Physics Chapter 9 Answers

Chapter 9: Linear Momentum and Collisions Answers to Even-Numbered Conceptual Questions 2.

Doubling an object’s speed increases its kinetic energy by a factor of four, and its momentum by a factor of two.

4.

No. Consider, for example, a system of two particles. The total momentum of this system will be zero if the particles move in opposite directions with equal momentum. The kinetic energy of each particle is positive, however, and hence the total kinetic energy is also positive.

6.

Yes. Just point the fan to the rear of the boat. The resulting thrust will move the boat forward.

8.

(a) The force due to braking—which ultimately comes from friction with the road—reduces the momentum

of the car. The momentum lost by the car does not simply disappear, however. Instead, it shows up as an increase in the momentum of the Earth. (b) As with braking, the ultimate source of the force accelerating the car is the force of static friction between the tires and the road. 10.

Yes. For example, we know that in a one-dimensional elastic collision of two objects of equal mass the objects “swap” speeds. Therefore, if one object is at rest before the collision, it is possible for one object to be at rest after the collision as well. See Figure 9-7(a).

12.

No. Any collision between cars will be at least partially inelastic, due to denting, sound production, heating, and other effects.

14.

The center of mass of the hourglass starts at rest in the upper half of the glass and ends up at rest in the lower half. Therefore, the center of mass accelerates downward when the sand begins to fall—to get it moving downward—and then accelerates upward when most of the sand has fallen—to bring it to rest again. It follows from equation 9-18 that the weight read by the scale is less than Mg when the sand begins falling, but is greater than Mg when most of the sand has fallen.

16.

(a) Assuming a very thin base, we conclude that the center of mass of the glass is at its geometric center of the glass. (b) In the early stages of filling, the center of mass is below the center of the glass. When the glass is practically full, the center of mass is again at the geometric center of the glass. Thus, as water is added, the center of mass first moves downward, then turns around and moves back upward to its initial position.

18.

As this jumper clears the bar, a significant portion of his body extends below the bar due to the extreme arching of his back. Just as the center of mass of a donut can lie outside the donut, the center of mass of the jumper can be outside his body. In extreme cases, the center of mass can even be below the bar at all times during the jump.

Solutions to Problems and Conceptual Exercises 1.

travel in a straight line at constant constant speed. The data given in the Picture the Problem: The car and the baseball each travel Exercise 9-1 include pcar = 15,80 15,800 0 kg kg ⋅ m/s m/s and m ball = 0.142 kg. Strategy: Use equation 9-1 to set the magnitude of the momentum of the baseball equal to the magnitude of the momentum of the car, and solve for m all . Solution: Set p all = pcar

and solve for v

all

:

p ball = mball vball = pcar v ball =

pcar m ball

=

15, 800 kg ⋅ m/s 0.142 kg

=

1.11 ×10 6 m s

×

3600 s h

×

1 mi 1609 m

= 2.49 × 105 mi/h

Insight: The huge required speed for the baseball (3,240 times the speed of sound!) is due to the huge mass difference between the two two objects. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–1

James S. Walker, Physics Walker, Physics,, 4th Edition

Chapter 9: Linear Momentum and Collisions

2.

Picture the Problem: The ducks approach each other and the goose recedes as indicated in the figure at right. Strategy: Sum the velocity vectors from the three birds using the component method to find the total momentum. Solution: 1. Find the momentum of the left duck:

p d1 = md vd xˆ = ( 4.00 4.00 kg) (1.10 .10 m/s m/s) xˆ

2. Find the momentum of the top duck:

p d2 = md vd yˆ = ( 4.00 kg) ( −1.10 m/s ) yˆ

3. Find the momentum of the goose:

p g = mgv g yˆ = ( 9.00 kg) ( −1.30 m/s ) yˆ

4. Add the momenta:

p total = pd1 + pd 2 + p g

G

4.40 kg ⋅ m/s) xˆ = ( 4.40 G

.40 kg kg ⋅ m/s m/s ) yˆ = ( − 4.40 G

= ( −11. 11.7 kg kg ⋅ m/s) yˆ G

G

G

G

= 4.40 kg ⋅ m/s xˆ − 4.40 kg ⋅ m/s yˆ − 11.7 kg ⋅ m/s yˆ p total = ( 4.40 kg ⋅ m/s ) xˆ + ( −16.1 kg ⋅ m/s ) yˆ G

positive x axis. Insight: The total momentum vector now has a magnitude of 16.7 kg·m/s and points 74.7° below the positive x

3.

Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward and the dog runs northward. Strategy: Sum the momenta of the dog and cat using the component method. Use the known components of the total momentum to find its magnitude and direction. Let north be in the y the y direction, east in the x the x direction. Use the momentum together with the owner’s mass to find the velocity of the owner. Solution: 1. Use the component method of vector addition to find the owner’s momentum:

p total = p cat + pdog = mcat v cat + mdog v dog G

G

G

G

G

5.30 kg) ( 3.04 3.04 m/s xˆ ) + ( 26.2 26.2 kg) ( 2.70 .70 m/s yˆ ) = ( 5.30 p total = (16.1 kg ⋅ m/s ) xˆ + ( 70.7 kg ⋅ m/s ) yˆ G

powner = mowner v owner = p total G

2. Divide the owner’s momentum by his mass to to get the components of the owner’s velocity:

G

G

v owner =

p total

m0

=

G

(16.1 kg ⋅ m/s ) xˆ + ( 70.7 kg ⋅ m/s ) yˆ 74.0 kg

= ( 0.218 m/s ) xˆ + ( 0.955 m/s ) yˆ 3. Use the known components to find the direction and magnitude of the owner’s velocity:

vowner =

2 2 ( 0.218 m/s ) + ( 0.955 m/s) =

0.980 m/s

⎛ 0.955 ⎞ ⎟ = 77.1° north of east ⎝ 0.218 ⎠

θ = tan −1 ⎜

Insight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The owner is moving much slower than either the cat or the dog because of his larger mass.

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–2



4.

Picture the Problem: The two carts approach each other on a frictionless track at different speeds. Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting expression for v2 . Then use

equation 7-6 to find the total kinetic energy of the two-cart system. Let cart 1 travel in the positive direction. Solution: 1. (a) Set

∑ p = 0 and solve for v G

2

∑p = m v G

:

G

1

v2 = −

+ m2 v 2 = 0 G

1

m1 v1 m2

=

0.35 kg) (1.2 1.2 m/s m/s) ( 0.35 0.61 kg

= 0.69 m/s

always greater than than or equal to zero. zero. 2. (b) No, kinetic energy is always

∑ K =

3. (c) Use equation 7-6 to sum the kinetic energies of the two carts:

1 2

m1 v12 + 12 m2 v2 2 2

= 12 ( 0.35 kg ) (1.2 m/s ) + 12 ( 0.61 kg ) ( 0.69 m/s)

2

= 0.40 J 0.42 kg kg ⋅ m/s ) xˆ and the momentum of Insight: If cart 1 is traveling in the positive xˆ direction, then its momentum is ( 0.42 cart 2 is ( − 0.42 0.42 kg ⋅ m/s) xˆ . 5.

Picture the Problem: The baseball drops straight down, gaining momentum due to the acceleration of gravity. Strategy: Determine the speed of the baseball before it hits the ground, then use equation 2-12 to find the height from which it was dropped. Solution: 1. Use equation 9-1 to find the speed of the ball when it lands:

v=

p m

=

0.78 0.780 0 kg kg ⋅ m/s 0.150 kg

= 5.20 m/s

v 2 = v02 − 2 g ( y − y0 )

2. (b) Use equation 2-12 to solve for y0 . Let y = 0 and v0 = 0 :

y0 =

v2 2 g

=

( 5.20 m/s )

2

2 ( 9.81 9.81 m/s m/s2 )

= 1.38 m

Insight: Another way to find the initial height is to use conservation of energy, setting mgy0 = 12 mv 2 and solving for y0 .

6.

Picture the Problem: The ball falls vertically downward, landing with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. Strategy: Use equation 9-1 to find the change in momentum of the ball when it rebounds.

Δp = p f − pi = m ( vf − vi ) G

Solution: 1. (a) Use equation 9-1 to find Δp :

G

G

G

G

G

= ( 0.285 kg ) ⎡⎣( 2.0 m/s ) yˆ − ( −2.5 m/s ) yˆ ⎤⎦ = (1.3 kg ⋅ m/s) yˆ 1.3 kg kg ⋅ m/s Δp = 1.3 G

pf − pi = m ( vf − vi )

2. (b) Subtract the magnitudes of the momenta:

= ( 0.285 kg ) ( 2.0 m/s − 2.5 m/s ) pf − pi = − 0.14 0.14 kg kg ⋅ m/s

3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its collision with the

floor, first of all because

∑

F = Δ p Δt (equation 9-3) and as we can see from above that Δp ≠ pf − pi . Secondly, we G

G

G

expect the floor to exert an upward force on the ball but we calculated a downward (negative) value in part (b). the ball ball were were to rebo reboun und d at 2.5 2.5 m/s m/s upwa upward rd we woul would d find find Δp = 2 mv = 1.1 1.1 kg ⋅ m/s m/s and pf − pi = 0 . Such a Insight: If the G

collision with the floor would be called elastic, elastic, as discussed in section 9-6.


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7.

Picture the Problem: The individual momenta and final momentum vectors are depicted at right.

y

objects are perpendicular. perpendicular. Because of this we Strategy: The momenta of the two objects can say that the momentum of object 1 is equal to the x the x-component -component of the total momentum and the momentum of object 2 is equal to the y the y-component -component of the total momentum. Find the momenta of objects 1 and 2 in this this manner and divide by their speeds to determine the masses. Solution: 1. Find ptotal, x total, x and divide by v1 :

and divide by v2 : 2. Find ptotal, y total, y

G

p total

G

p2

x

p1 G

p1 = ptotal, x = ptotal cos θ = (17.6 kg ⋅ m/s ) ( cos 66.5° ) = 7.02 kg ⋅ m/s m1 =

p1

m2 =

p2

v1

v2

=

7.02 7.02 kg kg ⋅ m/s

=

2.80 m/s ptotal sin θ v2

=

= 2.51 kg 6.5° ) (17.6 kg ⋅ m/s ) ( sin 6 6. 3.10 m/s

= 5.21 kg

Insight: Note that object 2 has the larger momentum because the total momentum points mostly in the yˆ direction. The

two objects have similar speeds, so object 2 must have the larger mass in order to have the larger momentum. 8.

Picture the Problem: Your car rolls slowly in a parking lot and bangs into the metal base of a light pole. Strategy: Use the concept of impulse during a collision to answer the conceptual question. Solution: If the collision were inelastic the light pole gives your car only enough impulse to bring it to rest. If the collision were elastic, elastic, the impulse given to your car is about twice as much; the car must be brought to rest and then accelerated to its initial speed but in the opposite direction. We conclude that in terms of safety, it is better for your collision with the light pole to be inelastic. Insight: The additional impulse required for an elastic collision, which acts over a very short period of time, means the driver will experience a larger force that could cause injury.

9.

Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. Strategy: Use Newton’s Second Law in terms of momentum to answer the conceptual question. Solution: 1. (a) One way to write Newton’s Second Law is to equate the net force with the change in momentum per time (equation 9-3). Both the boulder and the pebble have the same rate of momentum change because the same force acts on both objects. We conclude that the change of the boulder’s momentum in one second is equal to the change of the pebble’s momentum in the same time period.

time. Statement I is false, false, 2. (b) The best explanation is III. Equal force means equal change in momentum for a given time. and statement II is partly true (the pebble’s speed is larger) but the pebble’s change in momentum is the same because it has a smaller mass than the boulder. Insight: Newton originally formulated his second law by asserting that the net force is equal to the rate of change of momentum. Later textbooks introduced introduced the more familiar F = F = ma equation.

10. Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. Strategy: Use Newton’s Second Law in terms of momentum to answer the conceptual question. Solution: 1. (a) Both the boulder and the pebble have the same rate of momentum change because the same force acts on both objects (equation 9-3). As a result, the rate of change in velocity for the less massive object (the pebble) must be greater than it is for the more massive object (the boulder). We conclude that the change of the boulder’s speed in one second is less than the change of the pebble’s speed in the same time period.

boulder results in a small acceleration. acceleration. Statement II is false false (the 2. (b) The best explanation is I. The large mass of the boulder same force results in the same change in momentum, not speed), and statement III is true but irrelevant. Insight: We know that the acceleration (rate of change of velocity) of an object is proportional to the force acting on it and inversely proportional to its mass. These objects experience the same force, and therefore the less massive object has the greater acceleration. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–4



11. Picture the Problem: A friend tosses a ball of mass m to you with a speed v. The force you exert on the ball when you catch it produces a noticeable sting in your hand. Strategy: Use the concept of impulse to answer the conceptual question. Solution: 1. (a) The two balls have the same momentum. If they are stopped in the same amount of time by your hand, your hand will exert an equal force on each ball. We conclude that the sting you feel feel is equal to the sting you felt when you caught the first ball. 2. (b) The best explanation is II. The two balls have the same momentum, and hence they produce the same sting. Statement I is true if the two balls are stopped in the same distance (see the Insight statement below) and statement III is false (the momentum, not just the mass, needs to be considered). Insight: The first ball has a kinetic energy equal to 1 2

1 2

mv 2 and the second ball has half that much kinetic energy:

2

force F will will be required to stop the ( 2m ) ( 12 v ) = 12 ( 12 mv 2 ) . If each ball is stopped in the same distance d , a smaller force F

second ball because Fd because Fd = = Δ K and K and Δ K is K is smaller for the second ball. A smaller force would produce a smaller “sting.”

12. Picture the Problem: Several forces of different magnitudes act for different time intervals. Strategy: Use the concept of impulse to rank the forces in terms of the impulses they produce. Solution: Multiply the force by the time interval in order to calculate the impulse. Using this approach we find that impulse A is F Δt , impulse B is 32 F Δt , impulse C is 12 F Δt , and impulse D is 101 F Δt . We arrive at the ranking

D < C < B < A. F acting over a time interval 2 Δt is t is exactly the same as a force F force F acting acting over a time Insight: The impulse of a force 3 F acting interval 6Δt .

13. Picture the Problem: The force of the kick acting over a period of time imparts an impulse to the soccer ball. Strategy: Multiply the force by the time of impact to find the impulse according to equation 9-5. Solution: Apply equation 9-5 directly:

I = Fav Δt = (1250 N ) ( 5.95 × 10−3 s ) = 7.44 kg ⋅ m/s

Insight: The 1250-N force is equivalent to 281 lb! The same impulse could be delivered to the ball with 7.44 N (1.67 lb) of force acting over a period of one second.

14. Picture the Problem: The golf club exerts an impulse on the ball, imparting momentum. Strategy: Find the change of momentum of the golf ball and use it to find the force exerted on it according to equation 9-3. Solution: Apply equation 9-3 directly:

F av =

Δ p m ( vf − vi ) ( 0.045 kg ) ( 67 m/s − 0 ) = = = 3.0 kN 0.0010 s Δt Δt

to 670 lb! A ball launched with with a speed of 67 m/s launched at 45° will land land 457 m Insight: This is a force equivalent to (500 yd) down range in the absence of air friction. friction. A par 5 hole in one?

15. Picture the Problem: The croquet mallet exerts an impulse on the ball, imparting momentum. Strategy: Find the change in momentum of the croquet ball and then use it to find Δt using equation 9-3. Solution: Solve equation 9-3 for Δt :

Δt =

Δ p Fav

=

m ( vf − vi ) F av

=

( 0.50 kg ) ( 3.2 m/s − 0) 230 N

= 0.0070 s = 7.0 ms

Insight: The large force (52 lb) is exerted over a very brief time to give the ball its small velocity (7.2 mi/h).


9–5



16. Picture the Problem: The hand of the volleyball player exerts an impulse on the ball, imparting momentum. Strategy: Use equation 9-6 together with the known impulse and change of velocity of the ball in order to find its mass. I = Δ p = mΔv

Solution: Solve equation 9-6 for mass:

m=

I

Δv

=

9.3 kg ⋅ m/s −9.3 = 0.33 kg m/s −24 m/s − 4.2 m/

Insight: A mass this large weighs about ¾ lb at sea level. This ball is 22% heavier than the 0.27 kg regulation ball.

17. Picture the Problem: The marble drops straight down from rest and rebounds from the floor. Strategy: Use equation 2-12 to find the speed of the marble just before it hits the floor. Use the same equation and the known rebound height to find the rebound speed. Then use equation 9-6 to find the impulse delivered to the marble by the floor. Let upward be the positive direction, so that the marble hits the floor with speed −vi and rebounds upward

with speed vf . Solution: 1. (a) Find vi

vi2 = v02 − 2 g Δy

using equation 2-12:

vi = − v02 − 2 g Δy = − 02 − 2 ( 9. 9.81 m/s 2 ) ( −1.44 m ) = −5.32 m/s

2. Find vf using equation 2-12 again:

v 2 = vf 2 − 2 g Δy vf =

v 2 + 2 g Δy =

02 + 2 ( 9.81 m/s 2 ) ( 0.640 m ) = 3.54 m/s

I = mΔv = m ( vf − vi )

find I : 3. Use equation 9-6 to find I

= ( 0.0150 kg ) ⎡⎣ 3.54 − ( − 5.32) m/s⎤⎦ = 0.133 kg ⋅ m/s 4. (b) If the marble had bounced to a greater height, its rebound speed would have been larger and the impulse would have been greater than the impulse found in part (a).

delivers a downward impulse on the floor. The Earth theoretically theoretically Insight: By Newton’s Third Law the marble also delivers moves in response to this impulse, but its change of velocity is tiny (2.2×10-26 m/s) due to its enormous mass.

18. Picture the Problem: The ball rebounds from the floor in the manner indicated by the figure at right. the y-component -component of the Strategy: The impulse is equal to the change in the y momentum of the ball. Use equation 9-6 in the vertical vertical direction to find the impulse. Solution: Apply equation 9-6 in the y the y direction:

65°

65°

vi G

v f G

I = Δp y = mΔv y

= m [v0 cos 65° − (−v0 cos 65° )] = ( 0.60 kg ) ( 5.4 m/s) ( 2 cos 65° ) = 2.7 kg⋅ m/s

the x direction because the ball does not change its horizontal speed or momentum. Insight: There is no impulse in the x

19. Picture the Problem: The ball rebounds from the bat in the manner indicated by the figure at right. Strategy: The impulse is equal to the vector change in the momentum. Analyze the x the x and y and y components of Δp separately, then use the components G

v f G

G

to find the direction and magnitude of I . Solution: 1. (a) Find Δ p x :

vi G

Δ p x = m ( vfx − vix ) = ( 0.14 kg ) ⎡⎣ 0 − ( −36) m/s ⎤⎦ = 5.0 kg ⋅ m/s


9–6



Δ p y = m ( vfy − viy ) = ( 0.14 kg) (18 − 0 m/s ) = 2.5 kg ⋅ m/s

2. Find Δ p y :

I = Δp = ( 5.0 kg ⋅ m/s ) xˆ + ( 2.5 kg ⋅ m/s ) yˆ G

G

G

3. Use equation 9-6 to find I :

⎛ I y ⎞ ⎛ 2.5 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 27° above the horizontal ⎝ 5.0 ⎠ ⎝ I x ⎠

G

4. Find the direction of I :

I =

G

5. Find the magnitude of I :

I x2 + I y2 =

2 2 ( 5.0 kg ⋅ m/s ) + ( 2.5 kg ⋅ m/s ) =

5.6 kg ⋅ m/s

6. (b) If the mass of the ball were doubled the impulse would double in magnitude. There would be no change in the direction. 7. (c) If Δp of the ball is unchanged, the impulse delivered to the ball would not change , regardless of the mass of the G

bat. speed. Verify for yourself that Insight: The impulse brings the ball to rest horizontally but gives it an initial horizontal speed. this ball will travel straight upward upward 16.5 m (54 feet) before falling back to Earth. An easy popup!

20. Picture the Problem: The ball rebounds from the player’s head in the manner indicated by the figure at right.

y

vector change in the momentum. momentum. Analyze the x the x Strategy: The impulse is equal to the vector and y and y components of Δp separately, then use the components to find the direction and G

x

G

v f G

magnitude of I .

vi G

Solution: 1. (a) Find Δ p x :

Δ p x = m ( vfx − vix ) = ( 0.43 kg ) ( 5.2 − 8.8 m/s ) 1.5 kg ⋅ m/s = − 1.5

Δ p y = m ( vfy − viy ) = ( 0.43 kg ) ⎡⎣ 3.7 − ( −2.3) m/s ⎤⎦ = 2.6 kg ⋅ m/s

2. Find Δ p y : G

3. Use equation 9-6 to find I :

I = Δp = ( −1.5 kg ⋅ m/s ) xˆ + ( 2.6 kg ⋅ m/s ) yˆ G

G

4. Find the direction of I :

⎛ I y ⎞ ⎛ 2.58 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ positive x axis ⎟ = −59° + 180° = 121° from the positive x ⎝ −1.55 ⎠ ⎝ I x ⎠

5. (b) Find the magnitude:

I =

G

I x2 + I y2 =

2 2 ( −1.55 kg ⋅ m/s ) + ( 2.58 kg ⋅ m/s ) =

3.0 kg ⋅ m/s

Insight: The ball delivers an equal and opposite impulse to the player’s head, which would exert a force of 300 N (67 lb) if the time of collision were 10 milliseconds.

21. Picture the Problem: The two canoes are pushed apart by the force exerted by a passenger. Strategy: By applying the conservation of momentum we conclude that the total momentum of the two canoes after the push is zero, just just as it was before the push. Set the total momentum of the system to zero zero and solve for m2 . Let the

velocity v1 point in the negative negative direction, v 2 in the positive direction. G

Solution: Set ptotal = 0 and solve for m2 :

G

p1 x + p2 x = 0 = m1 v1x + m2v 2x − m1 v1 x − ( 320 kg ) ( − 0.58 m/s) m2 = = = 440 kg v2 x 0.42 m/s

Insight: An alternative way to find the mass is to use the equations of kinematics in a manner similar to that described in Example 9-3.


9–7



22. Picture the Problem: The two skaters push apart and move in opposite directions without friction. Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the push is zero, just just as it was before the push. Set the total momentum of the system to zero zero and solve for m2 . Let the

velocity v1 point in the negative negative direction, v 2 in the positive direction. G

G

p1 x + p2 x = 0 = m1 v1x + m2v 2x −m1 v1 x − ( 45 kg ) ( − 0.62 m/s ) m2 = = = 31 kg v2 x 0.89 m/s

Solution: Set ptotal = 0 and solve for m2 :

Insight: An alternative way to find the mass is to use the equations of kinematics in a manner similar to that described in Example 9-3.

23. Picture the Problem: The bee and the stick move in opposite directions without friction. Strategy: By applying the conservation of momentum we conclude that the total momentum of the bee and the stick is zero, just as it was before the bee started walking. Set the total momentum of the system to zero and solve for the speed of the stick. Let the velocity v s point in the negative negative direction, v in the positive direction. G

Solution: Set ptotal = 0 and solve for vs :

G

p1 x + p2 x = 0 = mbv b + ms vs vs =

− m bv b ms

=

0.175 5 g)( 14.1 mm/s m/s ) − ( 0.17 ) (14.1 4.75 g

= − 0.519 mm/s

Insight: The larger mass of the stick means that it must have a slower speed in order to have the same momentum as the bee. At scales this this small the surface surface tension of the water plays an an important role role and the stick’s speed would be even smaller than 0.512 mm/s.

24. Picture the Problem: The two pieces fly in opposite directions at different speeds. Strategy: As long as there is no friction the total momentum of the two pieces must remain zero, as it was before the explosion. Combine the conservation of momentum with with the given kinetic energy ratio to determine the ratio of the masses. Let m1 represent the piece with the smaller kinetic energy. Solution: 1. Set

∑p = 0 G

and solve for m1 m2 : p1 + p2 = 0 = m1 v1 + m2v 2 m1 m2

2. Set K 2 K 1 = 2 :

K 2 K1

=−

=2=

2

⎛m ⎞ ⎛v ⎞ ⇒ ⎜ 1 ⎟ =⎜ 2 ⎟ ⎝ m2 ⎠ ⎝ v1 ⎠

v2 v1

1 2

m2 v22

1 2

m1 v12

2

2

2

2

⎛v ⎞ m ⇒ ⎜ 2⎟ =2 1 m2 ⎝ v1 ⎠

⎛v ⎞ ⎛ m ⎞ m 3. Combine the expressions from steps 1 and 2: ⎜ 2 ⎟ = ⎜ 1 ⎟ = 2 1 ⇒ m2 ⎝ v1 ⎠ ⎝ m2 ⎠

m1 m2

=2

4. The piece with the smaller kinetic energy has the larger mass. Insight: The smaller mass carries the larger kinetic energy because kinetic energy increases with the square of the velocity but is linear with mass. Its higher speed more than compensates for its smaller mass.

25. Picture the Problem: The astronaut and the satellite move in opposite directions after the astronaut astronaut pushes off. The astronaut travels at constant speed a distance d before d before coming in contact with the space shuttle. Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must remain zero, as it was before the astronaut pushed off. Use the conservation of momentum to determine the speed of the astronaut, and then multiply the speed by the time to find the distance. Assume the satellite’s motion is in the negative x negative x-direction. -direction. Solution: 1. Find the speed of the astronaut using conservation of momentum:

pa + ps = 0 = ma v a + ms vs mv va = − s s ma


9–8



d = va t = −

2. Find the distance to the space shuttle:

ms vs ma

t = −

(1200 kg ) ( − 0.14 m/s ) ( 7.5 s ) = ( 92 kg )

14 m

Insight: One of the tricky things about spacewalking is that whenever you push on a satellite or anything else, because of Newton’s Third Law you yourself get pushed ! Conservation of momentum makes makes it easy to predict your speed.

26. Picture the Problem: The lumberjack moves to the right while the log moves to the left. Strategy: As long as there is no friction the total momentum of the lumberjack and the log remains zero, as it was before the lumberjack lumberjack started trotting. trotting. Combine vector vector addition for relative motion motion (equation 3-8) with with the expression from the conservation of momentum to find vL, s = speed of lumberjack lumberjack relative to to the shore. Let vL, log = speed of

lumberjack relative to the log, and vlog, s = speed of the log relative to the shore. Solution: 1. (a) Write out the equation for relative motion. motion. Let the log travel travel in the negative direction: 2. Write out the conservation of momentum with respect to the shore:

v L, s = v L, log + vlog, s G

G

G

vL, s = vL, log − vlog, s vlog, s = vL, log − vL, s

∑p = 0 = m v G

L L, s

− mlog vlog, s

mL vL, s = mlog vlog, s = mlog ( vL, log − vL, s )

3. Substitute the expression from step 1 into step 2 and solve for vL, s :

vL, s ( mL + mlog ) = mlog vL, log vL, s =

mlo gvL, l og og

(m

L

+ mlog )

=

kg ) ( 2.7 m/ m/s ) ( 380 kg = ( 85 + 380 kg )

2.2 m/s

4. (b) If the mass of the log had been greater, the lumberjack’s speed relative to the shore would have been greater than that found in part (a), because the log would have moved slower in the negative direction. 5. (c) Use the expression from step 3 to find the new speed of the lumberjack:

vL, s =

mlo gvL , lo g

(m

L

+ mlog )

=

kg ) ( 2.7 m/ m/s ) ( 450 kg = ( 85 + 450 kg )

2.3 m/s

Insight: Taking the argument in (b) to its extreme, if the mass of the log equaled the mass of the Earth the lumberjack’s speed would be exactly 2.7 m/s relative to the Earth (and the log). If the mass of the log were the same as the mass of the lumberjack, the speed of each relative to the Earth would be half the lumberjack’s walking speed. y

27. Picture the Problem: The vector diagram at right indicates the momenta of the three pieces.

G

the xy plane is Strategy: Because the plate falls straight down its momentum in the xy zero. That means the the momenta of all all three pieces must must sum to zero. zero. Choose the motion of the two pieces at right angles to one another to be in the xˆ and momentum equal to zero and solve for v 3 . yˆ directions. Set the total momentum

p2

225° p1 G

G

Solution: 1. Set

∑p = 0 G

and solve for v 3 : G

∑ p = mv xˆ + mv yˆ + mv G

G

3

( −mv ) xˆ + ( −mv ) yˆ m

2 2 (− v ) + ( −v ) =

2. Find the speed v3 :

v3 =

3. Find the direction of v3 :

θ = tan 1 ⎜ ⎜

−

p3 G

=0

G

v3 =

x

= ( −v ) xˆ + ( −v ) yˆ

2v

⎛ v3, y ⎞ −1 ⎛ − v ⎞ ⎟⎟ = tan ⎜ ⎟ = 45° + 180° = 225° ⎝ −v ⎠ ⎝ v3, x ⎠

Insight: As long as the first two pieces have equal masses the direction of v 3 will always be the same. For instance, if G

the third piece has four times the mass of either piece 1 or 2, its speed would be v

8 but θ would remain 225°.


9–9



28. Picture the Problem: The two carts collide on a frictionless track and stick together. Strategy: The collision is completely inelastic because the two carts stick together. Momentum is conserved during the collision because the track track has no friction. The two carts move as if they were were a single object after after the collision. Use the conservation of momentum to find the final speed of the carts and final kinetic energy of the system.

pi = pf

Solution: 1. Conserve momentum to find the final speed of the carts:

mv + m(0) = 2mvf

2. Use equation 7-6 to find the final kinetic energy:

⇒ vf = 2

K f = 12 (2m)vf 2 = m ( 12 v ) =

mv 2m 1 4

=

v 2

mv 2

Insight: Half of the initial kinetic energy is gone, having been converted to heat, sound, and permanent deformation of material during the inelastic collision.

29. Picture the Problem: The car and the minivan collide and stick together, as indicated in the figure at right. The data given in Example 9-6 include m1 = 950 kg and m2 = 1300 1300 kg .

Strategy: If we ignore frictional forces the total momentum of the two vehicles before the crash equals equals the momentum momentum after the collision. The vehicles stick stick together so they behave as a single object object with the sum of their masses. masses. Use conservation of momentum to find the initial speed v2 of the minivan and the

final speed vf of the two vehicles after the crash.

the x Solution: 1. Conserve momentum in the x direction:

∑ p

= m1 v1 + 0 = ( m1 + m2 ) vf cos θ

the y direction: 2. Conserve momentum in the y

∑ p

= m2v 2 + 0 = ( m1 + m2 ) vf sin θ

3. Divide the equation from step 2 by the equation for step 1 and solve for v2 :

m2 v2

=

x

y

m1v1 v2

4. Solve the equation in step 1 for vf :

vf =

( m1 + m2 ) vf sin θ = tan θ ( m1 + m2 ) vf cos θ ( 950 kg ) m tan 40.0° = = v1 1 tan θ = ( 20.0 m/s ) m2 (1300 kg ) m1 v1

( m1 + m2 ) cosθ

=

kg ) ( 20.0 m/s ) ( 950 kg = 40.0° ( 950 kg + 1300 kg) cos 40

12 m/s

11 m/s

Insight: In real life the assumption that there is no friction is quite incorrect unless the collision occurs on very slick, icy pavement. Substantial deviations from the conservation conservation of momentum will be observed if the pavement is dry.


9 – 10



30. Picture the Problem: The initial and final momentum vectors for this collision are depicted at right.

y

Strategy: Assuming there is no friction between the players’ skates and the ice, we can use conservation of momentum together with the fact that the players stick together after the collision to find the final velocity. Let the motion of player 1 be in the positive x positive x-direction -direction and the motion of player 2 be at an an angle of 115° measured measured counterclockwise counterclockwise from the positive x positive x-axis. -axis.

p f G

G

p 2i

115° x

p1i G

p1i + p 2i = p f ( mvxˆ ) + ( mv cosθ xˆ + mv sin θ yˆ ) = 2mv f G

Solution: 1. Write out the conservation of momentum and solve for v f : G

G

G

G

1 2 1 2

cos θ ) xˆ + 12 v ( sin θ ) yˆ = v f v (1 + co G

( 5.45 m/s ) (1 + cos115°) xˆ + 12 ( 5.45 m/s) ( sin 115°) yˆ = vf (1.57 m/s ) xˆ + ( 2.47 m/s ) yˆ = vf G

G

2 2 (1.57 m/s ) + ( 2.47 m/s ) =

vf = vf2 x + vf 2y =

2. Determine the magnitude of v f : G

2.93 m/s

positive x axis. Player 2’s initial momentum momentum Insight: The two players slide away from the collision at 57.5° above the positive x provides the y the y component of the final momentum, but his x his x momentum in the − xˆ direction is smaller than player 1’s momentum in the xˆ direction, and so the two players have a positive x positive x component of their their total momentum. momentum. Note that because the mass mass cancels out, in this particular case only (the case case of two identical identical masses colliding) colliding) the final final velocity is 1 the average of the two initial velocities: mv1i + mv 2i = 2 mv f ⇒ v f = 2 ( v1i + v 2i ) . G

G

G

G

G

G

31. Picture the Problem: A 1200-kg car moving at 2.5 m/s is struck in the rear by a 2600-kg truck moving at 6.2 m/s. The vehicles stick together after the collision and move together with a speed of 5.0 m/s. Strategy: In an inelastic collision we expect a loss of kinetic energy. Use equation 7-6 to find the kinetic energies before and after the collision and verify the loss. loss. Solution: 1. (a) The final kinetic energy of the car and truck together is less than the sum of their initial kinetic energies. Some of the initial kinetic energy is converted to heat, sound, and the permanent deformations deformations in the materials of the car and truck. 2. (b) Use equation 7-6 to find the initial kinetic energy:

Ki = 12 mcvc2 + 12 mt vt2 =

3. Use equation 7-6 to find the final kinetic energy:

K f =

1 2

1 2

2 2 (1200 kg ) ( 2.5 m/s ) + 12 ( 2600 kg) ( 6.2 m/s) =

2 ( mc + mt ) vf2 = 12 ( 1200 kg + 2600 kg) ( 5.0 m/s) =

54 kJ

48 kJ

initially. If Insight: The kinetic energy loss is only 11% because the two vehicles are traveling in the same direction initially. this were a head-on collision, the final speed would be 3.5 m/s in the direction the truck was traveling, and K f = 23 kJ , a loss of 58%. In fact, two identical objects traveling at the same speed and colliding together will lose 100% of their kinetic energy if they stick together after the collision.


9 – 11



32. Picture the Problem: The bullet collides with the block, passing right through it and continuing in a straight line but at a slower speed than it had initially. Afterwards the block moves with constant constant speed in the same direction as the bullet. Strategy: Use conservation of momentum to find the speed of the bullet after the collision. Because this is an inelastic collision we expect a loss of kinetic energy. Use equation 7-6 to find the initial and final kinetic energies and confirm the energy loss. Let the subscripts b and B denote the bullet and the block, respectively. Solution: 1. (a) Let pi = p f and solve for v f : G

m b vbi + 0 = mb vbf + mBv Bf m v − mBvBf v bf = b bi m b ( 0.0040 kg ) ( 650 m/s) − ( 0.095 kg) ( 23 m/s)

G

=

0.0040 kg

= 1.0 × 10 m/s 2

2. (b) The final kinetic energy is less than the initial kinetic energy because energy is lost to the heating and deformation of the bullet and block. 2 ( 0.0040 kg ) ( 650 m/s ) =

3. (c) Use equation 7-6 to find K i :

K i = 12 mbv bi 2 =

4. Use equation 7-6 to find K f :

K f = 12 mbv bf 2 + 12 mBvBF 2

1 2

850 J

2

2

= 12 ( 0.0040 kg ) (104 m/s ) + 12 ( 0.095 kg) ( 23 m/s) = 47 J Insight: In this collision 94% of the bullet’s initial kinetic energy is converted to heat, sound, and permanent deformation of materials.

33. Picture the Problem: The putty is thrown straight upward, strikes the bottom of the block, and sticks to it. The putty and the block move together in the upward direction immediately after the collision. Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the collision, and then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximum height above the collision point to which the conglomerate will rise. Solution: 1. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of the putty will be converted to heat, sound, and permanent deformation of material during the inelastic collision. 2. (b) Set pi = p f and solve for vf : G

3. Set E

ottom

G

m p vp = ( mb + mp ) vf

⎞ ⎟⎟ v p ⎠

K bottom + 0 = 0 + U top

= E top after the collision: 1 2

4. Solve the resulting expression for ytop

⎛ m p ⇒ vf = ⎜ ⎜ m b + mp ⎝

⎛ m ( m + m p ) ⎜⎜ m + pm p ⎝ b

ytop

⎛ m p =⎜ ⎜ m b + mp ⎝

⎞ ⎟⎟ ⎠

2

2

⎞ 2 ⎟⎟ vp = ( mb + mp ) g y top ⎠

2 2 ⎛ vp 2 ⎞ ⎛ ⎞ ⎡ ( 5.74 m/s ) ⎤ 0.0750 kg ⎥ ⎜⎜ ⎟⎟ = ⎜ ⎟ ⎢ 9.81 m/s m/s 2 ) ⎥ ⎝ 2 g ⎠ ⎝ 0.420 + 0.0750 kg ⎠ ⎢⎣ 2 ( 9.81 ⎦ cm = 0.0386 m = 3.86 cm

Insight: The putty-block conglomerate will rise even higher if either v p is larger or m p is larger.

34. Picture the Problem: The putty is thrown horizontally, strikes the side of the block, and sticks to it. The putty and the block move together together in the horizontal horizontal direction direction immediately after after the collision, collision, compressing the the spring. Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the collision, then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximum compression of the spring after the collision. Solution: 1. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of the putty will be converted to heat, sound, and permanent deformation of material during the inelastic collision. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 – 12



G

G

⎞ ⎟⎟ v p ⎠

K after + 0 = 0 + U rest

3. Set Eafter = E rest after the collision: 1 2

4. Solve the resulting expression for xmax :

⎛ m p ⇒ vf = ⎜ ⎜ m b + mp ⎝

m p vp = ( mb + mp ) vf

2. Set pi = p f and solve for vf :

2

⎛ m ( m + m p ) ⎜⎜ m + pm p ⎝ b

xmax

⎞ 2 1 2 ⎟⎟ vp = 2 kxmax ⎠

2 2 ⎡ m p2 vp2 ⎤ ⎡ ( 0.0500 kg ⎤ kg ) ( 2.30 m/ m/s ) ⎥=⎢ = ⎢ ⎥ N/m ) ( 0.430 + 0.0500 kg kg ) ⎥ ⎢⎣ k ( m p + mb ) ⎥⎦ ⎢⎣ ( 20.0 N/ ⎦ = 0.0371 m = 3.71 cm

2

Insight: The putty-block conglomerate will compress the spring even farther if v p is larger or if m p is larger.

35. Picture the Problem: Two objects with different masses travel in opposite directions with the same speed, collide, and stick together. Strategy: Use equation 7-6 to find the ratio of the kinetic energies before and after the collision. Use conservation of momentum to find the ratio of the masses.

K f

Solution: 1. (a) Use equation 7-6 to find K i and K f :

K i

( m1 + m2 ) ( v 4 ) 1 2

m1 v + 12 m2v 2

2

2

( m1 + m2 ) v 2 = 2 1 m m v + ( ) 1 2 2

1 32

1 16

m1v + m2 ( −v ) = ( m1 + m 2 ) ( v 4 )

2. (b) Use pi = pf to find m1 m2 : G

=

1 2

G

m1 − m2 = 14 ( m1 + m2 ) 4m1 − 4m2 = m1 + m2 3m1 = 5m 2 ⇒

m1 m2 = 5 3

Insight: Most of the kinetic energy is lost in the inelastic collision. If the same two masses collide and stick together except they are traveling in the same direction, with m1 traveling at speed v and m2 at v 4 , their final speed will be

( 23 32) v

and the energy loss will be only 17%.

36. Picture the Problem: The hammer strikes the nail with an elastic collision. collision. The collision drives the nail in the forward direction and slows down the speed of the hammer. collision where one of the objects (the nail) is initially at rest. Therefore, Strategy: This is a one-dimensional, elastic collision equation 9-12 applies and can be used to find the final speed of the nail. Once the nail speed is found, its kinetic energy is determined determined by equation 7-6. Let m1 be the mass of the hammer, m2 be the mass of the nail, and v0 be the initial speed of the hammer.

⎡ 2 ( 550 g ) ⎤ ⎛ 2m1 ⎞ ⎟ v0 = ⎢ ⎥ ( 4.5 m/s ) = 8.8 m/s = vnail ⎝ m1 + m2 ⎠ ⎣ 550 + 12 g ⎦

Solution: 1. Use equation 9-12 to find v2,f :

v2,f = ⎜

2. Use equation 7-6 to find K nail :

K nail = 12 mvn2ail =

1 2

2

( 0.012 kg ) ( 8.8 m/s) =

0.46 J

Insight: If the collision were instead considered to be inelastic (the hammer remains in contact with the nail and they move together) the final velocity would be 4.4 m/s and the nail’s kinetic energy would be 0.12 J, a reduction by a factor of four.


9 – 13



37. Picture the Problem: The truck strikes the car from behind. The collision sends the car lurching forward and slows down the speed of the truck. collision where one of the objects (the car) is initially at rest. Therefore, Strategy: This is a one-dimensional, elastic collision equation 9-12 applies and can be used used to find the final speeds of the vehicles. vehicles. Let m1 be the mass of the truck, m2 be the mass of the car, and v0 be the initial speed of the truck.

⎛ m1 − m2 ⎞ ⎛ 1720 − 732 kg ⎞ ⎟ v0 = ⎜ ⎟ (15.5 m/s ) = 6.25 m/s = vtruck m m 1 7 2 0 7 3 2 k g + + ⎝ ⎠ ⎝ 1 2 ⎠

Solution: 1. Use equation 9-12 to find v1,f :

v1,f = ⎜

2. Use equation 9-12 to find v2,f :

v2,f = ⎜

1720 kg) ⎤ ⎡ 2 (1720 ⎛ 2m1 ⎞ ⎟ v0 = ⎢ ⎥ (15.5 m/s ) = 21.7 m/s = vcar m m 1 7 2 0 7 3 2 k g + + 1 2 ⎝ ⎠ ⎣ ⎦

Insight: The elastic collision produces a bigger jolt for the car. If the collision were instead inelastic and the two vehicles stuck together, the final speed of the car (and the truck) would be 10.9 m/s.

38. Picture the Problem: You throw a rubber ball at an elephant that is charging directly toward you. Strategy: Use the principle that the speed of separation after a head-on elastic collision is always equal to the speed of approach before the collision to answer the conceptual question.

+ve. The speed of approach Solution: Suppose the ball is thrown at velocity −v b and the elephant is charging at velocity +v of the two objects is ve −(− v b) = ve + v b. After the collision collision the elephant’s speed will will be essentially unchanged because because it has so much more inertia and momentum than the ball. In order to maintain the same speed of separation (v ( v bf − ve) as the speed of approach, the ball’s speed must now be v bf = v b + 2 ve. We conclude that the speed of the ball ball after bouncing off the elephant will be greater than the speed it had before before the collision. collision. Insight: The situation is similar to that discussed in Conceptual Checkpoint 9-4, except that the ball has a nonzero speed before the collision. From an energy standpoint the elephant exerts a force on the ball, doing work on it and increasing its kinetic energy. The elephant loses a little kinetic energy in the process, but because its mass is so large its speed decreases by only a tiny bit. If the elephant were at rest, the ball would rebound with its initial speed.

39. Picture the Problem: The ball and the elephant move horizontally toward each other, and then an elastic collision occurs, sending the ball back along the direction it came from. Strategy: This problem can be very difficult because both the ball and the elephant are moving before and after the collision (see problem 88). However, if we adopt a frame of reference in which the observer is moving with the ball as it approaches the elephant, then in that frame of reference the ball is initially initially at rest, and equation 9-12 applies. We’ll take that approach, finding v 0 of the elephant in the ball’s frame of reference, determining the ball’s final velocity in G

that frame, and then switching back to the Earth frame of reference to report the ball’s final final speed. Let the elephant initially travel in the positive direction so v eg = 4.45 m/s xˆ and v bg = − 7.91 7.91 m/s xˆ . G

G

Solution: 1. (a) Find v 0 using

v eg,i = v eb,i + v bg,i

equation 3-8:

v eb,i = v eg,i − v bg,i = ( 4.55 m/s ) xˆ − ( −7.81 m/s ) xˆ = (12.36 m/s ) xˆ

G

2. Apply equation 9-12 to find v b,f for

G

G

G

G

G

G

⎛ 2me ⎝ me + mb

kg ) ⎤ ⎡ 2 ( 5240 kg ⎞ m/s ) = 24.7 m/ m/s ⎟ veb,i = ⎢ ⎥ (12.36 m/ 5 2 4 0 0 . 1 5 0 k g + ⎠ ⎣ ⎦

the ball in this frame of reference:

v bb,f = ⎜

3. Apply equation 3-8 again to find v g,f :

v bg,f = v bb,f + v bg,i = ( 24.7 m/s ) xˆ + ( −7.81 m/s ) xˆ = (16.9 m/s ) xˆ G

G

G

v bg,f = 16.9 m/s 4. (b) The ball’s kinetic energy has increased because kinetic energy has been transferred from the elephant to the ball as a result of the collision. Insight: The ball’s kinetic energy has increased from 4.57 J to 21.4 J, almost a factor of five! The elephant’s speed and kinetic energy hardly changes at all. This is the basic idea behind the gravitational slingshot effect effect (passage problems 97-100). Note that the approach speed of 12.36 m/s is essentially the same as the separation speed of 16.9 − 4.55 = 12.4 m/s , as discussed in section 9-6 of the text. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 – 14



40. Picture the Problem: The neutron collides elastically and head-on with a stationary particle and is slowed down. Strategy: If we assume the target particle is initially stationary, we can use equation 9-12 to determine the final speed of the neutron after the collision, and then use that speed to determine the ratio of kinetic energies in each case. Let m be the mass of the neutron and M and M be be the mass of the target particle.

⎛ m − M ⎞ ⎟ vi ⎝ m + M ⎠

vf = ⎜

Solution: 1. (a) Use equation 9-12 to find vf for the neutron:

K f

2. Calculate the ratio K f K i :

K i −4

electron, M = = 5.49×10 3. Apply the expression in step 2 for an electron, M

=

1 2

2

m ( m − M m + M ) vi2 1 2

mvi2

⎛ m − M ⎞ =⎜ ⎟ ⎝ m + M ⎠

2

2

⎛ 1.009 u − 5.49 × 10−4 u ⎞ =⎜ ⎟ = 0.9978 K i ⎝ 1.009 u + 5.49 × 10−4 u ⎠

K f

u:

2

⎛ 1.009 u − 1.007 u ⎞ −6 =⎜ ⎟ = 1× 10 K i ⎝ 1.009 u + 1.007 u ⎠

K f

proton, M = = 1.007 u: 4. (b) Apply the expression in step 2 for a proton, M

2

⎛ 1.009 u − 207.2 u ⎞ =⎜ ⎟ = 0.9807 K i ⎝ 1.009 u + 207.2 u ⎠

K f

electron, M = = 207.2 u: 5. (c) Apply the expression in step 2 for an electron, M

ahead and is hardly slowed by the tiny electron. electron. In case (c) the neutron bounces Insight: In case (a) the neutron plows ahead elastically from the much heavier nucleus and hardly slows down. In case (b) the neutron transfers nearly all its kinetic energy to the similarly sized proton and almost comes to rest. You might notice this phenomenon when a pool ball collides with a second identical ball at rest, or with the apparatus of five suspended metal balls depicted on page 275. 41. Picture the Problem: The apple and orange collide in the manner depicted in the figure at right. right. The mass of the apple is 0.130 0.130 kg and the mass of the orange is 0.160 kg. the y direction to find the Strategy: Use conservation of momentum in the y component of the apple’s speed, and conservation of momentum in the x direction to find the x the x component of the apple’s speed. The known components can then be used to find the total speed and direction of the apple after the collision. Solution: 1. Conserve momentum in the y the y direction to find v1f, y y :

the x 2. Conserve momentum in the x direction to find v1f, x x :

∑ p

y

= 0 = m1 v1 f,y − m2v 2f,y v1f,y =

∑ p

x

m2v 2f y, m1

=

0.160 0 kg kg ) (1.03 1.03 m/s m/s ) ( 0.16 = 1.268 m/s ( 0.130 kg )

= m1 v1i,x + m2v 2i2i,x = m1v1f,x + 0

v1i x, +

m2 m1

⎛ 0.160 kg ⎞ ⎟ ( −1.21 m/s ) ⎝ 0.130 kg ⎠

v2i,x = v1f,x = (1.11 m/s ) + ⎜

= − 0.38 m/s 3. Find the apple’s speed:

v1f = v12f x, + v12f,y =

4. Find the apple’s direction:

θ = tan −1 ⎜

⎛ v1f, y ⎜v ⎝ 1f, x

2 2 (1.268 m/s ) + ( − 0.38 m/s ) =

1.32 m/s

⎞ −1 ⎛ 1.27 m/s ⎞ 1 80° = 107° ⎟⎟ = tan ⎜ ⎟ = −73° + 18 0.38 m/s − ⎝ ⎠ ⎠

This angle is measured counterclockwise from the positive x positive x axis. Insight: Because the collision is elastic you can also set K f = 12 m1 v12f + 12 m2 v22f = K i and use the K i = 0.197 J given in

the example to find v1f = 1.31 m/s , which is correct to within rounding error. Note that we bent the rules of significant figures in steps 1 and 3 to avoid such rounding rounding error. This approach leaves an ambiguity ambiguity in the x the x component of the apple’s final velocity, however, and you still need conservation of momentum in the x the x direction to resolve it. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 – 15



42. Picture the Problem: The cart 4m 4m collides with the cart 2 2m m, which is given kinetic energy as a result and later collides with the cart m. Strategy: In each case a moving cart collides with a cart that is at rest, so application of equation 9-12 will yield the final velocities of all the carts. First apply equation 99-12 12 to the collision between carts carts 4m 4m and 2m, then to the collision between 2m 2m and m. Let the 4m 4m cart be called cart 4, the 2m 2 m cart be called cart 2, and the m cart be called cart 1:

⎛ m4 − m2 ⎞ ⎛ 4m − 2m ⎞ 1 ⎟ v4 , i = ⎜ ⎟ v0 = 3 v0 m m 4 m 2 m + + ⎝ ⎠ ⎝ 4 2 ⎠

Solution: 1. (a) Apply equation 9-12 to the first collision:

v4,f = ⎜

⎡ 2 ( 4m ) ⎤ ⎛ 2m4 ⎞ 4 ⎟ v 4, i = ⎢ ⎥ v0 = 3 v0 ⎝ m4 + m2 ⎠ ⎣ 4m + 2m ⎦

v2,f = ⎜ 2. Apply equation 9-12 to the second collision. In this case cart 2 has an initial speed of 43 v0 :

⎛ m2 − m1 ⎞ ⎛ 2m − m ⎞ 4 ⎟ v2,i = ⎜ ⎟ ( 3 v0 ) = ⎝ 2m + m ⎠ ⎝ m2 + m1 ⎠

4 9

v0

⎡ 2 ( 2m ) ⎤ 4 ⎛ 2m2 ⎞ ⎟ v2,i = ⎢ ⎥ ( 3 v0 ) = ⎝ m2 + m1 ⎠ ⎣ 2m + m ⎦

16 9

v0

v2,f = ⎜ v1,f = ⎜

?

3. (b) Verify that K i = K f by writing using

1 2

2

2

( 4m ) v02 = 12 ( 4m ) ( 13 v0 ) + 12 ( 2m ) ( 49 v0 ) + 12 ( m ) ( 169 v0 )

equation 7-6 and dividing both sides by mv02 :

2=

2 9

+

16 81

+

256 162

=

36 162

+

32 162

+

256 162

=

324 162

2

=2

Insight: Note that due to the transfer of kinetic energy via collisions, the cart with the smallest mass ends up with the largest speed.

43. Picture the Problem: The balls collide elastically along a single direction. Strategy: Set the initial momentum equal to the final momentum and solve for the final speed of the two balls. Then set the initial kinetic energy equal to the final kinetic energy and solve for the final speed of the two balls again. Solution: 1. (a) Set pi = p f and solve for vf :

mv0 = ( 2m ) vf

2. (b) Set K i = K f and solve for vf :

1 2

G

G

mv02 =

1 2

⇒ vf =

( 2m ) vf2 ⇒

v0 2

vf =

v0 2

Insight: If you conserve the momentum and kinetic energy during the collision, assuming the balls that are going into the collision come to rest afterwards, you can show that min = mout no matter how many balls are sent into the collision.

44. Picture the Problem: A stalactite in a cave has drops of water falling from it to the cave floor below. The drops are equally spaced in time and come in rapid succession, so that at any given moment there are many drops in midair. Strategy: Consider the center of mass as the “average” location of the system’s mass in order to answer the conceptual question. Solution: 1. (a) The center of mass is a weighted average of the location of the mass of a system. As the drops fall, their separations increase (see Conceptual Checkpoint 2-5). With the drops more closely spaced on the upper half of their fall, the center of mass is shifted above the halfway mark. We conclude that the center of mass of the midair drops is higher than the halfway distance between the tip of the stalactite and the cave floor.

time, the drops are closer together together higher up. Statements I 2. (b) The best explanation is III. Though equally spaced in time, and II are each false. Insight: It does not matter whether you place the origin of the coordinate system at the tip of the stalactite or at the floor of the cave. In either case there are more droplets closer to the stalactite tip and the center of mass is higher than the midpoint between the tip of the stalactite and the cave floor.


9 – 16



45. Picture the Problem: The bricks are configured as shown at right. the x-coordinate -coordinate of the center of Strategy: Use equation 9-14 to calculate the x mass. Assume that the mass of each brick is m and that the mass of each brick is distributed distributed uniformly. Reference all positions from the the left side of the bottom brick, as indicated in the figure.

∑ mx = m x + m x

X cm =

Solution: Apply equation 9-14 directly:

+ m3 x3 m ( L2 + L + 5 L4 ) 1 ⎛ 11L ⎞ 11 L = = ⎜ ⎟= m1 + m2 + m3 3m 3 ⎝ 4 ⎠ 12

1 1

M

2

2

Insight: The bricks will be balanced as long as the center of mass remains over the point of support. That means the table could end at 11 12 the length of the bottom brick and the stack would still be stable.

46. Picture the Problem: The cereal cartons are at the left end of the basket and the milk will be placed toward the right end.

y

Strategy: Place the origin at the left end of the L = 0.75 m basket. Then, the two cartons of cereal are at xc = 0 m . Put the milk at position x position x and set the center of mass

equal to L 2 , then solve for x. x. That will be the correct position position of the milk relative relative to

l a e r e C

the left end of the basket to put the center of mass at the center of the basket. Solution: Use equation 9-14 and set X cm = L 2 :

k l i M x

∑ mx = 2m x

+ mm xm 0 + mm x L = = M 2 mc + mm 2 mc + mm 2 L ( 2mc + mm ) ( 0.71 m ) ⎡⎣ 2 ( 0.56 kg ) + 1.8 kg ⎤⎦ x = = = 0.58 m X cm =

c

c

2 (1.8 kg )

2mm

Insight: Another approach is to put the origin at the center of the basket and find the position of the milk that puts the center of mass at zero. You should get 0.22 m to the right of center, center, or about 0.58 m from the left left end.

47. Picture the Problem: The Earth and Moon lie along a line connecting their centers of mass. the x-axis -axis along the line connecting the centers of the Earth and Moon, and put the origin at the center Strategy: Place the x of the Earth. Use equation 9-14 to find the location of the center of mass of the Earth-Moon system. Solution: 1. Find the distance between the center center of mass of the Earth and the center of mass of the Earth-Moon system: 2. Subtract X cm from REarth = 6.37 × 10 m : 6

X cm

∑ mx = m =

E

M

xE + mm xm

mE + mm

=

0 + ( 7.35 ×10 22 kg ) (3.85 ×10 8 m ) 5.98 × 10

24

kg + 7.35 ×10

22

kg

= 4.67 ×106 m

6.37 × 106 m − 4.67 × 10 6 m = 1.70 ×10 6 m below the surface of the Earth .

Insight: In Chapter 12 we’ll explore how the Earth and Moon orbit about their common center of mass, producing a “wobble” in the Earth’s path around the Sun.

48. Picture the Problem: A triangular piece of sheet metal of mass M mass M is is marked with a vertical dashed line at the point where the mass to the left of the line ( M M /2) /2) is equal to the mass to the right of the line (also M (also M /2). /2). Strategy: Use the definition of the center of mass to answer the question.

mass as the “average” location of the system’s system’s mass. The mass M mass M /2 /2 on Solution: 1. (a) You can think of the center of mass the right, on average, is located closer to the fulcrum than the mass M mass M /2 /2 on the left. The center of mass must must therefore be to the left of the dashed line, line, and we conclude conclude that the metal metal sheet will tip to the the left. 2. (b) The best explanation is II. The metal sheet extends for a greater distance to the left, which shifts the center of mass to the left of the dashed line. line. Statements I and III III are each false. Insight: The center of mass of a system is a combination of how much mass it contains and where it is located. If the sheet metal were cut into a rectangular shape, the center of mass would be located right at the dashed line. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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49. Picture the Problem: A pencil standing upright on its eraser end falls over and lands on a table. As the pencil falls, its eraser does not slip. Strategy: Consider the location of the pencil’s center of mass when answering the conceptual question. Let the positive x direction be in the direction the pencil falls, and the positive y positive y direction be vertically upward.

positive x direction and in the negative y negative y Solution: 1. (a) As the pencil falls its center of mass moves along an arc in the positive x direction (downward). (downward). By Newton’s Second Second Law the net force must point point in the direction of the acceleration. acceleration. We conclude that the x the x component of the contact force is positive. 2. (b) Because the center of mass of the pencil has a nonzero, downward acceleration, it follows that the net vertical force acting on it is downward. The net vertical force is the vector sum of the upward contact force and the downward weight of the pencil. We conclude that the y the y component of the contact force is less than the weight of the pencil. Insight: When the vertical contact force equals the weight the pencil is in equilibrium in the vertical direction.

50. Picture the Problem: The box with no top rests on a flat surface. surface. Its center of mass is measured measured relative to the geometric center of the box, a distance L 2 above the bottom surface. positive z -axis. -axis. Strategy: Place the origin at the center of the box with the plane of the missing top perpendicular to the positive z Due to symmetry, X symmetry, X cm written in the z the z direction direction to determine the location of cm = Y cm cm = 0. Use a version of equation 9-14 written the center of mass. Assume each side of the box has mass mass m. Let z Let z = = 0 correspond to the center of the box. Solution: Apply equation 9-14 directly:

Z cm =

Σmz M

=

mz1 + mz2 + mz3 + mz 4 + mz b ottom 5m

=

m ( 0 + 0 + 0 + 0 − L2 ) 5m

=−

L 10

The center of mass is L is L/10 /10 units below the center of the box. Insight: Placing some other masses at the bottom of the box will lower the center of mass even farther.

51. Picture the Problem: The partially eaten pizza is depicted at right. Strategy: We expect the center of mass of the second quadrant to be at a negative x negative x value that is greater in magnitude than −1.4 in, and a positive y positive y value. Adding the third and fourth quadrants quadrants back in would shift the center center of mass to the given ( −1.4 in,−1.4 in) position. By symmetry the positions of the the centers of mass of each quadrant must be the same distances x distances x and y and y from the middle of the pizza. Use this fact to find those positions. positions. Solution: 1. Use equation 9-14 to set X cm = − 1.4 in:

X cm =

=

mx2 + mx3 + mx4 3m

( − x ) + ( − x ) + x

= − 1.4 in 3 x = −3 ( −1.4 in ) = 4.2 in 2. Use equation 9-15 to set Y cm = − 1.4 in:

Y cm =

my2 + my3 + my4 3m

=

y + ( − y ) + ( − y ) 3

= −1.4 in

y = −3 ( −1.4 in ) = 4.2 in

quadrant x must be 3. This tells us the distances from the centers of mass of each quadrant to the origin. For the second quadrant x negative and y and y must be positive: ( x, y ) = ( − 4.2 in, 4.2 in ) Insight: We can therefore conclude the center of mass of the third quadrant is ( − 4.2 in, − 4.2 in) and the center of mass of the fourth quadrant is (4.2 in, − 4.2 in).


9 – 18



52. Picture the Problem: The geometry of the sulfur dioxide molecule is shown at right. Strategy: The center of mass of the molecule will lie somewhere along the axis because it is symmetric in the x the x direction. Find Y cm using equation

9-15. Both oxygen atoms will be the same vertical distance yO from the origin. Let m represent the mass of an oxygen atom, ms the sulfur atom.

Solution: 1. Use equation 9-15 to find Y cm

−9

2. Recalling that 1 nm = 1×10

∑ my = my

+ myO + ms ys 2 myO + 0 = 2m + ms 2 m + ms 2 (16 u ) ( 0.143 nm ) sin 30 30° = = 0.036 nm 2 (16 u ) + 32 u

Y cm =

O

m, we can write ( X cm , Y cm ) = ( 0, 3.6 ×10 −11 m )

Insight: If the angle were to decrease from 120° the center of mass would move upward. For instance, if the bond angle were only 90°, the center of mass would be located at (0, 5.1×10−11 m).

53. Picture the Problem: The sticks are arranged as depicted at right.

y

Strategy: Use equations 9-14 and 9-15 to find the position of the center of mass. Solution: 1. (a) Apply equation 9-14 directly: X

cm

X cm 2. Apply equation 9-15 directly:

=

Mx1 + Mx2 + Mx3

0.5 m

3 M 1 = 3 ( 0 + 0.50 m + 1.5 m ) = 0.67 m

Y cm =

My1 + My2 + My3 3 M

Center of Mass x 0.50 m

1.5 m

1

= (0.50 m + 0 + 0) = 0.17 m 3

3. The center of mass of the meter sticks is therefore located at ( X cm , Y cm ) = ( 0.67 m, 0.17 m ) . 4. (b) The location of the center of mass would not be affected. The mass drops out of the equations. Insight: If three more meter sticks were added to make a rectangle, the center of mass would be at (1.00 m, 0.50 m).


9 – 19



54. Picture the Problem: One end of the rope is lifted at constant speed. Strategy: Use equation 9-15 to find the position of the center of mass and equation 9-16 to find the velocity velocity of the center of mass of the rope. When writing subscripts for the variables, let f = on the floor, nf = not on the floor, and L and L = length of the rope. The graphs should all end at t = L v = ( 2.00 m ) 0.710 m/s = 2.82 s, at which time the rope will be

entirely off the floor and the velocity of its center of mass will be 0.710 m/s upward. The top of the rope rope is at position position vt during the lift, so the center of mass of the above floor portion of the rope is halfway between zero and vt , and the fraction of the rope that is above the floor is mnf = ( vt L ) M , where is the total mass of the rope. Solution: 1. Use equation 9-15 to find the position of the center of mass of the rope:

Y cm =

mnf yn f + mf yf mnf + mf

=

mnf ( 12 vt ) + 0 M

⎡( vt L ) M ⎤⎦ ( 12 vt ) v 2 t 2 =⎣ = =

M 2 ( 0.710 m/s ) t 2

2L

2 ( 2.00 m)

Ycm = ( 0.126 m/s 2 ) t 2 , 0 < t < 2.82 s

2. Use equation 9-16 to find the velocity of the center of mass of the rope:

V cm =

=

mnf vnf + mf v f nf mnf + mf v 2 t L

=

⎡( vt L) M ⎤⎦ v + 0 = ⎣ M

( 0.710 m/s ) 2.00 m

2

t

Vcm = ( 0.252 m/s 2 ) t 3. The plots of Y cm and V cm as a function of time are shown above. Insight: The position of the center of mass varies as the square of the time at first but would become linear with a slope of 0.710 m/s as soon as the last bit of rope left the floor (at t = t = 2.82 s).


9 – 20



55. Picture the Problem: One end of the rope is lowered at constant speed. Strategy: Use equation 9-15 to find the position of the center of mass and equation 9-16 to find the velocity velocity of the center of mass of the rope. When writing subscripts for the variables, let f = on the floor, nf = not on the floor, and L and L = length of the rope. The time t is t is zero the instant the first part of the rope touches the floor. The graphs should all end at t = L v = ( 2.00 m ) 0.710 m/s = 2.82 s, at which time the rope will be entirely

on the floor and the velocity of its its center of mass will be zero. zero. The top of the rope is at position L − vt during the drop, so the center of mass of the above floor portion of the rope is halfway between zero and L − vt , and the fraction of the rope that is above the floor is mnf = ⎡⎣( L − vt ) L ⎤⎦ M , where M where M is is the total mass of the rope. Solution: 1. Use equation 9-15 to find the position of the center of mass of the rope:

Y cm =

=

mnf yn f + mf yf mnf + mf

{⎡⎣( L − vt )

=

mnf ⎡⎣ 12 ( L − vt ) ⎤⎦ + 0 M

}

L ⎤⎦ M ⎡⎣ 12 ( L − vt ) ⎤⎦ M

⎡ 2.00 m − ( 0.710 m/s ) t ⎤ ⎥ =⎢ ⎢⎣ ⎥⎦ 2 ( 2.00 m)

=

( L − vt )

2

2L

2

2

Ycm = ⎡1.00 − ( 0.355 s−1 ) t ⎤ m, 0 < t < 2.82 s

⎣

2. Use equation 9-16 to find the velocity of the center of mass of the rope:

V cm =

⎦

mnf vnf + mf v f nf mnf + mf v t − vL 2

=

L

=

=

{⎡⎣( L − vt )

L ⎤⎦ M

} ( −v ) + 0

M

( 0.710 m/s ) 2.00 m

2

t

− 0.710 m/s

Vcm = ( 0.252 m/s 2 ) t − 0.710 m/s 3. The plots of Y cm and V cm as a function of time are shown above. Insight: The position of the center of mass varies linearly as the rope is lowered until the end of the rope hits the floor (at t = t = 0), at which time it begins to vary with the square of the time.

56. Picture the Problem: The physical arrangement for this problem is depicted in the figure at right. Strategy: Before the string breaks, the reading on the scale is the total weight of the ball and the liquid. After the string breaks, the ball falls with constant speed, so that the center of mass of the ball–liquid system does not accelerate. Solution: 1. (a) Find the total weight of the ball and the liquid:

Mg = (1.20 + 0.150 kg ) ( 9.81 m/s2 ) = 13.2 N .

2. (b) After the string breaks, the reading is 13.2 N. Because the center of mass of the ball-liquid system undergoes no net acceleration, the scale must not exert any net force on it (Newton’s Second Law), and by Newton’s Third Law it exerts no net force on the scale. Therefore, the reading will not change. Insight: If the ball were to accelerate, the center of mass of the ball-liquid system would accelerate downward, and the scale force would decrease. Another way of looking at the problem is to realize that the water must exert exactly the same upward force on the ball (1.47 N) when it is at rest as when it is falling at constant speed.


9 – 21



57. Picture the Problem: The two blocks are connected by a string and hang vertically from a spring. determine the force exerted by the spring before the string is cut. Then sum the Strategy: Use Newton’s Second Law to determine forces on the two blocks immediately after the string is cut to find the net force. Finally, use Newton’s Second Law again to find the acceleration of the two-block system just after the string is cut. Solution: 1. (a) Before the string is cut, the force of gravity is countered by the force of the spring, so the spring force at this stretch distance is 2mg . Just after the string is cut, the upper block experiences a force of

Fs + Fg = 2mg − mg = mg , and the lower block experiences a force of Fg = − mg . The net force acting on the two-block system is Fnet,ext = mg + ( −mg ) = 0 . 2. (b) Because Fnet,ext = MAcm = 0, we conclude that Acm = 0 . Insight: While the top block moves upward it experiences a smaller and smaller force from the spring. When it reaches the unstretched position of the spring, the force on it is − mg , so that the net force on the two-block system is −2mg

and the acceleration of the two blocks is − g . 58. Picture the Problem: The helicopter hovers in midair. Strategy: Use Newton’s Second Law to set the upward lift of the helicopter blades equal to the downward weight of the aircraft because the acceleration acceleration of the helicopter is zero. Solve for the rate of mass ejection by the blades. Solution: Use Newton’s Second Law and equation 9-19 to find Δm Δt :

∑ F

y

= thrust − mg = 0 ⎛ Δm ⎞ ⎟ v = mg ⎝ Δt ⎠

thrust = ⎜

5300 kg ) ( 9.81 9.81 m/s Δm mg ( 5300 = = v 62 m/s Δt

2

)

= 840 kg/s = 8.4 × 102 kg/s

Insight: 840 kg/s is about 1100 cubic meters or 39,000 cubic feet of air every second! That is why helicopter rotors need to be so large.

59. Picture the Problem: The child throws rocks in the backward direction, propelling the wagon in the forward direction. Strategy: Use Newton’s Second Law to set the forward thrust equal to the backward friction force because the acceleration of the wagon is zero. Solve for the rate of mass ejection by the rock-throwing girl. Solution: Use Newton’s Second Law and equation 9-19 to find Δm Δt :

∑ F

x

= thrust − f k = 0 ⎛ Δm ⎞ ⎟ v = f k ⎝ Δt ⎠

thrust = ⎜

kg 1 rock 60 s Δm f k 3.4 N = = = 0.31 × × = 29 rocks/min v 11 m/s s 0.65 kg min Δt Insight: If the girl had a 40-lb (18-kg) pile of rocks she could keep the wagon moving for about 1 minute.

60. Picture the Problem: The person throws the bricks in the backward direction, propelling the skateboard in the forward direction. determine the recoil speed. The total momentum is initially zero and Strategy: Use conservation of momentum to determine remains zero after the bricks are thrown because there is no friction. The speed of the bricks is given relative to the person ( v p ), not the ground ( v g ), so we must first use equation equation 3-8 to reference the speed speed relative to the ground. Let the bricks be thrown in the negative direction and the person be propelled in the positive direction. Solution: 1. Use equation 3-8 to relate the speed of the bricks relative to the ground:

v G

= v bp + v pg G

g

G


9 – 22



0 = 2m b vbg + ( mp + ms ) vpg

2. Set pi = p f , substitute the expression G

G

from step 1, and solve for the recoil speed v pg of the person relative to the ground:

= 2m b ( vbp + vpg ) + ( mp + ms ) v pg = m bv bp + ( m p + ms + 2m b ) v pg v pg = −

m bv bp

(m

p

+ ms + 2mb )

=−

( 2 × 0.880 kg ) ( −17.0 m/s ) = kg ) ( 57.8 + 2.10 + 2 × 0.880 kg

0.485 m/s

Insight: The concept of rocket thrust is more useful when mass is being continually ejected, and when the problem asks about forces and acceleration. In this case the mass is ejected all at once, and recoil speed is requested, so conservation of momentum is the way to go.

61. Picture the Problem: The person throws the bricks in the backward direction, propelling the skateboard in the forward direction. determine the recoil speed. This must be done separately for each brick, Strategy: Use conservation of momentum to determine and care must be taken to ensure the speeds are measured relative to the ground, not the person on the skateboard. The momentum is initially zero and remains zero after the first brick is thrown because because there is no friction. The speed of the bricks is given relative to the person ( v p ), not the ground ( v g ), so we must use equation 3-8 to reference the speed relative to the ground. Let the bricks be thrown in the negative direction and the person be propelled in the positive direction. Solution: 1. Use equation 3-8 to relate the speed of the bricks relative to the ground: 2. For the first brick, set pi = p f , G

G

v G

= v bp + v pg G

g

0 = m b vbg + ( mp + ms + mb ) v pg1

substitute the expression from step 1, and solve for the recoil speed v pg1 of the

= m ( v bp + vpg1 ) + ( mp + ms + mb ) v pg1 = m bv bp + ( m p + m s + 2m b ) v pg1

person relative to the ground after after throwing one brick:

v pg1 = −

3. If we concentrate only on the person and the remaining brick, the initial momentum is not zero. Repeat step 2, setting pi = p f and substituting the

(m

G

G

m bv bp

(m

p

p

+ ms + 2mb )

=−

( 0.880 kg ) ( −17.0 m/s ) kg ) ( 57.8 + 2.10 + 2 × 0.880 kg

= 0.243 m/s

+ ms + mb ) vpg1 = mbv bg + ( mp + ms ) v pg2 = m b ( vbp + vpg2 ) + ( mp + ms ) v pg2 = m b vbp + ( mp + ms + mb ) v pg2

G

expression from step 1to find v pg2 :

v pg2 = vpg1 −

m bv bp

(m

p

+ ms + mb )

= ( 0.243 m/s ) −

( 0.880 kg ) ( −17.0 m/s) = ( 57.8 + 2.10 + 0.880 kg )

0.489 m/s

Insight: The person ends up going slightly faster if the bricks are thrown one at a time instead of all at once. This is because when the the second brick is thrown thrown the person person has a smaller initial mass (he is holding only only one brick, not not two) so that the momentum of the second brick represents a larger fraction of the initial momentum and he gets a slightly larger boost in his speed. speed.


9 – 23



62. Picture the Problem: Sand is poured at a constant rate into a bucket that is sitting on a scale. Strategy: The reading of the scale is given by the total weight of the sand and bucket plus the force of impact due to the pouring sand. Use the concept of thrust (equation 9-19) to determine determine the total force force F S on the scale.

Δm v Δt

Solution: 1. (a) Add the sand thrust to the weight:

FS = ( mb + ms ) g +

2. (b) Calculate W :

W = ( m b + ms ) g = ( 0.540 + 0.750 kg ) ( 9.81 m/s2 ) = 12.7 N

= ( 0.540 + 0.750 kg ) ( 9.81 m/s2 ) + ( 0.0560 kg/s) ( 3.20 m/s) = 12.8 N

Insight: The thrust of the sand is 0.179 N, but it appears to make only a 0.1-N difference in the scale reading because of the rounding required by proper attention to significant figures.

63. Picture the Problem: The rope is lowered onto the scale, continuously adding to the mass that is resting on the scale. Strategy: Calculate the thrust due to the mass per time of rope being added to the scale at the given speed. The mass per time of the the rope is related related to the mass per length and the speed speed by: Δm Δt = ( Δm Δx ) ( Δx Δt ) = ( Δm Δx ) v . Add the

thrust to the weight to find the total force on the scale. thrust =

Solution: 1. (a) Use equation 9-19 to find the thrust:

Δm 2 ⎛ Δm ⎞ v=⎜ v ⎟ v = ( 0.13 kg/m ) (1.4 m/s ) = 0.25 N Δt ⎝ Δx ⎠

2. (b) The scale reads more than 2.5 N . It reads the weight of the rope on the scale plus the thrust due to the falling rope. 3. (c) Add the thrust to the weight:

Scale reading = thrust + mg = 0.25 N + 2.5 N = 2.8 N

Insight: The thrust in part (c) is only 10% of the weight. It represents 100% of the scale reading at the instant the rope first touches the scale, but becomes less and less significant as more of the rope’s mass rests on the scale.

64. Picture the Problem: Three objects have different masses but the same momenta. Strategy: Write the kinetic energy of an object in terms of its momentum: K = 12 mv 2 = m 2v 2 2 m = p 2 2 m . Use this

equation to determine the ranking of the kinetic energies. constant. The smallest mass Solution: The kinetic energy is inversely proportional to the mass when the momentum is constant. will therefore have the the highest kinetic energy. We therefore arrive at the the ranking B < A < C. Insight: The smaller mass will have a higher speed in order to have the same momentum as a larger mass. This will give the smaller mass a larger kinetic energy because K because K is is proportional to the square of the speed.

65. Picture the Problem: Three objects have different masses but the same kinetic energy. Strategy: Write the momentum of an object in terms of its kinetic energy: p

2

2m = K ⇒ p =

2mK . Use this

equation to determine the ranking of the momenta. Solution: The momentum is proportional to the square root of the mass when the kinetic energy is constant. The smallest mass will will therefore have the smallest smallest momentum. We therefore arrive at the the ranking C < A < B. Insight: Object A has four times more mass and twice the momentum of object C.


9 – 24



66. Picture the Problem: A block of wood is struck by a bullet. The bullet either embeds itself in the wood or it rebounds. Strategy: Consider the change in the momentum of the bullet for the two scenarios. The larger the change in momentum of the bullet, the larger the impulse the wood must deliver to the bullet, and by Newton's Third Law the larger the impulse the bullet delivers to the wood. Solution: 1. (a) If the bullet bounces off the wood then its change in momentum is greater than if it embeds itself into the wood and comes to rest. We conclude that block is more more likely to be knocked over if the bullet is rubber and bounces off the wood. 2. (b) The best explanation is I. The change in momentum when a bullet rebounds is larger than when it is brought to rest. Statement II is true, true, but irrelevant, and statement statement III is false. Insight: If the bullet were to bounce elastically, with the same rebound speed as its initial speed, the impulse delivered to the wood is exactly twice the impulse delivered if the bullet embeds itself in the wood.

67. Picture the Problem: A juggler performs a series of tricks with three bowling balls while standing on a bathroom scale. Strategy: Consider the forces on the juggler and the scale while he performs the tricks. Solution: 1. When the juggler exerts an u pward force on a bowling ball, by Newton's Third Law the ball exerts a downward force on the juggler and the scale reading is higher than when the juggler simply holds the ball at rest. A similar effect occurs when a ball lands in the juggler’s hand. However, when one or more balls are in the air, the scale supports a weight that is less than the weight of the juggler plus the balls. One way to think about the problem is to realize that the center of mass of the juggler plus balls remains more or less at rest. Therefore, we conclude that the average reading of the scale is equal to the weight of the juggler plus the weight of the three balls. Insight: If the juggler were tossing only two balls in the air with one hand, the scale force would be large when a ball is either being launched or caught by the juggler’s hand (requiring an upward scale force to accelerate it) but equal to the uggler’s weight during those instants when both balls are simultaneously simultaneously in the air. The average reading of the scale would equal the weight of the juggler plus both balls because the center of mass of the juggler-ball system does not accelerate on average.

68. Picture the Problem: The tourist climbs the staircase, placing her center of mass farther from the center of the Earth. internal force to this system) Strategy: Because the net external force on the person-Earth system is zero (gravity is an internal force the center of mass must remain stationary. That means that as the tourist moves away a distance xt from the center of mass, the Earth must move a distance xE to keep the center of mass stationary. stationary. Set the center of mass mass of the personEarth system equal to zero and determine the displacement of the Earth. Solution: Set X cm = 0 and solve for xE :

X cm = 0 = xE = −

mt xt + mE xE

mt mE

mt + mE

⎛

⎞ −21 ⎟ ( 555 ft ) = 6.73 × 10 ft ⎝ 5.97 × 10 kg ⎠

xt = ⎜

72.5 kg 24

Insight: The distance the Earth moves is 2.05×10 −21 m or about one-millionth of the width of the nucleus of a single atom inside the Earth!


9 – 25



69. Picture the Problem: The figure shows two blocks connected by a string that runs over a pulley. The table is frictionless, and the two blocks accelerate in a clockwise direction direction when they are are released. The initial and final centers of mass are marked in the diagram. Strategy: Consider Newton's Second Law in terms of the entire system of blocks when answering answering the conceptual conceptual question.

blocks—including gravity, the normal force, Solution: 1. (a) The net force acting on the system of the two connected blocks—including and the force exerted by the pulley—is constant in both magnitude and direction. Therefore, because the center of mass starts at rest, it will accelerate in a straight line in the direction of the net force. Only the green path satisfies this condition. 2. (b) The best explanation is II. The center of mass starts at rest, and moves in a straight line in the direction of the net force. Statement I is true but does not distinguish between the paths. Statement III is false.

consider all the the forces on a system in order to determine the net force and apply Newton’s Second Insight: You must consider all Law.

70. Picture the Problem: The moving car collides with a stationary car on a frictionless surface. Strategy: Conserve momentum in order to find the speed of the second car. Then calculate the initial and final kinetic energies of the entire system to determine if any energy is lost. If energy is lost the collision is inelastic.

mv1i + 0 = mv1f + ( 12 m ) v2f

Solution: 1. (a) Set pi = p f and solve for v2f : G

G

v2f =

m ( v1i − v1f ) m 2

= 2 ( v − v 3) =

2. (b) Use equation 7-6 to find K i :

K i = 12 mv 2

3. Use equation 7-6 to find K f :

K f = 12 mv12f + 12 ( 12 m ) v22f 2

4 3

v

2

= 12 m ( 13 v ) + 12 ( 12 m ) ( 43 v ) = 12 mv 2 ( 19 + 16 = 12 mv 2 18 ) 4. Because K i = K f we conclude the collision is elastic. Insight: The cars bounce off each other and separate because the final speed of the second car ( 4 3) v is greater than the

final speed of the first car v 3. The first car transfers transfers 8 9 of its initial kinetic energy energy to the the second car.

71. Picture the Problem: The bullet strikes the block and together they are launched horizontally off the edge of the table. Strategy: Use conservation of momentum to determine the horizontal speed of the bullet and block, then apply equation 4-9 to find the landing site of the bullet-block bullet-block combination. combination. Let m = the mass of the bullet, = the mass of the block, v0 = the initial speed of the bullet, and h = the

height of the table. Solution: 1. Set pi = p f and solve for vf : G

G

mv0 + M ( 0 ) = ( m + M ) vf

⎛ 0.0105 kg ⎞ ⎛ m ⎞ v0 = ⎜ ⎟ ( 715 m/s ) ⎟ 0.010 05 + 1.35 1.35 kg ⎠ ⎝ m + M ⎠ ⎝ 0.01

vf = ⎜

= 5.52 m/s Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 – 26



2h

x = vf

2. Now apply equation 4-9:

g

= ( 5.52 m/s )

2 ( 0.782 m) 9.81 m/s 2

= 2.20 m

Insight: If the block were twice as massive it would land about half as far, 1.11 m. In this way the landing spot is a measure of the mass of the block, as long as the block is much more massive than the bullet.

72. Picture the Problem: The egg carton’s dimensions are indicated in the figure at right. Either egg 1 or egg 2 is removed. Strategy: The center of mass changes when an egg is removed because the center of mass is an average mass weighted by the distance each egg is from the origin. Use equations 9-14 and 9-15 to calculate calculate the center of mass position. In each case we expect the center of mass to to move down and to the left. Solution: 1. (a) Because the center of mass is an average quantity weighted by the distance each egg is from the origin, its location will change more if egg 2 is removed. 2. (b) Apply equation 9-14 with egg 1 removed:

X cm =

3. Apply equation 9-15 with egg 1 removed:

Y cm =

2m ( −15.0 − 9.0 − 3.0 + 9.0 + 15.0 cm) + m ( 3.0 cm) 11m

6m ( −3.5 cm ) + 5m ( 3.5 cm) 11m

=

−3.5 cm 11

=

−3.0 cm 11

= − 0.27 cm

= − 0.32 cm

4. When egg 1 is removed the center of mass moves to ( X cm , Y cm ) = ( − 0.27 cm, − 0.32 cm ) 5. (c) Apply equation 9-14 with egg 2 removed:

X cm =

6. Apply equation 9-15 with egg 2 removed:

Y cm =

2m ( −15.0 − 9.0 − 3.0 + 3.0 + 9.0 cm ) + m (15.0 cm) 11m

6m ( −3.5 cm ) + 5m ( 3.5 cm) 11m

=

−3.5 cm 11

=

−15.0 cm 11

= −1.36 cm

= − 0.32 cm

7. When egg 2 is removed the center of mass moves to ( X cm , Y cm ) = ( −1.36 cm, − 0.32 cm ) Insight: In Chapter 11 we will learn how the weight of egg 2 produces a larger torque about the center of mass than egg 1 because of its greater moment arm. It’s another way to think about how removing egg 2 produces a larger effect.

73. Picture the Problem: The falling raindrops are stopped by the upward force from the ground. Strategy: Calculate the rate Δm Δt at which water is delivered to a square meter of the ground, and then use the thrust

equation (9-19) to estimate the force. Solution: 1. Find the rate at which the water fell over 1.0 m2:

Δm ⎛ 31 in ⎞ ⎛ 0.0254 m ⎞ ⎛ 1 h ⎞ ⎛ 1000 kg ⎞ =⎜ × (1.0 m2 ) ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ = 0.024 kg/s 3 in Δt ⎝ 9.0 h ⎠ ⎝ ⎠ ⎝ 3600 s ⎠ ⎝ m ⎠

2. Find the thrust from equation 9-19:

thrust = ⎜

⎛ Δm ⎞ ⎟ v = ( 0.024 kg/s ) (10 m/s ) = 0.24 N ⎝ Δt ⎠

Newton is about ¼ lb, lb, this force is about 1 16 lb or about one ounce on a square meter. meter. However, if Insight: Because a Newton you allow all 31 inches to pool on top of the square meter it will weigh 7.7 kN = 1700 lb = 0.87 ton!


9 – 27



74. Picture the Problem: The apple falls vertically downward due to gravity. Strategy: The rate of change of the apple’s momentum is the net force acting on it, according to equation 9-3. The only force acting on the apple is its weight mg . Calculate the change in momentum produced by gravity using equation 9-3. Solution: 1. (a) Solve equation 9-3 for Δ p Δt :

Δ p = F = 2.7 N Δt

2. (b) Solve equation 9-3 for Δ p :

Δ p = F Δt = ( 2.7 N ) (1.4 s) = 3.8 kg ⋅ m/s

Insight: We could also write the units of the answer to part (a) as 2.7 ( kg ⋅ m/s ) s .

75. Picture the Problem: The placement of the lead weight moved the center of mass of the tire from a certain location to the geometric center. Strategy: Let the coordinate system be placed so that the origin is at the geometric center of the wheel. Treat the entire mass of the wheel as a point mass at some small distance d from d from the origin. Together with with the lead weight the center of mass of the system is exactly at zero, so write out equation 9-14 and solve for d for d .

for d : Solution: 1. Solve equation 9-14 for d

X cm = 0 =

mlw xlw + mwh d mlw + mwh

d =−

mlw mwh

⎛ 0.0502kg ⎞ ⎟ ( 25.0 cm × 10 mm/cm ) = − 0.354 mm 35.5 kg ⎝ ⎠

xlw = − ⎜

2. Before the lead weight was added, the center of mass was 0.354 mm from the center of the wheel. Insight: Treating the entire wheel as if it were a point mass located at its center of mass is an important concept in mechanics. The attempt to prove this concept led Isaac Newton to develop the mathematical mathematical method we call calculus!

76. Picture the Problem: The locations of the hoop and the ball in each configuration are represented in the figure at right. Strategy: Because there are no external forces acting on the system in the x the x-direction, -direction, X cm of the system will not change.

Use equation 9-14 to set X cm before the ball moves equal to X cm after the ball moves and solve for x for x . Note that after the ball moves it is directly under under the center of the hoop, so that both the hoop and the ball are at position x . for x : Solution: Set X cm,before = X cm,after and solve for x

M ( 0 ) + 2M ( R − r )

=

Mx b + 2Mxb

M + 2 M M + 2M 2 ( R − r ) = 3x b x b =

2 3

( R − r ) = 23 ( R − 14 R ) =

1 2

R

Insight: If there were an unbalanced external force on the system, such as a child pushing on the hoop, the center of mass of the system would accelerate in accordance with Newton’s Second Law.


9 – 28



77. Picture the Problem: The locations of the canoeist and the canoe relative to shore in each configuration are depicted in the figure at right.

3.0 m

Strategy: Because there are no external forces, the center of mass of the canoe-person system remains remains stationary relative relative to the shore. Because the cano canoee is 3.0 3.0 m long long,, its its cent center er is 2.5 + 1.5 1.5 m = 4.0 4.0 m from the shore. Put the shore at the origin of the coordinate coordinate system. Force the center of mass to remain at 4.0 m and find the location of the person when the center of the canoe is 1.5 m farther from shore than the person.

2.5 m

4.0 m 3.6 m

Solution: 1. (a) When the canoeist walks toward the shore, the canoe will move away from the shore according to conservation of momentum. Therefore, Therefore, the end of the canoe closest to the shore will be a distance greater than 2.5 m from shore when the canoeist arrives at that end.

xc = xp + 1.5 m

2. (b) By inspection of the diagram the position of the canoe will be 1.5 m farther from shore than the person: 3. Use equation 9-14 to set X cm = 4.0 m.

X cm =

Substitute the expression for xc from step 2,

mc xc + mp xp mc + mp

= 4.0 m

mc ( xp + 1.5 m ) + mp xp = ( 4.0 m ) ( mc + mp )

and solve for x p :

x p =

( 4.0 m ) ( mc + mp ) − mc (1.5 m ) mc + mp

= ( 4.0 m ) −

( 22 kg ) (1.5 m) = ( 22 + 63 kg )

3.6 m

Insight: Another way to solve this problem is to set the origin of the coordinate system at the center of mass of the canoe (set X cm = 0 ), and solve for x p = 0.39 m when she has walked to the end of the canoe. canoe. She is then a distance

4.0 − 0.39 = 3.6 m from the shore.

78. Picture the Problem: The locations of the canoeist and the canoe relative to shore in each configuration are depicted in the figure at right.

3.0 m

Strategy: Because there are no external forces the center of mass of the canoe-person system remains remains stationary relative relative to the shore. Because the cano canoee is 3.0 3.0 m long long,, its its cent center er is 2.5 + 1.5 1.5 m = 4.0 4.0 m from the shore. Put the shore at the origin of the coordinate coordinate system. The center of mass of the canoe-person system remains at 4.0 m, the person is at 3.4 m, and the center of the canoe is 3.4 m + 1.5 m = 4.9 4.9 m from shore. Use these data to solve equation 9-14 for the mass of the canoe mc . Solution: Solve eq. 9-14 for mc :

X cm =

mc xc + mp xp mc + mp

2.5 m

4.0 m 3.4 m

= 4.0 m

mc xc + mp xp = ( 4.0 m ) ( mc + mp ) mc =

m p ( 4.0 m − xp )

( xc − 4.0 m )

=

( 63 kg ) ( 4.0 − 3.4 m) = ( 4.9 − 4.0 m )

42 kg

Insight: The heavier canoe moves less than the lighter canoe of problem 77, so the person can get slightly closer to shore while keeping the center of mass of the person-canoe system at 4.0 m. There is only one significant figure in the answer because of the rules of subtraction, in this case for both numerator and denominator. However, it is awkward to report the answer as 4×10 1 kg, so we bent the rules a little bit and reported 42 kg.


9 – 29



79. Picture the Problem: The physical arrangement for this problem is depicted in the figure at right. Strategy: Before the string breaks, the reading on the scale is the total weight of the ball and the liquid. liquid. After the string breaks, breaks, the ball falls with constant acceleration, so that the center of mass of the ball–liquid system also accelerates. accelerates. Use equation 9-17 to find the acceleration acceleration of the center of mass and then Newton’s Newton’s Second Law to find find the scale force. Let m be the mass of the ball, mf be the mass of the fluid and bucket, bucket, and M and M

be the combined mass mass of 1.20 + 0.150 kg = 1.35 kg. kg . Solution: 1. (a) Find the total weight of the ball and the liquid:

Mg = (1.35 kg ) ( 9.81 m/s2 ) = 13.2 N . mf ( 0 ) + mb ab

2. (b) Apply equation 9-17 to find Acm :

Acm =

3. Use Newton’s Second Law to find the scale force F s :

∑ F = F − Mg = MA

M

=

( 0.150 kg ) ( − 0.250 × 9.81 m/s2 ) 1.35 kg

= − 0.273 m/s 2

G

s

cm

Fs = M ( g + Acm ) = (1.35 kg ) ( 9.81 − 0.273 m/s2 ) = 12.9 N

Insight: The scale force must decrease as the ball accelerates downward because if it did not, the forces on the center of mass would be balanced and it could not accelerate.

80. Picture the Problem: The hockey player tosses the helmet in one direction and recoils in the other. Strategy: The total horizontal momentum of the hockey player m p and helmet mh is conserved because there is no

friction. The total horizontal momentum momentum remains zero both both before and after the helmet is tossed. tossed. Use this fact to find the mass of the player. Let the helmet be tossed tossed in the positive direction. Solution: Set pi x, = pf,x and solve for m p :

∑ p

x

= 0 = m pv p, x + mh vh, x

m p = −

mh vh, x v p, x

=−

kg ) ( 6.5 m/ m/s ) cos 11 11° (1.3 kg

− 0.25 m/s

= 33 kg

Insight: This must be a junior player (a 33 kg player weighs 73 lb). If a 70-kg adult tossed the helmet in the same manner, the recoil velocity would be − 0.118 m/s. Note that the momentum in the vertical direction is not conserved; conserved; the normal force from the ice is an external force that allows the vertical momentum to change from zero to mh vh sin 11° = 1.6 kg kg ⋅ m/s .

81. Picture the Problem: The carts move without friction in the manner indicated by the figure at right. Strategy: Use equation 9-16 and the known velocity of the center of mass to find the mass of cart 2.

Solution: Solve equation 9-16 for m2 :

Vcm = 2 3

2

v0 =

m1v0 + m2 ( v 0 2 )

3 m1 + m2 ( m1 + m2 ) = m1 + 12 m2

( 32 − 12 ) m2 = (1 − 23 ) m1 m2 = 2m1 = 2m 4 m, the center of mass of the system would move even slower, at Insight: If cart 2 were even more massive, say, 4m increases even more, the center of mass mass speed approaches v0 2, the speed of cart 2. ( 3 5) v0 . As the mass of cart 2 increases


9 – 30



82. Picture the Problem: One end of the rope is lifted at constant speed. Strategy: The force you need to exert on the end of the rope equals the weight of the rope that is off the floor plus the thrust required to add mass to the suspended portion of the rope at the given rate. No additional force is required because the rope is not accelerating. accelerating. Use equation 9-19 9-19 to find the thrust thrust and the given given information to find the mass and weight of the rope that is above the floor. 2 ⎛ Δm ⎞ ⎛ Δm Δx ⎞ ⎛ Δm ⎞ 2 v=⎜ v=⎜ v = ( 0.135 kg/m ) (1.13 m/s ) = 0.172 N ⎟ ⎟ ⎟ ⎝ Δt ⎠ ⎝ Δx Δt ⎠ ⎝ Δx ⎠

Solution: 1. Use equation 9-19 to find the required thrust:

thrust = ⎜

2. Find the total force required:

F = mg + thrust = ( x × Δm Δx ) g + thrust

= ( 0.525 m × 0.135 kg/m) ( 9.81 m/s2 ) + 0.172 N = 0.868 N Insight: If you were to stop lifting when the end of the rope was 0.525 m above the floor you would still need to exert a force of 0.695 N to support the weight of the 70.9 grams of rope suspended above the floor.

83. Picture the Problem: The geometry of the water molecule is shown at right.

the x axis Strategy: The center of mass of the molecule will lie somewhere along the x because it is symmetric symmetric in the y the y direction. Find X cm using equation 9-14. Both hydrogen atoms will be the same horizontal distance xH from the the origin. Let m represent the mass of a hydrogen atom, mO the oxygen atom.

Solution: 1. Use equation 9-14 to find X cm

−9

2. Recalling that 1 nm = 1×10

∑ mx = mx

+ mxH + mO xO 2myO + 0 = 2m + mO 2m + ms M 1 2 (1.0 u ) ( 0.096 nm) cos ( 2 104.5° ) = = 0.0065 nm 2 (1.0 u ) + 16 u

X cm =

H

m, we can write ( X cm , Y cm ) = ( 6.5 × 10−12 m, 0 ) .

Insight: If the angle were to increase from 104.5° the center of mass would move to the left. For instance, if the bond angle were 135° the center of mass would be located at (0, 4.1×10 −12 m).

84. Picture the Problem: The two moving carts collide with the 4m cart that is at rest. Strategy: Use conservation of momentum to find the speed of the carts before and after the the two collisions. Because the momentum always remains the same, it is not n ecessary to calculate the intermediate speed of carts 2 and 3 before cart 1 collides with them. them. Finally, use equation 7-6 to find find the ratio of the final kinetic energy to initial kinetic energy. Solution: 1. (a) Set pi = p f and solve for vf : G

G

mv0 + 2mv0 + 0 = ( m + 2m + 4m ) vf 3v0 = 7vf

2. (b) Calculate K f K i :

K f K i

2

( 7m ) ( 73 v0 ) = 1 = 3m ) v02 2( 1 2

⇒ vf =

3 7

v0

3 7

4m carts after they collide, you should get v0 3 . In part Insight: If you calculate the intermediate speed of the 2m and 4m (b) we learn that that 4 7 or 57% of the initial initial energy is dissipated dissipated as heat, sound, and permanent deformation of material. material.


9 – 31



85. Picture the Problem: The coasting rocket explodes into two pieces of equal mass that are ejected at 45.0° to the vertical. only force acting on the rocket after it it is launched. Find its Strategy: Assume gravity is the only speed after rising for 2.50 s, then use conservation of momentum in the vertical direction and the principles of center-of-mass motion to answer the questions.

45° 45° vf

vf

vi, y = v0 y − gt

Solution: 1. (a) Use equation 4-6 to find the speed of the rocket before the explosion: explosion:

= ( 44.2 m/s ) − ( 9.81 m/s 2 ) ( 2.50 s )

vi, y y

vi, y = 19.68 m/s 2mvi, y = 2mvf sin 45.0°

2. Set pi, y = pf,y and solve for vf

of each piece:

vf =

vi, y sin 45 45.0°

=

19.68 m/s sin 45 45.0°

= 27.8 m/s

3. (c) Before the explosion V cm = (19.7 m/s ) yˆ Because the momentum of the system is the same after the explosion,

and the total mass has not changed, V cm = (19.7 m/s ) yˆ after the explosion too. 4. (d) The only force acting on the system before and after the explosion is gravity. Therefore, Acm = ( −9.81 m/s 2 ) yˆ

the x direction, as well, as guaranteed by the fact that each piece has the same Insight: Momentum is conserved in the x speed and the same angle from vertical. The momentum in the x the x direction is zero both before and after the explosion. 86. Picture the Problem: The momenta of the two cars as they approach the intersection, as well as the total momentum, are depicted at right.

G

p2

Strategy: Use the component method of vector addition to determine the components of p 2 .

G

p1

G

p1 + p 2 = p total (11, 000 kg ⋅ m/s ) xˆ + ( −370 kg ⋅ m/s) yˆ + p2 = (15, 000 kg ⋅ m/s) xˆ + ( 2100 kg ⋅ m/s) yˆ G

Solution: 1. (a) Use the component method of vector addition to find p 2 :

G

p total

G

G

G

G

p 2 = ( 4000 kg ⋅ m/s ) xˆ + ( 2470 kg ⋅ m/s) yˆ G

matter which car is closer closer to the intersection intersection because momentum depends only only on mass and 2. (b) No, it does not matter velocity. It is independent of position. Insight: The position of the cars is important if you are attempting to predict whether the momenta of the two cars will produce a collision collision at the intersection! intersection!

87. Picture the Problem: The books are arranged in a stack as depicted at right, with book 1 on the bottom and book 4 at the top of the stack. this problem from the top down. The Strategy: It is helpful to approach this center of mass of each set of books must be above or to the left of the point of support. Find the positions of the centers centers of mass for successive successive stacks of books to determine d . Measure the positions of the books from the right edge of book 1 (right-hand dashed line in the figure). Solution: 1. The center of mass of book 4 needs to be above the right end of book 3:

d 3 =

L 2

2. The result of step 1 means that the center of mass of book 3 is located at L 2 + L 2 = L from the right edge of book 1. 3. The center of mass of books 4 and 3 needs to be above the right end of book 2:

d 2 = X cm,43 =

m ( L 2) + m ( L ) 2m

=

3 4

L

The res result ult of step tep 3 means eans that hat the center nter of mass ass of book ook 2 is loca ocated ted at 3 L 4 + L 2 = 5L 4 . 4. The Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 – 32


Chapter 9: Linear Momentum and Collisions 5. The center of mass of books 4, 3, and 2 needs to be above the right end of book 1:

d1 = X cm,432 =

m ( L 2) + m ( L ) + m ( 5L 4) 3m

=

11 12

L

The res resul ultt of of ste step p 3 means eans that hat the the center nter of mass ass of of boo book k 1 is loca ocated ted at at 11 L 12 + L 2 = 17 L 12. 6. The 7. The center of mass of all four books needs to be above the right edge of the table:

d = X cm,4321 =

m ( L 2 ) + m ( L ) + m (5L 4) + m (17L 12) 4m

=

25 24

L

problems such as these in Chapter 11. If you examine the Insight: We will explore more about static equilibrium problems L L L L 25 overhang of each book you find an interesting series: d = + + + = L . The series gives you a hint about how 2 4 6 8 24 to predict the overhang of even larger stacks of books!

88. Picture the Problem: The mass m1 has an initial velocity v1 and collides with m2 that has initial velocity v2 . Strategy: Combine the equations of momentum conservation and energy conservation to find the final speeds of the two masses. Solution: 1. Set pi = p f :

m1v1 + m2v 2 = m1v1f + m2v 2 f

2. Set Ki = K f :

1 2

G

G

3. Rearrange the equation from step 1:

4. Rearrange the equation from step 2:

5. Set the expressions in steps 3 and 4 equal to each other:

m1v12 + 12 m2v22 = 12 m1v12f + 12 m 2v 22f

m1 ( v1 − v1f ) m2 ( v2 f − v2 ) m1 ( v12 − v12f ) m2 ( v − v 2 2f

2 2

)

m1 ( v1 − v1f ) m2 ( v2f 2f − v2 )

=1

=1=

=

m1 ( v1 − v1f ) ( v1 + v1f )

m2 ( v2f − v2 ) ( v2 f + v2 )

m1 ( v1 − v1f ) ( v1 + v1f )

m2 ( v 2f 2f − v 2 ) ( v 2 f + v 2 )

v2 f + v2 = v1 + v1f

v2f = v1 + v1f − v 2 m1v1 + m2v 2 = m1v1f + m2 ( v1 + v1f − v 2 )

6. Substitute the expression for v2f in step 5

into the equation from step 1:

( m1 − m2 ) v1 + 2m2v 2 = ( m1 + m 2 ) v1f

⎛ m1 − m2 ⎞ ⎛ 2m2 ⎞ ⎟ v1 + ⎜ ⎟ v2 m m m m + + 2 ⎠ 2 ⎠ ⎝ 1 ⎝ 1

v1f = ⎜

7. Now substitute the expression for v1f into

⎡⎛ m1 − m2 ⎞ ⎛ 2m2 ⎞ ⎤ ⎟ v1 + ⎜ ⎟ v2 ⎥ − v 2 ⎢⎣⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠ ⎥⎦

v2f = v1 + ⎢⎜

the last equation of step 5 to find v2f :

⎛ m − m2 ⎞ ⎛ 2m2 ⎞ =⎜ 1 + 1⎟ v1 + ⎜ − 1 ⎟ v2 ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠ ⎛ 2m1 ⎞ ⎛ m2 − m1 ⎞ ⎟ v1 + ⎜ ⎟ v2 ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠

v2f = ⎜

predict the outcome of any one-dimensional elastic collision. collision. The equations get even Insight: These complex formulae predict more complex if you include two dimensions!


9 – 33



89. Picture the Problem: Cart 2 moves with speed v2i and collides with cart 1, which which is at rest. The collision imparts imparts kinetic energy to cart 1, but the collision is elastic so the total energy is conserved. conservation of momentum alone can Strategy: In the case where carts 1 and 2 have the same speed after the collision, conservation be used to determine determine the speeds of the carts at the instant their their speeds are equal. equal. Because cart 1 is initially initially at rest we can can use equations 9-12 and 9-13, except all the index labels must be switched because in this problem it is cart 1, not cart 2, that is initially at rest. Solution: 1. (a) Set pi = p f and G

G

solve for vf

m1v1 + m2v2 = m1v f + m 2v f 0 + ( 0.42 kg ) ( 0.68 m/s ) = ( 0.84 + 0.42 kg ) vf vf = 0.23 m/s

2. (b) The energy stored in the spring bumper is equal to the loss of kinetic energy at that time.

2

Δ K = K f − K i = 12 ( 0.84 + 0.42 kg ) ( 0.227 m/s ) − 12 ( 0.42 kg )( ) ( 0.68 m/s) = − 0.065 J

2

The “missing” energy must have gone into the energy stored in the bumper, or 0.065 J. 3. (c) Use equation 9-12 with the indices switched:

⎛ 2m2 v1,f = ⎜ ⎝ m1 + m2

kg ) ⎤ ⎡ 2 ( 0.42 kg ⎞ ⎟ v0 = ⎢ 0.84 + 0.42 kg ⎥ ( 0.68 m/s ) = 0.45 m/s ⎠ ⎣ ⎦

4. Use equation 9-12 with the indices switched:

⎛ m − m1 v2,f = ⎜ 2 ⎝ m1 + m2

⎡ 0.42 − 0.84 kg ⎤ ⎞ ⎟ v0 = ⎢ 0.84 + 0.42 kg ⎥ ( 0.68 m/s ) = − 0.23 m/s ⎣ ⎦ ⎠

Insight: The equations from problem 88 can also be used, but in this case v1i = 0 so that the first term of each equation

drops out and you end up with equation 9-12 again. 90. Picture the Problem: The geometry of the earring is indicated in the figure at right. Strategy: In order for the earring to be balanced at point P, the center of mass must reside there. The center of mass must lie lie along a vertical line through through P because it is symmetric to the left left and right of that line. Find the vertical vertical coordinate of the center of mass by subtracting the mass of a circle of diameter d diameter d from from the mass of a circle of diameter D. D. By inspection of the diagram we can see that the location of the center of the circular hollow must be a distance d 2 from the edge of the

earring, but we can’t assume anything about the relative sizes of d of d and and D D.. Place the origin of the coordinate system at the top of the figure and let y let y be positive in the downward direction in the figure. Use equation 9-15 to find the center of mass and set it equal to d . From that equation equation we can find an an expression for d for d and and D D and use it to solve for φ . mgold ( D 2 ) − mcircle ( d 2 )

Solution: 1. Write out equation 9-15 and set it equal to d :

Ycm =

2. The masses of the gold and the circle cutout are proportional to their areas. In these expressions the density density of gold ρ and the thickness t are t are constants:

mgold = ρ A t = ρ ( π 4 D 2 ) t

mgold − mcircle

= d

mcircle = ρ A t = ρ ( π 4 d 2 ) t

( D ) ( D 2 ) − ( d ) ( d 2 ) 2

3. Substituting these expressions into equation 9-15, the constants cancel:

set

2

D 2 − d 2

= d

D3 − 12 d 3 = d D 2 − d 3

1 2

D 3 + d 3 = 2d D 2 D d + d 2 D 2 = 2

4. Divide both sides by d D 2 and introduce φ = D d :

φ + 1 φ 2 = 2 φ 3 + 1 = 2φ 2


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several directions we could could go. We could simply test φ = 5. From this point there are several

1 2

(1 + 5 ) and verify that it is a

solution, or we could look up the roots of a cubic equation using a math reference book, or we could factor out a term and solve a quadratic equation. I will do the latter, with help from a math computer program that that suggested the factor. 6. Factor out (φ − 1) :

φ 3 − 2φ 2 + 1 = 0 = (φ − 1) (φ 2 − φ − 1 )

7. The solutions are φ = 1 and the roots of the quadratic term:

φ =

− b ± b 2 − 4 ac

=

2a

1 ± 12 − 4 (1) ( − 1) 2 (1)

=

1± 5 2

8. The solution φ = 1 is unphysical because it makes the hole the same size as the earring.

The solution φ =

1 2

(1 − 5 ) doesn’t work because it is negative, so the only valid solution is φ = (1 + 5 ) . 1 2

density ρ in in Chapter 15. Although this is a long and complex complex solution it shows that the Insight: We will discuss the density ρ basic concept of center-of-mass can can help you solve difficult difficult questions questions like this one without using calculus. calculus.

91. Picture the Problem: The mass m1 has an initial velocity v1 and collides with m2 that has initial velocity v2 . Strategy: Subtract the expressions for v2f and v1f found in problem 88 to prove the indicated relation. Solution: Use the results of problem 88 to write an expression for v2f − v1f :

⎡⎛ 2m1 ⎞ ⎛ m1 − m2 ⎟−⎜ ⎢⎣⎝ m1 + m2 ⎠ ⎝ m1 + m2

⎡⎛ m2 − m1 ⎞ ⎛ 2m2 ⎞⎤ ⎟ ⎥ v1i + ⎢⎜ ⎟− ⎜ m m + ⎥ ⎢ 2 ⎠ ⎠⎦ ⎝ m1 + m 2 ⎣⎝ 1 ⎛ m + m2 ⎞ ⎛ −m1 − m2 ⎞ =⎜ 1 ⎟ v1i + ⎜ ⎟ v2 i ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠

v2 f − v1f = ⎢⎜

⎞⎤ ⎟⎥ v 2 i ⎠⎥⎦

v2 f − v1f = v1i − v 2i difference in speeds remains constant. If m2 were at rest Insight: This shows that for a head-on, elastic collision the difference initially, then after the collision its speed relative to m1 will be v1i . If the two masses masses are equal, and m2 is initially at rest, then m1 stops completely and m2 leaves the collision with speed v1i .

92. Picture the Problem: The large ball of mass M mass M is is moving upward with speed v and collides elastically at floor level with the smaller ball of mass m that is moving with velocity −v. The small ball ball rebounds and rises rises to height hm . Strategy: Use the results of problem 88 to find the speed of the smaller ball after the collision. In that equation let m1 = m, m2 = M , v1 = − v, and v2 = v Let upward be the positive positive direction. Then use the speed of the smaller ball

together with the conservation of mechanical energy to find the height to which it will rise. Solution: 1. Find the speed of the small ball after the collision using the equation from problem 88:

2. Now use the conservation of energy for the drop and the rebound to relate the heights and speeds:

3. Substitute the expression from step 1 into step 2:

⎛ m1 − m2 ⎞ ⎛ 2m2 ⎟ v1 + ⎜ ⎝ m1 + m2 ⎠ ⎝ m1 + m2

⎞ ⎟ v2 ⎠ ⎛ m−M ⎞ ⎛ 2M ⎞ ⎛ 3M − m ⎞ v=⎜ =⎜ ( −v ) + ⎜ ⎟ ⎟ ⎟v ⎝ m+M ⎠ ⎝m+M ⎠ ⎝ m + M ⎠

v1f = ⎜

K i + U i = K f + U f 0 + ( m + M ) gh =

1 2

K i + U i = K f + U f

(m + M )v + 0 2

1 2

mv + 0 = 0 + mghm 2 1f

v2 2g = h

2 v1f 2 g = hm 2

2

2 ⎛ 3M − m ⎞ v ⎛ 3M − m ⎞ hm = =⎜ = ⎜ ⎟ ⎟ h 2 g ⎝ M + m ⎠ 2 g ⎝ M +m ⎠ 2 v1f

M = 0.600 kg) and a Insight: Note the impressive result of the transfer of kinetic energy. For instance, for a basketball ( M = tennis ball (m ( m = 0.0577 kg) the tennis ball will rise to a height hm = 7.02 h . Try it! Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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93. Picture the Problem: The child and the sled are acted upon by a 40.0 N force but each object undergoes a different acceleration. system. Then use Strategy: Use Newton’s Second Law to find the acceleration of the center of mass of the sled-child system. equation 9-17 to determine the acceleration of the child. Let mc be the mass of the child, ms be the mass of the sled, F sled, F be the pulling force, force, and ac and as be the accelerations of the child and sled, respectively. Solution: 1. Use Newton’s Second Law to find Acm relative to the ice:

∑ F = F = ( m

2. Use equation 9-17 and the result from step 1 to solve for ac :

Acm =

G

c

ac =

+ ms ) Acm ⇒ Acm = cm

mc ac + ms as mc + ms F − ms as mc

=

=

F mc + ms

F mc + ms

40.0 N − ( 9.75 kg ) ( 2.32 m/s2 ) 21.0 kg

= 0.828 m/s 2

Insight: If the child were also to accelerate at 2.32 m/s 2 with the sled, the pulling force would have to be 71.3 N.

94. Picture the Problem: The object of mass m has a speed v0 and collides

1

v0

elastically with another object of mass m that is at rest. rest. The diagrams at right indicate the directions the masses go after the collision and the relationship between the initial initial and final momenta.

v1f

1

φ

2

v2f

before

Strategy: Set the kinetic energies before and after the collision equal to each other because the collision collision is elastic. Then use conservation of momentum momentum in two dimensions and the law of cosines to find an expression for the angle between the two two final velocity velocity vectors.

2

φ

G

Solution: 1. Set K i = K f and find a

1 2

p1f

mv02 + 0 = 12 mv12f + 12 mv22f v02 = v12f + v22f

relation between the velocities:

after

p 2f G

α

pi G

2. Set pi = p f in two dimensions:

pi = p1f + p 2f

3. Apply the law of cosines to the triangle indicated in the lower diagram.

pi2 = p12f + p22f − 2 p1f p 2f2f co s α

G

G

4. From step 1, v02 − v12f − v22f = 0 . Use this

together with step 3 to find α and α and φ :

G

G

cos α =

G

pi2 − p12f − p22f 2 p1f p2f

=

m2 v02 − m 2v12f − m 2v22f 2 ( mv1f ) ( mv2f )

=

v02 − v12f − v 22f 2v1f v 2f

cos α = 0 ⇒ α = 90° φ = 180° − α = 180° − 90° = 90°

Insight: The law of cosines is mentioned in most geometry books but is not included in Appendix A. There is an algebraic way to solve this problem without using the law of cosines, but it is very lengthy.

95. Picture the Problem: The mass m2 slides down the incline as the reading on the scale is observed. Strategy: The reading on the scale is less than it would be if the two masses were at rest because the center of mass of the two masses is accelerating downward. Find the acceleration of the center of mass and use it together with Newton’s Second Law to find the reading on the scale. The acceleration of the block is g sin θ in θ in the direction indicated indicated by the vector. Let upward be the positive

direction in this problem. Solution: 1. (a) Find the vertical component of a :

a y = − a sin θ

G

= − ( g sin θ ) sin θ = − g sin 2 θ


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2. Use equation 9-17 to find Acm :

Newton’s Second Law to 3. Now use Newton’s find the scale force F force F :

Acm =

∑ F

y

m1 ( 0 ) + m2 ( − g sin 2 θ ) m1 + m2

=−

m2 g sin 2 θ m1 + m2

= F − ( m1 + m2 ) g = ( m1 + m 2 ) Acm

⎡ m2 g sin 2 θ ⎤ F = ( m1 + m2 ) g + ( m1 + m2 ) ⎢ − ⎥ ⎣ m1 + m2 ⎦ = m1 g + m2 g (1 − sin 2 θ ) = m1 g + m2 g cos 2 θ = ( m1 + m2 cos 2 θ ) g

4. (b) The scale force must decrease as the block accelerates downward because if it did not, the forces on the center of mass would be balanced and it could not accelerate. The net force on the two masses must point downward, so that the scale force upward must be less than the weight of the masses, ( m1 + m2 ) g . 5. (c) For the case where

θ =0°,

cos 2 θ = 1 and F = ( m1 + m2 ) g . That is, the block m2 does not accelerate because it is

on a level surface. The center of mass does not accelerate, either, and the reading on the scale is just the weight of both blocks. For the case where where θ = 90°, cos 2 θ = 0 and F = m1 g . In this case the block m2 is in freefall and does not exert any force on the scale, so the scale reads the weight of the wedge only. Insight: This problem is analogous to problem 79, where the ball accelerates downward through the fluid.

96. Picture the Problem: The rope is lifted by one end at constant speed until it is completely off the table, held steady for a moment, then lowered at constant speed back onto the table. Strategy: Use equation 9-16 to find the velocity of the center of mass of the rope. When writing subscripts for the variables, let f = on the floor, nf = not on the floor, and L and L = length of the rope. The expression for V cm can then be used

to determine determine the acceleration acceleration of the center of mass. The top of the rope is at position position vt during the lift, so the center of mass of the above floor portion of the rope is halfway between zero and vt , and the fraction of the rope that is above the floor is mnf = ( vt L ) M , where M where M is is the total mass of the rope. When the rope is being lowered its its top is at position L − vt , so the center of mass of the above floor portion of the rope is halfway between zero and L − vt , and the fraction

of the rope that is above the floor is mnf = ⎡⎣( L − vt ) L ⎤⎦ M . mnf vnf + mf vf nf

Solution: 1. (a) Use equation 9-16 to find V cm :

Vcm =

t graph: 2. Acm is the slope of the V cm vs. t graph:

Acm = v 2 L upward

mnf + mf

⎡( vt L) M ⎤⎦ v + 0 = ⎣ = (v2 L ) t M

3. (b) The rope being lowered has downward momentum. Its downward momentum is decreasing as more and more of its mass comes to rest. That means the rate of change of the momentum is upward. Therefore, there must be a net upward force acting on the rope, resulting in an upward acceleration of the rope’s center of mass. 4. (c) Use equation 9-16 again for the case when the rope is lowered to the table:

V cm =

= t graph: 5. Acm is the slope of the V cm vs. t graph:

mnf vnf + mf vf nf mnf + mf v t − vL 2

L

=

{⎡⎣( L − vt )

L ⎤⎦ M

} ( −v ) + 0

M

= ( v2 L ) t − v

Acm = v 2 L upward the same result as in part (a).

Insight: See problems 45 and 46 for plots of the position and velocity of the rope’s center of mass.


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97. Picture the Problem: The spacecraft approaches the planet with speed vi , interacts (“collides”) with it, and recedes in the opposite direction with speed vf , as depicted in the figure at right. Strategy: Subtract the initial velocities in order to find the speed of approach. Solution: The speed of approach can be found by subtracting the two

initial velocities: v app = u − v i ⇒ vapp = u − ( −vi ) = u + vi G

G

G

Insight: The spacecraft appears to be approaching at a high rate of speed due to the motion of the planet itself. If the spacecraft were at rest relative to the Sun, it would still approach the observer at speed u.

98. Picture the Problem: The spacecraft approaches the planet with speed vi , interacts (“collides”) with it, and recedes in the opposite direction with speed vf , as depicted in the figure at right. Strategy: Subtract the final velocities in order to find the speed of departure. Solution: The speed of departure can be found by subtracting the two

final velocities: v dep = v f − u ⇒ vdep = vf − u = vf − u G

G

G

Insight: The spacecraft appears to be receding at a relatively low rate of speed due to the motion of the planet itself. If the spacecraft were at rest relative to the Sun, it would actually approach the observer at speed u.

99. Picture the Problem: The spacecraft approaches the planet with speed vi , interacts (“collides”) with it, and recedes in the opposite direction with speed vf , as depicted in the figure at right. Strategy: Set the speed of approach from problem 97 to the speed of departure from problem 98 and solve for vf . Solution: Set vapp = vdep

vapp = u + vi = vf − u = vdep vf = vi + 2u

and solve for vf :

2 u at the expense of a tiny bit of the planet’s kinetic energy. Insight: The spacecraft has gained velocity 2u

100. Picture the Problem: The spacecraft approaches the planet with speed vi , interacts (“collides”) with it, and recedes in the opposite direction with speed vf , as depicted in the figure at right. Strategy: Use the results of problem 99, setting vi = u and calculating

the initial and final kinetic energies. Solution: 1. Find the initial and final speeds: 2. Calculate the ratio of the kinetic energies:

vf = vi + 2u = u + 2u = 3u K f K i

=

1 2

mvf 2

1 2

mvi2

=

( 3u ) u2

2

= 9

Insight: The kinetic energy of the spacecraft has increased nine-fold at the expense of a tiny fraction of the planet’s kinetic energy.


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101. Picture the Problem: The bullet enters the bob and together they rise to a height h. Strategy: Use the expression from Example 9-5, which was derived from conservation of momentum during the collision and conservation of mechanical energy after the collision, to find the speed of the bullet. Then use conservation of momentum momentum to find the speed of the bullet-bob combination after impact. 2 ⎛ m ⎞ ⎛ v0 ⎞ h=⎜ ⎟ ⎜ ⎟ ⎝ M + m ⎠ ⎝ 2 g ⎠ 2

Solution: 1. (a) Solve the expression from Example 9-5 for v0 :

2

⎛ 0.67 0.675 5 + 0.00 0.006 675 kg ⎞ ⎛ M + m ⎞ v0 = 2 gh ⎜ = 2 ( 9.81 m/s 2 ) ( 0.128 m ) ⎜ ⎟ ⎟ 0.00675 kg ⎝ m ⎠ ⎝ ⎠

2

m/s = 0.160 km km/s = 160 m/ 2. (b) Set pi = pf and solve for vf :

mv0 = ( m + M ) vf

⎛ ⎞ 0.00675 kg ⎛ m ⎞ v0 = ⎜ ⎟ (160 m/s ) = 1.58 m/s ⎟ 0.00 0. 0067 675 5 0.67 0. 675 5 kg k g + ⎝ m + M ⎠ ⎝ ⎠

vf = ⎜

Insight: Another way to get the answer to part (b) is to use conservation of energy again for the bullet-bob combination.

You’ll find that vf =

2 gh = 1.58 m/s for h for h = 0.128 m.

102. Picture the Problem: The bullet enters the bob and together they rise to a height h. Strategy: Use the expression from Example 9-5, which was derived from conservation of momentum during the collision and conservation of mechanical energy after the collision, to find the appropriate mass of the bob. 2 ⎛ m ⎞ ⎛ v0 ⎞ h=⎜ ⎟ ⎜ ⎟ ⎝ M + m ⎠ ⎝ 2 g ⎠ 2

Solution: Solve the expression from Example 9-5 for M M :

2 gh v02 = M =

m M + m mv0 2 gh

−m=

0.0081 81 kg) ( 320 320 m/s m/s) ( 0.00 2 ( 9.81 m/s 2 ) ( 0.125 m )

kg − 0.0081 kg = 1.6 kg

Insight: If the bob mass is made smaller, the speed of the bullet-bob combination after the collision will be greater and they will rise to a larger height h.


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103. Picture the Problem: Cart 1 has initial speed v0 and collides with cart 2, which which is at rest. The two carts stick together together and move with speed vf . determine the speed of the two carts after the collision. Then use equation Strategy: Use conservation of momentum to determine 7-6 to find the kinetic energy before and after the collision to calculate Δ K . Solution: 1. (a) Set pi = p f and solve for vf : G

G

m1v0 + 0 = ( m1 + m2 ) vf vf =

2. (b) Use equation 7-6 to find Δ K :

m1 m1 + m2

v0 =

3.0 kg 3.0+1.0 kg

( 0.25 m/s ) =

0.19 m/s

Δ K = 12 ( m1 + m2 ) vf2 − 12 m1v02 2

= 12 ( 4.0 kg ) ( 0.19 m/s ) − 12 ( 3.0 kg ) ( 0.25 m/s )

2

Δ K = −0.022 J = −22 mJ Insight: The kinetic energy that is lost represents 25% of the initial kinetic energy of 94 mJ. There are some rounding issues here; if we work with fractions instead we find vf = 3 16 m/s, /s, K i = 3 32 J, J, and K f = 9 128 128 J. J. Thus the loss of

energy energy Δ K = − 3 128 J or −23 mJ.

104. Picture the Problem: The carts move without friction in the manner indicated by the figure at right. Strategy: Use equation 9-16 and the given information to find the velocity of the center of mass. mass. After the collision collision the two carts will move with this same same velocity because they stick together. together. Use that information together with equation 7-6 to find the percentage of kinetic energy lost during the collision. collision. Finally, use the expression expression from problem 88 to find find the speeds of the the two carts after after an elastic elastic collision.

mv0 + m ( v0 2 )

Solution: 1. (a) Apply equation 9-16 directly:

Vcm =

2. (b) Use equation 7-6 to find K i :

K i = 12 mv02 + 12 m ( v0 2 ) = 58 mv02

3. Use the result of step 1 to find K f :

K f =

K : 4. Find the percentage loss of K

m+m

= 34 v0 2

1 2

K f − K i K i

2

( 2m ) vf2 = 12 ( 2m ) ( 34 v0 ) = 196 mv02 × 100 =

9 16

mv02 − 85 mv02 5 8

2 0

mv

× 100 = −

8 80

× 100 = − 10%

or 10% is lost

⎛ m−m⎞ ⎛ 2m ⎞ v0 v0 v0 + ⎜ ⎟ ⎟ = 2 ⎝ m+m⎠ ⎝m+m⎠ 2

5. (c) Use the expression from problem 88 to find find v1,f :

v1,f = ⎜

6. Use the expression from problem 88 to find find v2,f :

v2,f = ⎜

⎛ 2m ⎞ ⎛ m − m ⎞ v0 v0 + ⎜ ⎟ ⎟ = v0 ⎝ m+m⎠ ⎝m+m⎠ 2

Insight: Note that the two carts trade speeds after the collision. This is consistent with the result of problem 91 where we asserted that the relative velocities of the two masses are unchanged for a head-on, elastic collision.


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Physics Chapter 9 Answers

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