Version 018/AABAC — midterm 02 — turner — (56970) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page — find all choices before answering. 001 (part 1 of 2) 10.0 points Which of the following choices corresponds to a system of two electrons that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)? Note that the horizontal and vertical axes in each plot are the separation between the particles and energy, respectively. K+U (I)
1
5, Figure VI 6. Figure V Explanation: When the two electrons are very far away their potential energy is 0, and since they have nonzero initial velocities, this means that they are unbounded and thus have an overall positive energy at r = co, which is also equal to the kinetic energy at that location. As the electrons get closer, due to their Coulomb repulsion their kinetic energies drop to 0 while the potential energy rises. Thus the correct answer is Figure (I).
K (II) K+U
002 (part 2 of 2) 10.0 points Which of the diagrams corresponds to a system of a proton and an electron that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)? 1. Figure I
(Iii K K+u
U ps7 u
2, Figure V 3. Figure II correct 4. Figure III 5. Figure VI
(V)
K
6. Figure IV
3. Figure II
Explanation: When the two particles are very far away, their potential energy is 0, and since they have nonzero initial velocities, this means that they are unbounded and thus have an overall positive energy at r = co, which is also equal to the kinetic energy at that location. As the electron and proton get closer, due to their Coulomb attraction their kinetic energies increase while the negative potential energy decreases even further. Thus the correct answer is Figure (II).
4. Figure IV
003 10.0 points
r K U
L Figure III 2. Figure I correct
Version 018/AABAC — midterm 02 — turner — (56970) Two identical objects are moving directly toward one another at the same speed v.
What is the total kinetic energy of the system of the two objects? 1. Zero since (11 — '02 = 0 2. m v2 since energy is a scalar correct
1 3. Still m v2 from the definition of kinetic 2 energy 4. 4 m v2 since (2 v)2 = 4 v2 5. None of these Explanation: Since kinetic energy is an additive scalar, the total kinetic energy of the system of two objects is
1
2
smaller because it is at a radius that is 200km larger than it would be at the surface. As it circles around the earth, the gravitational force accelerates the wrench, changing only the direction, not the magnitude of it's velocity. Likewise, the astronaut experiences a gravitational force about the same as he or she would at the surface, and an acceleration also approximately equal to g. Thus, the wrench appears to not accelerate at all with respect to the astronaut, i.e. appears weightless to the astronaut. 005 10.0 points An object of volume V is floating at the interface between water and a denser liquid.
of the object's volume is displacing water, 4 1 while — of the object's volume is displacing 4 the heavier liquid.
m V 2 ± - m v2 = M V 2 . 2
004 10.0 points An astronaut in a space shuttle at a height 200 km above the surface of the earth (radius 6380 km) drops a wrench that has mass M. The wrench floats nearby the astronaut. What is the net force on the wrench?
1. Approximately Mg (g = 9.8 In/s2 ) correct 2. Impossible to determine without more information 3. Much less than Mg (g = 9.8 m/s2 ) 4. Zero 5, None of the above Explanation: The wrench experiences approximately the same gravitational force as it would on the surface of the earth Mg, though slightly
If the density of the heavier liquid is p and the density of water is pw , what is the total buoyant force on the object? 1,
pV g
4
2. (p
3.
pw )V g
pu,V g
4 4. (3 /3 +100 17 9 4 5. (13 — POI/ g
6.
3 p„17 g
4 (P+3Pw)Vg 7. correct 4
Explanation:
Version 018/AABAC — midterm 02 — turner — (56970) The buoyant force is the density of the fluid displaced times the volume displaced, so 3 FB = A. V g 1 pVg= (P+3Pw)Vg 4 4 4
006 10.0 points A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram.
At what time t in the period shown does the particle achieve its maximum negative acceleration? 1. t = 2 s
2. t = 0s 3. t = 4s 4. t = 5 s 5, t = 3 s correct 6. None of these; the acceleration is constant. 7. t = 1s Explanation: This oscillation is described by y(t) = — sin ( 7 t 2 ) dy 7F ( v(t) = = cos dt 2 d2 y a(t) = dt2 =
212 sin ( t ) 2 The maximum negative acceleration will oc7F t cur when sin (- = 1, or at t = 3 s . 2
3
From a non-calculus perspective, the velocity is positive just before t = 3 s but decreasing since the particle is slowing down. At t = 3 s, the particle is momentarily at rest and v = 0. Just after t = 3 s , the velocity is negative since the slope of the position versus time plot has a negative slope. Remember that a = acceleration is a negative maximum because the velocity is changing from a positive to a negative value.
007 10.0 points Consider a car of mass 1200 kg moving without slipping on an inclined circular track of radius R = 18 m. The track is inclined at an angle 0 = 25 ° with respect to the horizontal direction. Assume g = 9.8 m/s2 . Assuming that the track is frictionless, what is the speed at which the car must move to avoid slipping up or down the track. 1. 7.88271 2. 10.2841 3, 10.1455 4. 7.37037 5. 9.06955 6. 8.43808 7. 11.8235 8. 9.58536 9, 6.76449 10. 7.5545 Correct answer: 9.06955 m/s. Explanation: Take the radially inward direction as the x-axis and the vertical direction orthogonal to the x-axis as the y-axis. With the above choice for axes, the free body diagram leads to the following equations, Fcentripetal — max = mv2
IR
This force must be supplied by the xcomponent of the normal force so mv2 IR = Nsin9 Since the car only accelerates in the radially inward direction, the y-component of the normal force must be equal and opposite to the
Version 018/AABAC - midterm 02 - turner - (56970) gravitational force giving may = N cosB - mg = 0 Eliminating N from the above equations, we get v = •\./ gRtanh. 008 10.0 points A body oscillates with simple harmonic motion along the x-axis, Its displacement varies with time according to the equation x(t) = A cos(w t). Given the maximum acceleration amax 2 m/s2 , the period T = 3 s, determine the amplitude of oscillation, A. 1. 0.797904 2, 1.24118 3. 0.633257 4. 0.569932 5. 0.911891 6. 1.86178 7. 2.2164 8, 0.455945 9. 3.72355 10, 0.683918
4. 3.86461 5. 2.59284 6. 2.41475 7. 2.22463 8. 2.92496 9. 2.10699 10, 3.32923 Correct answer: 2.92496 m/s. Explanation: The system is the rock. In order for the rock to move in a circle of radius 0.873 m, it experiences a net force equal to the centripetal force necessary to maintain the circular motion. A free body diagram of the system includes only the force of gravity and the normal force exerted by the bottom of the bucket, so F„t = FN + F9 . When moving with the minimum velocity, the rock is just about to fall out, so the normal force is zero and the centripetal acceleration is supplied by gravity alone.
FN
Correct answer: 0.455945 m. Explanation: The acceleration is the rate of change of the velocity and so it's also the second time derivative of the position of the object, i.e. a = dv/dt = d2x/dt2. Thus we have a = -w2A cos(w t), which has a maximum amplitude of amax = Awe. We need only the angular frequency w which is related to the period by w = 2rr/T so in the end: A (27) 2 „ amax T giving A = amax 112 () 27r 009 10.0 points A bucket containing a rock of mass m is rotated in a vertical circle of radius 0.873 m. What must be the minimum speed of the pail at the top of the circle so that the rock won't fall out? 1. 3.5028 2. 2.82959 3. 2.34056
4
= rag = - rn acentripetal v2 =m .
Then, mg = m
V2
and so, v = gr
2.92496 m/s
010 10.0 points A force acting on a particle has the potential energy function U(x), shown by the graph. The particle is moving in one dimension under the influence of this force and has kinetic energy 1.0 Joule when it is at position xi.
Version 018/AABAC — midterm 02 — turner — (56970)
5
Potential Energy vs Position
Potential Energy (J)
011 10.0 points
■■■ E■ Two marbles, one twice as massive as the I MMEMIIIMAINEMMINUMINI ••••••••••••••••••• are dropped from the same height. ••••••••••••••••••11 other, 1 ■ When they strike the ground, how does the ■■ • u•••••••••••••••••• kinetic energy of the more massive marble 0
II■■ ■ ■ 11■■ compare to that of the other marble? Ell•••••••■ rn-m--11 111'••••••rAll•••••••••• ••••••IF.A■■ ■ •■ 1. It has 4 times the KE. il■■■ • •■■ •11•••,/••••••••••••11 ■■ 2. It has one-half the KE. ■ ► I■ IA■■ IIIMIWAMM•11••••IMMEME 11•111••••• • EMMEN= 3. It has the same KE. ■■ ■■ ■■ •■■ 4. It has twice the KE. correct
r• 1
xo
X1 X2 X3
Position Which of the following is a correct statement about the motion of the particle? 1. It comes to rest at either xo or x2 and remains at rest. 2. It moves to the left of xo and does not return. 3. It oscillates with maximum position x2 and minimum position xo. correct 4. It moves to the right of x3 and does not return. 5. It cannot reach either xo or x2. Explanation: In this case, the total energy of the particle is conserved so at any point on the axis, V(x )
u(x) = v-(x1) + (x1) = 1.0 J (-1.0 = 0
Explanation: KE = 1/27nv2 The two marbles experience the same acceleration while falling and so hit the ground with the same velocity. Therefore the marble with twice the mass has twice the kinetic energy. 012 10.0 points Object A with mass 4 kg moves at an initial speed of 2 m/s along a frictionless horizontal plane. A horizontal force F is applied opposite to the direction of the motion and brings object A to a stop in a distance AxA. Object B with mass 2 kg moves at an initial speed of 4 m/s along the frictionless horizontal plane. The same horizontal force F is applied opposite to the direction of the motion of B and brings object B to a stop in a distance AxB. What is the relation between AxA and Axi3? 1. AxA=40xB 2, Axj3=41AxA
V(x) = —U(x) 3. AxB=8AxA where V is the kinetic energy of the particle. Since kinetic energy > 0 , V(x) = —U(x) > 0
4. AxA=AKB 5, AxA=60,TB
U(x) < 0 J , 6. A,TB=20xA correct so the particle oscillates between position xo and x2.
7. Ax13-6AxA
Version 018/AABAC — midterm 02 — turner — (56970) 8, ,AxA=2AxB
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8. -489.868 9. -726.864 10. -266.496
9. AxA=8AxB Explanation: Consider each ball and the earth as three separate systems. Using the Energy Principle with the rest mass not changing, we know AK = W. The displacement is only in the x-direction and the only force is F, which is opposite to the direction of motion so
Correct answer: —87.0848 N. Explanation: The force exerted on an object is given by F = —dU/dr so taking the derivative of the potential energy function gives F = —dU I dr = —(2ar 4b r3)
WA = -FAxA
and inserting r = 1.1 m gives
WB = -F.AxB
F = —[2(27J/m2 )(1.1 m)+4(5.2 J/m4)(1,1 m)31 =
and For both blocks, Kf = 0 so the Energy Principle says: 2 FAXA = 0 — — 7nAvA 2 and
1 0 FAxB = 0 — — mBq 2 Taking the ratio of the two equations gives:
014 10.0 points Two electrons with charge q and mass m are held at a distance d apart. After they are let go, what would the final speed of each electron approach as time goes to infinity? (k = 471 €0 ) 1, v f = \I
(F.AxA)/(FAxB) = (mAv2A)/(mBv2B) (AxA)/(AxB) = (TnAv2A)/(mBv2B) = [4kg(2m/s)2]/[2kg(4m/8)2] = 1/2 so
2. vf = \I 3. vf
AxE, = 2.AxA
k (12
2d rn k q2
dm
correct
2 k q2 d2 m
4. vf = co 013 10.0 points Suppose the potential energy of an object is given by:
5, vf = 0 6. vf =
U = a r2 b r 4 where a = 27 J/m2 and b = 5.2 J/m4. What is the force on the object when it is at r = 1.1 m? 1. -353.69 2. -187.734 3. -47.454 4. -87.0848 5. -83.034 6. -2473.6 7. -1.0304
7. ti
k q2 d2 m
f = ,\/ k q2 2 d2 rrt
8. vf =
2 k q2 dm
Explanation: We take the initial state to be when the electrons are held at a distance d apart, and the final state to be when they are infinitely far apart, since there is a repulsive electric
Version 018/AABAC - midterm 02 - turner - (56970) force between them. Then, energy conservation gives: = Ef i +Ui = Kf Uf k (12 1 mv+ 0 0+ =2 -2 ik q2 of = dm
7
so that the frictional force is given by Ff = ma - F cos 0 The total work done by friction in moving the block from xi to xf is given by Xf
Wf = f
Xi
Ff
= f
(ma - F cos 9)dx
015 10.0 points Sally applies a total force of 115 N with a rope to drag a wooden crate of mass 100 kg across a floor with a constant acceleration in the x-direction of a = 0.1 m/s2. The rope tied to the crate is pulled at an angle of 8 = 54° relative to the floor.
T4 = (ma - F cos 0)d.
016 10.0 points The buoyant force exerted on a solid object immersed in a fluid can be: A. less than the weight of the object; B, greater than the weight of the object; C. the same as the weight of the object. 1. B only 2. C only 3. A and C only
Calculate the total work done by friction as the block moves a distance 45.4 m over the horizontal surface. The acceleration due to gravity is 9.8 m/s2 1. -10932.7 2. -2614.83 3. -4498.2 4. -14178.8 5, -17580.5 6. -4803.4 7. -2334.92 8. -8505.25 9. -7851.92 10. -12064.3 Correct answer: -2614.83 J.
Explanation: The system is the block. We know that Fnet x = ma because the block is accelerating in the x-direction. A free body diagram gives Ff
= ma
4. A only 5. All are true. correct 6. A and B only 7. None is true. 8. B and C only
Explanation: Imagine an object made of the fluid with the same shape as the solid object. When immersed into the fluid, the net force on this fluid object would be zero, which means the pressure of the surrounding liquid must just balance the gravitational force. Now replace the fluid object with the solid object that has the same geometry; the pressure of the surrounding fluid would be the same. Hence the force upward on the object (the buoyant force) equals the weight of the displaced fluid.
Version 018/AABAC — midterm 02 — turner — (56970) That force depends only on the geometric shape of the object and not its weight, i,e, gravitational force it experiences. Therefore all three statements A, B, and C could be true depending on the density of the object relative to the density of the fluid.
017 10.0 points The escape speed from a very small asteroid is only 16 m/s. If you throw a rock away from the asteroid at a speed of 42 m/s, what will be its final speed? G = 6.7 x 10-11 Nkg2 m2 1. 40.2989 2. 24.1868 3, 29,7489 4. 34.9857 5. 44.1928 6. 38.833 7. 33.9853 8. 42.72 9, 27,1293 10. 13.0767 Correct answer: 38.833 m/s. Explanation:
8
018 10,0 points The following graph represents a hypothetical potential energy curve for a particle of mass m, U(r) 3 U0 2 U0 Uo ro
0
2 ro
If the particle is released from rest at position ro, its speed ff at position 2 To is most nearly L =
U0 2m
2. =
Uo 6m
3- I =
U0 m
4. 111111 =
Uo 8m 4 U0
m
correct
ese Use the Energy Principle. = Ef + = Uf + Kf —GMm 1 2 1 2 2 7TIVi = + 2Tiv f ri Where Uf is zero. ,,2 2 f
= vi
,,2
2GM
2 2 V- — Vese
uf
Vf =
2 V 65C
9.
=
Uo 4m
Explanation: The total energy of the particle is conserved. So the change of the potential energy is converted into the kinetic energy of the particle, which gives
= \,/(42 m /s)2 — (16 111/0 2 VI
=
38,833 m/s
1 — m v' = 3 U0 — U0 2
Version 018/AABAC — midterm 02 — turner — (56970)
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Explanation: Consider each ball and the earth as three separate systems. There is no work done on these systems by the surroundings so AE,y, =
8.
Wsurr = 0.
Next, consider the system that consists of the ball with mass 'az and the earth. Initially, it has a total energy (ignoring rest mass energy) of
9. Esys , z =
Explanation: The normal force of the wall on the rider provides the centripetal acceleration necessary to keep her going around in a circle. The downward force of gravity is equal and opposite to the upward frictional force on her. Note: Since this problem states that it is viewed by a bystander, we assume that the free-body diagrams are in an inertial frame.
1 =mvi2 +mgyi 2
and a final energy of Esys,f
1 = —Trt
2
Three balls with masses: m 2 m and im re2 spectively are thrown from the top of a building, all with the same initial speed vi. The first ball is thrown horizontally, the second at some angle 9j above the horizontal, and the third with some angle 02 < 01 below the horizontal, Neglecting air resistance, rank the speeds of the balls as they reach the ground, from the slowest to the fastest: 1. 2, 3, 1 2. 2, 1, 3 3. 3, 1, 2 4. 3, 2, 1 5. 1, 3, 2 6. 1, 2, 3 7. All three balls strike the ground with the same speed. correct
2 rn g yf
SO
AE„ Y
021 10.0 points
= K f + Uf
—i m (v1 2 — 71,2 ) =2
+ My —
yi) =0
Solving this equation for vf gives
Vf
9
yi)
The result does not depend on the mass of the ball so the same expression can be derived for each one. All three balls begin with the same speed vi and experience the same change in height (yf — yi) so the final speed for all three is the same .
