1


2008 F = ma Exam

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23. Consider two uniform spherical planets of equal density but unequal radius. Which of the following quantities is the same for both planets? (a) The escape velocity from the planet’s surface. (b) The acceleration due to gravity at the planet’s surface. (c) The orbital period of a satellite in a circular orbit just above the planet’s surface. (d) The orbital period of a satellite in a circular orbit at a given distance from the planet’s center. (e) None of the above. 24. A ball is launched upward from the ground at an initial vertical speed of v0 and begins bouncing vertically. Every time it rebounds, it loses a proportion of the magnitude of its velocity due to the inelastic nature of the collision, such that if the speed just before hitting the ground on a bounce is v, then the speed just after the bounce is rv, where r < 1 is a constant. Calculate the total length of time that the ball remains bouncing, assuming that any time associated with the actual contact of the ball with the ground is negligible.

(a) (b) (c) (d) (e)

2v0 1 g 1−r v0 r g 1−r 2v0 1−r g r 2v0 1 g 1−r 2 2v0 1 g 1+(1−r)2

25. Two satellites are launched at a distance R from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed v0 and enters a circular orbit. The second satellite, however, is launched at a speed 12 v0 . What is the minimum distance between the second satellite and the planet over the course of its orbit? (a) (b) (c) (d) (e)

√1 R 2 1 2R 1 3R 1 4R 1 7R


2008 F = ma Exam

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ANSWERS TO 2008 Fnet = ma CONTEST A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 A12 A13 A14 A15 A16 A17 A18 A19 A20 A21 A22 A23 A24 A25

D C A D C A A C B B A D B B D E E C B D B E C A E

United States Physics Team Quarter Final Contest 2008

2008 Quarter-final Exam

AAPT AIP

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2008 Quarter-Final Exam 4 QUESTIONS - 60 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Show all your work. Partial credit will be given. • Start each question on a new sheet of paper. Put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie Prob. 1 - P. 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. • Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty. Good luck! • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after March 10, 2008.



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1. A charged particle with charge q and mass m is given an initial kinetic energy K0 at the middle of a uniformly charged spherical region of total charge Q and radius R. q and Q have opposite signs. The spherically charged region is not free to move. Throughout this problem consider electrostatic forces only.

R

(a) Find the value of K0 such that the particle will just reach the boundary of the spherically charged region. (b) How much time does it take for the particle to reach the boundary of the region if it starts with the kinetic energy K0 found in part (a)?

2. A uniform pool ball of radius r and mass m begins at rest on a pool table. The ball is given a horizontal impulse J of fixed magnitude at a distance βr above its center, where −1 ≤ β ≤ 1. The coefficient of kinetic friction between the ball and the pool table is µ. You may assume the ball and the table are perfectly rigid. Ignore effects due to deformation. (The moment of inertia about the center of mass of a solid sphere of mass m and radius r is Icm = 52 mr 2 .)

J βr r

(a) Find an expression for the final speed of the ball as a function of J, m, and β. (b) For what value of β does the ball immediately begin to roll without slipping, regardless of the value of µ?



3

3. A block of mass m slides on a circular track of radius r whose wall and floor both have coefficient of kinetic friction µ with the block. The size of the block is small compared to the radius of the track. The floor lies in a horizontal plane and the wall is vertical. The block is in constant contact with both the wall and the floor. The block has initial speed v0 .

(a) Let the block have kinetic energy E after traveling through an angle θ. Derive an expression for dE dθ in terms of g, r, µ, m and E. (b) Suppose the block circles the track exactly once before coming to a halt. Determine v0 in terms of g, r, and µ.

4. Two beads, each of mass m, are free to slide on a rigid, vertical hoop of mass mh . The beads are threaded on the hoop so that they cannot fall off of the hoop. They are released with negligible velocity at the top of the hoop and slide down to the bottom in opposite directions. The hoop remains vertical at all times. What is the maximum value of the ratio m/mh such that the hoop always remains in contact with the ground? Neglect friction. beads released with negligible velocity, slide on hoop wire vertical circular hoop

Ground


2008 Quarter-final Exam - Solutions

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2008 Quarter-Final Exam Solutions 1. A charged particle with charge q and mass m starts with an initial kinetic energy K at the middle of a uniformly charged spherical region of total charge Q and radius R. q and Q have opposite signs. The spherically charged region is not free to move. (a) Find the value of K0 such that the particle will just reach the boundary of the spherically charged region. (b) How much time does it take for the particle to reach the boundary of the region if it starts with the kinetic energy K0 found in part (a)? Solution: Assume that q is negative and Q is positive. (a) Apply Gauss’s Law to a spherical shell of radius r where r < R. Then, 4πρr3 0

E4πr2 = where the charge density ρ =

3Q 4πR3

Solving for the Electric Field at a distance r from the center, we find E=

ρr 30

We now find the potential difference between the center of the sphere and the outer boundary of the charged cloud. R

Z ∆V = −

Edr 0

Substituting in our expression for E, we now find ∆V = −

ρR2 60

Using conservation of energy, Klost = Ugained

K0 = −q∆V =

−qQ 8πR0

(b) Apply Newton’s Second Law to the object of charge −q when it is located a distance r away from the center of the sphere. c Copyright 2008 American Association of Physics Teachers


2

Fnet = ma −(−q)E = m

d2 r dt2

d2 r −qρr =− 2 dt 30 m We recognize that this is the differential equation for simple harmonic motion d2 r = −ω 2 r dt2 where ω =

q

−qρ 30 m .

Since the charge has the minimum kinetic energy needed to reach the surface, the trip from the center of the sphere to the outer boundary is one-fourth of a cycle of SHM. Therefore, π π T = t= = 4 2ω 2

r

30 m −qρ

. Substituting, in for rho, we find π t= 2

s

4π0 mR3 −qQ

. 2. A uniform pool ball of radius r begins at rest on a pool table. The ball is given a horizontal impulse J of fixed magnitude at a distance βr above its center, where −1 ≤ β ≤ 1. The coefficient of kinetic friction between the ball and the pool table is µ. You may assume the ball and the table are perfectly rigid. Ignore effects due to deformation. (The moment of inertia about the center of mass of a solid sphere of mass m and radius r is Icm = 52 mr2 .) (a) Find an expression for the final speed of the ball as a function of J, m, and β. (b) For what value of β does the ball immediately begin to roll without slipping, regardless of the value of µ. . Solution 1: (a) Consider an axis perpendicular to the initial impulse and coplanar with the table. (Throughout this solution we consider only torques and angular momenta with respect to this axis.) After the initial impulse, the torque about this axis is always zero, so angular momentum is conserved. The initial impulse occurs a perpendicular distance (β + 1) r from the axis, so the angular momentum is L = (β + 1) rJ After the ball has skidded along the table for a certain distance, it will begin to roll without v slipping. At that point, ω = rf . c Copyright 2008 American Association of Physics Teachers


3

Meanwhile, its moment of inertia about this axis is (by the parallel-axis theorem) I = Icm + mr2 = 75 mr2 , so that its final angular velocity ω is given by L = Iω 7 (β + 1) rJ = mr2 ω 5 7 (β + 1) J = mvf 5 . Therefore, vf =

5J (1 + β) 7m

. (b) The ball acquires linear momentum J as a result of the horizontal impulse, so its initial velocity v is given by mv = J We want the initial angular velocity and initial velocity to satisfy the no-slip condition v = ωr; thus 7 (β + 1) J = J 5 2 β= 5 . Solution 2:. Consider torques and angular momenta about the center of mass. If the horizontal impulse is large compared with the horizontal impulse from friction during the time that the cue stick is in contact with the ball, then angular impulse = change in angular momentum becomes: 2 Jβr = mr2 ω0 5 . Linear impulse = change in linear momentum yields: J = mv0 . (b) We want the initial angular velocity and initial velocity to satisfy the no-slip condition v0 = ω0 r; thus 2 2 Jβ = mv0 = J 5 5 . 2 β= 5 .



4

(a) In the case that the ball does not immediately begin to roll without slipping, friction will exert an angular impulse, f tr, about the center of mass of the ball as the ball skids along the surface: 2 f tr = mr2 (ωf − ω0 ) 5 . Friction will also exert a linear impulse −f t that will cause a change in linear momentum: −f t = m(vf − v0 ) . Combining the last two equations: 2 −m(vf − v0 ) = mr(ωf − ω0 ) 5 After the ball has skidded along the table for a certain distance, it will begin to roll without v slipping. At that point, ω = rf . 2 2 −vf + v0 = vf − rω0 5 5 Now, using the relationships between the impulse J and the initial angular and linear momentum, −vf +

J 2 Jβ = vf − m 5 m

. J 7 (1 + β) = vf m 5 . vf =

5J (1 + β) 7m

. 3. A block of mass m slides on a circular track of radius r whose wall and floor both have coefficient of kinetic friction µ with the block. The floor lies in a horizontal plane and the wall is vertical. The block is in constant contact with both the wall and the floor. The block has initial speed v0 . (a) Let the block have kinetic energy E after traveling through an angle θ. Derive an expression for dE dθ in terms of g, r, µ, m and E. (b) Suppose the block circles the track exactly once before coming to a halt. Determine v0 in terms of g, r, and µ.



5

Solution Let v be the speed of the block after it has traveled through an angle θ. When we apply Newton’s Second Law to the block as it is traveling around the circular path of radius r at a speed v, we find that the force of the wall on the block, Fw , is Fw =

mv 2 r

and the force of the floor on the block, FN , is FN = mg Therefore, the total force of friction on the block is FF ric = µm(

v2 + g) r

. Let dE denote the loss of kinetic energy as a result of the work W that friction does on the block when it has traveled through an angle dθ. The block will travel a distance ds = rdθ as it travels through an angle dθ. Then, dE = W = −FF ric ds = −FF ric rdθ . Now, substituting in the expressions for FF ric , we find dE = −µm(v 2 + gr)dθ .

dE = −µ(mv 2 + mgr) = −µ(2E + mgr) dθ

where we have simply used the fact that the kinetic energy E is E =

mv 2 2 .

(b) Now, we separate variables and solve the differential equation. Z Z dE = −µdθ 2E + mgr .

1 ln |2E + mgr| = −µθ + C 2

. Exponentiating both sides, yields 2E + mgr = Ce−2µθ Let E0 denote the initial kinetic energy. Then, using initial conditions, we find that C = 2E0 + mgr. Now, using the fact that E = 0 when θ = 2π, we obtain mgr = (2E0 + mgr)e−4πµ c Copyright 2008 American Association of Physics Teachers


6

. Solving for E0 ,

1 E0 = mgr(e4µπ − 1) 2

. Then, using E0 =

mv02 2 ,

and we find that v0 =

p gr(e4µπ − 1)

. 4. Two beads, each of mass m, are free to slide on a rigid, vertical hoop of mass mh . The beads are threaded on the hoop so that they cannot fall off of the hoop. They are released with negligible velocity at the top of the hoop and slide down to the bottom in opposite directions. What is the maximum value of the ratio m/mh such that the hoop always remains in contact with the ground? Neglect friction. Solution 1: Draw a free-body diagram for each bead; let FN be the (inward) normal force exerted by the hoop on the bead. Let θ be the angular position of the bead, measured from the top of the hoop, and let the hoop have radius r. We see that FN + mg cos θ = m

v2 r

v2 − mg cos θ r is given by

FN = m The (downward) vertical component FN y

FN y = FN cos θ From Newton’s third law, the two beads together exert an upwards vertical force on the hoop given by Fu = 2FN y 2 v Fu = 2m cos θ − g cos θ r noting that the beads clearly reach the same position at the same time. Meanwhile, when each bead is at a position θ it has moved through a vertical distance r (1 − cos θ). Thus from energy conservation, 1 mv 2 = mgr (1 − cos θ) 2 v2 = 2g (1 − cos θ) r Inserting this into the previous result, Fu = 2m cos θ (2g (1 − cos θ) − g cos θ) Fu = 2mg 2 cos θ − 3 cos2 θ c Copyright 2008 American Association of Physics Teachers


7

If the beads ever exert an upward force on the hoop greater than mh g, the hoop will leave the ground; i.e., the condition for the hoop to remain in contact with the ground is that for all θ, Fu ≤ mh g We can replace the left side by its maximum value. Letting s = cos θ, Fu = 2mg 2s − 3s2 d Fu = 2mg (2 − 6s) ds The derivative is zero at s = 13 , where 2 Fu(max) = mg 3 So our condition is

2 mg ≤ mh g 3 3 m ≤ mh 2

Solution 2: As before, we apply energy conservation to find the speed of the beads: p v = 2gr (1 − cos θ) The vertical (downward) component of the beads’ velocity is thus vy = v sin θ p vy = sin θ 2gr (1 − cos θ) While the hoop is in contact with the ground, the beads are the only part of the system in motion, so the momentum of the system is simply the beads’ momentum: py = 2mvy p py = 2m sin θ 2gr (1 − cos θ) The net (downward) force on the system is dpy dpy dθ dpy v = = dt dθ dt dθ r 1p p √ d Fnet = 2m 2gr sin θ 1 − cos θ · 2gr (1 − cos θ) dt r √ √ 1 Fnet = 4mg cos θ 1 − cos θ + sin θ √ sin θ 1 − cos θ 2 1 − cos θ 1 2 2 Fnet = 4mg cos θ − cos θ + sin θ 2 2 Fnet = 2mg 2 cos θ − 3 cos θ + 1 Fnet =



8

This downward force is provided by gravity and the normal force upward on the hoop: Fnet = 2mg + mh g − FN If the hoop is to remain in contact with the ground, the normal force can never be negative: FN ≥ 0 2mg + mh g − Fnet ≥ 0 2mg + mh g − 2mg 2 cos θ − 3 cos2 θ + 1 ≥ 0 2mg 2 cos θ − 3 cos2 θ ≤ mh g and we proceed as above.



2008 Semifinal Exam

1

AAPT AIP

UNITED STATES PHYSICS TEAM 2008 Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of three parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • Part C has one question and is allowed 20 minutes. The answer for Part C will not be used for team selection, but will be used for special recognition from the Optical Society of America. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for all three parts of the exam. • The three parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B or Part C. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students look at Part C or go back to Part A. Collect the answers to part B before allowing the examinee to begin Part C. Examinees are allowed a 10 to 15 minutes break between Parts B and C. • Allow 20 minutes to complete Part C. This part is optional; scores on Part C will not be used to select the US Team. Examinees may not go back to Part A or B. • Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-7), Part B (pages 8-10), and Part C (page 11). Examinees should be provided the parts individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2008. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A.


2008 Semifinal Exam

Cover Sheet

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Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Parts B or C during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Parts A or C during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your school ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, School ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • Part C is an optional part of the test. You will be given 20 additional minutes to complete Part C. Your score on Part C will not affect the selection for the US Team, but can be used for special prizes and recognition to be awarded by the Optical Society of America. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2008. Possibly Useful Information. You may use this sheet for all three parts of the exam. g = 9.8 N/kg G = 6.67 × 10−11 N · m2 /kg2 9 2 2 k = 1/4πǫ0 = 8.99 × 10 N · m /C km = µ0 /4π = 10−7 T · m/A 8 c = 3.00 × 10 m/s kB = 1.38 × 10−23 J/K 23 −1 NA = 6.02 × 10 (mol) R = NA kB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1eV = 1.602 × 10−19 J h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s −31 2 me = 9.109 × 10 kg = 0.511 MeV/c (1 + x)n ≈ 1 + nx for |x| ≪ 1 1 3 cos θ ≈ 1 − 21 θ2 for |θ| ≪ 1 sin θ ≈ θ − 6 θ for |θ| ≪ 1 c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

Part A

3

Part A Question A1 Four square metal plates of area A are arranged at an even spacing d as shown in the diagram. (Assume that A ≫ d2 .) Plate 1 Plate 2 Plate 3 d Plate 4 d d

Plates 1 and 4 are first connected to a voltage source of magnitude V0 , with plate 1 positive; plates 2 and 3 are then connected together with a wire. The wire is subsequently removed. Finally, the voltage source attached between plates 1 and 4 is replaced with a wire. The steps are summarized in the diagrams below.

Step 1

Step 2

Step 3

Find the resulting potential difference ∆V12 between plates 1 and 2; like wise find ∆V23 and ∆V34 , defined similarly. Assume, in each case, that a positive potential difference means that the top plate is at a higher potential than the bottom plate.


2008 Semifinal Exam

Part A

4

Question A2 A simple heat engine consists of a moveable piston in a cylinder filled with an ideal monatomic gas. Initially the gas in the cylinder is at a pressure P0 and volume V0 . The gas is slowly heated at constant volume. Once the pressure reaches 32P0 the piston is released, allowing the gas to expand so that no heat either enters or escapes the gas as the piston moves. Once the pressure has returned to P0 the outside of the cylinder is cooled back to the original temperature, keeping the pressure constant. For the monatomic ideal gas you should assume that the molar heat capacity at constant volume is given by CV = 32 R, where R is the ideal gas constant. You may express your answers in fractional form or as decimals. If you choose decimals, keep three significant figures in your calculations. The diagram below is not necessarily drawn to scale.

Pressure

32P0

P0

V0

Vmax

Volume

a. Let Vmax be the maximum volume achieved by the gas during the cycle. What is Vmax in terms of V0 ? If you are unable to solve this part of the problem, you may express your answers to the remaining parts in terms of Vmax without further loss of points. b. In terms of P0 and V0 determine the heat added to the gas during a complete cycle. c. In terms of P0 and V0 determine the heat removed from the gas during a complete cycle. d. What is the efficiency of this cycle?


2008 Semifinal Exam

Part A

5

Question A3 A certain planet of radius R is composed of a uniform material that, through radioactive decay, generates a net power P . This results in a temperature differential between the inside and outside of the planet as heat is transfered from the interior to the surface. The rate of heat transfer is governed by the thermal conductivity. The thermal conductivity of a material is a measure of how quickly heat flows through that material in response to a temperature gradient. Specifically, consider a thin slab of material of area A and thickness ∆x where one surface is hotter than the other by an amount ∆T . Suppose that an amount of heat ∆Q flows through the slab in a time ∆t. The thermal conductivity k of the material is then k=

∆Q 1 ∆x . ∆t A ∆T

It is found that k is approximately constant for many materials; assume that it is constant for the planet. For the following assume that the planet is in a steady state; temperature might depend on position, but does not depend on time. a. Find an expression for the temperature of the surface of the planet assuming blackbody radiation, an emissivity of 1, and no radiation incident on the planet surface. You may express your answer in terms of any of the above variables and the Stephan-Boltzmann constant σ. b. Find an expression for the temperature difference between the surface of the planet and the center of the planet. You may express your answer in terms of any of the above variables; you do not need to answer part (a) to be able to answer this part.


2008 Semifinal Exam

Part A

6

Question A4 A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0 s is the time at which the tape recorder was dropped. t (s) 2.0 4.0 6.0 8.0 10.0

f (Hz) 581 619 665 723 801

The acceleration due to gravity is g = 9.80 m/s2 and the speed of sound in air is vsnd = 340 m/s. Ignore air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, vsnd ± vobs f = f0 vsnd ± vsrc where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer, and vsnd , vsrc , and vobs are the speed of sound in air, the speed of the source, and the speed of the observer. The positive and negative signs are dependent upon the relative directions of the motions of the source and the observer. a. Determine the frequency measured on the ground at time t, in terms of f0 , g, h, and vsnd . Consider only the case where the falling tape recorder doesn’t exceed the speed of sound vsnd . b. Verify graphically that your result is consistent with the provided data. c. What (numerically) is the frequency played by the tape recorder? d. From what height h was the tape recorder dropped?


2008 Semifinal Exam

Part A

7

STOP: Do Not Continue to Part B If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.


2008 Semifinal Exam

Part B

8

Part B Question B1 A platform is attached to the ground by an ideal spring of constant k; both the spring and the platform have negligible mass; assume that your mass is mp . Sitting on the platform is a rather large lump of clay of mass mc = rmp , with r some positive constant that measures the ratio mc /mp . You then gently step onto the platform, and the platform settles down to a new equilibrium position, a vertical distance D below the original position. Throughout the problem assume that you never lose contact with the platform.

h D

a. You then slowly pick up the lump of clay and hold it a height h above the platform. Upon releasing the clay you and the platform will oscillate up and down; you notice that the clay strikes the platform after the platform has completed exactly one oscillation. Determine the numerical value of the ratio h/D. b. Assume the resulting collision between the clay and the platform is completely inelastic. Find the ratio of the amplitude of the oscillation of the platform after the collision (Af ) to the amplitude of the oscillations of the platform before the collision (Ai ). Determine Af /Ai in terms of the mass ratio r and any necessary numerical constants. c. Sketch a graph of the position of the platform as a function of time, with t = 0 corresponding to the moment when the clay is dropped. Show one complete oscillation after the clay has collided with the platform. It is not necessary to use graph paper. d. The above experiment is only possible if the mass ratio r is less than some critical value rc . Otherwise, despite the clay having been dropped from the height determined in part (a), the oscillating platform will hit the clay before the platform has completed one full oscillation. On your graph in part (c) sketch the position of the clay as a function of time relative to the position of the platform for the mass ratio r = rc .


2008 Semifinal Exam

Part B

9

Question B2 Consider a parallel plate capacitor with the plates vertical. The plates of the capacitor are rigidly supported in place. The distance between the plates is d. The plates have height h and area A ≫ d2 . Assume throughout this problem that the force of air resistance may be neglected; however, the force of gravity cannot be neglected. Neglect any edge effects as well as any magnetic effects. d/2

L

String

Rigid Support

d

h h h/2

a. A small metal ball with a mass M and a charge q is suspended from a string of length L that is tied to a rigid support. When the capacitor is not charged, the metal ball is located at the center of the capacitor— at a distance d/2 from both plates and at a height h/2 above the bottom edge of the plates. If instead a constant potential difference V0 is applied across the plates, the string will make an angle θ0 to the vertical when the metal ball is in equilibrium. i. Determine θ0 in terms of the given quantities and fundamental constants. ii. The metal ball is then lifted until it makes an angle θ to the vertical where θ is only slightly greater than θ0 . The metal ball is then released from rest. Show that the resulting motion is simple harmonic motion and find the period of the oscillations in terms of the given quantities and fundamental constants. iii. When the ball is at rest in the equilibrium position θ0 , the string is cut. What is the maximum value for V0 so that the ball will not hit one of the plates before exiting? Express your answer in terms of the given quantities and fundamental constants. b. Suppose instead that the ball of mass M and charge q is released from rest at a point halfway between the plates at a time t = 0. Now, an AC potential difference V (t) = V0 sin ωt is also placed across the capacitor. The ball may hit one of the plates before it falls (under the influence of gravity) out of the region between the plates. If V0 is sufficiently large, this will p range of angular p only occur for some frequencies ωmin < ω < ωmax . You may assume that ωmin ≪ g/h and ωmax ≫ g/h. Making these assumptions, find expressions for ωmin and ωmax in terms of the given quantities and/or fundamental constants. c. Assume that the region between the plates is not quite a vacuum, but instead humid air with a uniform resistivity ρ. Ignore any effects because of the motion of the ball, and assume that the humid air doesn’t change the capacitance of the original system. i. Determine the resistance between the plates. ii. If the plates are originally charged to a constant potential source V0 , and then the potential is removed, how much time is required for the potential difference between the plates to decrease to a value of V0 /e, where ln e = 1? iii. If the plates are instead connected to an AC potential source so that the potential difference across the plates is V0 sin ωt, determine the amplitude I0 of the alternating current through the potential source.


2008 Semifinal Exam

Part B

10

STOP: Do Not Continue to Part C If there is still time remaining for Part B, you should review your work for Part B, but do not continue to Part C until instructed by your exam supervisor. You may not return to Part A


2008 Semifinal Exam

Part C

11

Optical Society of America Bonus Question Researchers have developed a lens made of liquid. The spherical lens consists of a droplet of transparent liquid resting on an electrically controllable surface. When the voltage of the surface is changed, the droplet itself changes shape; it either tries to “ball-up” more strongly or it becomes flatter. The figure below is a sketch of the liquid lens and several parameters that describe it, including the thickness of the lens (t), the radius of curvature of the top surface (R) and the contact angle (θ), which represents the angle between the flat surface beneath the droplet and the tangent to the curved surface at the point of contact.

t θ R

R

a. When a certain voltage is applied, both the contact angle and lens thickness increase (and the lens becomes more curved). In this case, is the liquid attracted or repelled by the surface? b. Express the contact angle as a function of R and t. c. The total volume of the liquid lens is an important parameter because as the liquid lens changes shape, its volume is conserved. Calculate the volume of the lens as a function of R and t. d. Use your result to part (b) to eliminate the variable t from your expression for the volume and find V (R, θ). e. By changing the voltage on the control surface, the contact angle, θ, can be changed, which in turn changes the focal length of the lens, f . The lensmaker’s formula can be used to calculate the focal length and is given by 1 1 1 , − = (nliquid − nair ) f R1 R2 where nliquid and nair are the refractive indices of the liquid in the lens and air around it, and R1 and R2 are the radii of curvature of the two surfaces of the lens. In figure 1, R1 is the curved face and R2 is the flat face. Use the lensmaker’s formula to calculate the focal length of the lens in terms of the total volume of the liquid, the contact angle, and the relevant refractive indices. Sidenote: liquid lenses are interesting because they are electrically controllable, variable focus lenses that can be very compact. People are working on putting them into cell phone cameras for ultracompact zoom lenses. For more information on this type of liquid lens, see T. Krupenkin, S. Yang, and P. Mach, “Tunable liquid microlens,” Appl. Phys. Lett. 82, 316-318 (2003).


2008 Semifinal Exam

1

AAPT AIP

UNITED STATES PHYSICS TEAM 2008 Semifinal Exam 6 QUESTIONS - Several MINUTES INSTRUCTIONS

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DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Show all your work. Partial credit will be given.

• Start each question on a new sheet of paper. Put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie

Prob. 1 - P. 1/3

• A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. • Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Each of the four questions in part A are worth 25 points. Each of the two questions in part B are worth 50 points. The questions are not necessarily of the same difficulty. Good luck! • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after April 10, 2008.



2008 Semifinal Exam

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Part A Question A1 Four square metal plates of area A are arranged at an even spacing d as shown in the diagram. (Assume that A >> d2 .) Plate 1 Plate 2 Plate 3 d Plate 4 d d

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Plates 1 and 4 are first connected to a voltage source of magnitude V0 , with plate 1 positive. Plates 2 and 3 are then connected together with a wire, which is subsequently removed. Finally, the voltage source attached between plates 1 and 4 is replaced with a wire. The steps are summarized in the diagram below.

(a)

(b)

(c)

What is the resulting potential difference between a. Plates 1 and 2 (Call it V1 ),

b. Plates 2 and 3 (Call it V2 ), and c. Plates 3 and 4 (Call it V3 ).

Assume, in each case, that a positive potential difference means that the top plates is at a high potential than the bottom plate. Solution

There are two fairly easy ways to do this problem, one rather straightforward application of capacitors, the other a more elegant, and much, much short, application of boundary conditions in electric fields. The first method involves treating the problem as three series capacitors. Each has an identical capacitance C. The figure below then show the three steps.

C1

C1

C1

C2

C2

C2

C3

C3

C3

Since C2 is shorted out originally, then effectively there are only two capacitors in series, so the voltage drop across each is V0 /2, where the a positive potential difference means that the top plate of any given capacitor is positive. The top plate of C1 will then have a positive charge of q0 = CV0 /2. Note that this means that the bottom plate of the top capacitor will have a negative charge of −q0 . Removing the shorting wire across C2 will not change the charges or potential drops across the other two capacitors. Removing the source V0 will also make no difference. c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

3

Shorting the top plate of C1 with the bottom plate of C3 will make a difference. Positive charge will flow out of top plate of C1 into the bottom plate of C3 . Also, negative charge will flow out of the bottom plate of C1 into the top plate of C2 . The result is that C1 will acquire a potential difference of V1 , C2 a potential difference of V2 , and C3 a potential difference of V3 . Let the final charge on the top plate of each capacitor also be labeled as q1 , q2 , and q3 . The last figure implies that V1 + V2 + V3 = 0. By symmetry, we have V1 = V3 . so 2V1 = −V2 . By charge conservation between the bottom plate of C1 and the top plate of C2 we have

But q = CV , so

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−q0 = −q1 + q2 . 1 − V0 = −V1 + V2 2

Combining the above we get

1 − V0 2 1 − V0 3

=

1 V2 + V2 , 2

=

V2 .

Finally, solving for V1 , we get V1 = V0 /6. Alternatively, we could focus on the plate arrangement and the fact that across a boundary |∆E⊥ | = |σ/ǫ0 |, a consequence of Gauss’s Law. Also, we have, for parallel plate configurations, |∆V | = |Ed|. Since ǫ0 and d are the same for each of the three regions, it is sufficient to simply look at the behavior of the electric fields.

E0

E0

E1

E2

E1

In the first picture we require that 2E0 = V0 /d. The charge density on the second plate requires that ∆E = E0 . In the last picture we have 2E1 +E2 = 0, since the potential between the top plate and the bottom plate is zero. But we also have, on the second plate, ∆E = E1 − E2 . Combining, E0 = − 12 E2 − E2 = − 23 E2 , and therefore V2 = − 31 V0 , and V1 = V0 /6.


2008 Semifinal Exam

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Question A2 A simple heat engine consists of a piston in a cylinder filled with an ideal monatomic gas. Initially the gas in the cylinder is at a pressure P0 and volume V0 . The gas is slowly heated at constant volume. Once the pressure reaches 32P0 the piston is released, allowing the gas to expand so that no heat either enters or escapes the gas as the piston moves. Once the pressure has returned to P0 the the outside of the cylinder is cooled back to the original temperature, keeping the pressure constant. For the monatomic ideal gas you should assume that the specific heat capacity at constant volume is given by CV = 32 nR, where n is the number of moles of the gas present and R is the ideal gas constant. You may express your answers in fractional form or as decimals. If you choose decimals, keep three significant figures in your calculations. The diagram below is not necessarily drawn to scale.

Pressure

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32P0

P0

V0

Vmax

Volume

a. Let Vmax be the maximum volume achieved by the gas during the cycle. What is Vmax in terms of V0 ? If you are unable to solve this part of the problem, you may expressyour answers to the remaining parts in terms of Vmax without further loss of points. b. In terms of P0 and V0 determine the heat added to the gas during a complete cycle. c. In terms of P0 and V0 determine the heat removed from the gas during a complete cycle. d. Defining efficiency e as the net work done by the gas divided by the heat added to the gas, what is the efficiency of this cycle? e. Determine the ratio between the maximum and minimum temperatures during this cycle. Solution It is convenient to construct two tables and solve this problem in a manner similar to a Sudoku puzzle. Defining point 1 to be the initial point, and measuring P , V , and T in terms of P0 , V0 , and T0 , while measuring Q, W , and ∆U in terms of nRT0 = P0 V0 , we have initially Point 1 2 3

P 1 32 1

V 1 1

T 1

Process 1→2 2→3 3→1 net

Q

W 0

∆U

0 0


2008 Semifinal Exam

5

The “obvious” values have been filled in: the initial conditions, and zeroes corresponding to Q = 0 along an adiabat, W = 0 along a constant volume process, and finally ∆U = 0 for a net process. The convention that will be used here is Q + W = ∆U . The ideal gas law, P V /T = nR, can be used to quickly determine T2 , since P/T is a constant for that process. One can then use Q = CV ∆T to find Q1→2 . The table values will now read Point 1 2 3

P 1 32 1

V 1 1


T 1 32

Q 3 2 31

W 0

∆U

0

0

For the adiabatic process we have P V γ is a constant, where γ = CP /CV . Students who don’t know this can derive it, although it will take some time.

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The derivation is straightforward enough. Along an adiabatic process, Q = 0, so from Q + W = δU −P dV =

Rearranging,

3 3 nR dT = (P dV + V dP ) 2 2

0 = 5P dV + 3V dP

or

0=

Integrating,

dP 5 dV + 3 V P

Constant =

5 ln V + ln P 3

which can be written in the more familiar form

P V γ = Constant.

The factor of 32 was chosen so that the results are nice answers.

One can then find V3 (and, for that matter, Vmax ) by using this, and get V3 = V2

P2 P3

1/γ

= (32)3/5 = 8

Putting this in the table, and then quickly applying the ideal gas law to find T3 then enables the finding of Q3→1 , since along this process 5 Q = CP ∆T = nR∆T. 2 The tables now look like Point 1 2 3

P 1 32 1

V 1 1 8

T 1 32 8


Q

3 2 31

W 0

∆U

0 − 25 7 0

It is now possible to determine Qnet and, from Q + W = ∆U , Wnet . So the tables now look like Point 1 2 3

P 1 32 1

V 1 1 8

T 1 32 8


Q 3 2 31

0 − 25 7 29

W 0

∆U

−29

0


2008 Semifinal Exam

6

The negative net work reflects that the gas does work on the outside world. The efficiency of the process is then 58 29 = e= 93/2 93 It isn’t much more work to fill in all of the values for both tables, a task that each student ought be able to do. One key point will be process 3 → 1, where W = −P ∆V . the the rest are filled in by applications of Q + W = ∆U . P 1 32 1

V 1 1 8

T 1 32 8


Q 3 2 31

0 − 25 7 29

W 0 -36 7 −29

∆U 3 2 31 -36 − 32 7 0

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Point 1 2 3


2008 Semifinal Exam

7

Question A3 A certain planet of radius R is composed of a uniform material that, through radioactive decay, generates a net power P . This results in a temperature differential between the inside and outside of the planet as heat is transfered from the interior to the surface. The rate of heat transfer is governed by the thermal conductivity. The thermal conductivity of a material is a measure of how quickly heat flows through that material in response to a temperature gradient. Specifically, consider a thin slab of material of area A and thickness ∆x where one surface is hotter than the other by an amount ∆T . Suppose that an amount of heat ∆Q flows through the slab in a time ∆t. The thermal conductivity k of the material is then ∆Q 1 ∆x . ∆t A ∆T

k=

It is found that k is approximately constant for many materials; assume that it is constant for the planet. For the following assume that the planet is in a steady state; temperature might depend on position, but does not depend on time.

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a. Find an expression for the temperature of the surface of the planet assuming blackbody radiation, an emissivity of 1, and no radiation incident on the planet surface. You may express your answer in terms of any of the above variables and the Stephan-Boltzmann constant σ. b. Find an expression for the temperature difference between the surface of the planet and the center of the planet. You may express your answer in terms of any of the above variables; you do not need to answer part (a) to be able to answer this part. Solution

For the first question, apply the Boltzmann equation, and

P = σATs4

where A is the surface area of the planet, and Ts the temperature at the center. Then Ts =

P 4πσR2

1/4

For the second question, it is reasonable to assume that the temperature depends on the distance form the center only. Then the definition of k gives for a spherical shell of thickness dr k=

∆Q 1 dr . ∆t 4πr2 dT

The heat through the shell depends on the power radiated from within the shell. Since the planet is uniform, this depends on the volume according to 4 πr3 r3 ∆Q = P 43 3 = P 3 ∆t R 3 πR

so that rearrangement yields dT =

P r dr 4πkR3

Integrating between the center and the surface, ∆T =

P , 8πkR

which could be used to find the temperature of the interior. c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

8

Question A4 A tape recorder playing a single tone of frequency f0 is dropped from rest at a height h. You stand directly underneath the tape recorder and measure the frequency observed as a function of time. Here t = 0s is the time at which the tape recorder was dropped. t (s) 2.0 4.0 6.0 8.0 10.0

f (Hz) 581 619 665 723 801

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The acceleration due to gravity is g = 9.80 m/s2 and the speed of sound in air is va = 340 m/s. Ignore air resistance. You might need to use the Doppler shift formula for co-linear motion of sources and observers in still air, va ± vo f = f0 va ± vs where f0 is the emitted frequency as determined by the source, f is the frequency as detected by the observer, and va , vs , and vo are the speed of sounds in air, the speed of the source, and the speed of the observer. The positive and negative signs are dependent upon the relative directions of the source and the observer. a. Determine the frequency measured on the ground at time t, in terms of f0 , g, h, and va . b. Verify graphically that your result is consistent with the provided data. c. What (numerically) is the frequency played by the tape recorder? d. From what height h was the tape recorder dropped?

Solution

The position of the tape recorder above the ground at a time t is given by 1 y = h − gt2 2

and the speed of the tape recorder is given by

vs = −gt

The observer “hears” the sound emitted from the tape recorder a time δt earlier, since it takes time for the sound to travel to the listener. In this case, y = va δt So at time t the listener is hearing the tape recorder when it had emitted at time t′ = t − δt, or t′ = t −

h g ′ 2 + (t ) va 2va

Solve this for t′ , first by rearranging, g ′ 2 (t ) − va t′ + (va t − h) = 0 2 and the by applying the quadratic formula ′

t =

va ±

p va 2 + 2gh − 2gva t . g


2008 Semifinal Exam

9

This might not look right, but in the limit of small h and large va , it does reduce to the expected t′ = t if one keeps the negative result. Consequently, p vs = va 2 + 2gh − 2gva t − va

gives the velocity of that source had when it emitted the sound heard at time t. This result is negative, indicating motion down, and toward the observer, so one must use the positive sign in the denominator of the Doppler shift formula. Applying the Doppler shift formula, va f = f0 p 2 va + 2gh − 2gva t

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which, in the limit of large va and small h, reduces to g f = f0 1 + t va

Keeping to the correct expression, we can rearrange it as 2gh 2g 1 1 1 + = − t f2 f0 2 va 2 va

which would graph as a straight line by plotting t horizontally and 1/f 2 vertically. The slope of the line would yield 2g − va f0 2 while the vertical intercept would yield

1 f0 2

2gh 1+ 2 va

Using this hint, a quick table of data to consider graphing would be t (s) 2.0 4.0 6.0 8.0 10.0

The slope is −1.75 × 10−7 s. Then

f0 =

s

f (Hz) 581 619 665 723 801

1/f 2 (×10−6 s2 ) 2.96 2.61 2.26 1.91 1.56

2(9.8) Hz = 574Hz (1.75 × 10−7 )(340)

The intercept is 3.31 × 10−6 s2 . This yields a height in meters given by h = (340)

(340)2 (3.31) − = 533 (0.175) 2(9.8)

Clearly, an impressive building; and a more impressive tape player, that it could be heard from such a distance!


2008 Semifinal Exam

10

Part B Question B1

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A platform is attached to the ground by an ideal spring of constant k; both the spring and the platform have negligible mass. Sitting on the platform is a rather large lump of clay of mass mc . You then gently step onto the platform, and the platform settles down to a new equilibrium position, a vertical distance D below the original position. Assume that your mass is mp .

h

D

a. You then pick up the lump of clay and hold it a height h above the platform. Upon releasing the clay the you and the platform will oscillate up and down; you notice that the clay strikes the platform after the platform has completed exactly one oscillation. Determine h in terms of any or all of k, D, the masses mp and mc , the acceleration of free fall g, and any necessary numerical constants. You must express your answer in the simplest possible form. b. Assume the resulting collision between the clay and the platform is completely inelastic. Find the ratio of the amplitude of the oscillation of the platform before the collision (Ai ) and the amplitude of the oscillations of the platform after the collision (Af ). Determine Af /Ai in terms of any or all of k, D, the masses mp and mc , the acceleration of free fall g, and any necessary numerical constants. c. Sketch a graph of the position of the platform as a function of time, with t = 0 corresponding to the moment when the clay is dropped. Show one complete oscillation after the clay has collided with the platform. d. The above experiment is only possible if the ratio mc /mp is smaller than some critical value rc , otherwise the clay will hit the platform before one complete oscillation. An estimate for the value of the critical ratio rc can be obtained by assuming the clay hits the platform after exactly one-half of an oscillation. Assume that h is the same as is determined by part (a), and use this technique to determine rc in terms of any or all of k, D, the mass mp , the acceleration of free fall g, and any necessary numerical constants. e. Is this estimate for rc too large or too small? You must defend your answer with an appropriate diagram. Solution Stepping on the platform will lower it a distance D. This means that the spring constant of the platform spring is given by kD = mp g. c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

11

If the lump of clay is removed, then the equilibrium position of the platform would rise a distance A given by kA = mc g. This would also be the amplitude of the oscillations after the clay is released, so mc g Ai = k The frequency of oscillation of the plate without the clay is s 1 k f= 2π mp The time for a complete oscillation is T = 2π If the clay falls a distance h, then

r

mp k

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mp 1 2 gT = 2π 2 g = 2π 2 H. 2 k When the plate is at the stating point it is at rest. The clay will hit it with a speed given by h=

v0 = gT.

Conservation of momentum in an inelastic collision will then result in a final speed of the clay + platform system of mc vf = v0 . mc + mp The kinetic energy just after collision will be

1 (mc + mp )vf 2 . 2 So the amplitude of the resulting oscillations will be given by K=

or

1 1 kAf 2 = (mc + mp )vf 2 2 2 r mc + mp Af = vf k

Gluing stuff together

Af Ai

r vf mc + mp , Ai k r mc + mp v0 mc = . , Ai mc + mp k r mc gT k mc + mp . , = mc g mc + mp k s k = T mp + mc =

And then, combining with our previous expression for T , s r mp k Af = 2π , Ai k mp + mc r mp = 2π mp + mc c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

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If, instead, the clay manages to hit the platform at the top of an oscillation, then the distance the clay would fall would only be h − 2A and the time required would be r

mp . k

1 g 2

2 T , 2

T =π 2 Then h − 2A = where that complicated looking thing is actually

1 h. 4 So =

1 h + 2A, 4

3 h 8

=

A,

3 2 π D 4

=

A.

So lut ion s h

But from the very first two equations,

A 3 mc = = π2 . mp D 4

In this scenario the clay passes the platform three times: once at the highest point, once some distance further on, and once when the at the lowest point. That means that if A were smaller, it might be possible to find a value of A such that the clay just barely touches the platform once before hitting the platform at the bottom. Consequently, this is an over estimate for A, and an overestimate for mc , and therefore, and overestimate for the ratio. A larger ratio would guarantee collision, the actual critical ratio would be smaller than this.


2008 Semifinal Exam

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Question B2 Consider a parallel plate capacitor with the plates vertical. The plates of the capacitor are rigidly supported in place. The distance between the plates is d. The plates have height h and area A >> d2 . Assume throughout this problem that air resistance may be neglected; however, the force of gravity cannot be neglected. d/2

L

String

Rigid Support

d

h h

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h/2

a. A metal ball with a mass M and a charge q is suspended from a string that is tied to a rigid support. When the capacitor is not charged, the metal ball is located at the center of the capacitor (at a distance d/2 from both plates and at a height h/2 above the bottom edge of the plates. If instead a constant potential difference V0 is applied across the plates, the string will an angle θ0 to the vertical when the metal ball is in equilibrium. i. Determine θ0 in terms of the given quantities and fundamental constants. ii. The metal ball is then lifted until it makes an angle θ to the vertical where θ is only slightly greater than θ0 . The metal ball is released from rest. Show that the resulting motion is simple harmonic motion and find the period of the oscillations in terms of the given quantities and fundamental constants. iii. When the ball is at rest in its equilibrium position, the string is cut. What is the maximum value for V0 so that the ball will not hit one of the plates before exiting? Express your answer in terms of the given quantities and fundamental constants. b. Suppose instead that the ball of mass M and charge q is released from rest at a point halfway between the plates at a time t = 0. Now, an AC potential difference V (t) = V0 sin ωt is also placed across the capacitor. For what range of angular frequencies will the ball not hit either plate before it falls, under the influence of gravity, out of the region between the plates? Consider only two scenarios: either g >> hω 2 or g << hω 2 . Express your answer in terms of the given quantities and fundamental constants. Solution The electric field between the plates is given by E = V0 /d There are then three forces on the hanging ball: electrostatic, gravitational, and the tension in the string. They are FE = qE = qV0 /d and Fg = M g and then tension FT . c Copyright 2008 American Association of Physics Teachers

2008 Semifinal Exam

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In equilibrium, the string must make an angle such that the horizontal and vertical components of the tension balance the electrostatic and gravitational forces. Consequently, tan θ0 =

FE qV0 = Fg M gd

The tension in the string is given by FT 2 = Fg 2 + FE 2 . A simple pendulum oscillates with a period according to s L T = 2π g

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where g is the acceleration of free fall. In this case, if the string were cut, the instantaneous acceleration of the ball would be FT /M , so the period of small oscillations would be r ML T = 2π FT This is not a particularly pretty thing to substitute for, but yields s ML T = 2π p 2 2 M g + V0 2 q 2 /d2

which can be simplified, somewhat, to give

T = 2π

s

L g

1+

V0 q M gd

2 !− 14

An astute reader would recognize a few trigonometric identities and write s L cos θ0 T = 2π g

which is certainly much more compact. Upon cutting the string the ball will move in a straight line, tangent to the angle the string originally made. It won’t move with constant speed. So it will leave the region between the plates a distance x1 = (L + h/2) tan θ0

away from the center line. Set this equal to d/2, the condition that it hits one of the plates, and solve for V0 : d 2

=

(L + h/2)

V0

=

M gd2 (2L + h)q

qV0 M gd

If the ball is instead released from rest in the oscillating field then it will experience an oscillating force in the x direction given by V0 sin ωt. FE = d Hence, the x component of the acceleration will be given by ax =

qV0 sin ωt. Md


2008 Semifinal Exam

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The ball is released from rest, so this can be directly integrated to give the x component of the velocity vx =

qV0 (1 − cos ωt) . M dω

This can be directly integrated to find the position relative to the center, which we will define as x = 0: 1 qV0 t − sin ωt x= M dω ω The ball hits one of the plates if this value exceeds d/2 while it is still in the region between the plates. Solving for equality, sin ωt M d2 ω =t− 2qV0 ω is the concern. t is given by the equation of a falling object, or

So lut ion s

h 1 = gt2 , 2 2

p so t = h/g. Unfortunately, this can’t be solved for ω. p Consider first the condition that g << hω 2 . Then it is reasonable to neglect 1/ω compared to g/h, and the equation simplifies to M d2 ω p = g/h 2qV0 with solution

2qV0 p g/h M d2 At higher frequencies the ball will miss, falling out of the plates before it hits either side. Consider instead the condition that g >> hω 2 . This is a slowly oscillating field, and one can then approximate the sine function as 1 sin ωt ≈ ωt − ω 3 t3 . 6 yielding M d2 ω 1 = ω 2 t3 2qV0 6 or 3M d2 p 3 3 h /g ω2 = qV0 ω1 =

as the second critical frequency. At lower frequencies the ball will not hit either side before it falls out of the region between the plates. Note that the product of these two angular frequencies is θ1 θ2 = 6

h g


2008 Semifinal Exam

16

Optical Society of America Bonus Question Researchers have a developed a lens made of liquid. The spherical lens consists of a droplet of transparent liquid resting on a electrically controllable surface. When the voltage of the surface is changed, the droplet changes itself shape; it either tries to “ball-up” more strongly or it becomes flatter. Figure 1 is a sketch of the liquid lens and several parameters that describe it, including the thickness of the lens (t), the radius of curvature of the top surface (R) and the contact angle (θ), which represents the angle between the flat surface beneath the droplet and the tangent to the curved surface at the point of contact. a. When a certain voltage is applied, both the contact angle and lens thickness increase (and the lens becomes more curved). In this case, is the liquid attracted or repelled by the surface? b. Express the contact angle as a function of R and t. c. The total volume of the liquid lens is an important parameter because as the liquid lens changes shape, its volume is conserved. Calculate the volume of the lens as a function of R and t.

So lut ion s

d. Use your result to part (b) to eliminate the variable t from your expression for the volume and find V (R, θ). e. By changing the voltage on the control surface, the contact angle, θ, can be changed, which in turn changes the focal length of the lens, f . The lensmaker’s formula can be used to calculate the focal length and is given by 1 1 1 , − = (nliquid − nair ) f R1 R2 where nliquid and nair are the refractive indices of the liquid in the lens and air around it, and R1 and R2 are the radii of curvature of the two surfaces of the lens. In figure 1, R1 is the curved face and R2 is the flat face. Use the lensmaker’s formula to calculate the focal length of the lens in terms of the total volume of the liquid, the contact angle, and the relevant refractive indices. Sidenote: liquid lenses are interesting because they are electrically controllable, variablefocus lenses that can be very compact. People are working on putting them into cell phone cameras for ultracompact zoom lenses. For more information on this type of liquid lens, see T. Krupenkin, S. Yang, and P. Mach, “Tunable liquid microlens,” Appl. Phys. Lett. 82, 316-318 (2003).


United States Physics Team

F = ma Contest 2009

2009 F = ma Exam

1

AAPT AIP

UNITED STATES PHYSICS TEAM 2009 2009 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Use g = 10 N/kg throughout this contest. • You may write in this booklet of questions. However, you will not receive any credit for anything written in this booklet. • Your answer to each question must be marked on the optical mark answer sheet. • Select the single answer that provides the best response to each question. Please be sure to use a No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer, the previous mark must be completely erased. • Correct answers will be awarded one point; incorrect answers will result in a deduction of There is no penalty for leaving an answer blank.

1 4

point.




2009 F = ma Exam

2

1. A 0.3 kg apple falls from rest through a height of 40 cm onto a flat surface. Upon impact, the apple comes to rest in 0.1 s, and 4 cm2 of the apple comes into contact with the surface during the impact. What is the average pressure exerted on the apple during the impact? Ignore air resistance. (A) 67,000 Pa (B) 21,000 Pa← CORRECT (C) 6,700 Pa (D) 210 Pa (E) 67 Pa The following information is used for questions 2 and 3. Three blocks of identical mass are placed on a frictionless table as shown. The center block is at rest, whereas the other two blocks are moving directly towards it at identical speeds v. The center block is initially closer to the left block than the right one. All motion takes place along a single horizontal line.

2. Suppose that all collisions are instantaneous and perfectly elastic. After a long time, which of the following is true? (A) The center block is moving to the left. (B) The center block is moving to the right. (C) The center block is at rest somewhere to the left of its initial position. (D) The center block is at rest at its initial position.← CORRECT (E) The center block is at rest somewhere to the right of its initial position. 3. Suppose, instead, that all collisions are instantaneous and perfectly inelastic. After a long time, which of the following is true? (A) The center block is moving to the left. (B) The center block is moving to the right. (C) The center block is at rest somewhere to the left of its initial position. (D) The center block is at rest at its initial position. (E) The center block is at rest somewhere to the right of its initial position.← CORRECT 4. A spaceman of mass 80 kg is sitting in a spacecraft near the surface of the Earth. The spacecraft is accelerating upward at five times the acceleration due to gravity. What is the force of the spaceman on the spacecraft? (A) 4800 N← CORRECT (B) 4000 N (C) 3200 N (D) 800 N (E) 400 N


2009 F = ma Exam

3

5. Three equal mass satellites A, B, and C are in coplanar orbits around a planet as shown in the figure. The magnitudes of the angular momenta of the satellites as measured about the planet are LA , LB , and LC . Which of the following statements is correct? A B

C

(A) LA > LB > LC ← CORRECT (B) LC > LB > LA

(C) LB > LC > LA (D) LB > LA > LC (E) The relationship between the magnitudes is different at various instants in time. 6. An object is thrown with a fixed initial speed v0 at various angles α relative to the horizon. At some constant height h above the launch point the speed v of the object is measured as a function of the initial angle α. Which of the following best describes the dependence of v on α? (Assume that the height h is achieved, and assume that there is no air resistance.) (A) v will increase monotonically with α. (B) v will increase to some critical value vmax and then decrease. (C) v will remain constant, independent of α.← CORRECT (D) v will decrease to some critical value vmin and then increase. (E) None of the above. 7. A bird is flying in a straight line initially at 10 m/s. It uniformly increases its speed to 15 m/s while covering a distance of 25 m. What is the magnitude of the acceleration of the bird? (A) 5.0 m/s2 (B) 2.5 m/s2 ← CORRECT

(C) 2.0 m/s2

(D) 0.5 m/s2 (E) 0.2 m/s2


2009 F = ma Exam

4

The following information is used for questions 8 and 9.

Angular Velocity (rad/s)

A flat disk rotates about an axis perpendicular to the plane of the disk and through the center of the disk with an angular velocity as shown in the graph below.

4 2 Time (s)

0 1

2

3

−2

8. Determine the angular acceleration of the disk when t = 2.0 s. (A) -12 rad/s2. (B) -8 rad/s2 . (C) -4 rad/s2 . (D) -2 rad/s2 . ← CORRECT (E) 0 rad/s2 .

9. Through what net angle does the disk turn during the 3 seconds? (A) 9 rad. (B) 8 rad. (C) 6 rad. (D) 4 rad. (E) 3 rad. ← CORRECT 10. A person standing on the edge of a fire escape simultaneously launches two apples, one straight up with a speed of 7 m/s and the other straight down at the same speed. How far apart are the two apples 2 seconds after they were thrown, assuming that neither has hit the ground? (A) 14 m (B) 20 m (C) 28 m ← CORRECT

(D) 34 m

(E) 56 m


2009 F = ma Exam

5

11. A 2.25kg mass undergoes an acceleration as shown below. How much work is done on the mass?

Acceleration (m/s/s)

4 2 Position (m)

0 2

4

6

8

10

12

−2

(A) 36 J ← CORRECT (B) 22 J

(C) 5 J (D) -17 J (E) -36 J 12. Batman, who has a mass of M = 100 kg, climbs to the roof of a 30 m building and then lowers one end of a massless rope to his sidekick Robin. Batman then pulls Robin, who has a mass of m = 75 kg, up the roof of the building. Approximately how much total work has Batman done after Robin is on the roof? (A) 60 J (B) 7 × 103 J

(C) 5 × 104 J ← CORRECT

(D) 600 J

(E) 3 × 104 J 13. Lucy (mass 33.1 kg), Henry (mass 63.7 kg), and Mary (mass 24.3 kg) sit on a lightweight seesaw at evenly spaced 2.74 m intervals (in the order in which they are listed; Henry is between Lucy and Mary) so that the seesaw balances. Who exerts the most torque (in terms of magnitude) on the seesaw? Ignore the mass of the seesaw. (A) Henry (B) Lucy← CORRECT (C) Mary (D) They all exert the same torque. (E) There is not enough information to answer the question.


2009 F = ma Exam

6

14. A wooden block (mass M ) is hung from a peg by a massless rope. A speeding bullet (with mass m and initial speed v0 ) collides with the block at time t = 0 and embeds in it. Let S be the system consisting of the block and bullet. Which quantities are conserved between t = −10 s and t = +10 s?

(A) The total linear momentum of S. (B) The horizontal component of the linear momentum of S. (C) The mechanical energy of S. (D) The angular momentum of S as measured about a perpendicular axis through the peg. (E) None of the above are conserved.← CORRECT 15. A 22.0 kg suitcase is dragged in a straight line at a constant speed of 1.10 m/s across a level airport floor by a student on the way to the 40th IPhO in Merida, Mexico. The individual pulls with a 1.00 x 102 N force along a handle with makes an upward angle of 30.0 degrees with respect to the horizontal. What is the coefficient of kinetic friction between the suitcase and the floor? (A) µk = 0.013 (B) µk = 0.394 (C) µk = 0.509← CORRECT (D) µk = 0.866 (E) µk = 1.055 16. Two identical objects of mass m are placed at either end of a spring of spring constant k and the whole system is placed on a horizontal frictionless surface. At what angular frequency ω does the system oscillate? p k/m (A) p (B) 2k/m← CORRECT p (C) k/2m p (D) 2 k/m p (E) k/m/2 17. You are given a standard kilogram mass and a tuning fork that is calibrated in Hz. You are also provided with a complete collection of laboratory equipment, but none of it is calibrated in SI units. You do not know the values of any fundamental constants. Which of the following quantities could you measure in SI units? (A) The acceleration due to gravity. (B) The speed of light in a vacuum. (C) The density of room temperature water. (D) The spring constant of a given spring.← CORRECT (E) The air pressure in the room. c Copyright 2009 American Association of Physics Teachers

2009 F = ma Exam

7

18. A simple pendulum of length L is constructed from a point object of mass m suspended by a massless string attached to a fixed pivot point. A small peg is placed a distance 2L/3 directly below the fixed pivot point so that the pendulum would swing as shown in the figure below. The mass is displaced 5 degrees from the vertical and released. How long does it take to return to its starting position?

Fixed Pivot Point

L Small Peg

Point Object of mass m q q L 2 1 + g 3 q π Lg 2 + √23 q π Lg 1 + 31 q √ π Lg 1 + 3 q π Lg 1 + √13 ← CORRECT

(A) π (B) (C) (D) (E)

19. A certain football quarterback can throw a football a maximum range of 80 meters on level ground. What is the highest point reached by the football if thrown this maximum range? Ignore air friction. (A) 10 m (B) 20 m ← CORRECT

(C) 30 m

(D) 40 m (E) 50 m 20. Consider a completely inelastic collision between two lumps of space goo. Lump 1 has mass m and originally moves directly north with a speed v0 . Lump 2 has mass 3m and originally moves directly east with speed v0 /2. What is the final speed of the masses after the collision? Ignore gravity, and assume the two lumps stick together after the collision. (A) 7/16 v0 √ (B) 5/8 v0 √ (C) 13/8 v0 ← CORRECT

(D) 5/8 v0 p (E) 13/8 v0


2009 F = ma Exam

8

The following information is used for questions 21 and 22. Two stars orbit their common center of mass as shown in the diagram below. The masses of the two stars are 3M and M . The distance between the stars is d.

21. What is the value of the gravitational potential energy of the two star system? (A) − GM d (B)

2

3GM 2 d

(C) − GM d2

2

2

(D) − 3GM d ← CORRECT (E) − 3GM d2

2

22. Determine the period of orbit for the star of mass 3M . q d3 (A) π GM ← CORRECT q d3 (B) 3π 4 GM q d3 (C) π 3GM q d3 (D) 2π GM q d3 (E) π4 GM 23. A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T . At what time is the power delivered to the mass by the spring first a maximum? (A) t = 0 (B) t = T /8 (C) t = T /4 (D) t = 3T /8 ← CORRECT (E) t = T /2


2009 F = ma Exam

9

24. A uniform rectangular wood block of mass M , with length b and height a, rests on an incline as shown. The incline and the wood block have a coefficient of static friction, µs . The incline is moved upwards from an angle of zero through an angle θ. At some critical angle the block will either tip over or slip down the plane. Determine the relationship between a, b, and µs such that the block will tip over (and not slip) at the critical angle. The box is rectangular, and a 6= b. b

a

θ

(A) µs > a/b (B) µs > 1 − a/b

(C) µs > b/a← CORRECT (D) µs < a/b (E) µs < b/a − 1 25. Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii r1 and r2 . The discs are given angular velocities of magnitudes ω1 and ω2 , respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? Ignore effects associated with the vertical rods.

r1

r2

(A) ω1 2 r1 = ω2 2 r2 (B) ω1 r1 = ω2 r2 (C) ω1 r1 2 = ω2 r2 2 (D) ω1 r1 3 = ω2 r2 3 ← CORRECT (E) ω1 r1 4 = ω2 r2 4


United States Physics Team Quarter Final Contest 2009


AAPT AIP

1


Quarterfinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of one part. • The first page that follows is a cover sheet. Examinees may keep the cover sheet during the exam. • Allow 60 minutes to complete the exam. Examinees may read the cover sheet before beginning the exam, but may not look at the examination questions until the 60 minute time period begins. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 8, 2009. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • The examinees will need to use a ruler for one of the questions on this exam. They may not share rulers with other examinees. • AAPT must receive the students answer papers no later than Thursday, March 5, 2009. Marking of papers will occur that weekend, and the semifinalists will be selected by March 8, 2009. It will not be possible to mark late papers.



AAPT AIP

2


2009 Quarter-Final Exam 4 QUESTIONS - 60 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Show all your work. Partial credit will be given. • Start each question on a new sheet of paper. Put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie Prob. 1 - P. 1/3 • A ruler will be required on this exam. • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. • Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty. Good luck! • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after March 10, 2009.



3

1. Below is an image of Fomalhaut b, the first extrasolar planet to be observed directly by visible light, obtained by the Hubble Space Telescope.

Astronomy Picture of the Day 2008 November 14 http://antwrp.gsfc.nasa.gov/apod/ap081114.html Image Credit: NASA, ESA, P. Kalas, J. Graham, E. Chiang, E. Kite (Univ. California, Berkeley), M. Clampin (NASA/Goddard), M. Fitzgerald (Lawrence Livermore NL), K. Stapelfeldt, J. Krist (NASA/JPL)

The scale of the larger diagram is shown on the lower left; the 13” refers to the angle, in arc-second, subtended by 100 AU at the distance of Fomalhaut. The scale of the inset can be determined by size of the small box. (a) A planet is in a circular orbit of radius R and period T . Derive an expression for the mass of a star in terms of R, T , and the gravitational constant G. You may assume that the mass of the planet is much, much less than the mass of the star. (b) From the image, estimate the mass of the star Fomalhaut in solar masses. You may assume that the orbit of Fomalhaut b is circular and ignore any errors associated with the fact that the plane of the image is not coincident with the plane of the orbit. Recall that the radius of the Earth’s orbit around the Sun is 1 AU. You do not need to do an error analysis, but the number of significant digits in your answer ought reflect the accuracy of your answer.



4

Solution The force between the star of mass M and the planet of mass m is given by F =

GM m R2

Assuming that M m, then the center of mass of the system is located at the star, and therefore the orbital radius of the planet is effectively R. For a circular orbit, F =m

v2 (2πR/T )2 R =m = 4π 2 m 2 . R R T

Equating, GM 4π 2 R = 2 , 2 T R or the more familiar Kepler’s law GM R3 = 2 T 4π 2 Rearranging, M = 4π 2

GR3 T2

To find the mass of Fomalhaut b, Mb in terms of solar masses Ms , we simply require a ratio: Rb 3 /Tb 2 Mb = , Ms Re 3 /Te 2 where b refers to the planet around Fomalhaut b, while e refers to Earth. Then Mb = Ms

Rb Re

3

Te Tb

2 .

From the diagram we can obtain the ratios. Note that the 13 arcsecond notation is not needed for this computation, it simply is a comparison of the size of the object as seen from Earth. As such, seen from Earth, the orbital dust ring subtends about the same angle as does the planet Jupiter! A quick estimate yields Rb /Re ≈ 110, while slightly more work is required for the periods. The inset box is 10 times the scale of the main image. The planet moves 1.4 AU during the two year period. Since the orbital circumference is 690 AU, that means it takes 980 years to complete an orbit. Then Tb /Te = 980, and Mb /Ms ≈ 1.4. Allowing for 10% error in each measurement, the results would be between 0.87 and 2.2. Finally, after marking the quarter-final answers, it was mildly entertaining to consider the range of values obtained, from a minimum of 10−19 (about 1011 kg, or the mass of a pile of rocks some 300 meters tall) to a maximum of 1025 (About 1000 times greater than the mass of the universe!).



5

2. A ball of mass m is thrown vertically upward with a speed of v0 . The ball is subject to an air resistance force that is proportional to the velocity, F = −kv. The ball rises up to height of h and then returns to the starting point after some total time tf . The acceleration of free fall is g. Determine the following: (a) An expression for the velocity as a function of time in terms of any or all of the constants. (b) Sketch the graph of velocity vs time. On the sketch indicate the time to rise to the highest point, tr , and the total time of flight, tf . You do not need to find either tr or tf at this point, but your graph should reflect relative positions of both. (c) Find the time to rise, tr , to the highest point, h, in terms of any or all of the constants.

Solution Two forces act on the ball: gravity, and air friction. They are not necessarily equal, so the net force, and hence an expression for the acceleration, would be given by ma = −mg − kv where up is positive. We can’t simply rearrange this to solve for v, since a is a variable. We can, however, rearrange to prepare for integration, since a = dv/dt: dv k =− g+ v dt m or

dv = −dt. k g+m v

Integrating both sides m ln k Assume t0 is zero. Then

mg/k + v mg/k + v0

= −(t − t0 ). k

v = (mg/k + v0 )e− m t − mg/k Note that as t → ∞ the ball will approach a constant speed of vt = mg/k. Of course, it will probably hit the ground first, but if it were thrown up in the air over the edge of an infinitely deep cliff.... The ball will rise until the velocity is zero, or when vt vt = tr . − ln g vt + v0 or, if you prefer, v0 vt tr = ln 1 + g vt To find the height we need to integrate one more time. Since v = dx/dt we can write −gt dx = (vt + v0 )e vt − vt dt c Copyright 2009 American Association of Physics Teachers


6

which integrates quickly to

vt h = (vt + v0 ) − g

− vg tr

e

t

− 1 − (vt tr )

Gluing together with the expression for tr , v0 vt v0 vt 2 − ln 1 + . h= g g vt Showing that it reduces to h = v02 /2g in the limit as k → 0 is left as an exercise for the reader. 3. A parallel plate √ capacitor is made of two square parallel plates of area A, and separated by a distance d A. The capacitor is connected to a battery with potential V and allowed to fully charge. The battery is then disconnected. A square metal conducting slab also with area A but thickness d/2 is then fully inserted between the plates, so that it is always parallel to the plates. How much work has been done on the metal slab while it is being inserted?

Solution The capacitance of a thin rectangular capacitor is given by C = 0 A/d If you forgot this, then derive it. The electric field between the plates is given by E=

Q 0 Ad

so the potential difference between the plates must be V = Ed =

Q 0 A

and the capacitance is C = Q/V = 0 A/d For the original capacitor we’ll call this C0 . The energy stored in a capacitor is U = 12 QV = 12 CV0 2 , so once charged, the initial energy is U0 =

1 0 A 2 V 2 d 0

Note, we are going to need to know the initial charge sooner or later, which is Q0 = C0 V0 =

0 AV0 d

Inserting the conducting plate can be treated two ways: it results in the creation of two capacitors, in series, each with area A but differing separation. The net charge Q0 won’t change, that’s why we needed it above, but the resulting capacitance of the system will be 1/Ce = 1/C1 + 1/C2 c Copyright 2009 American Association of Physics Teachers


7

For our capacitors, the only thing that changes is d, which we’ll call x1 and x2 . Then 1/Ce = x1 /0 A + x2 /0 A which gives us Ce =

0 A = 2C0 d/2

Since the new total separation x1 + x2 is just d − d/2. The energy in the system after inserting the plate is 1 1 U = Q0 V = Q20 /Ce 2 2 since V is not constant! Finally, 1 1 U = Q20 /2C0 = U0 2 2 The work done on the plate is the same as the work done on the system, which is given by W = U − U0 or

1 0 A 2 V . 4 d 0 This means that the plate is “sucked” into the original capacitor, and we have to fight to hold it out. W =−

4. A certain electric battery is not quite ideal. It can be thought of as a perfect cell with constant output voltage V0 connected in series to a resistance r, but there is no way to remove this internal resistance from the battery. R Battery V0

r

N parallel bulbs (4 are shown)

The battery is connected to N identical lightbulbs in parallel. The bulbs each have a fixed resistance R, independent of the current through them. (a) Derive an expression for the total power dissipated by the N bulbs in terms of r, R, and V0 . c Copyright 2009 American Association of Physics Teachers


8

(b) It is observed that the configuration dissipates more total power through the bulbs when N = 5 than it does for any other value of N . In terms of R, what range of values is possible for r?

Solution The effective resistance of N identical resistors in parallel is Re = R/N The total resistance of the circuit is the r + R/N . The current drawn is then I=

V0 . r + R/N

The power dissipated by the parallel resistance is given by P = I 2 Re , so P =

R V0 2 V0 2 = N R (r + R/N )2 N (N r + R)2

One can simply start with the fact that the maximum power is dissipated when the external resistance is equal to the internal resistance. Or, take the derivative of P with respect to R and set it equal to zero. In either case, you will get r = Re =

R N

Don’t assume, however, that the answer is then r = R/5. Since we can’t have half a bulb, we don’t actually know that the power is a maximum for the circuit, only that it is the maximum configuration. It could be that a larger power output could occur with a resistance between R/4 and R/5, or maybe between R/5 and R/6. A rough guess would be to assume R R ≤r≤ 5.5 4.5 A better estimate would be to assume a quadratic function of 1/N with respect to P in the vicinity of Pmax (think Taylor expansion of P ), and then the important point is the halfway point between 1/N and 1/(N ± 1), or 1 1 N ± 1/2 1/N + 1/(N ± 1) = + = 2 2N 2(N ± 1) N (N ± 1) So the bounds are

4.5R 5.5R ≤r≤ 30 20 The correct approach is to assume the power outputs at N and N ± 1 are equal, and solve for the necessary value of r. Then N N ±1 = 2 (N r + R) (N r ± r + R)2 c Copyright 2009 American Association of Physics Teachers


9

or N 3 r2 + 2N 2 (R ± r)r + N (R ± r)2 = N 3 r2 + 2N 2 Rr + N R2 ± N 2 r2 ± 2N Rr ± R2 Thankfully, we loose the cubic term ±2N 2 r2 ± 2N Rr + N r2 = ±N 2 r2 ± 2N Rr ± R2 . To save space, flip the ± to the one term that doesn’t have it. N 2 r 2 ± N r 2 = R2 Only positive answers have meaning, so 1 R r=p N (N ± 1) which give the correct bounds as R R √ ≤r≤ √ 30 20 If you knew about geometric means, you might have been able to just write this down.



2009 Semifinal Exam

1

AAPT AIP


Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2009. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A.


2009 Semifinal Exam

AAPT AIP

Cover Sheet

2


Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your school ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, School ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2009. Possibly Useful Information. You may g = 9.8 N/kg k = 1/4π0 = 8.99 × 109 N · m2 /C2 c = 3.00 × 108 m/s NA = 6.02 × 1023 (mol)−1 σ = 5.67 × 10−8 J/(s · m2 · K4 ) 1eV = 1.602 × 10−19 J me = 9.109 × 10−31 kg = 0.511 MeV/c2 sin θ ≈ θ − 16 θ3 for |θ| 1

use this sheet for both parts of the exam. G = 6.67 × 10−11 N · m2 /kg2 km = µ0 /4π = 10−7 T · m/A kB = 1.38 × 10−23 J/K R = NA kB = 8.31 J/(mol · K) e = 1.602 × 10−19 C h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s (1 + x)n ≈ 1 + nx for |x| 1 cos θ ≈ 1 − 21 θ2 for |θ| 1


2009 Semifinal Exam

Part A

3

Part A Question A1 A hollow cylinder has length l, radius r, and thickness d, where l r d, and is made of a material with resistivity ρ. A time-varying current I flows through the cylinder in the tangential direction. Assume the current is always uniformly distributed along the length of the cylinder. The cylinder is fixed so that it cannot move; assume that there are no externally generated magnetic fields during the time considered for the problems below.

I r

l

a. What is the magnetic field strength B inside the cylinder in terms of I, the dimensions of the cylinder, and fundamental constants? b. Relate the emf E developed along the circumference of the cylinder to the rate of change of the current dI dt , the dimensions of the cylinder, and fundamental constants. c. Relate E to the current I, the resistivity ρ, and the dimensions of the cylinder. d. The current at t = 0 is I0 . What is the current I(t) for t > 0?

Solution The magnetic field through the inside of the cylinder is given by B = µ0 I/l so the magnetic flux is ΦB = BA = πµ0 r2 I/l The inductance is then L = ΦB /I = πµ0 r2 /l Induced emf as a function of changing current is then E = −L

dI πµ0 r2 dI =− dt l dt

But that induced emf will be driving the current, so E = IR


2009 Semifinal Exam

Part A

4

where R is the resistance, given by Length Area Here the legnth is the circumference, 2πr, while the area is the cross sectional area of the conductor, ld. Therefore, 2πr E = Iρ ld Combining the above, we get a differential equation, R=ρ

Iρ

2πr πµ0 r2 dI =− ld l dt

which can be written more simply as −αI = where

dI dt

2ρ µ0 rd

α= and the solution is then

I(t) = I(0)e−αt

Question A2 A mixture of 32 P and 35 S (two beta emitters widely used in biochemical research) is placed next to a detector and allowed to decay, resulting in the data below. The detector has equal sensitivity to the beta particles emitted by each isotope, and both isotopes decay into stable daughters. You should analyze the data graphically. Error estimates are not required. Day 0 5 10 20 30

Activity 64557 51714 41444 27020 18003

Day 40 60 80 100 150

a. Determine the half-life of each isotope. b. Determine the ratio of the number of sample.

Activity 12441 6385 3855 2734 1626

35 S

32 P

Day 200 250 300

Activity 1121 673 467

has a significantly longer half-life than

atoms to the number of

35 S

32 P.

atoms in the original

Question A3 Two stars, each of mass M and separated by a distance d, orbit about their center of mass. A planetoid of mass m (m M ) moves along the axis of this system perpendicular to the orbital plane.


2009 Semifinal Exam

Part A

5

axis perpendicular to plane of orbit

z

d

Let Tp be the period of simple harmonic motion for the planetoid for small displacements from the center of mass along the z-axis, and let Ts be the period of motion for the two stars. Determine the ratio Tp /Ts . This problem was adapted from a problem by French in Newtonian Mechanics.

Solution This problem starts out similar to the quarterfinal exam. The center of mass of the two stars is given by R = d/2. The force of attraction between the two stars is given by F = GM 2 /d2 The orbital period T will be given by the centripetal force law 4π 2 R M2 = G T2 d2 or r r d3 R3 Ts = 2π = 4π 2GM GM √ The planetoid is a distance z above the plane. The distance to either star is then R2 + z 2 . Only the part of the force that is perpendicular to the plane survives the vector addition, so the net force is given by GmM z √ F =2 2 . 2 2 R +z R + z2 This expression is exact. But if z d, then we can approximate this as F =M

GmM z R3 This is a restoring force; the planetoid will execute simple harmonic oscillations with period r m Tp = 2π k F ≈2

where k is the effective force constant, or r Tp = 2π

R3 = 2π GM

r

d3 8GM

so Tp /Ts = 1/2 c Copyright 2009 American Association of Physics Teachers

2009 Semifinal Exam

Part A

6

Question A4 A potato gun fires a potato horizontally down a half-open cylinder of cross-sectional area A. When the gun is fired, the potato slug is at rest, the volume between the end of the cylinder and the potato is V0 , and the pressure of the gas in this volume is P0 . The atmospheric pressure is Patm , where P0 > Patm . The gas in the cylinder is diatomic; this means that Cv = 5R/2 and Cp = 7R/2. The potato moves down the cylinder quickly enough that no heat is transferred to the gas. Friction between the potato and the barrel is negligible and no gas leaks around the potato. potato

L

closed end

open end

The parameters P0 , Patm , V0 , and A are fixed, but the overall length L of the barrel may be varied. a. What is the maximum kinetic energy Emax with which the potato can exit the barrel? Express your answer in terms of P0 , Patm , and V0 . b. What is the length L in this case? Express your answer in terms of P0 , Patm , V0 , and A.

Solution So long as the pressure inside the cylinder is greater than the external air pressure, the potato will accelerate. Therefore, maximum energy will be transferred to the potato if the cylinder is exactly long enough so that the final pressure inside the volume of the cylinder is Patm . The energy of an ideal diatomic gas is given by Cv nRT but, by the ideal gas law, P V = nRT so the energy in the gas is Cv P V for any volume and pressure. Maximum energy is delivered to the potato when the final pressure is atmospheric, so the work done by the gas on the potato is Cv (P0 V0 − Patm Vf ) But the potato is moving against air, so the actual energy given to the potato is Emax = Cv (P0 V0 − Patm Vf ) − Patm (Vf − V0 ) The relationship determining Vf is that of adiabatic expansion, Vf = V0

P0 Patm

1/γ ,


2009 Semifinal Exam

Part A

7

where γ = Cp /Cv . Things don’t simplify much, unless we use the numerical values of Cv and γ. Then 5 7 Emax = P0 V0 − Patm Vf + Patm V0 , 2 2 or 5 7 2/7 5/7 V0 Emax = P0 + Patm − Patm P0 2 2 The length of the tube is simply V0 L = Vf /A = A

P0 Patm

5/7 .


2009 Semifinal Exam

Part A

8

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.


2009 Semifinal Exam

Part B

9

Part B Question B1 A bowling ball and a golf ball are dropped together onto a flat surface from a height h. The bowling ball is much more massive than the golf ball, and both have radii much less than h. The bowling ball collides with the surface and immediately thereafter with the golf ball; the balls are dropped so that all motion is vertical before the second collision, and the golf ball hits the bowling ball at an angle α from its uppermost point, as shown in the diagram. All collisions are perfectly elastic, and there is no surface friction between the bowling ball and the golf ball.

α

h

l After the collision the golf ball travels in the absence of air resistance and lands a distance l away. The height h is fixed, but α may be varied. What is the maximum possible value of l, and at what angle α is it achieved? You may present your results as decimals, but remember that you are not allowed to use graphical or algebraic functions of your calculator.

Solution Both balls arrive at the surface with a speed v0 , which we can determine as usual via conservation of energy: 1 mv 2 = mgh 2 0 p v0 = 2gh After the bowling ball collides with the surface, it is traveling upwards at v0 and the golf ball is traveling downwards at v0 . The subsequent collision is most easily understood in the reference c Copyright 2009 American Association of Physics Teachers

2009 Semifinal Exam

Part B

10

frame of the bowling ball; in this frame, the golf ball is traveling downwards at 2v0 and the bowling ball is stationary. Since the bowling ball is very massive compared to the golf ball, the golf ball will rebound at the same speed 2v0 . Since there is no surface friction, the angle of reflection will be equal to the angle of incidence. Therefore the golf ball will emerge at an angle 2α to the vertical. In the bowling ball’s reference frame, then, the golf ball emerges with horizontal velocity 2v0 sin 2α and upward vertical velocity 2v0 cos 2α. Transforming these back to the original frame of reference, we obtain the initial conditions for the golf ball’s projectile motion: vx = 2v0 sin 2α vy = 2v0 cos 2α + v0 Meanwhile, the time of flight t of the golf ball is given by 1 vy t − gt2 = 0 2 t=

2vy g

and therefore the range is l = vx t 2vx vy l= g (This is a well-known result, and students may quote it directly.) Combining with our previous result, l=

2 (2v0 sin 2α)(2v0 cos 2α + v0 ) g

l=

1 8v0 2 sin 2α (cos 2α + ) g 2

and inserting the result for v0 ,

1 l = 16h sin 2α (cos 2α + ) 2 As a convenience set β = 2α. Then 1 l = 16h sin β (cos β + ) 2 dl 1 2 = 16h cos β (cos β + ) − sin β dβ 2

Setting this to zero to find the maximum, cos2 β − sin2 β + 2 cos2 β +

cos β =

1 cos β = 0 2

1 cos β − 1 = 0 2 q − 12 ± 14 − 8 4


2009 Semifinal Exam

Part B

Taking the positive root,

√ cos β =

33 − 1 = 0.593 8

and sin β = So at maximum

p 1 − cos2 β = 0.805

1 l = 16h · 0.805 · (0.593 + ) 2 l = 14.08 h

This occurs at cos 2α = 0.593 α = 0.468 = 26.8◦


11

2009 Semifinal Exam

Part B

12

Question B2 An electric dipole consists of two charges of equal magnitude q and opposite sign, held rigidly apart by a distance d. The dipole moment is defined by p = qd. Now consider two identical, oppositely oriented electric dipoles, separated by a distance r, as shown in the diagram.

A

B

d

d r

a. It is convenient when considering the interaction between the dipoles to choose the zero of potential energy such that the potential energy is zero when the dipoles are very far apart from each other. Using this convention, write an exact expression for the potential energy of this arrangement in terms of q, d, r, and fundamental constants. b. Assume that d r. Give an approximation of your expression for the potential energy to lowest order in d. Rewrite this approximation in terms of only p, r, and fundamental constants. c. What is the force (magnitude and direction) exerted on one dipole by the other? Continue to make the assumption that d r, and again express your result in terms of only p, r, and fundamental constants. d. What is the electric field near dipole B produced by dipole A? Continue to make the assumption that d r and express your result in terms of only p, r, and fundamental constants.

Solution a. There are two +q - −q pairs separated by a distance d, each having potential energy −

q2 4π0 d

There are two +q - −q pairs separated by a distance r, each having potential energy −

q2 4π0 r

√ There are a +q - +q pair and a −q - −q pair separated by a distance r2 + d2 , each having potential energy q2 √ 4π0 r2 + d2 Note that the latter two terms go to zero as r becomes large, whereas the first term is not dependent on r. Thus the given zero convention will include only the latter two terms: q2 −2 2 U= +√ 4π0 r r2 + d2 c Copyright 2009 American Association of Physics Teachers

2009 Semifinal Exam

b. We have

Part B

13

 2q 2  1 q U= 4π0 r 1+

 d 2 r

− 1

Using the binomial approximation (1 + x)n ≈ 1 + nx, 2q 2 U≈ 4π0 r

1 1− 2

! 2 d −1 r

U ≈−

q 2 d2 4π0 r3

U ≈−

p2 4π0 r3

or, in terms of p,

c. We can infer by symmetry that the force must be in the direction along the line separating the dipoles. Since the potential energy decreases with decreasing distance, the force is attractive. Its magnitude can be determined by taking the derivative of the potential energy: F =−

dU p2 = −3 dr 4π0 r4

with the negative sign confirming that the force is attractive. One can, of course, also use an approach analogous to the previous one, i.e. write down the force exactly and use a binomial approximation as above. One must take care to account for the fact that the force between like-signed charges is not exactly in the same direction as that between opposite-signed charges.


United States Physics Team F = ma Contest 2010

2010 F = ma Exam

1

AAPT AIP

UNITED STATES PHYSICS TEAM 2010 2010 F = ma Contest

Entia non multiplicanda sunt praeter necessitatem

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25 QUESTIONS - 75 MINUTES INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Use g = 10 N/kg throughout this contest.

• You may write in this booklet of questions. However, you will not receive any credit for anything written in this booklet. • Your answer to each question must be marked on the optical mark answer sheet.

de

• Select the single answer that provides the best response to each question. Please be sure to use a No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer, the previous mark must be completely erased. 1 4

point. There is

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• Correct answers will be awarded one point; incorrect answers will result in a deduction of no penalty for leaving an answer blank.

• A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.

Co

• This test contains 25 multiple choice questions. Your answer to each question must be marked on the optical mark answer sheet that accompanies the test. Only the boxes preceded by numbers 1 through 25 are to be used on the answer sheet. • All questions are equally weighted, but are not necessarily the same level of difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers or solutions) on this contest until after February 8, 2010. • The question booklet and answer sheet will be collected at the end of this exam. You may not use scratch paper.



2010 F = ma Exam

2

Questions 1 to 3 refer to the figure below which shows a representation of the motion of a squirrel as it runs in a straight-line along a telephone wire. The letters A through E refer to the indicated times.

time

A

B

C

D

E

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1. If the graph is a graph of POSITION vs. TIME, then the squirrel has the greatest speed at what time(s) or during what time interval(s)? (A) From A to B (B) From B to C only (C) From B to D← CORRECT (D) From C to D only (E) From D to E

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2. If, instead, the graph is a graph of VELOCITY vs. TIME, then the squirrel has the greatest speed at what time(s) or during what time interval(s)? (A) at B (B) at C (C) at D

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(D) at both B and D ← CORRECT (E) From C to D

3. If, instead, the graph is a graph of ACCELERATION vs. TIME and the squirrel starts from rest, then the squirrel has the greatest speed at what time(s) or during what time interval? (A) at B

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(B) at C ← CORRECT (C) at D

(D) at both B and D (E) From C to D

4. Two teams of movers are lowering a piano from the window of a 10 floor apartment building. The rope breaks when the piano is 30 meters above the ground. The movers on the ground, alerted by the shouts of the movers above, first notice the piano when it is 14 meters above the ground. How long do they have to get out of the way before the piano hits the ground? (A) 0.66 sec← CORRECT (B) 0.78 sec (C) 1.67 sec (D) 1.79 sec (E) 2.45 sec


2010 F = ma Exam

3

5. Two projectiles are launched from a 35 meter ledge as shown in the diagram. One is launched from a 37 degree angle above the horizontal and the other is launched from 37 degrees below the horizontal. Both of the launches are given the same initial speed of v0 = 50 m/s. Projectile 1

35 m

Projectile 2

nt ial

The difference in the times of flight for these two projectiles, t1 − t2 , is closest to (A) 3 s (B) 5 s (C) 6 s ← CORRECT (D) 8 s (E) 10 s

6. A projectile is launched across flat ground at an angle θ to the horizontal and travels in the absence of air resistance. It rises to a maximum height H and lands a horizontal distance R away. What is the ratio H/R?

(B) 2 tan θ (D) (E)

2 tan θ 1 2 tan θ 1 4 tan θ←

CORRECT

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(C)

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(A) tan θ

7. Harry Potter is sitting 2.0 meters from the center of a merry-go-round when Draco Malfoy casts a spell that glues Harry in place and then makes the merry-go-round start spinning on its axis. Harry has a mass of 50.0 kg and can withstand 5.0 g’s of acceleration before passing out. What is the magnitude of Harry’s angular momentum when he passes out? (A) 200 kg·m2 /s

Co

(B) 330 kg·m2 /s (C) 660 kg·m2 /s

(D) 1000 kg·m2 /s ← CORRECT (E) 2200 kg·m2 /s

8. A car attempts to accelerate up a hill at an angle θ to the horizontal. The coefficient of static friction between the tires and the hill is µ > tan θ. What is the maximum acceleration the car can achieve (in the direction upwards along the hill)? Neglect the rotational inertia of the wheels. (A) g tan θ (B) g(µ cos θ − sin θ) ← CORRECT (C) g(µ − sin θ) (D) gµ cos θ (E) g(µ sin θ − cos θ)


2010 F = ma Exam

4

9. A point object of mass M hangs from the ceiling of a car from a massless string of length L. It is observed to make an angle θ from the vertical as the car accelerates uniformly from rest. Find the acceleration of the car in terms of θ, M , L, and g.

L

θ

(A) M g sin θ (C) g tan θ ← CORRECT (D) g cot θ (E) M g tan θ

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(B) M gL tan θ

10. A block of mass m1 is on top of a block of mass m2 . The lower block is on a horizontal surface, and a rope can pull horizontally on the lower block. The coefficient of kinetic friction for all surfaces is µ. What is the resulting acceleration of the lower block if a force F is applied to the rope? Assume that F is sufficiently large so that the top block slips on the lower block. 1

de

F

2

(A) a2 = (F − µg(2m1 + m2 ))/m2 ← CORRECT (B) a2 = (F − µg(m1 + m2 ))/m2

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(C) a2 = (F − µg(m1 + 2m2 ))/m2 (D) a2 = (F + µg(m1 + m2 ))/m2 (E) a2 = (F − µg(m2 − m1 ))/m2

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11. The three masses shown in the accompanying diagram are equal. The pulleys are small, the string is lightweight, and friction is negligible. Assuming the system is in equilibrium, what is the ratio a/b? The figure is not drawn to scale! a b

(A) 1/2 (B) 1 √ (C) 3 (D) 2 √ (E) 2 3← CORRECT


2010 F = ma Exam

5

12. A ball with mass m projected horizontally off the end of a table with an initial kinetic energy K. At a time t after it leaves the end of the table it has kinetic energy 3K. What is t? Neglect air resistance. p (A) (3/g) K/m p (B) (2/g) K/m ← CORRECT p (C) (1/g) 8K/m p (D) (K/g) 6/m p (E) (2K/g) 1/m

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13. A ball of mass M and radius R has a moment of inertia of I = 25 M R2 . The ball is released from rest and rolls down the ramp with no frictional loss of energy. The ball is projected vertically upward off a ramp as shown in the diagram, reaching a maximum height ymax above the point where it leaves the ramp. Determine the maximum height of the projectile ymax in terms of h.

h

(C) (D) (E)

25 49 h 2 5h 5 7 h← 7 5h

CORRECT

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(B)

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(A) h

14. A 5.0 kg block with a speed of 8.0 m/s travels 2.0 m along a horizontal surface where it makes a head-on, perfectly elastic collision with a 15.0 kg block which is at rest. The coefficient of kinetic friction between both blocks and the surface is 0.35. How far does the 15.0 kg block travel before coming to rest? (A) 0.76 m

(B) 1.79 m← CORRECT

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(C) 2.29 m (D) 3.04 m (E) 9.14 m


2010 F = ma Exam

6

The following figure is used for questions 15 and 16.

m

M

v0

A small block of mass m is moving on a horizontal table surface at initial speed v0 . It then moves smoothly onto a sloped big block of mass M . The big block can also move on the table surface. Assume that everything moves without friction.

(A) h =

v02 2g

(B) h =

2 1 M v0 g m+M

(C) h =

2 1 M v0 2g m+M

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15. A small block moving with initial speed v0 moves smoothly onto a sloped big block of mass M . After the small block reaches the height h on the slope, it slides down. Find the height h.

← CORRECT

mv02

(D) h =

1 2g m+M

(E) h =

v02 g

(A) v = v0 m m+M v0

(C) v =

M m+M v0

(D) v =

M −m m v0 M −m m+M v0 ←

(E) v =

CORRECT

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(B) v =

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16. Following the previous set up, find the speed v of the small block after it leaves the slope.

17. Four masses m are arranged at the vertices of a tetrahedron of side length a. What is the gravitational potential energy of this arrangement? 2

(B) −3 Gm a

2

(C) −4 Gm a

2

Co

(A) −2 Gm a

2

(D) −6 Gm ← CORRECT a (E) −12 Gm a

2


2010 F = ma Exam

7

nt ial

The following graph of potential energy is used for questions 18 through 20.

(B)

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(A)

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18. Which of the following represents the force corresponding to the given potential?

(C)

(D)

(E)← CORRECT


2010 F = ma Exam

8

19. Consider the following graphs of position vs. time.

I.

II.

III.

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Which of the graphs could be the motion of a particle in the given potential? (A) I (B) III (C) I and II (D) I and III← CORRECT (E) I, II, and III

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20. Consider the following graph of position vs. time, which represents the motion of a certain particle in the given potential.

What is the total energy of the particle? (A) -5 J← CORRECT (B) 0 J (C) 5 J

(D) 10 J (E) 15 J


2010 F = ma Exam

9

21. The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self potential energy of a ball of mass density ρ and radius 2R? (A) 2E (B) 4E (C) 8E (D) 16E (E) 32E ← CORRECT

A B

D

de

C

(A) A (B) B (C) C

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(D) D ← CORRECT (E) Remains vertical

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22. A balloon filled with helium gas is tied by a light string to the floor of a car; the car is sealed so that the motion of the car does not cause air from outside to affect the balloon. If the car is traveling with constant speed along a circular path, in what direction will the balloon on the string lean towards?

23. Two streams of water flow through the U-shaped tubes shown. The tube on the left has cross-sectional area A, and the speed of the water flowing through it is v; the tube on the right has cross-sectional area A0 = 1/2A. If the net force on the tube assembly is zero, what must be the speed v 0 of the water flowing through the tube on the right?

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Neglect gravity, and assume that the speed of the water in each tube is the same upon entry and exit.

(A) 1/2v

(B) v √ (C) 2v ← CORRECT (D) 2v (E) 4v


2010 F = ma Exam

10

24. A uniform circular disk of radius R begins with a mass M ; about an axis through the center of the disk and perpendicular to the plane of the disk the moment of inertia is I0 = 21 M R2 . A hole is cut in the disk as shown in the diagram. In terms of the radius R and the mass M of the original disk, what is the moment of inertia of the resulting object about the axis shown?

R/2

R

R

axis of rotation

(A) (15/32)M R2

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(B) (13/32)M R2 ← CORRECT (C) (3/8)M R2 (D) (9/32)M R2 (E) (15/16)M R2

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25. Spaceman Fred’s spaceship (which has negligible mass) is in an elliptical orbit about Planet Bob. The minimum distance between the spaceship and the planet is R; the maximum distance between the spaceship and the planet is 2R. At the point of maximum distance, Spaceman Fred is traveling at speed v0 . He then fires his thrusters so that he enters a circular orbit of radius 2R. What is his new speed?

2R

Co

R

p 3/2v0 ← CORRECT √ (B) 5v0 p (C) 3/5v0 √ (D) 2v0 (A)

(E) 2v0



2010 Semifinal Exam

1

AAPT AIP


Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2010. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.


2010 Semifinal Exam

AAPT AIP

Cover Sheet

2


Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your school ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, School ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2010. Possibly Useful Information. You may g = 9.8 N/kg k = 1/4π0 = 8.99 × 109 N · m2 /C2 c = 3.00 × 108 m/s NA = 6.02 × 1023 (mol)−1 σ = 5.67 × 10−8 J/(s · m2 · K4 ) 1eV = 1.602 × 10−19 J me = 9.109 × 10−31 kg = 0.511 MeV/c2 sin θ ≈ θ − 16 θ3 for |θ| 1



2010 Semifinal Exam

Part A

3

Part A Question A1 An object of mass m is sitting at the northernmost edge of a stationary merry-go-round of radius R. The merry-go-round begins rotating clockwise (as seen from above) with constant angular acceleration of α. The coefficient of static friction between the object and the merry-go-round is µs . a. Derive an expression for the magnitude of the object’s velocity at the instant when it slides off the merry-go-round in terms of µs , R, α, and any necessary fundamental constants. b. For this problem assume that µs = 0.5, α = 0.2 rad/s2 , and R = 4 m. At what angle, as measured clockwise from north, is the direction of the object’s velocity at the instant when it slides off the merry-go-round? Report your answer to the nearest degree in the range 0 to 360◦ .

Solution The object will begin to slide when the force required to keep it accelerating according to the motion of the merry-go-round exceeds the maximum static force of friction. The angular speed ω of the merry-go-round is ω = αt. Since F = ma, we need consider the magnitude of the acceleration of the object. There are two components, the tangential and radial. at = αR ar = ω 2 R p a = ar 2 + at 2 We also have for the static frictional force Ff ≤ µs N , where N is the normal force and N = mg, where g is the acceleration of gravity. Consequently, ar 2 + at 2 = µs 2 g 2 is the condition for slipping. Combining the above expressions, α 2 R 2 + ω 4 R 2 = µs 2 g 2 , α 2 R 2 1 + α 2 t 4 = µs 2 g 2 , s r 1 µs 2 g 2 −1 t = α α2 R 2 Since the object moves off with a tangential velocity once it begins to slip, we have q p vt = Rω = Rαt = R µs 2 g 2 − α2 R2 The angular position made by the merry-go-round is then given by θ = 12 αt2 , so 1 p 2 2 θ= µs g − α 2 R 2 2αR For the numbers given, p 1 θ= (0.5)2 (9.8)2 − (0.2)2 (4)2 = 3.021rad, 2(0.2)(4) or 173 degrees. Add 90 for the angle as measured from north, or 263 degrees. c Copyright 2010 American Association of Physics Teachers

2010 Semifinal Exam

Part A

4

Question A2 A spherical shell of inner radius a and outer radius b is made of a material of resistivity ρ and negligible dielectric activity. A single point charge q0 is located at the center of the shell. At time t = 0 all of the material of the shell is electrically neutral, including both the inner and outer surfaces. What is the total charge on the outer surface of the shell as a function of time for t > 0? Ignore any effects due to magnetism or radiation; do not assume that b − a is small.

Solution The material of the shell will remain electrically neutral, although a charge −Q will build up on the inner surface while a charge of +Q will build up on the outer surface. By spherical symmetry and Gauss’s law we can conclude that the electric field in the material of the shell will be given by E=

1 q0 − Q(t) . 4π0 r2

This will cause a current density J=

E 1 q0 − Q(t) = . ρ 4π0 ρ r2

and therefore a current I = JA =

1 (q0 − Q) 0 ρ

where Q is still a function of time. But I = dQ/dt, so dt dQ = q0 − Q 0 ρ which can easily be integrated to yield ln or

q0 q0 − Q

=

t 0 ρ

Q = 1 − e−t/0 ρ q0


2010 Semifinal Exam

Part A

5

Question A3 A cylindrical pipe contains a movable piston that traps 2.00 mols of air. Originally, the air is at one atmosphere of pressure, a volume V0 , and at a temperature of T0 = 298 K. First (process A) the air in the cylinder is compressed at constant temperature to a volume of 14 V0 . Then (process B) the air is allowed to expand adiabatically to a volume of V = 15.0 L. After this (process C) this piston is withdrawn allowing the gas to expand to the original volume V0 while maintaining a constant temperature. Finally (process D) while maintaining a fixed volume, the gas is allowed to return to the original temperature T0 . Assume air is a diatomic ideal gas, no air flows into, or out of, the pipe at any time, and that the temperature outside the remains constant always. Possibly useful information: Cp = 72 R, Cv = 52 R, 1 atm = 1.01 × 105 Pa. a. Draw a P-V diagram of the whole process. b. How much work is done on the trapped air during process A? c. What is the temperature of the air at the end of process B?

Solution a. Isotherm, adiabat, isotherm, isochoric

b. 1 V1 = V0 4 For an ideal gas along an isotherm: PV = nRT P1 V1 = P0 V0 =⇒

P0 V1 = V0 P1

W = nRT ln

V1 V0

J W = 2 mols · 8.31 · 298 K · ln mol·K

1 4

W = −6866 J (work done by the gas) W = 6866 J (work done on the gas) c. Adiabatic process T1 V1γ−1 = T2 V2γ−1 T1 = T0 = 298 K – from isothermal process in part b γ−1 V1 T2 = T0 V2 γ−1=

Cp −1= Cv

7 2 5 2

R 7 2 −1= −1= 5 5 R


2010 Semifinal Exam

Part A

6

T2 = 298 K

V1 V2

2 5

P0 V0 = nRT0 V0 =

nRT0 P0

P0 = 1 atm = 1.01 · 105 Pa J · 298 K mol·K 1.01 · 105 Pa

2 mols · 8.31 V0 =

V0 = 0.0490 m3 1 1 V1 = V0 = · 0.0490 m3 = 0.0123 m3 4 4 1 m3 103 mL 1 cm3 · · 6 = 0.0150 m3 1L 1 mL 10 cm3 2 0.0123 5 T2 = 298 K 0.0150

V2 = 15.0 L = 15.0 L ·

T2 = 275 K d. Lowest Pressure From the P-V diagram, the lowest pressure occurs at the end of the second isothermal process. P3 V3 = nRT3 P3 =

nRT3 V3

T3 = T2 = 275 K V3 = V0 = 0.0490 m3 J · 275 K mol·K 0.0490 m3

2.00 mols · 8.31 P3 =

P3 = 9.33 · 104 Pa


2010 Semifinal Exam

Part A

7

Question A4 The energy radiated by the Sun is generated primarily by the fusion of hydrogen into helium-4. In stars the size of the Sun, the primary mechanism by which fusion takes place is the proton-proton chain. The chain begins with the following reactions: 2 p → X1 + e+ + X2 (0.42 MeV)

(A4-1)

p + X1 → X3 + γ (5.49 MeV)

(A4-2)

The amounts listed in parentheses are the total kinetic energy carried by the products, including gamma rays. p is a proton, e+ is a positron, γ is a gamma ray, and X1 , X2 , and X3 are particles for you to identify. The density of electrons in the Sun’s core is sufficient that the positron is annihilated almost immediately, releasing an energy x: e+ + e− → 2 γ (x)

(A4-3)

Subsequently, two major processes occur simultaneously. The “pp I branch” is the single reaction 2 X3 → 4 He + 2 X4 (y),

(A4-4)

which releases an energy y. The “pp II branch” consists of three reactions: X3 + 4 He → X5 + γ

(A4-5)

X5 + e− → X6 + X7 (z)

(A4-6)

X6 + X4 → 2 4 He

(A4-7)

where z is the energy released in step A4-6. a. Identify X1 through X7 . X2 and X7 are neutral particles of negligible mass. It is useful to know that the first few elements, in order of atomic number, are H, He, Li, Be, B, C, N, O. b. The mass of the electron is 0.51 MeV/c2 , the mass of the proton is 938.27 MeV/c2 , and the mass of the helium-4 nucleus is 3 727.38 MeV/c2 . Find the energy released during the production of one helium-4 nucleus, including the kinetic energy of all products and all energy carried by gamma rays. c. Find the unknown energies x and y above. d. Step (A4-6) does not proceed as follows because there is insufficient energy. X5 → X6 + e+ + X7 What constraint does this fact place on z? e. In which of the reaction steps is the energy carried by any given product the same every time the step occurs? Assume that the kinetic energy carried in by the reactants in each step is negligible, and that the products are in the ground state.


2010 Semifinal Exam

Part A

8

Solution a. We know that in all nuclear processes, total charge is conserved, lepton number (electrons plus neutrinos minus positrons minus antineutrinos), and baryon number (neutrons plus protons) are conserved. As X2 is a neutral particle of negligible mass, X1 must have charge +1 and contain two baryons. Thus it is 2 H. X2 is then a neutral particle with lepton number +1 and is an electron neutrino. Similar reasoning shows that X3 is 3 He, X4 is a proton (or 1 H, X5 is 7 Be, X6 is 7 Li, and X7 is an electron neutrino. b. The overall reaction (not including gamma rays) is 4 p + 2 e− →4 He + 2 νe as can be seen from conservation considerations or by combining the given reactions. The energy released is the difference in mass between the reactants and the products; using the given values, this is 26.72 MeV. c. x is simply twice the mass of the electron, 1.02 MeV. To compute y, note that we can sum the other known energies to obtain the result from the previous problem: 2(0.42 MeV + 5.49 MeV + 1.02 MeV) + y = 26.72 MeV where the factor of 2 arises because two 3 He are produced in the course of the combined reaction. Solving, y = 12.86 MeV. d. The forbidden reaction produces an energy of z − 2(0.51 MeV), as it differs from the naturally occurring one by the consumption of one fewer electron and the production of an additional positron. Since it is forbidden, we know that z − 2(0.51 MeV) < 0; in other words, z < 1.02 MeV. e. It is a well-known result that reactions with two products have a single set of product energies, whereas those with three or more products produce a spectrum of output energies. Students may quote this result; alternatively, observe that in a two-product reaction the conservation of momentum and of energy give two equations in the two unknowns, fixing their values, whereas there are insufficient equations in the case of three or more products. The reactions with only two products are (A4-2), (A4-3), (A4-5), and (A4-7).


2010 Semifinal Exam

Part A

9




2010 Semifinal Exam

Part B

10

Part B Question B1 A thin plank of mass M and length L rotates about a pivot at its center. A block of mass m M slides on the top of the plank. The system moves without friction. Initially, the plank makes an angle θ0 with the horizontal, the block is at the upper end of the plank, and the system is at rest. Throughout the problem you may assume that θ 1, and that the physical dimensions of the block are much, much smaller than the length of the plank. L/2

x

Let x be the displacement of the block along the plank, as measured from the pivot, and let θ be the angle between the plank and the horizontal. You may assume that centripetal acceleration of the block is negligible compared with the linear acceleration of the block up and down the plank. a. For a certain value of θ0 , x = kθ throughout the motion, where k is a constant. What is this value of θ0 ? Express your answer in terms of M , m, and any fundamental constants that you require. b. Given that θ0 takes this special value, what is the period of oscillation of the system? Express your answer in terms of M , m, and any fundamental constants that you require. c. Determine the maximum value of the ratio between the centripetal acceleration of the block and the linear acceleration of the block along the plank, writing your answer in terms of m and M , therefore justifying our approximation.

Solution Eventually we will need to compute the rotational inertia of a long plank. The answer is I=

1 M L2 12

and it is acceptable to just write it down with no work. Assume the block is at point x away from the central pivot. The magnitude of the torque on the plank is then given by τ = ymg cos θ ≈ mgx so the angular acceleration is α=−

mg x I


2010 Semifinal Exam

Part B

11

It is fair to neglect the rotational inertia of the block, since m M . A similar expression exists for the linear acceleration of the board a = −g sin θ ≈ gθ Assume that the motion of the block is given by x = L2 cos ωt, then the acceleration is given by a = − L2 ω 2 cos ωt. Similarly, θ = θ0 cos ωt, so α = −θ0 ω 2 cos ωt. Note that ω is not the angular velocity of the plank! Combine the equations and get two equation of the form θ0 ω 2 = and

mg L I 2

L 2 ω = gθ0 2

which can be combined to yield θ0 2 L/2 θ02

mL , I 2 m = 3 M =

and 2g ω = L 2

r m 3 . M

We neglected centripetal acceleration of the block. Is this reasonable? It would be on the order of 2 dθ ac = x = xθ0 2 ω 2 sin 2 ωt. dt Now a = −xω 2 , so by considering the ratio of ac /a, we get ac = θ0 2 sin2 ωt, a an expression which is always small. The kinetic energy of the system is given by the sum of the energy of the block and the rotating plank, 2 1 L 1 1 2 m ω sin ωt + M L (θ0 ω sin ωt)2 2 2 2 12 which can be simplified to 1 (Lω sin ωt)2 2

m M 2 + θ0 4 12

using the above results, the kinetic energy reduces to m mgLθ0 (Lω sin ωt)2 = sin2 ωt 4 2 The potential energy is given by the location of the block only, so mgh = mgxθ = mg

L mgLθ0 cos ωtθ0 cos ωt = cos2 ωt. 2 2


2010 Semifinal Exam

Part B

12

Thankfully, this does result in the expected behavior of a system undergoing simple harmonic motion. The kinetic energy equals the potential energy when ωt is equal to π/4, so the position of the block is √ L 2 x= 2 2 and √ 2 θ = θ0 2 so r L m L y = xθ = θ0 = 3 4 4 M of course, if we had thought about this, we could have simply written down this last line without worrying about computing the kinetic or potential energy at all!


2010 Semifinal Exam

Part B

13

Question B2 These three parts can be answered independently. a. One pair of ends of two long, parallel wires are connected by a resistor, R = 0.25 Ω, and a fuse that will break instantaneously if 5 amperes of current pass through it. The other pair of ends are unconnected. A conducting rod of mass m is free to slide along the wires under the influence of gravity. The wires are separated by 30 cm, and the rod starts out 10 cm from the resistor and fuse. The whole system is placed in a uniform, constant magnetic field of B = 1.2 T as shown in the figure. The resistance of the rod and the wires is negligible. When the rod is released is falls under the influence of gravity, but never loses contact with the long parallel wires.

Resistor

Fuse

Sliding Rod

The magnetic field is directed into the page

i. What is the smallest mass needed to break the fuse? ii. How fast is the mass moving when the fuse breaks? b. A fuse is composed of a cylindrical wire with length L and radius r L. The resistivity (not resistance!) of the fuse is small, and given by ρf . Assume that a uniform current I flows through the fuse. Write your answers below in terms of L, r, ρf , I, and any fundamental constants. i. What is the magnitude and direction of the electric field on the surface of the fuse wire? ii. What is the magnitude and direction of the magnetic field on the surface of the fuse wire? iii. The Poynting vector, ~S is a measure of the rate of electromagnetic energy flow through ~ × B, ~ a unit surface area; the vector gives the direction of the energy flow. Since ~S = 1 E µ0

~ and B ~ are the electric and magnetic where µ0 is the permeability of free space and and E field vectors, find the magnitude and direction of the Poynting vector associated with the current in the fuse wire. c. A fuse will break when it reaches its melting point. We know from modern physics that a hot object will radiate energy (approximately) according to the black body law P = σAT 4 , where T is the temperature in Kelvin, A the surface area, and σ is the Stefan-Boltzmann constant. If Tf = 500 K is the melting point of the metal for the fuse wire, with resistivity ρf = 120 nΩ · m, and If = 5 A is the desired breaking current, what should be the radius of the wire r? c Copyright 2010 American Association of Physics Teachers

2010 Semifinal Exam

Part B

14

Solution a. Magnetic flux ΦB = BA B = 1.2T A = 0.1 m · 0.3 m = 0.03 m2 ΦB = 1.2 T · 0.03 m2 ΦB = 3.6 · 10−2 Wb b. Mass needed to break fuse F = iL × B = iLB sin (θ = 90◦ ) = iLB F = 5 amps · 0.3 m · 1.2 T F = 1.8 N To cause a current of 5 amperes, a force of 1.8 N is required. F = ma m=

1.8 N 9.8 m s2

m = 0.18 kg c. A released mass of m = 0.18 kg will produce a 5 ampere current once it reaches a certain velocity. That velocity is: F = iLB i= F = E=

E R

E LB R

dΦB dBA dBLx dx = = = BL = BLv dt dt dt dt i=

v=

BLv R

F =

B 2 L2 v R

v=

FR B 2 L2

1.8 N · 0.25Ω m = 3.47 2 2 (1.2 T) · (0.3 m) s


2010 Semifinal Exam

Part B

15

A quick application of E = δV /L, where L is the length of the fuse wire, followed by ∆V = IRf , yields IRf E= . L The resistance of the fuse wire is Rf = ρf L/πr2 , so E=

Iρf πr2

The magnetic field is found from Ampere’s law, B=

µ0 I 2π r

and then the Poynting vector is I 2 ρf 2πr3 It is directed inward toward the wire, making it hotter. The total power radiated inward is then found by multiplying by the surface area of a cylinder, or P = I 2 Rf . S=

What a surprise. Finally, the temperature increases until there is a power balance, so σTf 4 =

If 2 ρ 2π 2 r3

This can be solved to find the radius needed for the fuse wire, s (5A)2 (120 × 10−9 Ω · m) r= 3 = 0.35 mm 2π 2 (5.67 × 10−8 J/(s · m2 · K4 ))(500K)4


United States Physics Team F = ma Contest Papers 2011

2011 F = ma Exam

1

AAPT AIP


DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Use g = 10 N/kg throughout this contest. • You may write in this booklet of questions. However, you will not receive any credit for anything written in this booklet. • Your answer to each question must be marked on the optical mark answer sheet. • Select the single answer that provides the best response to each question. Please be sure to use a No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer, the previous mark must be completely erased. • Correct answers will be awarded one point; incorrect answers will result in a deduction of no penalty for leaving an answer blank.

1 4

point. There is




2011 F = ma Exam

2

1. A cyclist travels at a constant speed of 22.0 km/hr except for a 20 minute stop. The cyclist’s average speed was 17.5 km/hr. How far did the cyclist travel? (A) 28.5 km (B) 30.3 km (C) 31.2 km (D) 36.5 km (E) 38.9 km The correct answer is A Questions 2 to 4 refer to the three graphs below which show velocity of three objects as a function of time. Each object is moving only in one dimension.

+2 0 −2 0

2

4 6 time (s)

Object I

8

10

+4 velocity (m/s)

+4 velocity (m/s)

velocity (m/s)

+4

+2 0 −2 0

2

4 6 time (s)

8

+2 0 −2

10

0

2

Object II

2. Rank the magnitudes of the average acceleration during the ten second interval. (A) I > II > III (B) II > I > III (C) III > II > I (D) I > II = III (E) I = II = III The correct answer is E 3. Rank the magnitudes of the maximum velocity achieved during the ten second interval. (A) I > II > III (B) II > I > III (C) III > II > I (D) I > II = III (E) I = II = III The correct answer is D 4. Rank the magnitudes of the distance traveled during the ten second interval. (A) I > II > III (B) II > I > III (C) III > II > I (D) I = II > III (E) I = II = III The correct answer is B


4 6 time (s)

Object III

8

10

2011 F = ma Exam

3

5. A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit? (A) exactly 0 m/s2 (B) 0.006 m/s2 (C) 0.6 m/s2 (D) 6 m/s2 (E) 10 m/s2 The correct answer is B 6. A child is sliding out of control with velocity vc across a frozen lake. He runs head-on into another child, initially at rest, with 3 times the mass of the first child, who holds on so that the two now slide together. What is the velocity of the couple after the collision? (A) 2vc (B) vc (C) vc /2 (D) vc /3 (E) vc /4 The correct answer is E 7. An ice skater can rotate about a vertical axis with an angular velocity ω0 by holding her arms straight out. She can then pull in her arms close to her body so that her angular velocity changes to 2ω0 , without the application of any external torque. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy? √ (A) 2 (B) 2

√ (C) 2 2 (D) 4 (E) 8 The correct answer is B 8. When a block of wood with a weight of 30 N is completely submerged under water the buoyant force on the block of wood from the water is 50 N. When the block is released it floats at the surface. What fraction of the block will then be visible above the surface of the water when the block is floating? (A) 1/15 (B) 1/5 (C) 1/3 (D) 2/5 (E) 3/5 The correct answer is D


2011 F = ma Exam

4

9. A spring has an equilibrium length of 2.0 meters and a spring constant of 10 newtons/meter. Alice is pulling on one end of the spring with a force of 3.0 newtons. Bob is pulling on the opposite end of the spring with a force of 3.0 newtons, in the opposite direction. What is the resulting length of the spring? (A) 1.7 m (B) 2.0 m (C) 2.3 m (D) 2.6 m (E) 8.0 m The correct answer is C 10. Which of the following changes will result in an increase in the period of a simple pendulum? (A) Decrease the length of the pendulum (B) Increase the mass of the pendulum (C) Increase the amplitude of the pendulum swing (D) Operate the pendulum in an elevator that is accelerating upward (E) Operate the pendulum in an elevator that is moving downward at constant speed. The correct answer is C

water level

time (B)

water level

time (A)

water level

water level

water level

11. A large metal cylindrical cup floats in a rectangular tub half-filled with water. The tap is placed over the cup and turned on, releasing water at a constant rate. Eventually the cup sinks to the bottom and is completely submerged. Which of the following five graphs could represent the water level in the sink as a function of time?

time (D)

time (E)

The correct answer is C


time (C)

2011 F = ma Exam

5

12. You are given a large collection of identical heavy balls and lightweight rods. When two balls are placed at the ends of one rod and interact through their mutual gravitational attraction (as is shown on the left), the compressive force in the rod is F . Next, three balls and three rods are placed at the vertexes and edges of an equilateral triangle (as is shown on the right). What is the compressive force in each rod in the latter case?

(A) (B)

√1 F 3 √ 3 2 F

(C) F √ (D) 3F (E) 2F The correct answer is C 13. The apparatus in the diagram consists of a solid cylinder of radius 1 cm attached at the center to two disks of radius 2 cm. It is placed on a surface where it can roll, but will not slip. A thread is wound around the central cylinder. When the thread is pulled at the angle θ = 90 to the horizontal (directly up), the apparatus rolls to the right. Which below is the largest value of θ for which it will not roll to the right when pulling on the thread?

θ

(A) θ = 15◦ (B) θ = 30◦ (C) θ = 45◦ (D) θ = 60◦ (E) None, the apparatus will always roll to the right The correct answer is D


2011 F = ma Exam

6

14. You have 5 different strings with weights tied at various point, all hanging from the ceiling, and reaching down to the floor. The string is released at the top, allowing the weights to fall. Which one will create a regular, uniform beating sound as the weights hit the floor?

(A)

(B)

(C)

(D)

(E)

The correct answer is D 15. A vertical mass-spring oscillator is displaced 2.0 cm from equilibrium. The 100 g mass passes through the equilibrium point with a speed of 0.75 m/s. What is the spring constant of the spring? (A) 90 N/m (B) 100 N/m (C) 110 N/m (D) 140 N/m (E) 160 N/m The correct answer is D


2011 F = ma Exam

7

Questions 16 and 17 refer to the information and diagram below. Jonathan using rope to lift a box with Becky in it; the box is hanging off the side of a bridge, Jonathan is on the top. A rope is hooked up from the box and passes a fixed railing; Jonathan holds the box up by pressing the rope against the railing with a massless, frictionless physics textbook. The static friction coefficient between the rope and railing is µs ; the kinetic friction coefficient between the rope and railing is µk < µs ; the mass of the box and Becky combined is M ; and the initial height of the bottom of the box above the ground is h. Assume a massless rope.

Loose rope

fixed hard railing

Jonathan, pushes on book against rope

Floor

Becky

16. What magnitude force does Jonathan need to exert on the physics book to keep the rope from slipping? (A) M g (B) µk M g (C) µk M g/µs (D) (µs + µk )M g (E) M g/µs The correct answer is E 17. Jonathan applies a normal force that is just enough to keep the rope from slipping. Becky makes a small jump, barely leaving contact with the floor of the box. Upon landing on the box, the force of the impact causes the rope to start slipping from Jonathan’s hand. At what speed does the box smash into the ground? Assume Jonathan’s normal force does not change. √ (A) 2gH(µk /µs ) √ (B) 2gH(1 − µk /µs ) p √ (C) 2gH µk /µs p √ (D) 2gH (1 − µk /µs ) √ (E) 2gH(µs − µk ) The correct answer is D 18. A block of mass m = 3.0 kg slides down one ramp, and then up a second ramp. The coefficient of kinetic friction between the block and each ramp is µk = 0.40. The block begins at a height h1 = 1.0 m above the horizontal. Both ramps are at a 30◦ incline above the horizontal. To what height above the horizontal does the block rise on the second ramp? (A) 0.18 m (B) 0.52 m (C) 0.59 m (D) 0.69 m (E) 0.71 m The correct answer is A


2011 F = ma Exam

8

Questions 19 and 20 refer to the following information A particle of mass 2.00 kg moves under a force given by ~ = −(8.00 N/m)(xî + yˆj) F where î and ˆj are unit vectors in the x and y directions. The particle is placed at the origin with an initial velocity ~v = (3.00 m/s)î + (4.00 m/s)ˆj. 19. After how much time will the particle first return to the origin? (A) 0.785 s (B) 1.26 s (C) 1.57 s (D) 2.00 s (E) 3.14 s The correct answer is C 20. What is the maximum distance between the particle and the origin? (A) 2.00 m (B) 2.50 m (C) 3.50 m (D) 5.00 m (E) 7.00 m The correct answer is B 21. An engineer is given a fixed volume Vm of metal with which to construct a spherical pressure vessel. Interestingly, assuming the vessel has thin walls and is always pressurized to near its bursting point, the amount of gas the vessel can contain, n (measured in moles), does not depend on the radius r of the vessel; instead it depends only on Vm (measured in m3 ), the temperature T (measured in K), the ideal gas constant R (measured in J/(K · mol)), and the tensile strength of the metal σ (measured in N/m2 ). Which of the following gives n in terms of these parameters?

(C) n =

2 Vm σ 3 RT √ 2 3 Vm σ 3 RT √ 2 3 Vm σ 2 3 RT

(D) n =

2 3

(E) n =

2 3

(A) n = (B) n =

√ 3

Vm 2 σ RT

q 3

Vm σ 2 RT

The correct answer is A


2011 F = ma Exam

9

Output Torque (Nm)

22. This graph depicts the torque output of a hypothetical gasoline engine as a function of rotation frequency. The engine is incapable of running outside of the graphed range.

30

20

10

0 1,000

0

I

2,000

II

III

Engine Revolutions per Minute At what engine RPM (revolutions per minute) does the engine produce maximum power? (A) I (B) At some point between I and II (C) II (D) At some point between II and III (E) III The correct answer is D 23. A particle is launched from the surface of a uniform, stationary spherical planet at an angle to the vertical. The particle travels in the absence of air resistance and eventually falls back onto the planet. Spaceman Fred describes the path of the particle as a parabola using the laws of projectile motion. Spacewoman Kate recalls from Kepler’s laws that every bound orbit around a point mass is an ellipse (or circle), and that the gravitation due to a uniform sphere is identical to that of a point mass. Which of the following best explains the discrepancy? (A) Because the experiment takes place very close to the surface of the sphere, it is no longer valid to replace the sphere with a point mass. (B) Because the particle strikes the ground, it is not in orbit of the planet and therefore can follow a nonelliptical path. (C) Kate disregarded the fact that motions around a point mass may also be parabolas or hyperbolas. (D) Kepler’s laws only hold in the limit of large orbits. (E) The path is an ellipse, but is very close to a parabola due to the short length of the flight relative to the distance from the center of the planet. The correct answer is E


2011 F = ma Exam

10

24. A turntable is supported on a Teflon ring of inner radius R and outer radius R+δ (δ R), as shown in the diagram. To rotate the turntable at a constant rate, power must be supplied to overcome friction. The manufacturer of the turntable wishes to reduce the power required without changing the rotation rate, the weight of the turntable, or the coefficient of friction of the Teflon surface. Engineers propose two solutions: increasing the width of the bearing (increasing δ), or increasing the radius (increasing R). What are the effects of these proposed changes?

R

(A) Increasing δ has no significant effect on the required power; increasing R increases the required power. (B) Increasing δ has no significant effect on the required power; increasing R decreases the required power. (C) Increasing δ increases the required power; increasing R has no significant effect on the required power. (D) Increasing δ decreases the required power; increasing R has no significant effect on the required power. (E) Neither change has a significant effect on the required power. The correct answer is A 25. A hollow cylinder with a very thin wall (like a toilet paper tube) and a block are placed at rest at the top of a plane with inclination θ above the horizontal. The cylinder rolls down the plane without slipping and the block slides down the plane; it is found that both objects reach the bottom of the plane simultaneously. What is the coefficient of kinetic friction between the block and the plane? (A) 0 (B) (C) (D)

1 3 1 2 2 3

tan θ tan θ tan θ

(E) tan θ The correct answer is C


United States Physics Team Semi Final Contest Papers 2011

2011 Semifinal Exam

1

AAPT AIP


Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2011. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A.


2011 Semifinal Exam

AAPT AIP

Cover Sheet

2


Semifinal Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, AAPT ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 1, 2011. Possibly Useful Information. You may g = 9.8 N/kg k = 1/4π0 = 8.99 × 109 N · m2 /C2 c = 3.00 × 108 m/s NA = 6.02 × 1023 (mol)−1 σ = 5.67 × 10−8 J/(s · m2 · K4 ) 1eV = 1.602 × 10−19 J me = 9.109 × 10−31 kg = 0.511 MeV/c2 sin θ ≈ θ − 16 θ3 for |θ| 1



2011 Semifinal Exam

Part A

3

Part A Question A1 Single bubble sonoluminescence occurs when sound waves cause a bubble suspended in a fluid to collapse so that the gas trapped inside increases in temperature enough to emit light. The bubble actually undergoes a series of expansions and collapses caused by the sound wave pressure variations. We now consider a simplified model of a bubble undergoing sonoluminescence. Assume the bubble is originally at atmospheric pressure P0 = 101 kPa. When the pressure in the fluid surrounding the bubble is decreased, the bubble expands isothermally to a radius of 36.0 µm. When the pressure increases again, the bubble collapses to a radius of 4.50 µm so quickly that no heat can escape. Between the collapse and subsequent expansion, the bubble undergoes isochoric (constant volume) cooling back to its original pressure and temperature. For a bubble containing a monatomic gas, suspended in water of T = 293 K, find a. the number of moles of gas in the bubble, b. the pressure after the expansion, c. the pressure after collapse, d. the temperature after the collapse, and e. the total work done on the bubble during the whole process. You may find the following useful: the specific heat capacity at constant volume is CV = 3R/2 and the ratio of specific heat at constant pressure to constant volume is γ = 5/3 for a monatomic gas.

Solution We consider the bubble to be filled with an ideal monatomic gas, so originally: P0 V0 = nRT0 . The bubble undergoes 3 processes: 1) isothermal expansion, 2) adiabatic collapse (no heat escapes), and 3) isochoric (constant volume) cooling. The final process is isochoric, so we know that the bubble’s collapsed volume is equal to its original volume, so V2 = V0 , and P0 V0 = P0 V2 = nRT0 . Rearranging, n= n=

P0 V 2 RT0

P0 43 πr23 RT0

101, 000 N2 · 34 π · (4.50 × 10−6 m)3 3.86 × 10−11 m n= = moles 2430 8.31 J · 293 K mol·K c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

a)

Part A

4

n = 1.58 × 10−14 moles.

Process 1: Isothermal expansion This process is isothermal, so T1 = T0 and P1 V1 = nRT1 = nRT0 P1 = b)

1.58 × 10−14 moles · 8.31 J · 293 K nRT0 mol·K = 4 −5 m)3 V1 π · (3.60 × 10 3

N = 197 Pa. P1 = 197 m 2 The work done by the bubble is: W1 = nRT0 ln W1 = 1.58 × 10−14 moles · 8.31

V1 V0

J · 293 K · ln mol·K

(3.60 × 10−5 )3 (4.50 × 10−6 )3

W1 = 2.40 × 10−10 J So, the work done on the bubble during the expansion is: W1 = −2.40 × 10−10 J.

Process 2: Adiabatic collapse For an adiabatic process P1 V1γ = P2 V2γ P2 =

P1 V1γ V2γ

For a monatomic gas γ = 5/3 so, N · (3.60 × 10−5 m)5 197 m 2 P2 = (4.50 × 10−6 m)5 c)

P2 = 6.46 × 106 Pa. And T2 =

P2 V 2 nR

6.46 × 106 Pa · 34 π · (4.50 × 10−6 m)3 1.58 × 10−14 moles · 8.31 J mol·K T2 = 18800 K. Lord, have mercy! That’s hot! T2 =

d)

The work done by the bubble during an adiabatic process is W2 = −∆Einternal = −nCv ∆T c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part A

5

where Cv = 3R/2. −14

W2 = −1.58 × 10

3 J moles · 8.31 · (18800 − 293) K 2 mol·K W2 = −3.64 × 10−9 J

The work done on the bubble is then W2 = 3.64 × 10−9 J

Process 3: Isochoric cooling The work done on the bubble during an isochoric process is zero, so W3 = 0 J. The total work is then the sum of the work on the bubble Wtotal = W1 + W2 + W3 Wtotal = −2.40 × 10−10 J + 3.64 × 10−9 J + 0 J e)

Wtotal = 3.4 × 10− 9 J.

Question A2 A thin, uniform rod of length L and mass M = 0.258 kg is suspended from a point a distance R away from its center of mass. When the end of the rod is displaced slightly and released it executes simple harmonic oscillation. The period, T , of the oscillation is timed using an electronic timer. The following data is recorded for the period as a function of R. What is the local value of g? Do not assume it is the canonical value of 9.8 m/s2 . What is the length, L, of the rod? No estimation of error in either value is required. The moment of inertia of a rod about its center of mass is (1/12)M L2 . R (m) 0.050 0.075 0.102 0.156 0.198

T (s) 3.842 3.164 2.747 2.301 2.115

R (m) 0.211 0.302 0.387 0.451 0.588

T (s) 2.074 1.905 1.855 1.853 1.900

You must show your work to obtain full credit. If you use graphical techniques then you must plot the graph; if you use linear regression techniques then you must show all of the formulae and associated workings used to obtain your result.

Solution The period of a physical pendulum is given by s s 1 2 L + R2 I T = 2π = 2π 12 mgR gR c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part A

6

A little math, and g

T 2R 1 − L2 = R2 . 2 4π 12

This is of the form mx + b = y if we let y = R2 and x=

T 2R 4π 2

R T T 2 R/4π 2 R2 0.050 3.842 0.0187 0.0025 0.075 3.164 0.0190 0.0056 0.102 2.747 0.0195 0.0104 0.156 2.301 0.0209 0.0243 Filling out a table of data, we get 0.198 2.115 0.0224 0.0392 The corresponding graph of 0.211 2.074 0.0230 0.0445 0.302 1.905 0.0278 0.0912 0.387 1.855 0.0337 0.1498 0.451 1.853 0.0392 0.2034 0.588 1.900 0.0538 0.3457 1 2 2 2 2 L . T /R/4π versus R ought yield a straight line such that the slope is g and the intercept is − 12

g = 9.7923 m/s2 and L = 1.470 m


2011 Semifinal Exam

Part A

7

Question A3 A light bulb has a solid cylindrical filament of length L and radius a, and consumes power P . You are to design a new light bulb, using a cylindrical filament of the same material, operating at the same voltage, and emitting the same spectrum of light, which will consume power nP . What are the length and radius of the new filament? Assume that the temperature of the filament is approximately uniform across its cross-section; the filament doesn’t emit light from the ends; and energy loss due to convection is minimal.

Solution Since the new bulb emits the same spectrum of light, the emitted power is simply proportional to the area: P ∝ 2πaL P ∝ aL If the resistivity of the filament is ρ, the resistance is R=

ρL ρL = A πa2

and therefore the power is also given by P =

V2 V 2 πa2 = R ρL P ∝

a2 L

Combining our conditions, a ∝ P 2/3 L ∝ P 1/3 So the new filament must have length n2/3 a and length n1/3 L.

Question A4 In this problem we consider a simplified model of the electromagnetic radiation inside a cubical box of side length L. In this model, the electric field has spatial dependence E(x, y, z) = E0 sin(kx x) sin(ky y) sin(kz z) where one corner of the box lies at the origin and the box is aligned with the x, y, and z axes. Let h be Planck’s constant, kB be Boltzmann’s constant, and c be the speed of light. a. The electric field must be zero everywhere at the sides of the box. What condition does this impose on kx , ky , and kz ? (Assume that any of these may be negative, and include cases where one or more of the ki is zero, even though this causes E to be zero.) b. In the model, each permitted value of the triple (kx , ky , kz ) corresponds to a quantum state. These states can be visualized in a state space, which is a notional three-dimensional space c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part A

8

with axes corresponding to kx , ky , and kz . How many states occupy a volume s of state space, if s is large enough that the discreteness of the states can be ignored? c. Each quantum state, in turn, may be occupied by photons with frequency ω = where q |k| = kx 2 + ky 2 + kz 2

f 2π

= c|k|,

In the model, if the temperature inside the box is T , no photon may have energy greater than kB T . What is the shape of the region in state space corresponding to occupied states? d. As a final approximation, assume that each occupied state contains exactly one photon. What is the total energy of the photons in the box, in terms of h, kB , c, T , and the volume of the box V ? Again, assume that the temperature is high enough that there are a very large number of occupied states. (Hint: divide state space into thin regions corresponding to photons of the same energy.) Note that while many details of this model are extremely inaccurate, the final result is correct except for a numerical factor.

Solution a. We require that sin(kx L) = 0, so that kx L = nx π for any integer nx , and similarly for ky and kz . π b. The occupied states are equally spaced a distance L apart. Each can therefore be thought of π3 as taking up volume L3 , and the number of states in the volume s is

L3 s π3 c. A photon’s energy is E = h ¯ω = h ¯ c|k|, where h ¯=

h 2π .

Thus the occupied states obey

¯hc|k| ≤ kB T |k| ≤ This corresponds to a ball of radius

kB T hc ¯

kB T ¯hc

in state space.

d. As we have seen, the energy of a photon is proportional to its distance |k| from the origin in state space. Thus consider the spherical shell in state space between radius k and radius k + dk. The volume of this region is ds = 4πk 2 dk Each photon in the region has energy h ¯ ck, and from above there are region. Therefore the photons in the region have total energy dE = h ¯ ck ·

L3 ds π3

L3 · 4πk 2 dk π3


photons in the

2011 Semifinal Exam

Part A

dE = From above, k ranges from zero to kmax = Z

4 ¯hcL3 k 3 dk π2 kB T hc , ¯

kmax

E= 0

9

so the total energy is

4 ¯hcL3 k 3 dk π2

1 4 E = 2 ¯hcL3 · π 4

kB T ¯hc

4

Since the volume of the box is V = L3 , and h = 2π¯h, this cleans up to E=

8πkB 4 4 T V h3 c3


2011 Semifinal Exam

Part A

10




2011 Semifinal Exam

Part B

11

Part B Question B1 An AC power line cable transmits electrical power using a sinusoidal waveform with frequency 60 Hz. The load receives an RMS voltage of 500 kV and requires 1000 MW of average power. For this problem, consider only the cable carrying current in one of the two directions, and ignore effects due to capacitance or inductance between the cable and with the ground. a. Suppose that the load on the power line cable is a residential area that behaves like a pure resistor. i. What is the RMS current carried in the cable? ii. The cable has diameter 3 cm, is 500 km long, and is made of aluminum with resistivity 2.8 × 10−8 Ω · m. How much power is lost in the wire? b. A local rancher thinks he might be able to extract electrical power from the cable using electromagnetic induction. The rancher constructs a rectangular loop of length a and width b < a, consisting of N turns of wire. One edge of the loop is to be placed on the ground; the wire is straight and runs parallel to the ground at a height h much less than the length of the wire. Write the current in the wire as I = I0 sin ωt, and assume the return wire is far away. i. Determine an expression for the magnitude of the magnetic field at a distance r from the power line cable in terms of I, r, and fundamental constants. ii. Where should the loop be placed, and how should it be oriented, to maximize the induced emf in the loop? iii. Assuming the loop is placed in this way, determine an expression for the emf induced in the loop (as a function of time) in terms of any or all of I0 , h, a, b, N , ω, t, and fundamental constants. iv. Suppose that a = 5 m, b = 2 m, and h = 100 m. How many turns of wire N does the rancher need to generate an RMS emf of 120 V? c. The load at the end of the power line cable changes to include a manufacturing plant with a large number of electric motors. While the average power consumed remains the same, it now behaves like a resistor in parallel with a 0.25 H inductor. i. Does the power lost in the power line cable increase, decrease, or stay the same? (You need not calculate the new value explicitly, but you should show some work to defend your answer.) ii. The power company wishes to make the load behave as it originally did by installing a capacitor in parallel with the load. What should be its capacitance?

Solution a.

i. Because the load is purely resistive, the average power is simply Pav = Vrms Irms c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part B

so

12

Pav = 2000 A Vrms

Irms =

ii. The cross-sectional area of the wire is A = πr2 = 7.07 × 10−4 m2 , so its resistance is ρL = 19.8 Ω A

R= The power loss is then

P = I 2 R = 79.2 MW b.

i. The field is perpendicular to the wire and to the radius, and from Ampere’s Law I B · ds = µ0 Iencl B · 2πr = µ0 I B=

µ0 I 2πr

ii. The induced emf is proportional to the rate of change of the flux through the loop. Since the time dependence of the magnetic field is uniform across space, the rate of change of flux is maximized by maximizing the flux itself. This in turn can be accomplished by maximizing the field in the loop and ensuring that it is normal to the loop. Because the field gets stronger closer to the wire, the loop should be directly below the wire, and since the field is horizontal and perpendicular to the wire at this location, the loop should be vertical and parallel to the wire. Finally, again because the field gets stronger closer to the wire, the long edge of the loop should be vertical. In summary, the loop should be placed vertically, parallel to the wire and directly beneath it, with the long edge vertical. iii. From Faraday’s law, d ΦB dt where we have dropped the sign and ΦB is the magnetic flux through a single loop. The flux, in turn, is defined as Z E =N

B · dA

ΦB =

Dividing the loop into strips of radial width dr and length b, Z h ΦB = B(r)b dr h−a

Z

h

ΦB = h−a

ΦB =

µ0 I b dr 2πr

h µ0 Ib ln 2π h−a

So, E =N

d µ0 Ib h ln dt 2π h−a


2011 Semifinal Exam

Part B

13

Since the loop is stationary, only I depends on t, and E =N E =N

h dI µ0 b ln 2π h − a dt

h µ0 b ln I0 ω cos ωt 2π h−a

iv. Note that the RMS value of I0 cos ωt is the same as the RMS value of I0 sin ωt, i.e. Irms . So, taking the RMS value of both sides of our previous result, Erms = N And (conveniently) the frequency f =

h µ0 b ln ωIrms 2π h−a

ω 2π ,

so

Erms = N µ0 bf ln

h Irms h−a

With the given numbers, µ0 bf ln

h Irms = 0.0155 V h−a

so that the required number of turns is N = 7757 c.

i. The inductor adds a new component of the current in the wire out of phase with the voltage; this component does not transmit power, so the in-phase component must remain unchanged. The total current is thus increased, and with it the power lost in the wire increases as well. ii. The resonant frequency of an LC circuit is given by ω=√ so that C=

1 LC

1 ω2L

Here ω = 2πf = 377 s−1 , so C = 28.1 µF


2011 Semifinal Exam

Part B

14

Question B2 A particle is constrained to move on the inner surface of a frictionless parabolic bowl whose crosssection has equation z = kr2 . The particle begins at a height z0 above the bottom of the bowl with a horizontal velocity v0 along the surface of the bowl. The acceleration due to gravity is g.

z z = kr2

r

a. For a particular value of horizontal velocity v0 , which we will name vh , the particle moves in a horizontal circle. What is vh in terms of g, z0 , and/or k? b. Suppose that the initial horizontal velocity is now v0 > vh . What is the maximum height reached by the particle, in terms of v0 , z0 , g and/or k? c. Suppose that the particle now begins at a height z0 above the bottom of the bowl with an initial velocity v0 = 0. i. Assuming that z0 is small enough so that the motion can be approximated as simple harmonic, find the period of the motion in terms any or all of the mass of the particle m, g, z0 , and/or k. ii. Assuming that z0 is not small, will the actual period of motion be greater than, less than, or equal to your simple harmonic approximation above? (You need not calculate the new value explicitly, but you should show some work to defend your answer.)

Solution a. Let the particle have mass m, let the radius of the bowl at height z0 be r0 , and let the angle made by the bowl’s surface to the horizontal at that height be θ. Two forces act on the particle: the normal force and gravity. If the particle moves in a horizontal circle the horizontal component of the net force must equal the centripetal force 2 Fc = mvr0h , whereas the vertical component must be zero. From the free body diagram [Diagram], these conditions are mvh 2 FN sin θ = r0 FN cos θ − mg = 0 Combining these, tan θ =

vh 2 gr0

c American Association of Physics Teachers Copyright 2011

2011 Semifinal Exam

Part B

15

dz dr

However, tan θ is simply the slope of the bowl

= 2kr0 , so that

2ar0 =

vh 2 gr0

p

2gz0

Using the fact that z0 = kr0 2 , vh =

b. Let the maximum height be z, let the radius of the bowl at this point be r, and let the speed of the particle at this point be v. From conservation of energy, 1 1 mv0 2 + mgz0 = mv 2 + mgz 2 2 Meanwhile, the two forces acting on the particle never exert a torque in the direction of the bowl’s axis, and so angular momentum about this axis is conserved. Furthermore, at the point of maximum height the velocity of the particle is entirely tangential to the axis, so the conservation condition is simply mv0 r0 = mvr r0 v = v0 r or, since z = kr2 and z0 = kr0 2 , r v = v0

z0 z

Combining our results, 1 z0 1 mv0 2 + mgz0 = mv0 2 + mgz 2 2 z 2 2 v0 v0 + z0 z + z0 = 0 z2 − 2g 2g v0 2 (z − z0 ) z − =0 2g The root z = z0 corresponds to our starting condition, so the desired root is z=

v0 2 2g

Note that we recover z = z0 if v0 = vh as we would expect; indeed, the analysis in this section is an alternative path to the previous result. c.

i. We present a force-based approach and an energy-based approach. In each case, let r be the radial position of the particle, so that z = kr2 is the height of the particle above the bottom of the bowl. The force-based approach begins with the free body diagram. [Diagram] Again, let the angle of the bowl’s surface to the horizontal be θ. Because z0 is small, sin θ ≈ θ ≈ tan θ =

dz = 2kr dr

and cos θ ≈ 1. c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part B

16

We can consider the component of force tangential to the bowl, which is mg sin θ; Newton’s third law then gives for the magnitude of the acceleration a ma = mg sin θ noting that the acceleration is entirely tangential because the particle is constrained to the surface of the bowl (and there is no centripetal force anymore). The radial acceleration ar is given by ar = −a cos θ where we have introduced the appropriate sign to match the sign of r. Thus ar = −g cos θ sin θ So in the small-z approximation, ar ≈ −g tan θ ar =

d2 r = −2krg dt2

The energy-based approach begins with the total energy 1 E = mv 2 + mgz 2 The velocity v is given by

2

v = Because z is small,

dz dt

dr dt ,

dr dt

2

+

dz dt

2

and we conclude that 1 E= m 2

From conservation of energy,

dE dt

dr dt

2

+ mgkr2

= 0:

0=m

dr d2 r dr + 2mgkr 2 dt dt dt

0=

d2 r + 2krg dt2

Both approaches lead to the standard SHM differential equation d2 x + ω2x = 0 dt2 with angular frequency ω =

√

2kg; since the period T =

2π ω ,

2π T =√ 2kg c Copyright 2011 American Association of Physics Teachers

2011 Semifinal Exam

Part B

17

ii. The period is greater than the simple harmonic period. We can see this using both approaches: In the force-based approach, we obtained the exact equation ar = −g cos θ sin θ and approximated it as ar = −g tan θ Since cos θ sin θ < tan θ, the exact radial acceleration is smaller than the approximate one, so that the particle takes longer to reach the origin in reality than it does in the approximation, meaning that the period is larger. In the energy-based approach, we dropped a (positive) term in the formula for the speed dr v as expressed in terms of dr dt . Therefore we overestimated dt , and again the particle takes longer to reach the origin in reality than it does in the approximation.



2012 F = ma Exam

1

AAPT AIP



1 4

point. There is


DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Contributors to this years exam include Jiajia Dong, Qiuzi Li, Paul Stanley, Warren Turner, former US Team members Marianna Mao, Andrew Lin, Steve Byrnes, Adam Jermyn, Ante Qu, Alok Saxena, Tucker Chan, Kenan Diab, Jason LaRue.


2012 F = ma Exam

2

1. Consider a dripping faucet, where the faucet is 10 cm above the sink. The time between drops is such that when one drop hits the sink, one is in the air and another is about to drop. At what height above the sink will the drop in the air be right as a drop hits the sink? (A) Between 0 and 2 cm, (B) Between 2 and 4 cm, (C) Between 4 and 6 cm, (D) Between 6 and 8 cm, ← CORRECT (E) Between 8 and 10 cm, 2. A cannonball is launched with initial velocity of magnitude v0 over a horizontal surface. At what minimum angle θmin above the horizontal should the cannonball be launched so that it rises to a height H which is larger than the horizontal distance R that it will travel when it returns to the ground? (A) θmin = 76◦ ← CORRECT (B) θmin = 72◦ (C) θmin = 60◦ (D) θmin = 45◦ (E) There is no such angle, as R > H for all range problems. 3. An equilateral triangle is sitting on an inclined plane. Friction is too high for it to slide under any circumstance, but if the plane is sloped enough it can “topple” down the hill. What angle incline is necessary for it to start toppling? (A) 30 degrees (B) 45 degrees (C) 60 degrees ← CORRECT (D) It will topple at any angle more than zero (E) It can never topple if it cannot slide 4. A particle at rest explodes into three particles of equal mass in the absence of external forces. Two particles emerge at a right angle to each other with equal speed v. What is the speed of the third particle? (A) v √ (B) 2v← CORRECT (C) 2v √ (D) 2 2v (E) The third particle can have a range of different speeds. 5. A 12 kg block moving east at 4 m/s collides head on with a 6 kg block that is moving west at 2 m/s. The two blocks move together after the collision. What is the loss in kinetic energy in this collision? (A) 36 J (B) 48 J (C) 60 J (D) 72 J← CORRECT (E) 96 J


2012 F = ma Exam

3

The following information applies to questions 6 and 7 Two cannons are arranged vertically, with the lower cannon pointing upward (towards the upper cannon) and the upper cannon pointing downward (towards the lower cannon), 200m above the lower cannon. Simultaneously, they both fire. The muzzle velocity of the lower cannon is 25m/s and the muzzle velocity of the upper cannon is 55m/s.

200 meters

6. How long after the cannons fire do the projectiles collide? (A) 2.2 s (B) 2.5 s← CORRECT (C) 3.6 s (D) 6.7 s (E) 8.0 s 7. How far beneath the top cannon do the projectiles collide? (A) 31 m (B) 67 m (C) 110 m (D) 140 m (E) 170 m← CORRECT 8. A block of mass m = 3.0 kg is moving on a horizontal surface towards a massless spring with spring constant k = 80.0 N/m. The coefficient of kinetic friction between the block and the surface is µk = 0.50. The block has a speed of 2.0 m/s when it first comes in contact with the spring. How far will the spring be compressed? (A) 0.19 m (B) 0.24 m← CORRECT (C) 0.39 m (D) 0.40 m (E) 0.61 m


2012 F = ma Exam

4

9. A uniform spherical planet has radius R and the acceleration due to gravity at its surface is g. What is the escape velocity of a particle from the planet’s surface? √ (A) 12 gR √ (B) gR √ (C) 2gR← CORRECT √ (D) 2 gR (E) The escape velocity cannot be expressed in terms of g and R alone. 10. Four objects are placed at rest at the top of an inclined plane and allowed to roll without slipping to the bottom in the absence of rolling resistance and air resistance. • Object A is a solid brass ball of diameter d. • Object B is a solid brass ball of diameter 2d. • Object C is a hollow brass sphere of diameter d. • Object D is a solid aluminum ball of diameter d. (Aluminum is less dense than brass.) The balls are placed so that their centers of mass all travel the same distance. In each case, the time of motion T is measured. Which of the following statements is correct? (A) TB > TC > TA = TD (B) TA = TB = TC > TD (C) TB > TA = TC = TD (D) TC > TA = TB = TD ← CORRECT (E) TA = TB = TC = TD 11. As shown below, Lily is using the rope through a fixed pulley to move a box with constant speed v. The kinetic friction coefficient between the box and the ground is µ < 1; assume that the fixed pulley is massless and there is no friction between the rope and the fixed pulley. Then, while the box is moving, which of the following statements is correct?

(A) The magnitude of the force on the rope is constant. (B) The magnitude of friction between the ground and the box is decreasing. ← CORRECT (C) The magnitude of the normal force of the ground on the box is increasing. (D) The pressure of the box on the ground is increasing. (E) The pressure of the box on the ground is constant.


2012 F = ma Exam

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12. A rigid hoop can rotate about the center. Two massless strings are attached to the hoop, one at A, the other at B. These strings are tied together at the center of the hoop at O, and a weight G is suspended from that point. The strings have a fixed length, regardless of the tension, and the weight G is only supported by the strings. Originally OA is horizontal.

B T2

T1 A

O G

Now, the outer hoop will start to slowly rotate 90◦ clockwise until OA will become vertical, while keeping the angle between the strings constant and keeping the object static. Which of the following statements about the tensions T1 and T2 in the two strings is correct? (A) T1 always decreases. (B) T1 always increases. (C) T2 always increases. (D) T2 will become zero at the end of the rotation.← CORRECT (E) T2 first increases and then decreases. 13. Shown below is a graph of the x component of force versus position for a 4.0 kg cart constrained to move in one dimension on the x axis. At x = 0 the cart has a velocity of −3.0 m/s (in the negative direction). Which of the following is closest to the maximum speed of the cart?

F (Newtons)

6

−4

4 2

−2

2

4 x (meters)

6

−2

(A) 1.6 m/s (B) 2.5 m/s (C) 3.0 m/s (D) 4.0 m/s (E) 4.2 m/s← CORRECT


8

2012 F = ma Exam

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14. A uniform cylinder of radius a originally has a weight of 80 N. After an off-axis cylinder hole at 2a/5 was drilled through it, it weighs 65 N. The axes of the two cylinders are parallel and their centers are at the same height. a

2a/5

A force T is applied to the top of the cylinder horizontally. In order to keep the cylinder is at rest, the magnitude of the force is closest to: (A) 6 N← CORRECT (B) 10 N (C) 15 N (D) 30 N (E) 38 N 15. A car of mass m has an engine that provides a constant power output P . Assuming no friction, what is the maximum constant speed vmax that this car can drive up a long incline that makes an angle θ with the horizontal? (A) vmax = P/(mg sin θ)← CORRECT (B) vmax = P 2 sin θ/mg p (C) vmax = 2P/mg/ sin θ (D) There is no maximum constant speed. (E) The maximum constant speed depends on the length of the incline. 16. Inside a cart that is accelerating horizontally at acceleration ~a, there is a block of mass M connected to two light springs of force constants k1 and k2 . The block can move without friction horizontally. Find the vibration frequency of the block.

1 2π

q

k1 +k2 M

(B)

1 2π

q

k1 k2 (k1 +k2 )M

(C)

1 2π

q

k1 k2 (k1 +k2 )M

q

|k1 −k2 | M

q

k1 +k2 M ←

(A)

(D)

1 2π

(E)

1 2π

+a

+a

CORRECT


2012 F = ma Exam

7

17. Shown below is a log/log plot for the data collected of amplitude and period of oscillation for certain non-linear oscillator.

log (T)

8

6

4

2 0

1

log (A)

2

According to the data, the relationship between period T and amplitude A is best given by (A) T = 1000A2 ← CORRECT (B) T = 100A3 (C) T = 2A + 3 √ (D) T = 3 A (E) Period is independent of amplitude for oscillating systems 18. A mass hangs from the ceiling of a box by an ideal spring. With the box held fixed, the mass is given an initial velocity and oscillates with purely vertical motion. When the mass reaches the lowest point of its motion, the box is released and allowed to fall. To an observer inside the box, which of the following quantities does not change when the box is released? Ignore air resistance. (A) The amplitude of the oscillation (B) The period of the oscillation← CORRECT (C) The maximum speed reached by the mass (D) The height at which the mass reaches its maximum speed (E) The maximum height reached by the mass 19. A 1,500 Watt motor is used to pump water a vertical height of 2.0 meters out of a flooded basement through a cylindrical pipe. The water is ejected though the end of the pipe at a speed of 2.5 m/s. Ignoring friction and assuming that all of the energy of the motor goes to the water, which of the following is the closest to the radius of the pipe? The density of water is ρ = 1000 kg/m3 . (A) 1/3 cm (B) 1 cm (C) 3 cm (D) 10 cm ← CORRECT (E) 30 cm


2012 F = ma Exam

8

45.0 kg

??? kg

12.0 kg

20. A container of water is sitting on a scale. Originally, the scale reads M1 = 45 kg. A block of wood is suspended from a second scale; originally the scale read M2 = 12 kg. The density of wood is 0.60 g/cm3 ; the density of the water is 1.00 g/cm3 . The block of wood is lowered into the water until half of the block is beneath the surface. What is the resulting reading on the scales?

??? kg

(A) M1 = 45 kg and M2 = 2 kg. (B) M1 = 45 kg and M2 = 6 kg. (C) M1 = 45 kg and M2 = 10 kg. (D) M1 = 55 kg and M2 = 6 kg. (E) M1 = 55 kg and M2 = 2 kg.← CORRECT 21. A spring system is set up as follows: a platform with a weight of 10 N is on top of two springs, each with spring constant 75N/m. On top of the platform is a third spring with spring constant 75N/m. If a ball with a weight of 5.0 N is then fastened to the top of the third spring and then slowly lowered, by how much does the height of the spring system change? The ball is placed here

Original height

(A) 0.033 m (B) 0.067 m (C) 0.100 m← CORRECT (D) 0.133 m (E) 0.600 m 22. The softest audible sound has an intensity of I0 = 10−12 W/m2 . In terms of the fundamental units of kilograms, meters, and seconds, this is equivalent to (A) I0 = 10−12 kg/s3 ← CORRECT (B) I0 = 10−12 kg/s (C) I0 = 10−12 kg2 m/s (D) I0 = 10−12 kg2 m/s2 (E) I0 = 10−12 kg/m · s3


2012 F = ma Exam

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23. Which of the following sets of equipment cannot be used to measure the local value of the acceleration due to gravity (g)? (A) A spring scale (which reads in force units) and a known mass. (B) A rod of known length, an unknown mass, and a stopwatch. (C) An inclined plane of known inclination, several carts of different known masses, and a stopwatch.← CORRECT (D) A launcher which launches projectiles at a known speed, a projectile of known mass, and a meter stick. (E) A motor with a known output power, a known mass, a piece of string of unknown length, and a stopwatch. 24. Three point masses m are attached together by identical springs. When placed at rest on a horizontal surface the masses form a triangle with side length l. When the assembly is rotated about its center at angular velocity ω, the masses form a triangle with side length 2l. What is the spring constant k of the springs? (A) 2mω 2 (B) (C) (D) (E)

√2 mω 2 3 2 2 3 mω ← √1 mω 2 3 1 2 mω 3

CORRECT

25. Consider the two orbits around the sun shown below. Orbit P is circular with radius R, orbit Q is elliptical such that the farthest point b is between 2R and 3R, and the nearest point a is between R/3 and R/2. Consider the magnitudes of the velocity of the circular orbit vc , the velocity of the comet in the elliptical orbit at the farthest point vb , and the velocity of the comet in the elliptical orbit at the nearest point va . Which of the following rankings is correct?

c

vc

orbit Q va

a

R

b vb

orbit P

(A) vb > vc > 2va (B) 2vc > vb > va (C) 10vb > va > vc ← CORRECT (D) vc > va > 4vb √ (E) 2va > 2vb > vc


2012 F = ma Exam

1 1d 2a 3c 4b 5d

10

Solutions 6b 7e 8b 9c 10d

11b 12d 13e 14a 15a

16e 17a 18b 19d 20e

21c 22a 23c 24c 25c



2012 Semifinal Exam

1

AAPT AIP


Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minute break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 1, 2012. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A.


2012 Semifinal Exam

AAPT AIP

Cover Sheet

2





2012 Semifinal Exam

Part A

3

Part A Question A1 A newly discovered subatomic particle, the S meson, has a mass M . When at rest, it lives for exactly τ = 3 × 10−8 seconds before decaying into two identical particles called P mesons (peons?) that each have a mass of αM . a. In a reference frame where the S meson is at rest, determine i. the kinetic energy, ii. the momentum, and iii. the velocity of each P meson particle in terms of M , α, the speed of light c, and any numerical constants. b. In a reference frame where the S meson travels 9 meters between creation and decay, determine i. the velocity and ii. kinetic energy of the S meson. Write the answers in terms of M , the speed of light c, and any numerical constants.

Solution a. Apply conservation of four momentum. For the S meson, we have pS c = (Es , 0) and for the two P mesons we have pP c = (Ep , ±p), where p is the magnitude of the (relativistic) three momentum of the P mesons. This yields ES = 2EP We must also satisfy the relation E 2 = p2 c2 + m2 c4 for each particle, so ES 2 = M 2 c4 and EP 2 = p2 c2 + α2 M 2 c4 . Therefore, the kinetic energy of each P meson is 1 KP = EP − αM c = M c2 − αM c2 = 2 2

1 − α M c2 . 2


2012 Semifinal Exam

Part A

4

Square the energy conservation expression, and combine with the momentum/energy/mass relations:

1 2 4 M c = p2 c2 + α2 M 2 c4 , 4

1 − α2 M 2 c4 = p2 c2 4 r 1 − α2 M c2 = pc. 4

The velocity of each P meson will be found from the relativistic three momentum, p = mγv and the relativistic energy, E = γmc2 so pc E

Putting in the values for the P meson, q v = c

1 4

mcγv γmc2 = β. =

− α2 M c2 1 2 2Mc

 p  = c 1 − 4α2

b. From relativistic kinematics, d = vt = vγτ, so

v d γ= c cτ

Call this k for now. Then k = βγ, k2 = β 2 γ 2 , β2 , k2 = 1 − β2 k 2 (1 − β 2 ) = β 2 , k 2 = (1 + k 2 )β 2 , k2 1 + k2

= β2.

Combine, r v=c

d2 c2 τ 2 + d2


2012 Semifinal Exam

Part A

so

r v=c

Then

c 92 =√ 92 + 92 2

p √ γ = 1/ 1 − 1/2 = 2

It isn’t much work to find the kinetic energy, √ K = (γ − 1)M c2 = ( 2 − 1)M c2 . c. This is a velocity addition problem, so v=

vS + vS 1 + vS 2 /c2

or, using the numbers from the first part of the problem, √ 1 − 4α2 v=c 1 − 2α2


5

2012 Semifinal Exam

Part A

6

Question A2 An ideal (but not necessarily perfect monatomic) gas undergoes the following cycle. • The gas starts at pressure P0 , volume V0 and temperature T0 . • The gas is heated at constant volume to a pressure αP0 , where α > 1. • The gas is then allowed to expand adiabatically (no heat is transferred to or from the gas) to pressure P0 • The gas is cooled at constant pressure back to the original state. The adiabatic constant γ is defined in terms of the specific heat at constant pressure Cp and the specific heat at constant volume Cv by the ratio γ = Cp /Cv . a. Determine the efficiency of this cycle in terms of α and the adiabatic constant γ. As a reminder, efficiency is defined as the ratio of work out divided by heat in. b. A lab worker makes measurements of the temperature and pressure of the gas during the adiabatic process. The results, in terms of T0 and P0 are Pressure Temperature

units of P0 units of T0

1.21 2.11

1.41 2.21

1.59 2.28

1.73 2.34

2.14 2.49

Plot an appropriate graph from this data that can be used to determine the adiabatic constant. c. What is γ for this gas?

Solution a. Label the end points as 0, 1, and 2. A quick application of P V = nRT requires that T1 = αT0 . It takes more work to do the process 1 → 2; it is acceptable to simply state the adiabatic law of P V γ = constant; if you don’t know this, you will need to derive it. In the case that you know the adiabatic process law, P1 V1γ = P2 V2γ = αP1 V2γ , so that

1

V2 = V1 (α) γ . 1

Another quick application of P V = nRT requires that T2 = (α) γ T0 . Heat enters the gas during isochoric process 0 → 1, so Qin = nCv ∆T = nCv (α − 1)T0 Heat exits the system during process 2 → 0, so Qout = nCp ∆T = nCp (α1/γ − 1)T0 c Copyright 2012 American Association of Physics Teachers

2012 Semifinal Exam

Part A

7

We only consider absolute values, and insert negative signs later as needed. The work done is the difference, so W = Qin − Qout = nCv (α − 1)T0 − nCp (α1/γ − 1)T0 and the efficiency is then e=

Cv (α − 1) − Cp (α1/γ − 1) Cv (α − 1)

This can be greatly simplified to e=1−γ

α1/γ − 1 α−1

b. Along an adiabatic path, the relationship between pressure and temperature is given by γ T γ P V = constant ∝ P P so

γ

P T 1−γ = constant As such,

γ

P ∝ T γ−1 Note that, for an ideal gas, Cp /Cv Cp γ = = γ−1 Cp /Cv − 1 R This means that we want to plot a log-log plot with log T horizontal and log P vertical. The slope of the graph will be Cp /R. For the data given, Cp = (7/2)R, so γ = 7/5.


2012 Semifinal Exam

Part A

8

Question A3 This problem inspired by the 2008 Guangdong Province Physics Olympiad Two infinitely long concentric hollow cylinders have radii a and 4a. Both cylinders are insulators; the inner cylinder has a uniformly distributed charge per length of +λ; the outer cylinder has a uniformly distributed charge per length of −λ. An infinitely long dielectric cylinder with permittivity = κ0 , where κ is the dielectric constant, has a inner radius 2a and outer radius 3a is also concentric with the insulating cylinders. The dielectric cylinder is rotating about its axis with an angular velocity ω c/a, where c is the speed of light. Assume that the permeability of the dielectric cylinder and the space between the cylinders is that of free space, µ0 .

a. Determine the electric field for all regions. b. Determine the magnetic field for all regions.

Solution a. Consider a Gaussian cylinder of radius r and length l centered on the cylinder axis. Gauss’s Law states that Z qencl E dA = 0 λencl l 2πrEl = 0 λencl E= 2πr0 where λencl is the enclosed linear charge density. The field due to the hollow cylinders alone is therefore   r 4a The field within the dielectric is reduced by a   0    λ    2πr0 λ E = 2πrκ 0   λ    2πr0   0

factor κ, so that in total r 4a


2012 Semifinal Exam

Part A

9

b. We can apply the results of the previous section to obtain the enclosed charge density λencl as a function of radius:   0 r < a     λ a < r < 2a λencl = λκ 2a < r < 3a    λ 3 < r < 4a     0 r > 4a Defining λi =

1 1− κ

λ

we conclude that a charge density −λi exists on the inner surface of the dielectric, a charge density λi on the outer surface, and no charge on the interior. As with the case of a very long solenoid, we expect the magnetic field to be entirely parallel to the cylinder axis, and to go to zero for large r. Consider an Amperian loop of length l extending along a radius, the inner side of which is at radius r and the outer side of which is at a very large radius. We have on this loop I B dl = µ0 Iencl Letting B now be the magnetic field at radius r, lB = µ0 Iencl µ0 Iencl l For r > 3a, Iencl = 0, since the charge on the hollow cylinders is not moving. For 2a < r < 3a, the loop now encloses the outer surface of the dielectric. In time 2π ω a charge λi l passes through the loop, so the current due to the outer surface is B=

Iout =

λi lω 2π

and thus this is Iencl for 2a < r < 3a. For r < 2a, the loop now encloses both surfaces of the dielectric; the inner surface contributes a current that exactly cancels the outer one, so again Iencl = 0. Putting this together,   r < 2a 0 µ0 ω B= 2π λi 2a < r < 3a   0 r > 3a or, using our expression for λi ,   0 B= 1 − κ1 µ02πωλ   0

r < 2a 2a < r < 3a r > 3a


2012 Semifinal Exam

Part A

10

Question A4 Two masses m separated by a distance l are given initial velocities v0 as shown in the diagram. The masses interact only through universal gravitation.

v0 l v0

a. Under what conditions will the masses eventually collide? b. Under what conditions will the masses follow circular orbits of diameter l? c. Under what conditions will the masses follow closed orbits? d. What is the minimum distance achieved between the masses along their path?

Solution a. In order for the masses to collide, the total angular momentum of the system must be zero, which only occurs if v0 = 0. b. In this case, the masses undergo uniform circular motion with radius

l 2

and speed v0 , so that

Gm2 mv0 2 = l 2 l 2 Gm =2 v0 2 l c. The masses follow closed orbits if they do not have enough energy to escape, i.e. if the total energy of the system is negative. The total energy of the system is Gm2 1 2 · mv0 2 − 2 l so that the condition required is Gm2 <0 l Gm >1 v0 2 l

mv0 2 −


2012 Semifinal Exam

Part A

11

d. Note that the masses will always move symmetrically about the center of mass. Thus, in order to be at minimum separation, their velocities must be perpendicular to the line joining them (and will be oppositely directed). Let the minimum separation be d, and let the speed of each mass at minimum separation be v. L = 2mv

d = mvd 2

The initial angular momentum is likewise mv0 l, and so by conservation of angular momentum mvd = mv0 l v = v0

l d

By conservation of energy 1 Gm2 1 Gm2 2 · mv0 2 − = 2 · mv 2 − 2 l 2 d v0 2 −

Gm Gm = v2 − l d

Combining these, l2 Gm Gm = v0 2 2 − l d d 2 Gm d Gm d 1− 2 + 2 −1=0 v0 l l v0 l l Gm d d −1 1− 2 +1 =0 l v0 l l v0 2 −

so that d=l

or d =

l Gm v0 2 l

−1

Gm The second root is only sensible if vGm 2 > 1, and is only smaller than the first if v 2 l > 2. 0 l 0 (Note that both of these results make sense in light of the previous ones.) Thus the minimum l otherwise. separation is l if vGm 2 ≤ 2 and Gm 0 l −1 v0 2 l


2012 Semifinal Exam

Part A

12




2012 Semifinal Exam

Part B

13

Part B Question B1 A particle of mass m moves under a force similar to that of an ideal spring, except that the force repels the particle from the origin: F = +mα2 x In simple harmonic motion, the position of the particle as a function of time can be written x(t) = A cos ωt + B sin ωt Likewise, in the present case we have x(t) = A f1 (t) + B f2 (t) for some appropriate functions f1 and f2 . a. f1 (t) and f2 (t) can be chosen to have the form ert . What are the two appropriate values of r? b. Suppose that the particle begins at position x(0) = x0 and with velocity v(0) = 0. What is x(t)? c. A second, identical particle begins at position x(0) = 0 with velocity v(0) = v0 . The second particle becomes closer and closer to the first particle as time goes on. What is v0 ?

Solution a. We have ma = mα2 x d2 x − α2 x = 0 dt2 As with the case of simple harmonic motion, we insert a trial function, in this case x(t) = Aert : d2 Aert − α2 Aert = 0 dt2 r2 Aert − α2 Aert = 0 r 2 − α2 = 0 r = ±α b. We have x(t) = Aeαt + Be−αt and therefore v(t) = αAeαt − αBe−αt Inserting our initial values, x(0) = A + B = x0 c Copyright 2012 American Association of Physics Teachers

2012 Semifinal Exam

Part B

14

v(0) = αA − αB = 0 These equations have solution A=B= and therefore x(t) =

x0 2

x0 αt x0 −αt e + e 2 2

c. This time our initial values are x(0) = A + B = 0 v(0) = αA − αB = v0 with solution

v0 2α v0 B=− 2α A=

Therefore x(t) =

v0 αt x0 −αt e − e 2α 2α

After a long time, the second (e−αt ) term will become negligible. Thus, the second particle will approach the first particle if the first term matches: x0 αt v0 αt e = e 2 2α v0 = αx0


2012 Semifinal Exam

Part B

15

Question B2 For this problem, assume the existence of a hypothetical particle known as a magnetic monopole. Such a particle would have a “magnetic charge” qm , and in analogy to an electrically charged particle would produce a radially directed magnetic field of magnitude B=

µ0 qm 4π r2

and be subject to a force (in the absence of electric fields) F = qm B A magnetic monopole of mass m and magnetic charge qm is constrained to move on a vertical, nonmagnetic, insulating, frictionless U-shaped track. At the bottom of the track is a wire loop whose radius b is much smaller than the width of the “U” of the track. The section of track near the loop can thus be approximated as a long straight line. The wire that makes up the loop has radius a b and resistivity ρ. The monopole is released from rest a height H above the bottom of the track. Ignore the self-inductance of the loop, and assume that the monopole passes through the loop many times before coming to a rest. a. Suppose the monopole is a distance x from the center of the loop. What is the magnetic flux φB through the loop? b. Suppose in addition that the monopole is traveling at a velocity v. What is the emf E in the loop? c. Find the change in speed ∆v of the monopole on one trip through the loop. d. How many times does the monopole pass through the loop before coming to a rest? e. Alternate Approach: You may, instead, opt to find the above answers to within a dimensionless multiplicative constant (like 23 or π 2 ). If you only do this approach, you will be able to earn up to 60% of the possible score for each part of this question. You might want to make use of the integral Z ∞ 1 3π du = 2 3 8 −∞ (1 + u ) or the integral Z

π

sin4 θ dθ =

0

3π 8

Solution Version 1 The magnetic field around a monopole is given by B=

µ0 qm 4π r2


2012 Semifinal Exam

Part B

16

The flux through the loop will then be Z Z ~ · dA ~ = 1 µ0 qm sin θ dθ ΦB = B 2 where θ is the angle between a line along the axis of the loop and a line drawn between the monopole and any point on the rim of the loop. It is easy to see that 1 dΦB = µ0 qm sin θ 2 From Faraday’s law, we have that a changing flux will induce a current I in the loop. dΦB = IR, dt where R is the resistance of the loop. We’ll figure R out later. The induced current will create a magnetic field that will oppose the monopole motion. We need to use the law of Biot & Savart to find that field. Along the axis of the loop, we have µ0 I d~l × ~r , 4π r3

~ = dB

where ~r is a vector connecting the monopole with some point on the rim of the loop. Only compo~ parallel to the axis of the loop will survive, so we can concern ourselves with nents of B µ0 I dl sin θ 4π r2

dB =

The integral is trivial; dl is around the circumference; nothing else changes, so B=

µ0 Ib sin θ 2r2

It is better to think in terms of b, the radius of the loop, than it is to deal with r, the distance from the rim of the loop to the monopole. In that case, b sin θ = , r so B=

µ0 I (sin θ)3 2b

The force on the monopole is then F = qm B = qm

µ0 1 dΦB µ0 I (sin θ)3 = qm (sin θ)3 2b 2b R dt

Note than multiplying through by dt gives an expression that is related to the change in momentum, dp = qm

µ0 1 µ0 2 (sin θ)3 dΦB = qm 2 (sin θ)4 dθ 2b R 4bR

Using the provided integral, ∆p =

qm 2 µ0 2 3π 4bR 8


2012 Semifinal Exam

Part B

17

as the monopole moves from one side to the other. If the monopole started from rest a distance H above the loop, then the initial energy of the system is mgH, and the initial momentum when passing through the loop (assuming there is no loop) is then p p0 = m 2gH The monopole will lose ∆p from the momentum on each pass through the loop, so the number of times it passes through the loop N is N=

p p0 = m 2gH ∆p

or

qm 2 µ0 2 3π 4bR 8

−1

√ 32bRm 2gH N= 3πµ0 2 qm 2

Oh, we still need to do R. Since a b, we can treat it as a long thin cylindrical wire, and R= so we finally get

ρ2πb πa2

√ 64b2 ρm 2gH N= 3πµ0 2 qm 2 a2

Version 2 Instead of force, focus on the power dissipated in the loop, which is P =

1 2 µ0 qm

P =

P =

E2 R b2 3

(b2 +x2 ) 2 2πbρ πa2

dx dt

1 µ0 2 qm 2 a2 b3 8ρ (b2 + x2 )3

2

dx dt

2

The energy is lost from the particle: d 1 mv 2 dt 2 dv P = −mv dt

P =−

Combining these, and since v =

dx dt ,

µ0 2 qm 2 a2 b3 dx 1 dv =− 2 2 3 dt 8ρm (b + x ) dt Use the provided integral (or do a trig subsitution that gives the other provided integral), and continue.



2013 F = ma Exam

1

AAPT AIP

UNITED STATES PHYSICS TEAM CEE 2013 2013 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS


1 4

point. There is


DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Contributors to this year’s exam include David Fallest, Jiajia Dong, Paul Stanley, Warren Turner, Qiuzi Li, and former US Team members Marianna Mao, Andrew Lin, Steve Byrnes.


2013 F = ma Exam

2

1. An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, it takes 5 seconds for the first car to pass the observer. How long will it take for the 10th car to pass? (A) 1.07s (B) 0.98s (C) 0.91s (D) 0.86s (E) 0.81s

Solution Start with

1 2 at + vi t 2 We have four times. t0 = 0 is when the train starts, and when the first car is aligned with the observer. t1 is when the end of the first car is aligned with the observer. Then ∆x =

L=

1 2 at1 2

We are assuming the car has a length L. t2 is when the tenth car is first aligned with the observer, so 9L =

1 2 at2 2

10L =

1 2 at3 2

and finally, t3 is when that car has passed,

From the equation for t1 we find 2L/a = 25 s2 so

√ t2 =

and

√ t3 =

9 · 25 s2 = 15 s

10 · 25 s2 = 15.81 s


2013 F = ma Exam

3

2. Jordi stands 20 m from a wall and Diego stands 10 m from the same wall. Jordi throws a ball at an angle of 30◦ above the horizontal, and it collides elastically with the wall. How fast does Jordi need to throw the ball so that Diego will catch it? Consider Jordi and Diego to be the same height, and both are on the same perpendicular line from the wall. (A) 11 m/s (B) 15 m/s (C) 19 m/s (D) 30 m/s (E) 35 m/s

Solution The wall acts like a mirror for a perfectly elastic collision. In that case, Jordi and Diego are effectively 30 meters apart. Using the range formula, v2 sin 2θ, R= g we get v2 =

gR 300 m2 /s2 ≈ √ sin 2θ 3/2 v ≈ 19 m/s


2013 F = ma Exam

4

3. Tom throws a football to Wes, who is a distance l away. Tom can control the time of flight t of the ball by choosing any speed up to vmax and any launch angle between 0◦ and 90◦ . Ignore air resistance and assume Tom and Wes are at the same height. Which of the following statements is incorrect? √ (A) If vmax < gl, the ball cannot reach Wes at all. (B) Assuming the ball can reach Wes, as vmax increases with l held fixed, the minimum value of t decreases. (C) Assuming the ball can reach Wes, as vmax increases with l held fixed, the maximum value of t increases. (D) Assuming the ball can reach Wes, as l increases with vmax held fixed, the minimum value of t increases. (E) Assuming the ball can reach Wes, as l increases with vmax held fixed, the maximum value of t increases.

Solution The greater the initial vertical velocity of the football, the longer it will take to fall back to the ground. Meanwhile, the initial horizontal velocity of the football increases with l and decreases with the time of flight. Thus the minimum time of flight is obtained by the minimum possible initial vertical velocity; it is limited by the increasing required horizontal velocity, and this limitation becomes more severe as l increases. The maximum time of flight is obtained by the maximum possible initial vertical velocity; it is limited by the required horizontal velocity component, and this limitation likewise becomes more severe as l increases. Thus increasing vmax expands the range of available time of flight in both directions; increasing l contracts it in both directions. (A) follows from the standard result for the maximum range of a projectile. For those who prefer the mathematical approach, we have for a launch angle θ and initial speed v0 2v0 sin θ = gt (v0 cos θ)t = l 2

Since sin θ + cos2 θ = 1, v0 2 =

gt 2

2 +

2 l t

g2 4 t − v0 2 t2 + l2 = 0 4 p v0 2 ± v0 4 − g 2 l 2 2 t =2 g2 (A) follows immediately from the requirement that the result be real. It is also clear that increasing l contracts the range of available times in both directions, and that increasing vmax increases the maximum available time. As for the minimum available time, we have p v0 2 − v0 4 − g 2 l 2 2 t− = 2 g2 ! ∂ t− 2 2 2v0 2 = 2 1− p ∂ (v0 2 ) g 2 v0 4 − g 2 l 2   ∂ t− 2 2  1  = 2 1− q 2 2 ∂ (v0 2 ) g 1− g l v0 4

The right hand side is always negative, so t− is always minimized by choosing the largest possible v0 .


2013 F = ma Exam

5

4. The sign shown below consists of two uniform legs attached by a frictionless hinge. The coefficient of friction between the ground and the legs is µ. Which of the following gives the maximum value of θ such that the sign will not collapse?

θ

L

(A) sin θ = 2µ (B) sin θ/2 = µ/2 (C) tan θ/2 = µ (D) tan θ = 2µ (E) tan θ/2 = 2µ

Solution For each leg, friction must balance the horizontal force from the other leg at the top. Balancing torques, L (M/2)g sin α = LFf cos α 2 where Ff is the force of friction on the leg, and α = 21 θ. The upward normal force on the leg is equal to the weight of the leg, so FN = (M/2)g Equating, µFN = Ff or

L (M/2)g sin α = Lµ(M/2)g cos α 2

or tan θ/2 = 2µ


2013 F = ma Exam

6

The following information applies to questions 5 and 6 A student steps onto a stationary elevator and stands on a bathroom scale. The elevator then travels from the top of the building to the bottom. The student records the reading on the scale as a function of time. 120

Scale reading (kg)

100 80 60 40 20 0

0

2

4

6

8

10

12

14

16

18

20

22

24

26

Time (s) 5. At what time(s) does the student have maximum downward velocity? (A) At all times between 2 s and 4 s (B) At 4 s only (C) At all times between 4 s and 22 s (D) At 22 s only (E) At all times between 22 s and 24 s 6. How tall is the building? (A) 50 m (B) 80 m (C) 100 m (D) 150 m (E) 400 m

Solution The bathroom scale does not directly measure the weight of the student; instead it measures the normal force FN supporting her feet, scaled by g. mscale g = FN The normal force acts upward on the student. Gravity, of course, always exerts a force mg downward. Thus her downward acceleration a is given by ma = mg − FN Combining these and rearranging,

mscale a=g 1− m

Since the elevator begins at rest, the student’s acceleration is initially zero, so her mass must be 80 kg. From


2013 F = ma Exam

7

the graph we see that there are two periods of uniform acceleration; the first is downward with magnitude 60 kg 2 a = (10 m/s ) 1 − = 2.5 m/s2 80 kg and the second is upward with the same magnitude. The maximum downward velocity occurs between the periods of uniform acceleration, when the car travels at a constant speed; this occurs between 4 s and 22 s. The downward acceleration lasts for 2 s, so the speed of the elevator after the acceleration is 5 m/s. The period between accelerations is approximately 20 s, and so the distance traveled is approximately 100 m. In fact, this calculation yields the exact answer, as can be seen by considering the exact graph of velocity vs. time: 0

Velocity (m/s)

−1 −2 −3 −4 −5

0

2

4

6

8

10

12

14

16

18

20

22

24

26

Time (s) The displacement is the area between the curve and the horizontal axis, which is exactly 100 m.

7. A light car and a heavy truck have the same momentum. The truck weighs ten times as much as the car. How do their kinetic energies compare? (A) The truck’s kinetic energy is larger by a factor of 100 (B) They truck’s kinetic energy is larger by a factor of 10 (C) They have the same kinetic energy (D) The car’s kinetic energy is larger by a factor of 10 (E) The car’s kinetic energy is larger by a factor of 100

Solution Express kinetic energy in terms of momentum, K= Then the ratio is

p2 2m

K1 m2 = K2 m1


2013 F = ma Exam

8

The following information applies to questions 8 and 9 A truck is initially moving at velocity v. The driver presses the brake in order to slow the truck to a stop. The brake applies a constant force F to the truck. The truck rolls a distance x before coming to a stop, and the time it takes to stop is t. 8. Which of the following expressions is equal the initial kinetic energy of the truck (i.e. the kinetic energy before the driver starts braking)? (A) F x (B) F vt (C) F xt (D) F t (E) Both (a) and (b) are correct

Solution Considering magnitudes only, Ki = δK = W = F x

9. Which of the following expressions is equal the initial momentum of the truck (i.e. the momentum before the driver starts braking)? (A) F x (B) F t/2 (C) F xt (D) 2F t (E) 2F x/v

Solution Considering magnitudes only, pi = ∆p = F t but x=

1 1 1 2 at = (at)t = vt 2 2 2

so pi = 2F x/v


2013 F = ma Exam

9

10. Which of the following can be used to distinguish a solid ball from a hollow sphere of the same radius and mass? (A) Measurements of the orbit of a test mass around the object. (B) Measurements of the time it takes the object to roll down an inclined plane. (C) Measurements of the tidal forces applied by the object to a liquid body. (D) Measurements of the behavior of the object as it floats in water. (E) Measurements of the force applied to the object by a uniform gravitational field.

Solution The measurement described in (B) is certainly appropriate; the solid ball has a smaller moment of inertia than the hollow sphere and will accelerate down the inclined plane faster. The measurements in (A) and (C) are unhelpful because they only probe the gravitational field outside the object; for a spherically symmetric object this depends only on the mass. The force measured in (E) depends only on the object’s total mass; the buoyant force applied in (D) depends only on the external shape of the object.


2013 F = ma Exam

10

11. A right-triangular wooden block of mass M is at rest on a table, as shown in figure. Two smaller wooden cubes, both with mass m, initially rest on the two sides of the larger block. As all contact surfaces are frictionless, the smaller cubes start sliding down the larger block while the block remains at rest. What is the normal force from the system to the table?

m

90

m

M α

β

(A) 2mg (B) 2mg + M g (C) mg + M g (D) M g + mg(sin α + sin β) (E) M g + mg(cos α + cos β)

Solution Two forces act on each cube: the normal force from the triangular block, and gravity. The normal force must balance the normal component of gravity, which in the case of the left cube is FN = mg cos α The vertical component of this normal force is transmitted through the triangular block to the ground, and is FN y = mg cos2 α A similar result holds for the other cube, and in addition the ground must support the weight of the triangular block; thus in total Ftot = mg cos2 α + mg cos2 β + M g However, because the block is a right triangle, cos2 α + cos2 β = 1, so that Ftot = mg + M g Note that the horizontal component of the normal force due to the left cube is FN x = mg cos α sin α The right cube likewise applies a normal force with horizontal component mg cos β sin β in the other direction. But, once again, because the block is a right triangle, cos α sin α = cos β sin β, so the net horizontal force is zero! This justifies the given assumption that the triangular block does not slide on the table.


2013 F = ma Exam

11

12. A spherical shell of mass M and radius R is completely filled with a frictionless fluid, also of mass M . It is released from rest, and then it rolls without slipping down an incline that makes an angle θ with the horizontal. What will be the acceleration of the shell down the incline just after it is released? Assume the acceleration of free fall is g. The moment of inertia of a thin shell of radius r and mass m about the center of mass is I = 32 mr2 ; the moment of inertia of a solid sphere of radius r and mass m about the center of mass is I = 25 mr2 . (A) a = g sin θ (B) a = 43 g sin θ (C) a = 12 g sin θ (D) a = 38 g sin θ (E) a = 35 g sin θ

Solution One can use torque or energy to solve this problem. The torque about an axis through the point of contact is τ = RF sin θ = 2M gR sin θ The angular acceleration is given by τ = Iα where the moment of inertia is I=

2 8 M R2 + M R2 + M R2 = M R2 3 3

The acceleration is then a = αR =

2M gR sin θ 3 R = g sin θ 8 2 4 M R 3

Alternatively, the kinetic energy of the object is T =

1 1 2 (2M )v 2 + · M R2 ω 2 2 2 3

4 M v2 3 The potential energy is related to the vertical position y by T =

U = −(2M )gy and so by conservation of energy d (T + U ) = 0 dt 8 dv dy Mv = −(2M )g 3 dt dt But

and so

dy = −v sin θ dt 3 dv = g sin θ dt 4


2013 F = ma Exam

12

13. There is a ring outside of Saturn. In order to distinguish if the ring is actually a part of Saturn or is instead part of the satellites of Saturn, we need to know the relation between the velocity v of each layer in the ring and the distance R of the layer to the center of Saturn. Which of the following statements is correct? (A) If v ∝ R, then the layer is part of Saturn. (B) If v 2 ∝ R, then the layer is part of the satellites of Saturn. (C) If v ∝ 1/R, then the layer is part of Saturn. (D) If v 2 ∝ 1/R, then the layer is part of Saturn. (E) If v ∝ R2 , then the layer is part of the satellites of Saturn.

Solution If attached to Saturn, then ω = v/R is a constant, so v∝R If in orbit, then GM v2 = 2 , R r or v 2 ∝ 1/R

14. A cart of mass m moving at 12 m/s to the right collides elastically with a cart of mass 4.0 kg that is originally at rest. After the collision, the cart of mass m moves to the left with a velocity of 6.0 m/s. Assuming an elastic collision in one dimension only, what is the velocity of the center of mass (vcm ) of the two carts before the collision? (A) vcm = 2.0 m/s (B) vcm = 3.0 m/s (C) vcm = 6.0 m/s (D) vcm = 9.0 m/s (E) vcm = 18 m/s

Solution In the center of mass frame, the carts initially have equal and opposite momenta, so that the total momentum is zero. After the elastic collision in this frame the momenta simply reverse direction; this clearly conserves total energy and momentum. The frame of reference in which the velocity of the cart of mass m is reversed travels to the right at 3 m/s, so that the initial velocity is 9 m/s to the right and the final velocity is 9 m/s to the left. Note that the mass and initial velocity of the other cart are not required at all!


2013 F = ma Exam

13

15. A uniform rod is partially in water with one end suspended, as shown in figure. The density of the rod is 5/9 that of water. At equilibrium, what portion of the rod is above water?

11111111111111111 00000000000000000 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 (A) 0.25 (B) 0.33 (C) 0.5 (D) 0.67 (E) 0.75

Solution Suppose a fraction α of the rod is above water. Let the length of the rod be l, the volume of the rod be V , the density of water be ρw , and the density of the rod be ρr . Consider torques about the pivot point. Gravity applies a torque 1 τg = ρr V g · l 2 A fraction 1 − α of the rod is submerged, and the center of the submerged portion is a distance (α + 12 (1 − α))l from the pivot. So the buoyant force applies a torque 1 τb = ρw (1 − α)V g · α + (1 − α) l 2 τb =

1 ρw (1 − α2 )V gl 2

These torques must balance: τg = τb 1 1 ρr = ρw (1 − α2 ) 2 2 ρr = 1 − α2 ρw We are given

ρr ρw

= 95 , so that 5 = 1 − α2 9 2 α= 3


2013 F = ma Exam

14

16. Inspired by a problem from the 2012 International Physics Olympiad, Estonia. A very large number of small particles forms a spherical cloud. Initially they are at rest, have uniform mass density per unit volume ρ0 , and occupy a region of radius r0 . The cloud collapses due to gravitation; the particles do not interact with each other in any other way. q How much time passes until the cloud collapses fully? (The constant 0.5427 is actually 3π 32 .) (A)

0.5427 √ r0 2 Gρ0

(B)

0.5427 √ r0 Gρ0

(C)

√ √0.5427 r0 Gρ0

(D)

0.5427 √ Gρ0

(E)

0.5427 √ r Gρ0 0

Solution This problem is a matter of dimensional analysis. All of the answers have the form −1/2

t = 0.527G−1/2 ρ0

r0 n

The dimensions of t are [T ], the dimensions of G are [L]3 [M ]−1 [T ]−2 , the dimensions of ρ0 are [L]−3 [M ], and the dimensions of r0 are [L]. So [T ] = ([L]3 [M ]−1 [T ]−2 )−1/2 ([L]−3 [M ])−1/2 [L]n [T ] = [T ][L]n n=0 The time to collapse does not depend on the size of the cloud at all!


2013 F = ma Exam

15

17. Two small, equal masses are attached by a lightweight rod. This object orbits a planet; the length of the rod is smaller than the radius of the orbit, but not negligible. The rod rotates about its axis in such a way that it remains vertical with respect to the planet. • Is there a force in the rod? If so, is it tension or compression? • Is the equilibrium stable, unstable, or neutral with respect to a small perturbation in the angle of the rod? (Assume this perturbation maintains the rate of rotation, so that in the co-rotating frame the rod is still stationary but at an angle to the vertical.)

(A) There is no force in the rod; the equilibrium is neutral. (B) The rod is in tension; the equilibrium is stable. (C) The rod is in compression; the equilibrium is stable. (D) The rod is in tension; the equilibrium is unstable. (E) The rod is in compression; the equilibrium is unstable.

Solution Because the gravitational force goes as 1/r2 , the inward force on the inner mass is greater than the inward force on the outer mass. On the other hand, both masses have the same angular velocity, so the centripetal acceleration goes as r, and the outer mass must be subject to a greater centripetal force than the inward one. So the rod must exert an inward force on the outer mass and an outward force on the inner mass. Thus the rod is in tension. The stability of the position is best understood in the corotating frame. In this frame the outer mass experiences an outward radial force and the inward tension of the rod, and the inner mass experiences an inward radial force and the outward tension of the rod. When the rod is rotated slightly these forces create a restoring torque, so the equilibrium is stable.


2013 F = ma Exam

16

18. Two point particles, each of mass 1 kg, begin in the state shown below. 3 2 1.0 m/s y (m)

1 0 1.0 m/s

−1 −2 −3 −3 −2 −1

0

1

2

3

x (m)

The system evolves through internal forces only. Which of the following could be the state after some time has passed? 3

3

2

2

1

1 1.0 m/s

0 1.0 m/s

(B)

−1

y (m)

(A)

y (m)

2.0 m/s 0 −1

−2

−2

−3 −3 −2 −1

0

1

2

−3 −3 −2 −1

3

x (m)

0

1

2

3

x (m)

3

3

2

2

1.0 m/s 1

1.0 m/s

0 −1

(D)

1.0 m/s

−2

y (m)

(C)

y (m)

1

0 −1 1.0 m/s

−2

−3 −3 −2 −1

0

1

2

−3 −3 −2 −1

3

x (m)

0

1

2

3

x (m)

3 2 1.0 m/s

(E)

y (m)

1 0 −1 1.0 m/s −2 −3 −3 −2 −1

0

1

2

3

x (m)

Solution First, linear momentum is conserved; as it happens, the system begins with zero linear momentum. This eliminates choice (B). c Copyright 2013 American Association of Physics Teachers

2013 F = ma Exam

17

Angular momentum about any point is also conserved. (It is specified that the particles are point particles to rule out their having spin angular momentum.) As it happens, the system has zero linear momentum, so the angular momentum is the same measured about any point; otherwise it would suffice to consider any one point. The total angular momentum must be 2 kg m2 /s counterclockwise. This eliminates choices (C) and (D). Finally, because the linear momentum of the system is zero, the center of mass does not move. This eliminates choice (A). The following information applies to questions 19, 20, and 21. A simple pendulum experiment is constructed from a point mass m attached to a pivot by a massless rod of length L in a constant gravitational field. The rod is released from an angle θ0 < π/2 at rest and the period of motion is found to be T0 . Ignore air resistance and friction. 19. At what angle θg during the swing is the tension in the rod the greatest? (A) The tension is the greatest at the point θg = θ0 . (B) The tension is the greatest at the point θg = 0. (C) The tension is the greatest at an angle θg with 0 < θg < θ0 . (D) The tension is constant. (E) None of the above is true for all values of θ0 with 0 < θ0 < π/2. 20. What is the maximum value of the tension in the rod? (A) mg (B) 2mg (C) mLθ0 /T0 2 (D) mg sin θ0 (E) mg(3 − 2 cos θ0 ) 21. The experiment is repeated with a new pendulum with a rod of length 4L, using the same angle θ0 , and the period of motion is found to be T . Which of the following statements is correct? (A) T = 2T0 regardless of the value of θ0 . (B) T > 2T0 with T ≈ 2T0 if θ0 1. (C) T < 2T0 with T ≈ 2T0 if θ0 1. (D) T > 2T0 for some values of θ0 and T < 2T0 for other values of θ0 . (E) T0 and T are undefined because the motion is not periodic unless θ0 1.

Solution The mass accelerates towards the pivot due to centripetal acceleration; if its speed is v, this acceleration is ac =

v2 L

Two forces act in the radial direction; the tension in the rod, inward, and the radial component of gravity, outward. At an angle θ to the vertical the radial component of gravity is Fg,rad = mg cos θ From Newton’s second law we thus have for the tension F F − Fg,rad = mac c Copyright 2013 American Association of Physics Teachers

2013 F = ma Exam

18

v2 L Both terms increase as the mass approaches the bottom of the swing, so the maximum tension certainly occurs there. At that point the mass has traveled a vertical distance L(1 − cos θ0 ), so from conservation of energy 1 mv 2 = mgL(1 − cos θ0 ) 2 and so at the bottom F = mg + 2mg(1 − cos θ0 ) F = mg cos θ + m

F = mg(3 − 2 cos θ0 ) The motion is certainly periodic (one argument considers the fact that energy is conserved and the particle eventually returns to rest). From dimensional analysis the period must take the form s l0 T = f (θ0 ) g Therefore, for any fixed θ0 , the period goes as

√

l0 .

22. A simplified model on the foot is shown. When a student of mass m = 60 kg stands on a single toe, the tension T in the Achilles Tendon is closest to Leg Bone Achilles Tendon

Ankle Joint Approximation of Foot Bone Outline of Foot Toe

0

4

8

12

16

20

24

28

32

36

40

Centimeter Rule

(A) T = 600 N (B) T = 1200 N (C) T = 1800 N (D) T = 2400 N (E) T = 3000 N

Solution The entire weight of the student, W = mg = 600 N, is supported by the toe. Balancing torques about the ankle, T · 5 cm = 600 N · 20 cm T = 2400 N


2013 F = ma Exam

19

The following information applies to questions 23 and 24 A man with mass m jumps off of a high bridge with a bungee cord attached to his ankles. The man falls through a maximum distance H at which point the bungee cord brings him to a momentary rest before he bounces back up. The bungee cord is perfectly elastic, obeying Hooke’s force law with a spring constant k, and stretches from an original length of L0 to a final length L = L0 + h. The maximum tension in the Bungee cord is four times the weight of the man. 23. Determine the spring constant k. (A) k = (B) k = (C) k = (D) k = (E) k =

mg h 2mg h mg H 4mg H 8mg H

24. Find the maximum extension of the bungee cord h. (A) h = 12 H (B) h = 41 H (C) h = 51 H (D) h = 52 H (E) h = 81 H

Solution At the moment of maximum extension, all of the gravitational potential energy has been converted to spring potential energy. 1 mgH = kh2 2 We are given that the maximum tension in the cord is four times the weight of the man; this occurs at the moment of maximum extension. kh = 4mg These can be solved to yield 8mg H 1 h= H 2

k=

as well as k=

4mg h

although the latter is not an available answer choice.


2013 F = ma Exam

20

25. A box with weight W will slide down a 30◦ incline at constant speed under the influence of gravity and friction alone. If instead a horizontal force P is applied to the box, the box can be made to move up the ramp at constant speed. What is the magnitude of P ? (A) P = W/2 √ (B) P = 2W/ 3 (C) P = W √ (D) P = 3W (E) P = 2W

Solution If block slides down at constant speed, then tan θ = µ If block is pushed by horizontal force P up ramp at constant speed, then FN = mg cos θ + P sin θ so friction is Ff = µ(mg cos θ + P sin θ) This is balanced by gravity and the push up the ramp, so P cos θ = µ(mg cos θ + P sin θ) + mg sin θ Divide through by cos θ, and use first equation P = µ(W + P µ) + W µ or P = or P =

2µ W 1 − µ2 2 √13 1−

1 3

W



2013 Semifinal Exam

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AAPT AIP

UNITED STATES PHYSICS TEAM CEE 2013

Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-9), and Part B (pages 11-17). Examinees should be provided parts A and B individually, although they may keep the cover sheet. • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 1, 2013. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas. • Please provide the examinees with graph paper for Part A. A straight edge or ruler could also be useful.


2013 Semifinal Exam

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Cover Sheet

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UNITED STATES PHYSICS TEAM CEE 2013




2013 Semifinal Exam

Part A

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Part A Question A1 The flow of heat through a material can be described via the thermal conductivity κ. If the two faces of a slab of material with thermal conductivity κ, area A, and thickness d are held at temperatures differing by ∆T , the thermal power P transferred through the slab is κA∆T d A heat exchanger is a device which transfers heat between a hot fluid and a cold fluid; they are common in industrial applications such as power plants and heating systems. The heat exchanger shown below consists of two rectangular tubes of length l, width w, and height h. The tubes are separated by a metal wall of thickness d and thermal conductivity κ. Originally hot fluid flows through the lower tube at a speed v from right to left, and originally cold fluid flows through the upper tube in the opposite direction (left to right) at the same speed. The heat capacity per unit volume of both fluids is c. The hot fluid enters the heat exchanger at a higher temperature than the cold fluid; the difference between the temperatures of the entering fluids is ∆Ti . When the fluids exit the heat exchanger the difference has been reduced to ∆Tf . (It is possible for the exiting originally cold fluid to have a higher temperature than the exiting originally hot fluid, in which case ∆Tf < 0.) P =

h d h

v l

v w

Assume that the temperature in each pipe depends only on the lengthwise position, and consider transfer of heat only due to conduction in the metal and due to the bulk movement of fluid. Under the assumptions in this problem, while the temperature of each fluid varies along the length of the exchanger, the temperature difference across the wall is the same everywhere. You need not prove this. Find ∆Tf in terms of the other given parameters.

Solution Suppose the temperature difference across the wall is ∆Tw . Since the total area of the wall is simply lw, the power transferred across the wall is P =

κlw ∆Tw d


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Part A

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In a time dt, the energy transferred is therefore dE = P dt =

κlw ∆Tw dt d

Meanwhile, suppose the red fluid enters at temperature Tr and the blue fluid at temperature Tb . The red fluid then exits at temperature Tb + ∆Tw , so the overall temperature change of the red fluid is ∆Tr = Tr − (Tb + ∆Tw ) = ∆Ti − ∆Tw In a time dt, a volume of red fluid vwh dt flows through the pipe; the heat capacity of this amount of fluid is vwhc dt, so the energy transferred out of it is therefore dE = vwhc dt ∆Tr = vwhc(∆Ti − ∆Tw ) dt (Note that the same result is obtained for the heat transferred into the blue fluid; if the flow rates or heat capacities were not the same, this would not hold, exposing the fact that ∆Tw is not constant in that case.) Equating, κlw ∆Tw = vwhc(∆Ti − ∆Tw ) d ∆Ti ∆Tw = κl 1 + dvhc Because the red fluid exits at Tb + ∆Tw , and the blue fluid at Tr − ∆Tw , ∆Tf = (Tb + ∆Tw ) − (Tr − ∆Tw ) = −∆Ti + 2∆Tw ! 2 −1 ∆Tf = ∆Ti κl 1 + dvhc κl ; as The performance of the heat exchanger is controlled entirely by the dimensionless parameter dvhc might be intuitive, a long tube and high conductivity are beneficial, whereas a thick wall, high flow rate, and high heat capacity is not. The poorest performance, unsurprisingly, reflects essentially no heat transfer, with the red fluid and blue fluid exiting with the same temperatures they started with. Interestingly, the limit of performance is a complete reversal in the temperatures of the two fluids, with ∆Tf → −∆Ti .


2013 Semifinal Exam

Part A

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Question A2 A solid round object of radius R can roll down an incline that makes an angle θ with the horizontal. Assume that the rotational inertia about an axis through the center of mass is given by I = βmR2 . The coefficient of kinetic and static friction between the object and the incline is µ. The object moves from rest through a vertical distance h. a. If the angle of the incline is sufficiently large, then the object will slip and roll; if the angle of the incline is sufficiently small, then the object with roll without slipping. Determine the angle θc that separates the two types of motion. b. Derive expressions for the linear acceleration of the object down the ramp in the case of i. Rolling without slipping, and ii. Rolling and slipping.

Solution As the object rolls down the incline, there is a torque about the center of mass given by τ = Rf where f is the force of friction. The angular acceleration is then α=

τ f = I βmR

or, as will be more useful, f = αβmR The object experience a linear acceleration down the incline given by ma = mg sin θ − f We have to cases to consider. Either the object rolls without slipping so that f ≤ µmg cos θ and a = αR, or the object rolls while slipping so that f = µmg cos θ and a > αR. Rolling without slipping Combining the equalities, we get ma = mg sin θ − βma or a=g

sin θ 1+β

Rolling while slipping Combining the equalities, we get ma = mg sin θ − µmg cos θ or a = g (sin θ − µ cos θ) c Copyright 2013 American Association of Physics Teachers

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Part A

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The motion is changes at an angle where the static friction is greatest, or when both conditions are equalities: f = µmg cos θ and a = αR In that case

sin θc = sin θc − µ cos θc 1+β 1 +1 µ tan θc = β

or

Question A3 A beam of muons is maintained in a circular orbit by a uniform magnetic field. Neglect energy loss due to electromagnetic radiation. The mass of the muon is 1.88 × 10−28 kg, its charge is −1.602 × 10−19 C, and its half-life is 1.523 µs. a. The speed of the muons is much less than the speed of light. It is found that half of the muons decay during each full orbit. What is the magnitude of the magnetic field? b. The experiment is repeated with the same magnetic field, but the speed of the muons is increased; it is no longer much less than the speed of light. Does the fraction of muons which decay during each full orbit increase, decrease, or stay the same? The following facts about special relativity may be useful: • The Lorentz factor for a particle moving at speed v is 1 γ=p 1 − v 2 /c2 • The Lorentz factor gives the magnitude of time dilation; that is, a clock moving at speed v in a given reference frame runs slow by a factor γ in that frame. • The momentum of a particle is given by p~ = γm~v where m does not depend on v. • The Lorentz force law in the form d~ p ~ + ~v × B) ~ = q(E dt continues to hold.


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Solution For brevity we simply present the full relativistic solution. The relationships for circular motion d~v |~v |2 = dt r 2πr = |~v | T are purely a matter of mathematics, and thus continue to hold under special relativity. Meanwhile, since |~v | is constant for circular motion, γ is constant as well. Thus we can take magnitudes in the ~ = 0) to find Lorentz force law (and set E d~v γm = q |~v | B dt Combining these relationships, 2π γm = qB T If half of the muons decay during each orbit, in the muons’ frame of reference each orbit takes one half-life T1/2 . In the lab frame, then, T = γT1/2 and so

2π γm = qB γT1/2 B=

2πm qT1/2

Numerically, B = 4.85 mT The speed of the muons is irrelevant to the fraction which decay per orbit, even in the relativistic case. (The orbits take longer, but the muons live longer, both by the same factor γ.)

Question A4 A graduated cylinder is partially filled with water; a rubber duck floats at the surface. Oil is poured into the graduated cylinder at a slow, constant rate, and the volume marks corresponding to the surface of the water and the surface of the oil are recorded as a function of time.


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Part A

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500 Volume mark at water surface Volume mark at oil surface

450 400

Volume (mL)

350 300 250 200 150 100 50 0

0

1

2

3

4 5 6 Time (min)

7

8

9

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Water has a density of 1.00 g/mL; the density of air is negligible, as are surface effects. Find the density of the oil.

Solution As the oil is poured in, more and more of the weight of the duck is supported by oil, and it rises out of the water, reducing the water level. Eventually this stops, either because the duck is fully submerged in oil or because it is floating entirely above the water. At all times, the weight of the water that is no longer displaced equals the weight of the newly displaced oil: ρo g∆Vo = ρw g∆Vw With this understanding many approaches are possible; we illustrate one. The change in the volume of displaced water is easily read off the graph as the distance between the dotted and dashed lines; it is 143 mL. Finding the volume of displaced oil requires us to take into account the increasing amount of oil in the cylinder. We know there is no oil at t = 0, because the oil level and water level coincide, and we know that the rate of change of the oil level for t > 6 min is the pour rate, because the water level is not changing. Extrapolating to t = 0 we conclude that the volume of oil in the container at any time is given by the height of the shaded region. The volume of displaced oil can then be read as the distance between the solid and dashed lines; it is 186 mL. The density of the oil is ∆Vw 143 mL ρo = ρw = (1.00 g/mL) ∆Vo 186 mL c Copyright 2013 American Association of Physics Teachers

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Part A

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ρo = 0.77 g/mL 500 Volume mark at water surface Volume mark at oil surface

450 400

Volume (mL)

350 300 250 200 150 100 50 0

0

1

2

3

4 5 6 Time (min)

7

8

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2013 Semifinal Exam

Part B

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Part B Question B1 Shown below is the Blackbird, a vehicle built in 2009. There is no source of stored energy such as a battery or gasoline engine; all of the power used to move the car comes from the wind. The only important mechanism in the car is a gearbox that can transfer power between the wheels and the propeller. The Blackbird was driven both directly downwind and directly upwind, as shown below. In each case the car remained exactly parallel (or anti-parallel) to the wind without turning. The tests were conducted on level ground, in steady, uniform wind, and continued long enough to reach the steady state. ~vw

~v

Source: fasterthanthewind.org

Downwind

~vw

~v

Upwind

When driving downwind, the builders claim that they were able to drive “faster than the wind”: that is, with |~v | > |~vw |, so that the car experienced a relative headwind while traveling. Commenters on the Internet claimed, often angrily, that this was physically impossible and that the Blackbird was a hoax. Some commenters also claimed that the upwind case was physically impossible. a. Consider first the downwind faster than the wind case. • Is the motion actually possible as claimed? If not, offer a brief explanation! • If the motion is possible, is power transferred from the propeller to the wheels or vice versa? • If the motion is possible, what ground speed is attained? For this question, suppose that when transferring power in either direction between the propeller and the wheels, a fraction α of the useful work is lost; let the wind speed be vw . Neglect all other losses of energy. b. Answer the previous questions for the upwind case.

Solution


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Part B

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Both modes are possible as claimed. When a force exists between two bodies in relative motion, the net power produced or consumed is proportional to the relative velocity between them: P = F vr This value is independent of the reference frame in which the power is measured, even though the power delivered to each body separately is not. This is a consequence of Newton’s third law; if the velocities of the bodies in a particular reference frame are v1 and v2 , the net power is P = F v1 + (−F )v2 P = F (v1 − v2 ) The relative speed of the Blackbird and the air is different from the relative speed of the Blackbird and the ground; if the ground speed is v, the airspeed is v − vw in the downwind mode and v + vw in the upwind mode. Thus, even though the force due to the air and the force due to the ground must balance, one force can produce more power than the other one absorbs. Power should always be produced by the force corresponding to the larger relative velocity. Thus in the downwind case, power is transferred from the wheels to the propeller, whereas in the upwind case, power is transferred from the propeller to the wheels. (In practice this reversal required a lengthy reconfiguration of the Blackbird between the two trials.) In the downwind case, the wheel power is Pw = F v and the propeller power is Pp = F (v − vw ) Power is transferred to the propeller, so Pp = (1 − α)Pw F (v − vw ) = (1 − α)F v vw v= α At high energy loss one moves close to the wind speed (as expected); with sufficiently low energy loss, any speed is possible. In the upwind case, the wheel power is still Pw = F v but the propeller power is Pp = F (v + vw ) Power is transferred to the wheels, so Pw = (1 − α)Pp F v = (1 − α)F (v + vw ) 1 v = vw −1 α At high energy loss one cannot make any progress at all (as expected); again, with sufficiently low energy loss, any speed is possible. c Copyright 2013 American Association of Physics Teachers

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Part B

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Question B2 This problem concerns three situations involving the transfer of energy into a region of space by electromagnetic fields. In the first case, that energy is stored in the kinetic energy of a charged object; in the second and third cases, the energy is stored in an electric or magnetic field. In general, whenever an electric and a magnetic field are at an angle to each other, energy is transferred; for example, this principle is the reason electromagnetic radiation transfers energy. The power transferred per unit area is given by the Poynting vector : ~= 1E ~ ×B ~ S µ0 In each part of this problem, the last subpart asks you to verify that the rate of energy transfer agrees with the formula for the Poynting vector. Therefore, you should not use the formula for the Poynting vector before the last subpart! a. A long, insulating cylindrical rod has radius R and carries a uniform volume charge density ρ. A uniform external electric field E exists in the direction of its axis. The rod moves in the direction of its axis at speed v. i. What is the power per unit length P delivered to the rod? ii. What is the magnetic field B at the surface of the rod? Draw the direction on a diagram. iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer. b. A parallel plate capacitor consists of two discs of radius R separated by a distance d R. The capacitor carries charge Q, and is being charged by a small, constant current I. i. What is the power P delivered to the capacitor? ii. What is the magnetic field B just inside the edge of the capacitor? Draw the direction on a diagram. (Ignore fringing effects in the electric field for this calculation.) iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer. c. A long solenoid of radius R has N turns of wire per unit length. The solenoid carries current I, and this current is increased at a small, constant rate dI dt . i. What is the power per unit length P delivered to the solenoid? ii. What is the electric field E just inside the surface of the solenoid? Draw its direction on a diagram. iii. Compute the Poynting vector, draw its direction on a diagram, and verify that it agrees with the rate of energy transfer.

Solution


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Part B

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i. A length l of the rod has charge q = πR2 lρ; the force on it is F = qE and the power delivered is P = F v. Combining these, P = πR2 lρEv P = πR2 ρEv ii. The length l of the rod moves past a point in a time t = vl , so the current carried by the rod is q I = = πR2 ρv t Applying Ampere’s law to a loop of radius R, I B dl = µ0 Ienc 2πRB = µ0 πR2 ρv 1 B = µ0 Rρv 2 The field is circumferential as given by the right-hand rule. iii. The electric and magnetic fields are perpendicular, so the Poynting vector has magnitude S=

1 EB µ0

1 S = RρvE 2 A quick application of the right hand rule indicates that it points inward along the surface of the cylinder, as it ought. The cylinder has area per unit length 2πr, so the rate of energy transfer per unit length is P = 2πrS = πR2 ρvE in agreement with the previous result. b.

i. The capacitance is given by the standard parallel-plate capacitor formula: C=

0 πR2 d

The voltage on the capacitor is thus V =

Q Qd = C 0 πR2

and the power is P = IV IQd P = 0 πR2 Students may choose instead to apply the formula for the volume energy density, 1 U = 0 E 2 2 c Copyright 2013 American Association of Physics Teachers

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Part B

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ii. Consider an Amperian loop encircling the edge of the capacitor, and use a flat Gaussian surface through the center of the capacitor. The electric field here is perpendicular to the surface and has magnitude E=

V Q = d 0 πR2

The electric flux through the surface is thus φE = πR2 E =

Q 0

This can also be determined directly using Gauss’s law and appropriate symmetries. There is no current through the surface, so from Ampere’s Law I dφE B dl = µ0 0 dt 2πRB = µ0

dQ dt

µ0 I 2πR The field is circumferential as given by the right-hand rule. Note that we could instead use a curved Gaussian surface that avoids the center of the capacitor and intersects one of the charging wires! In this case we have directly I B dl = µ0 I B=

and the calculation proceeds as before. iii. The electric and magnetic fields are perpendicular, so again S=

1 EB µ0

IQ 20 π 2 R3 A quick application of the right hand rule indicates that it points inward along the edge of the capacitor, as it ought. The area of this region is 2πRd, so the power delivered is S=

P = 2πRdS =

IQd 0 πR2

in agreement with the previous result. c.

i. Suppose that the solenoid has length l. The inductance is L = µ0 N 2 πR2 l Students may quote this formula directly, or derive it as follows. Consider an Amperian loop of length d intersecting the solenoid. This loop encloses N d turns of wire, so from Ampere’s law (remembering that the magnetic field exists entirely within the solenoid) I B dl = µ0 Ienc


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Part B

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Bd = µ0 N dI B = µ0 N I There are N l loops, so the total flux is Φ = N lBπR2 Φ = µ0 N 2 IπR2 l and since Φ = LI, L = µ0 N 2 πR2 l as quoted above. The voltage across the inductor is thus V =L

dI dt

V = µ0 N 2 πR2 l

dI dt

and the power delivered is P = IV P = µ0 N 2 πR2 lI

dI dt

P = µ0 N 2 πR2 I

dI dt

or, dividing by l,

Students may choose instead to apply the formula for the volume energy density, U=

1 2 B 2µ0

ii. Consider an Amperian loop just inside the surface of the solenoid. From above, the magnetic field through this loop is B = µ0 N I; thus (working in magnitudes) I dφB E dl = dt 2πRE = µ0 N πR2

dI dt

1 dI E = µ0 N R 2 dt The field is circumferential as given by Lenz’s law and the right-hand rule. iii. The electric and magnetic fields are perpendicular, so again S=

1 EB µ0

1 dI S = µ0 N 2 RI 2 dt c Copyright 2013 American Association of Physics Teachers

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Part B

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A quick application of the right hand rule indicates that it points inward towards the axis of the solenoid, as it ought. The area per unit length is just 2πR, so the power per unit length is P = 2πRS dI P = µ0 N 2 πR2 I dt in agreement with the previous result.



2014 F = ma Exam

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2014 F = ma Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Use g = 10 N/kg throughout this contest. • You may write in this booklet of questions. However, you will not receive any credit for anything written in this booklet. • Your answer to each question must be marked on the optical mark answer sheet. • Select the single answer that provides the best response to each question. Please be sure to use a No. 2 pencil and completely fill the box corresponding to your choice. If you change an answer, the previous mark must be completely erased. • Correct answers will be awarded one point; incorrect answers will result in a deduction of point. There is no penalty for leaving an answer blank.

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DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN Contributors to this year’s exam include David Fallest, David Jones, Jiajia Dong, Paul Stanley, Warren Turner, Qiuzi Li, and former US Team members Andrew Lin, Matthew Huang, Samuel Zbarsky.


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1. A car turning to the right is traveling at constant speed in a circle. From the driver’s perspective, the angular momentum vector about the center of the circle points X and the acceleration vector of the car points Y where (A) X is left, Y is left. (B) X is forward, Y is right. (C) X is down, Y is forward. (D) X is left, Y is right. (E) X is down, Y is right. ← CORRECT

Solution Use right hand rule to find angular momentum, and acceleration is toward center in uniform circular motion.


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2. A ball rolls without slipping down an inclined plane as shown in the diagram.

Which of the following vectors best represents the direction of the total force that the ball exerts on the plane? (A)

(B)

(C)

(D)

(E)

CORRECT answer is (E)

Solution Normal force into plane. Friction must be less than the parallel force, or it wouldn’t slide down incline! Correct answer is into plane and down plane; note that if the ball is rolling downward and accelerating, friction must be less than the parallel component of gravity.


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3. An object of uniform density floats partially submerged so that 20% of the object is above the water. A 3 N force presses down on the top of the object so that the object becomes fully submerged. What is the volume of the object? The density of water is ρH2 O = 1000 kg/m3 . (A) Vobject = 0.3 L (B) Vobject = 0.67 L (C) Vobject = 1.2 L (D) Vobject = 1.5 L ← CORRECT (E) Vobject = 3.0 L

Solution Submerging object requires pushing against the additional buoyant force of 0.2ρw V g, so V = 1.5 L.


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4. What are the correct values of the numbers in the following statements? Assume there are no external forces, and take N = 1 to mean that the statement cannot be made for any meaningful number of particles. • If a particle at rest explodes into N1 or fewer particles with known masses, and the total kinetic energy of the new particles is known, the kinetic energy of each of the new particles is completely determined. • If a particle at rest explodes into N2 or fewer particles, the velocities of the new particles must lie in a line. • If a particle at rest explodes into N3 or fewer particles, the velocities of the new particles must lie in a plane. (A) N1 = 2, N2 = 1, N3 = 1 (B) N1 = 1, N2 = 2, N3 = 3 (C) N1 = 2, N2 = 2, N3 = 3 ← CORRECT (D) N1 = 3, N2 = 2, N3 = 3 (E) N1 = 2, N2 = 3, N3 = 4

Solution Certainly N2 is at most 2 and N3 is at most 3; it is certainly possible for three particles to emerge with non-collinear velocities, or four to emerge with non-planar ones. (Consider for example the case where all of the particles have equal mass and they emerge at the corners of a triangle or tetrahedron.) However, note that the total momentum of the daughter particles must be zero; it is impossible for two non-collinear vectors to sum to zero, nor three non-coplanar vectors. Thus N2 = 2 and N3 = 3. Meanwhile, N1 is at most 2; consider three identical particles, where (among many other possibilities) any one could emerge at rest with the energy split equally among the other two. In the case of two particles, the velocities must be in opposite direction and (as argued) in a straight line; thus, up to overall direction, they are entirely determined by the kinetic energies. There are two independent equations governing the kinetic energies; the total momentum must be zero, and the total kinetic energy is fixed. Thus the kinetic energies are fully determined in this case, and N1 = 2.


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5. A unicyclist goes around a circular track of radius 30 m at a (amazingly fast!) constant speed of 10 m/s. At what angle to the left (or right) of vertical must the unicyclist lean to avoid falling? Assume that the height of the unicyclist is much smaller than the radius of the track. (A) 9.46◦ (B) 9.59◦ (C) 18.4◦ ← CORRECT (D) 19.5◦ (E) 70.5◦

Solution Force toward center is mv 2 /r. Force toward ground is mg. The force toward the center is from friction and acts at the point in contact with the ground. The normal force is equal I’m magnitude to the force of gravity and acts upward at the point of contact. We are then interested in the tangent angle as defined by tan θ =

v2 gr


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6. A cubical box of mass 10 kg with edge length 5 m is free to move on a frictionless horizontal surface. Inside is a small block of mass 2 kg, which moves without friction inside the box. At time t = 0, the block is moving with velocity 5 m/s directly towards one of the faces of the box, while the box is initially at rest. The coefficient of restitution for any collision between the block and box is 90%, meaning that the relative speed between the box and block immediately after a collision is 90% of the relative speed between the box and block immediately before the collision. 10 kg cube

2 kg block frictionless surface After 1 minute, the block is a displacement x from the original position. Which of the following is closest to x? (A) 0 m (B) 50 m ← CORRECT (C) 100 m (D) 200 m (E) 300 m

Solution The easiest way to consider this is the motion of the center of mass of the system, which, by conservation of momentum, must move with a speed of vcm =

(2 kg)(5 m/s) 5 = m/s. (2 kg) + (10 kg) 6

After one minute the center of mass has moved ( 65 m/s)(60 s) = 50 m. Now the center of mass must be bounded by the cubical box, so the position of the inside block must be somewhere near the 50 meter mark.


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7. A 1.00 m long stick with uniform density is allowed to rotate about a point 30.0 cm from its end. The stick is perfectly balanced when a 50.0 g mass is placed on the stick 20.0 cm from the same end. What is the mass of the stick? (A) 35.7 g (B) 33.3 g (C) 25.0 g ← CORRECT (D) 17.5 g (E) 14.3 g 8. An object of mass M is hung on a vertical spring of spring constant k and is set into vertical oscillations. The period of this oscillation is T0 . The spring is then cut in half and the same mass is attached and the system is set up to oscillate on a frictionless inclined plane making an angle θ to the horizontal. Determine the period of the oscillations on the inclined plane in terms of T0 . (A) T0 (B) T0 /2 (C) 2T0 sin θ √ (D) T0 / 2 ← CORRECT √ (E) T0 sin θ/ 2

Solution The spring constant of a spring is inversely proportional to the length of the un-stretched spring; cutting a spring in half will double the spring constant. Placing the system on an incline will change the equilibrium position of the mass, but will not affect the period of oscillation in any way. As such, the new period of oscillation is given by r r m 1 m T0 T = 2π = √ 2π =√ 2k k 2 2


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9. A 5.0 kg object undergoes a time-varying force as shown in the graph below. If the velocity at t = 0.0 s is +1.0 m/s, what is the velocity of the object at t = 7 s? 4

Force (N)

3 2 1

0

1

2

3

4 5 time (s)

6

7

(A) 2.45 m/s (B) 2.50 m/s (C) 3.50 m/s ← CORRECT (D) 12.5 m/s (E) 15.0 m/s

Solution The area under the graph is a measure of the change in momentum, so 1 1 ∆p = (3 s)(3 N) + (4 s)(3 N + 1 N) = 12.5 N · s. 2 2 The additional velocity is then ∆v = 2.5 m/s.


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2014 F = ma Exam

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10. A radio controlled car is attached to a stake in the ground by a 3.00 m long piece of string, and is forced to move in a circular path. The car has an initial angular velocity of 1.00 rad/s and smoothly accelerates at a rate of 4.00 rad/s2 . The string will break if the centripetal acceleration exceeds 2.43 × 102 m/s2 . How long can the car accelerate at this rate before the string breaks? (A) 0.25 s (B) 0.50 s (C) 1.00 s (D) 1.50 s (E) 2.00 s ← CORRECT

Solution The string breaks if ac exceeds the maximum tension, or 2 1 2 2 αt + ω0 ac = rω = r 2 Solve for the time, s r 2 ac t= − ω0 α r or t = 2s


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11. A point mass m is connected to an ideal spring on a horizontal frictionless surface. The mass is pulled a short distance and then released. Which of the following is the most correct plot of the kinetic energy as a function of potential energy? Ekinetic Ekinetic

(A)

(B)

Epotential

Ekinetic

(C)

Epotential

Ekinetic

(D)

Epotential

Epotential

Ekinetic

(E)

Epotential

← CORRECT

Solution Energy is conserved, so the total must be constant. So it is the descending diagonal line of slope magnitude one.


2014 F = ma Exam

12

The following information applies to questions 12 and 13 A paper helicopter with rotor radius r and weight W is dropped from a height h in air with a density of ρ. r

Assuming that the helicopter quickly reaches terminal velocity, a function for the time of flight T can be found in the form T = khα rβ ρδ W ω . where k is an unknown dimensionless constant (actually, 1.164). α, β, δ, and ω are constant exponents to be determined. 12. Determine α. (A) α = −1 (B) α = −1/2 (C) α = 0 (D) α = 1/2 (E) α = 1 ← CORRECT

Solution If the helicopter is at terminal velocity, then it is falling at constant speed. As such, h = vT , where v is the terminal velocity. In that case, α = 1.


2014 F = ma Exam

13

13. Determine β. (A) β = 1/3 (B) β = 1/2 (C) β = 2/3 (D) β = 1 ← CORRECT (E) β can not be uniquely determined without more information.

Solution Dimensional analysis on mass requires that δ = −ω. SInce only W has units of time (inverse p squared), then ω = −1/2. But ρ/W has units of length−2 time−1 , and we know α = 1, then β = 1.

14. A disk of moment of inertia I, mass M , and radius R has a cord wrapped around it tightly as shown in the diagram. The disk is free to slide on its side as shown in the top down view. A constant force of T is applied to the end of the cord and accelerates the disk along a frictionless surface. rope Force T

Solid disk

After the disk has accelerated some distance, determine the ratio of the translational KE to total KE of the disk, KEtranslational /KEtotal = (A)

I M R2

(B) (D)

M R2 I I 3M R2 I M R2 +I

(E)

M R2 M R2 +I

(C)

← CORRECT

Solution c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

14

The applied force T accelerates the center of mass of the disk at a rate a = T /M . The applied force produces a torque on the disk about the center of mass given by τ = RT , and therefore an angular acceleration of α = RT /I. After a time t the velocity of the disk will be v = at and the angular velocity will be ω = αt. Then KEtranslational

1 = M 2

T M

and KErotational

1 = I 2

RT I

2

2

t2 ,

t2 ,

so the desired ratio is then KEtranslational = KEtotal = =

1 T 2 2 M M t 2 , 2 T 2 + 1 I RT 2 t2 t M 2 I

1 M 2 1 M 2 1 + RI M

,

I . I + M R2

15. The maximum torque output from the engine of a new experimental car of mass m is τ . The maximum rotational speed of the engine is ω. The engine is designed to provide a constant power output P . The engine is connected to the wheels via a perfect transmission that can smoothly trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of static friction between the wheels and the road is µ. What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no frictional losses and assume µ is large enough so that the tires do not slip. (A) v = 2P/(mg) ← CORRECT √ (B) v = 2P/( 3mg) (C) v = 2P/(µmg) (D) v = τ ω/(mg) (E) v = τ ω/(µmg)

Solution The fundamental idea is P = F v where F is the component of the weight parallel to the incline. Then v = P/mg sin θ Since θ = 30◦ , the answer is v = 2P/mg c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

15

16. An object of mass m1 initially moving at speed v0 collides with an object of mass m2 = αm1 , where α < 1, that is initially at rest. The collision could be completely elastic, completely inelastic, or partially inelastic. After the collision the two objects move at speeds v1 and v2 . Assume that the collision is one dimensional, and that object one cannot pass through object two. After the collision, the speed ratio r1 = v1 /v0 of object 1 is bounded by (A) (1 − α)/(1 + α) ≤ r1 ≤ 1 (B) (1 − α)/(1 + α) ≤ r1 ≤ 1/(1 + α) ← CORRECT (C) α/(1 + α) ≤ r1 ≤ 1 (D) 0 ≤ r1 ≤ 2α/(1 + α) (E) 1/(1 + α) ≤ r1 ≤ 2/(1 + α)

Solution Conserving momentum and kinetic energy in the completely elastic collision yields the following quantities. 1−α v1 = , r1 = v0 (1 + α) r2 =

2 v2 = , v0 (1 + α)

Since α < 1, object 1 is tricking a less massive object, and therefore continues to move forward. Conserving momentum in a completely inelastic collision yields r1 = r2 =

1 , (1 + α)

Note that object 2 will always be moving forward, and since object 1 can’t pass through it, object 2 must always move with a more positive (or equal) velocity than object 1. Consequently, 1/(1 + α) ≤ r2 ≤ 2/(1 + α) Sorting out object 1 is a little harder, since under certain circumstances it can bounce backward. But in this case, since α < 1, it retains a forward velocity after the collision. Then (1 − α)/(1 + α) ≤ r1 ≤ 1/(1 + α)


2014 F = ma Exam

16

17. A spherical cloud of dust in space has a uniform density ρ0 and a radius R0 . The gravitational acceleration of free fall at the surface of the cloud due to the mass of the cloud is g0 . A process occurs (heat expansion) that causes the cloud to suddenly grow to a radius 2R0 , while maintaining a uniform (but not constant) density. The gravitational acceleration of free fall at a point R0 away from the center of the cloud due to the mass of the cloud is now (A) g0 /32 (B) g0 /16 (C) g0 /8 ← CORRECT (D) g0 /4 (E) g0 /2

Solution Newton’s Law of gravitation for a spherical object depends only on matter located closer to the center than the point of reference. g=

4π GM = Gρr, 2 r 3

where the last equality is true for uniform density. Doubling the size cuts the density to 1/8.


2014 F = ma Exam

17

18. Consider the following diagram of a box and two weight scales. Scale A supports the box via a massless rope. A pulley is attached to the top of the box; a second massless rope passes over the pulley, one end is attached to the box and the other end to scale B. The two scales read indicate the tensions TA and TB in the ropes. Originally scale A reads 30 Newtons and scale B reads 20 Newtons. TA

Scale A

Pulley

Box Frame

Scale B

TB If an additional force pulls down on scale B so that the reading increases to 30 Newtons, what will be the new reading on scale A? (A) 35 Newtons (B) 40 Newtons ← CORRECT (C) 45 Newtons (D) 50 Newtons (E) 60 Newtons Adapted from a demonstration by Richard Berg. c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

18

Solution As cute as the box is, the only thing that matter is that you pull down on scale B with 10 extra newtons, so scale A must increase by 10 to balance. 19. A helicopter is flying horizontally at constant speed. A perfectly flexible uniform cable is suspended beneath the helicopter; air friction on the cable is not negligible. Which of the following diagrams best shows the shape of the cable as the helicopter flies through the air to the right?

(A)

(B) ← CORRECT

(C)

(D)

(E)

Solution Since there is air friction on the cable, then there must be a horizontal component to the force where the cable attaches to the helicopter. Since the amount of air friction would be proportional to the length of the cable hanging beneath at any point on the cable, then the cable would be hanging in a straight diagonal line. This question generated a great deal of controversy. At least two test takers challenged the answer, one who even tried to do the experiment. I’m told that several different “Ph.D” physicists declared that the correct answer was X, but, interestingly enough, couldn’t agree on what X should be. Originally the problem was worded as a cable with a hanging mass. As such, the correct answer would have been D, as explained below. c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

19

Since air friction on the hanging object is negligible, the only forces to consider are gravity and the cable. So the cable must be pulling directly upward. Since there is air friction on the cable, then there must be a horizontal component to the force where the cable attaches to the helicopter. The correct answer would then be curving to the left then falling straight down. 20. A crew of scientists has built a new space station. The space station is shaped like a wheel of radius R, with essentially all its mass M at the rim. When the crew arrives, the station will be set rotating at a rate that causes an object at the rim to have radial acceleration g, thereby simulating Earth’s surface gravity. This is accomplished by two small rockets, each with thrust T newtons, mounted on the station’s rim. How long a time t does one need to fire the rockets to achieve the desired condition? p (A) t = gR3 M/(2T ) √ (B) t = gR M/(2T )← CORRECT √ (C) t = gR M/T p (D) t = gR/π M/T √ (E) t = gR M/(πT ) Adapted from a problem in Physics for Scientists and Engineers by Richard Wolfson

Solution p Desired acceleration: g = ω 2 R → ω = g/R The two rockets provide: 2T R = M R2 α → α = 2T /M R g 1/2 R1/2 M time needed: t = ω/α = 2T


2014 F = ma Exam

20

21. Two pulleys (shown in figure) are made of the same metal with density ρ. Pulley A is a uniform disk with radius R. Pulley B is identical except a circle of R/2 is removed from the center. When two boxes M = αm (α > 1) are connected over the pulleys through a massless rope and move without slipping, what is the ratio between the accelerations in system A and B? The mass of pulley A is M + m. Pulley B

Pulley A

m

m M

M

(A) aA /aB = 47/48 ← CORRECT (B) aA /aB = 31/32 (C) aA /aB = 15/16 (D) aA /aB = 9/16 (E) aA /aB = 3/4

Solution This is effectively a force balance problem with a generalized Atwood machine. For mass M , the equations are M a = M g − TM where TM is tension in rope on the M side. For mass m, the equations are ma = Tm − mg For the pulley, the equations are Iα = R(TM − tm ) with α = a/R and I the moment of inertia. Combining, Ia/R2 = (M g − M a) − (ma + mg) or a=g

M −m I/R2 + M + m

Let I = β(M + m)R2 for the two pulleys. Then the acceleration ratio is βB + 1 aA = aB βA + 1 c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

21

βA is easy enough, it is just 1/2, since it is a uniform disk. βB requires a little more work, it is 1/2(1 − (1/4)(1/4)) = 15/32 to account for the removal of the interior disk. The ratio of accelerations is then 47 aA = aB 48

22. A body of mass M and a body of mass m M are in circular orbits about their center of mass under the influence of their mutual gravitational attraction to each other. The distance between the bodies is R, which is much larger than the size of either body. A small amount of matter δm m is removed from the body of mass m and transferred to the body of mass M . The transfer is done in such a way so that the orbits of the two bodies remain circular, and remain separated by a distance R. Which of the following statements is correct? (A) The gravitational force between the two bodies increases. (B) The gravitational force between the two bodies remains constant. (C) The total angular momentum of the system increases. (D) The total angular momentum of the system remains constant. (E) The period of the orbit of two bodies remains constant. ← CORRECT

Solution The force between two bodies is given by F =G

Mm R2

The new force, F 0 , would be given by F0 = G

(M + δm)(m − δm) M m − (M − m)δm − (δm)2 = G r2 R2

This can be approximated to first order as δm 0 F = 1− F m For centripetal motion, mv 2 /r = F where r is the distance from m to the center of mass, which is r=

M R M +m


2014 F = ma Exam

22

In terms of angular momenta, Lm = mvr = or

s Lm =

p

m(mv 2 /r)r3 =

mM 3 RGmM = mM 2 (m + M )3

s

√

mr3 F

GR (m + M )3

By symmetry, s LM = m2 M

GR (m + M )3

and the sum is then

r

GR m+M and we already saw that the quantity M m decreases as δm is transferred from m to M . L = Lm + LM = mM

Finally, the period of motion can be found from v = 2πr/T , so r 2πr mr T = = 2π v F Combining the above, r T = 2π

mM R R2 = 2π m + M GmM

s

R3 G(m + M )

And that’s constant. The following information applies to questions 23 and 24 A 100 kg astronaut carries a launcher loaded with a 10 kg bowling ball; the launcher and the astronaut’s spacesuit have negligible mass. The astronaut discovers that firing the launcher results in the ball moving away from her at a relative speed of 50 m/s. 23. What is the impulse delivered to the astronaut when firing the launcher? (A) 455 N s ← CORRECT (B) 500 N s (C) 550 N s (D) 5000 N s (E) 5500 N s

Solution Work in the center of mass frame, where the impulse delivered is simply the final momentum c Copyright 2014 American Association of Physics Teachers

2014 F = ma Exam

23

of the astronaut. Let the astronaut have mass m1 and final velocity v1 , and let the launcher have mass m2 and final velocity v2 . Then from conservation of momentum m1 v1 + m2 v2 = 0 and the relative velocity is (taking v1 > 0) vr = v1 − v2 Substituting, m1 v1 + m2 (v1 − vr ) = 0 m1 m2 m1 v1 = vr m1 + m2 so J = m1 v1 = 455 N s

24. The astronaut in the previous situation is now moving at 10 m/s (as measured in a certain frame of reference). She wishes to fire the launcher so that her velocity turns through as large an angle as possible (in this frame of reference). What is this maximum angle? (Hint: a diagram may be useful.) (A) 24.4◦ (B) 26.6◦ (C) 27.0◦ ← CORRECT (D) 30.0◦ (E) 180.0◦

Solution The final momentum of the astronaut is the sum of her initial momentum and the impulse delivered to her. Her initial momentum has magnitude pi = (100 kg)(10 m/s) = 1000 kg m/s and the impulse available has the fixed, smaller magnitude found in the previous problem. (Note that the impulse is the same in all inertial reference frames.) Her final momentum is thus confined to the circle shown below:


2014 F = ma Exam

24

p~i θ J~

p~f

The maximum angle is obtained by the tangent to the circle.

p~i θ p~f

J~

The maximum angle is thus θ = arcsin

J = 27.0◦ pi

25. A block with mass m is released from rest at the top of a frictionless ramp. The block starts at a height h1 above the base of the ramp, slides down the ramp, and then up a second ramp. The coefficient of kinetic friction between the block and the second ramp is µk . If both ramps make an angle of θ with the horizontal, to what height h2 above the base of the second ramp will the block rise? (A) h2 = (h1 sin θ)/(µk cos θ + sin θ) ← CORRECT (B) h2 = (h1 sin θ)/(µk + sin θ) (C) h2 = (h1 sin θ)/(µk cos2 θ + sin θ) (D) h2 = (h1 sin θ)/(µk cos2 θ + sin2 θ) (E) h2 = (h1 cos θ)/(µk sin θ + cos θ)



2014 USA Physics Olympiad Exam

AAPT AIP

1

UNITED STATES PHYSICS TEAM 2014 USA Physics Olympiad Exam

DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • This examination consists of two parts. • Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes. • The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam. • The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B. • Allow 90 minutes to complete Part B. Do not let students go back to Part A. • Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-16), Part B (pages 17-23), and several answer sheets for two of the questions in part A (pages 25-28). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided! • The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2014. • Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.



AAPT AIP

Cover Sheet

2


USA Physics Olympiad Exam INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN • Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. • After you have completed Part A you may take a break. • Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. • Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. • Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, AAPT ID # Doe, Jamie A1 - 1/3 • A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. • Questions with the same point value are not necessarily of the same difficulty. • In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 15, 2014. Possibly Useful Information. You may g = 9.8 N/kg k = 1/4π0 = 8.99 × 109 N · m2 /C2 c = 3.00 × 108 m/s NA = 6.02 × 1023 (mol)−1 σ = 5.67 × 10−8 J/(s · m2 · K4 ) 1eV = 1.602 × 10−19 J me = 9.109 × 10−31 kg = 0.511 MeV/c2 sin θ ≈ θ − 16 θ3 for |θ| 1




Part A

3

Part A Question A1 Inspired by: http://www.wired.com/wiredscience/2012/04/a-leaning-motorcycle-on-a-vertical-wall/ A unicyclist of total height h goes around a circular track of radius R while leaning inward at an angle θ to the vertical. The acceleration due to gravity is g. a. Suppose h R. What angular velocity ω must the unicyclist sustain?

Solution Work in the rotating frame, where four forces act on the unicyclist: a normal and frictional force at the point of contact, gravity downwards at the center of mass, and a (fictitious) centrifugal force. If h R, all parts of the unicyclist are at a distance of approximately R from the center of the circle, so the centripetal acceleration of every part of the unicyclist is ω 2 R. The centrifugal force can then be taken to act at the center of mass for purposes of computing the torque. If the center of mass is a distance l from the point of contact, the torque about the point of contact is τ = mω 2 Rl cos θ − mgl sin θ Since the unicyclist is stationary in this frame, τ = 0: mω 2 Rl cos θ − mgl sin θ = 0 r g tan θ ω= R

b. Now model the unicyclist as a uniform rod of length h, where h is less than R but not negligible. This refined model introduces a correction to the previous result. What is the new expression for the angular velocity ω? Assume that the rod remains in the plane formed by the vertical and radial directions, and that R is measured from the center of the circle to the point of contact at the ground.

Solution The centripetal acceleration now varies meaningfully along the length of the unicyclist. In the rotating frame, the torque about the point of contact is given by Z τc = ω 2 rz dm where r is the distance from the center of the circle, z is the height above the ground, and dm is a mass element. Because the mass of the unicyclist is uniformly distributed along a length h, the mass element dm can be written as m h ds for a length element ds, and we have Z h m τc = ω 2 (R − s sin θ)(s cos θ) ds h 0 c Copyright 2014 American Association of Physics Teachers


Part A

2

τc = mω h cos θ

4

R h − sin θ 2 3

Gravity continues to act at the center of mass, a distance h2 from the point of contact, and in the opposite direction: h τg = −mg sin θ 2 Again, the total torque is zero, so R h h 2 mω h cos θ − sin θ − mg sin θ = 0 2 3 2 s −1 g 2h 1− tan θ sin θ ω= R 3R

Question A2 A room air conditioner is modeled as a heat engine run in reverse: an amount of heat QL is absorbed from the room at a temperature TL into cooling coils containing a working gas; this gas is compressed adiabatically to a temperature TH ; the gas is compressed isothermally in a coil outside the house, giving off an amount of heat QH ; the gas expands adiabatically back to a temperature TL ; and the cycle repeats. An amount of energy W is input into the system every cycle through an electric pump. This model describes the air conditioner with the best possible efficiency.

heating coil

pump

room

valve

cooling coil

Assume that the outside air temperature is TH and the inside air temperature is TL . The air-conditioner unit consumes electric power P . Assume that the air is sufficiently dry so that no condensation of water occurs in the cooling coils of the air conditioner. Water boils at 373 K and freezes at 273 K at normal atmospheric pressure. a. Derive an expression for the maximum rate at which heat is removed from the room in terms of the air temperatures TH , TL , and the power consumed by the air conditioner P . Your derivation must refer to the entropy changes that occur in a Carnot cycle in order to receive full marks for this part.



Part A

5

Solution From Carnot cycles, and by entropy conservation, we have QH TH = QL TL Also, by energy conservation, QH = QL + W So heat is removed at a rate QL /t. But QL = QH − W = QL or

W = QL

TH −W TL

TH −1 TL

Rearrange and divide by time, QL =P t

TL TH − TL

b. The room is insulated, but heat still passes into the room at a rate R = k∆T , where ∆T is the temperature difference between the inside and the outside of the room and k is a constant. Find the coldest possible temperature of the room in terms of TH , k, and P .

Solution Equate. k∆T = P

TL TH − ∆T =P ∆T ∆T

or k(∆T )2 = P TH − P ∆T, which is a quadratic that can be solved as ∆T =

−P ±

p P 2 + 4P kTH , 2k

but only the positive root has physical significance. Writing x = P/k, x p ∆T = 1 + 4TH /x − 1 2 That’s the amount the room is colder than the outside, so x p TL = TH − 1 + 4TH /x − 1 2



Part A

6

c. A typical room has a value of k = 173 W/◦ C. If the outside temperature is 40◦ C, what minimum power should the air conditioner have to get the inside temperature down to 25◦ C?

Solution Don’t forget to convert to Kelvin! From above, P =

k(∆T )2 , TL

so P = 130 W



Part A

7

Question A3 When studying problems in special relativity it is often the invariant distance ∆s between two events that is most important, where ∆s is defined by (∆s)2 = (c∆t)2 − (∆x)2 + (∆y)2 + (∆z)2 where c = 3 × 108 m/s is the speed of light.1 a. Consider the motion of a projectile launched with initial speed v0 at angle of θ0 above the horizontal. Assume that g, the acceleration of free fall, is constant for the motion of the projectile. i. Derive an expression for the invariant distance of the projectile as a function of time t as measured from the launch, assuming that it is launched at t = 0. Express your answer as a function of any or all of θ0 , v0 , c, g, and t. ii. The radius of curvature of an object’s trajectory can be estimated by assuming that the trajectory is part of a circle, determining the distance between the end points, and measuring the maximum height above the straight line that connects the endpoints. Assuming that we mean “invariant distance” as defined above, find the radius of curvature of the projectile’s trajectory as a function of any or all of θ0 , v0 , c, and g. Assume that the projectile lands at the same level from which it was launched, and assume that the motion is not relativistic, so v0 c, and you can neglect terms with v/c compared to terms without.

Solution The particle begins at ct = x = y = 0 and takes a path satisfying x = v0 t cos θ0 1 z = v0 t sin θ0 − gt2 2 Thus

1 s2 = (ct)2 − (v0 t cos θ0 )2 − (v0 t sin θ0 − gt2 )2 2 which can be simplified to 1 1 s2 = (c2 − v02 )t2 + gv0 sin θ0 t3 + g 2 t4 2 4

b. The particle reaches the ground again at tf = 2

v0 sin θ g

xf = v0 cos θtf zf = 0 1

We are using the convention used by Einstein



Part A

8

and so the invariant distance between the endpoints is s2 = (ctf )2 + (v0 cos θ tf )2 s ≈ 2c

v0 sin θ g

The maximum height above the ground is zmax =

(v0 sin θ)2 2g

Because zmax s, we have from similar triangles 1 s zmax ≈ 2 1 R 2s

R≈

1 s2 4 zmax

R≈2

c2 g

c. A rocket ship far from any gravitational mass is accelerating in the positive x direction at a constant rate g, as measured by someone inside the ship. Spaceman Fred at the right end of the rocket aims a laser pointer toward an alien at the left end of the rocket. The two are separated by a distance d such that dg c2 ; you can safely ignore terms of the form (dg/c2 )2 . i. Sketch a graph of the motion of both Fred and the alien on the space-time diagram provided in the answer sheet. The graph is not meant to be drawn to scale. Note that t and x are reversed from a traditional graph. Assume that the rocket has velocity v = 0 at time t = 0 and is located at position x = 0. Clearly indicate any asymptotes, and the slopes of these asymptotes.

Solution Since the rocket-ship can never exceed the speed of light, yet it is always accelerating (in the local frame), it must approach an asymptote that has a slope of one on the space-time diagram shown. There is a slight challenge to consider, however. Since the rocket ship is an extended object, do the two ends (represented by Fred and the Alien) approach the same asymptote, or two different asymptotes? At this point we must remember a consequence of special relativity for a ship moving at relativistic speeds: the ship will contract in length as measured in the original frame. As the speed of the ship approaches that of light, the length of the ship will approach zero. The only way for that to happen is for the two ends of the ship to have slightly different accelerations.



Part A

9

If you had assumed (somewhat incorrectly) that the two ends of the ship have the same acceleration, then the two trajectories would be approaching two different asymptotes separated by a constant horizontal distance. But this would mean the apparent length of the ship was constant, regardless of speed. In the instantaneous rest frame of the ship we then require that Fred and the Alien be moving apart. This means that the ship must be stretching, and eventually breaking.



Part A

10

ii. If the frequency of the laser pointer as measured by Fred is f1 , determine the frequency of the laser pointer as observed by the alien. It is reasonable to assume that f1 c/d.

Solution If the spaceship is uniformly accelerating then we can choose a reference frame which is instantaneously at rest with respect to the spaceship at t = 0. Consider two instantaneous flashes from the astronaut. Flash 1 is emitted by Fred at t = 0, flash 2 is emitted by the Fred at t = τ1 .



Part A

11

Flash 1 travels down towards the alien, who is accelerating upward. Let t1 be the time at which the alien sees flash 1. Equate the distances, 1 ct1 = d − gt1 2 . 2 Flash 2 is emitted at τ1 . Flash 2 travels down towards the alien, who is still accelerating upward. Let t2 be the time at which the alien sees flash 2. Equate the distances, 1 1 c(t2 − τ1 ) = gτ1 2 + d − gt2 2. 2 2 Notice that the pulse travel for a time t2 − τ1 ! Defining t2 =1 +τ2 , and then expanding to keep terms linear in τ t1 , 1 c(t1 + τ2 − τ1 ) = h − gt1 2 − gt1 τ2 . 2 Combining, c(τ2 − τ1 ) = −gt1 τ2 . but t1 ≈ d/c, so gh τ2 = τ1 1 − 2 c In terms of frequency this is f2 = f1

gd 1+ 2 c

Alternatively, we can follow the motion of two wave crests from Fred to the alien. We can work in the reference frame where Fred is stationary when the first crest is emitted. Because f1 c/d, we can assume that Fred remains stationary during the period between the first and second crests; then, because the first crest moves towards the alien at a speed c, they are separated by a distance c/f1 . Because dg c2 , the time taken for the crests to reach the alien is due almost entirely to the motion of the crests, and is d/c. In this time, the spaceship accelerates to a speed gd/c, and (in Fred’s frame of reference) the relative speed of the crests and the alien is c + gd c . The time between crests reaching the alien is thus 1 c/f1 = f2 c + gd c gd f2 = f1 1 + 2 c While this time interval is measured in Fred’s reference frame, time dilation effects are 2 and can thus be ignored. of the order gd c2



Part A

12

Question A4 A positive point charge q is located inside a neutral hollow spherical conducting shell. The shell has inner radius a and outer radius b; b − a is not negligible. The shell is centered on the origin. y

b a x

a. Assume that the point charge q is located at the origin in the very center of the shell. i. Determine the magnitude of the electric field outside the conducting shell at x = b.

Solution Conducting shell is neutral, so there is equal but opposite charge on surface r = a and r = b. The electric field inside of a static conductor is zero, so the charge on inner surface is equal but opposite to q by Gauss’s Law. Spherical symmetry requires a spherically symmetric electric field, so by Gauss’s law, outside the shell, we have E(r > b) =

q 4π0 r2

and then, at x = r = b, we have E(b) =

q 4π0 b2

ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer sheet provided.

Solution



Part A

13

iii. Determine the electric potential at x = a.

Solution The shell is a conductor, so it is an equipotential surface. This means potential at r = a is same as r = b. For points outside the shell, spherical symmetry (and Gauss’s Law) makes the problem reducible to a point charge at the origin, so V (r > b) =

q 4π0 r

But V (a) = V (b), so V (x = a) is V (a) =

q 4π0 b

iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.

Solution

b. Assume that the point charge q is now located on the x axis at a point x = 2a/3. c Copyright 2014 American Association of Physics Teachers


Part A

14

i. Determine the magnitude of the electric field outside the conducting shell at x = b.

Solution The problem (for r > a) maintains the spherical symmetry of above, so the answer is unchanged: q E(b) = 4π0 b2 ii. Sketch a graph for the magnitude of the electric field along the x axis on the answer sheet provided.

Solution

iii. Determine the electric potential at x = a.

Solution The problem (for r > a) maintains the spherical symmetry of above, so the answer is unchanged: q V (a) = 4π0 b iv. Sketch a graph for the electric potential along the x axis on the answer sheet provided.

Solution



Part A

15

v. Sketch a figure showing the electric field lines (if any) inside, within, and outside the conducting shell on the answer sheet provided. You should show at least eight field lines in any distinct region that has a non-zero field.

Solution



Part A

16





Part B

17

Part B Question B1 A block of mass M has a hole drilled through it so that a ball of mass m can enter horizontally and then pass through the block and exit vertically upward. The ball and block are located on a frictionless surface; the block is originally at rest.

M m

v

frictionless horizontal surface a. Consider the scenario where the ball is traveling horizontally with a speed v0 . The ball enters the block and is ejected out the top of the block. Assume there are no frictional losses as the ball passes through the block, and the ball rises to a height much higher than the dimensions of the block. The ball then returns to the level of the block, where it enters the top hole and then is ejected from the side hole. Determine the time t for the ball to return to the position where the original collision occurs in terms of the mass ratio β = M/m, speed v0 , and acceleration of free fall g.

Solution The ball will be ejected vertically relative to the block, so the collision is effectively inelastic. This means the horizontal velocity v1 of the block and ball after the collision will be given by v1 =

m v0 m+M

The ball will now have a vertical component to the velocity, v2 Since there are no frictional losses, we can use conservation of energy to determine this vertical velocity. Prior to the collision, the kinetic energy is 1 E0 = mv0 2 2 After the collision, the block carries away a kinetic energy of 1 E1 = M v1 2 2 so the ball must have kinetic energy 1 E2 = E0 − E1 = m v1 2 + v2 2 2



Part B

18

Equating, mv0 2 − M v1 2 2 m 2 mv0 − (M + m) v0 2 m+M m 1− v0 2 m+M r M v0 m+M

= m v1 2 + v2 2

= mv2 2 = v2 2 = v2

The time spent by the ball in the air is given by t2 = 2v2 /g The distance traveled horizontally by the ball in the air is given by x = v1 t2 = 2v1 v2 /g or m x=2 v0 m+M

r

M 1 2v0 2 v0 = m+M g g

s

m2 M (m + M )3

When the ball returns to the block it then is projected horizontally back toward the starting point (but it is now a distance x further away. It moves with a velocity v3 given by the equation for an elastic collision m−M v3 = v0 m+M The time for the ball to return to where first collision occurs is then given by t3 = or 2v0 t3 = g or 2v0 t3 = g

s

s

x |v3 |

m2 M m + M (m + M )3 M − m

m2 M (M + m)(M − m)2

The total time is the sum of t2 and t3 , or s ! r 2v0 M m2 M + g m+M (M + m)(M − m)2 which can be simplified to give 2v0 g

r

M m+M

s 1+

m2 (M − m)2

!



Part B

or 2v0 g

r

19

M m+M

M M −m

s

β β−1

Writing it in terms of β = M/m, 2v0 g

β 1+β

b. Now consider friction. The ball has moment of inertia I = 25 mr2 and is originally not rotating. When it enters the hole in the block it rubs against one surface so that when it is ejected upwards the ball is rolling without slipping. To what height does the ball rise above the block?

Solution Some of the previous work still holds true. The ball will be ejected vertically relative to the block, so the collision is effectively inelastic. This means the horizontal velocity v1 of the block and ball after the collision will be given by m v1 = v0 m+M The vertical velocity v4 of the ball, however, is now changed from v2 . Friction slows the ball down with an impulse given by ∆p = f ∆t all the while causing the ball to rotate with a new angular momentum given by L = τ ∆T = rf ∆t but L = Iω = Iv4 /r and m(v2 − v4 ) = ∆p, so Iv4 /r = mr(v2 − v4 ) or, writing I = αmr2 ,

v2 1+α The vertical velocity of the ball will take it to a height v4 =

h=

v4 2 2g

so

1 v0 2 M 2g (1 + α)2 m + M Thankfully, the problem reduces to what we expect if M m and α = 0. h=

Writing it in terms of β = M/m and α = 2/5, h=

v0 2 25 β 2g 49 1 + β



Part B

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Question B2 In parts a and b of this problem assume that velocities v are much less than the speed of light c, and therefore ignore relativistic contraction of lengths or time dilation. a. An infinite uniform sheet has a surface charge density σ and has an infinitesimal thickness. The sheet lies in the xy plane. ~ (magnitude and direction) i. Assuming the sheet is at rest, determine the electric field E above and below the sheet.

Solution From symmetry, the fields above and below the sheet are equal in magnitude and directed away from the sheet. From Gauss’s Law, using a cylinder of end area A, 2EA =

σA 0

E = σ/20 pointing directly away from the sheet in the z direction, or ~ = σ ˆ z E 20 above the sheet and

~ =− σ ˆ E z 20

below the sheet ii. Assuming the sheet is moving with velocity ~v = vˆ x (parallel to the sheet), determine ~ (magnitude and direction) above and below the sheet. the electric field E

Solution The motion does not affect the electric field, which is still directed away from the sheet and still has magnitude σ E= 20 ~ iii. Assuming the sheet is moving with velocity ~v = vˆ x, determine the magnetic field B (magnitude and direction) above and below the sheet.

Solution Application of the right-hand rule indicates that (for v > 0) there is a magnetic field in the −y direction for z > 0 and in the +y direction for z < 0. From Ampere’s law applied to a loop of length l normal to the x direction, 2Bl = µ0 σvl c Copyright 2014 American Association of Physics Teachers


Part B

21

(In a time t, an area vtl moves through the loop, so the charge that moves is σvtl and the current is σvl.) So µ0 σv B= 2 This can also be written as ~ = − µ0 σv y B ˆ 2 above the sheet and ~ = µ0 σv y B ˆ 2 below the sheet iv. Assuming the sheet is moving with velocity ~v = vˆ z (perpendicular to the sheet), deter~ mine the electric field E (magnitude and direction) above and below the sheet.

Solution Once again the motion does not affect the electric field, which is still directed away from the sheet and still has magnitude σ E= 20 ~ v. Assuming the sheet is moving with velocity ~v = vˆ z, determine the magnetic field B (magnitude and direction) above and below the sheet.

Solution There is no magnetic field above and below the sheet. Interestingly, there’s no magnetic field at the sheet either. Consider an Amperian loop of area A in the x-y plane as the sheet passes through. The loop experiences a current of the form Aσδ(t). But the loop also experiences an oppositely directed change in flux of the form A

σ δ(t), 0

so the right-hand side of Ampere’s law remains zero. ~ = Ex x b. In a certain region there exists only an electric field E ˆ + Ey y ˆ + Ez ˆ z (and no magnetic ~ 0 and B ~ 0 as meafield) as measured by an observer at rest. The electric and magnetic fields E ~ regardless sured by observers in motion can be determined entirely from the local value of E, of the charge configuration that may have produced it. ~ 0 as measured by an observer moving with i. What would be the observed electric field E velocity ~v = vˆ z?



Part B

22

Solution The electric field was unaffected by the motion of the sheet of charge, so the electric field in the frame of reference of the moving observer should be the same: ~0 = E ~ E

~ 0 as measured by an observer moving with ii. What would be the observed magnetic field B velocity ~v = vˆ z?

Solution No magnetic field was created by the motion of the sheet of charge in the direction of the electric field, so the magnetic field in the frame of reference of the moving observer should likewise not depend on the component of the electric field in the direction of motion. When the sheet of charge was moving in the +x direction, a magnetic field was created in the −y direction; the observer moving in the +x direction is equivalent to the sheet of charge moving in the −x direction, creating a magnetic field in the +y direction. That is, an electric field in the zˆ direction causes an observer moving in the x ˆ direction to observe a magnetic field in the yˆ = zˆ × x ˆ direction. Previously, furthermore, the electric and magnetic fields satisfied B=

µ0 0 E v

Combining this with the previous equation, ~ 0 = −µ0 0 ~v × E ~ B or, using c2 = 1/µ0 0 , ~ 0 = − 1 ~v × E ~ B c2 or in our case, where ~v = vˆ z, ~ 0 = v (Ey x B ˆ − Ex y ˆ) c2

c. An infinitely long wire wire on the z axis is composed of positive charges with linear charge density λ which are at rest, and negative charges with linear charge density −λ moving with speed v in the z direction. ~ (magnitude and direction) at points outside the wire. i. Determine the electric field E

Solution The wire as a whole is neutral, so there is no electric field outside the wire.



Part B

23

~ (magnitude and direction) at points outside the wire. ii. Determine the magnetic field B

Solution The current in the wire is λv, so Ampere’s Law quickly yields B = µ0

λv 2πr

in the tangential direction to a circle of radius r centered on and perpendicular to the wire. The actual direction is given by the right hand rule; since the current is in the −z direction, the circular B field lines would point clockwise looking in that direction. iii. Now consider an observer moving with speed v parallel to the z axis so that the negative charges appear to be at rest. There is a symmetry between the electric and magnetic fields such that a variation to your answer to part b can be applied to the magnetic field in this part. You will need to change the multiplicative constant to something dimensionally correct and reverse the sign. Use this fact to find and describe the electric field measured by the moving observer, and comment on your result. (Some familiarity with special relativity can help you verify the direction of your result, but is not necessary to obtain the correct answer.)

Solution Part b indicated that a velocity boost in an electric field resulted in a magnetic field given by ~ 0 = − 1 ~v × E. ~ B c2 By symmetry, and insisting on a dimensionally correct answer, we assume that ~ 0 = ±~v × B ~ E is correct, give or take a sign. We were told to switch the sign, so make it positive. Taking the cross product yields an electric field vector that points outward, with magnitude λv λ v2 E 0 = vµ0 = 2πr 2π0 r c2 This can be attributed to length contraction of the positive charges and inverse length contraction of the (now-stationary) negative charges.



Answer Sheets

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Answer Sheets

Following are answer sheets for some of the graphical portions of the test.



Answer Sheets

25

Answer for Part A, Question 3 Space-time graph for accelerated rocket. The positions of Fred and the Alien at t = 0 are shown.

ct

t=0 Fred

Alien


x


Answer Sheets

26

Answer for Part A, Question 4 E

−b

−a

0

a

b

x

a

b

x

Answer for Part A, Question 4 V

−b

−a



Answer Sheets

27

Answer for Part A, Question 4 E

−b

−a

0

a

b

x

a

b

x

Answer for Part A, Question 4 V

−b

−a



Answer Sheets

Answer for Part A, Question 4


28

USAPhO Problems (2007-2014)

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