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

4.

No. For example, suppose object 2 has four times the mass of object 1. In this case, the two objects have the same kinetic energy if object 2 has half the speed of object 1. On the other hand, it follows that the momentum of object 2 is twice the momentum of object 1.

6.

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.

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.

It is better if the collision is inelastic, because then the light pole gives your car only enough impulse to bring it to rest. If the collision is elastic, the impulse given to your car is about twice as much. This additional impulse – which acts over a very short period of time – could cause injury.

12.

The rate of change in momentum is the same for both objects. 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). Alternatively, we know that the acceleration (rate of change in 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.

14.

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

16.

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

18.

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 was at rest before the collision, it is possible for one object to be at rest after the collision as well. See Figure 97 (a).

20.

The speed of the ball when it leaves the tee is about twice the speed of the club. This follows for two reasons: (i) The collision is approximately elastic; and (ii) the mass of the club and the arms swinging the club is much greater than the mass of the ball. See Figure 9-7 (c).

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22.

Kinetic energy can be lost as objects rub against one another during a collision, doing negative, nonconservative work. These are internal forces, however, and hence they can have no effect on the total momentum of the system.

24.

The speed of the ball after bouncing off the elephant will be greater than the speed it had before the collision. The situation is similar to that shown in Figure 9-7 (c), except that the small mass has a nonzero speed before the collision.

26.

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.

28.

The center of mass is higher than the midway point between the tip of the stalactite and the cave floor. The reason is that 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.

30.

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.

32.

The scale supports the juggler and the three balls for an extended period of time. 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.

34.

(a) Assuming a very thin base, we conclude that the center of mass of the glass is at its geometric center. (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.

Solutions to Problems 1. Setting the momentum of the ball equal to that of the car gives Pb = mb vb = Pc vb =

Pc 15,800 kg ⋅ = mb 0.142 kg

m s

= 111.3

km 111.3 km 3600 s 1 mi 5 = = 2.49 × 10 mi/h s s h 1.609 km

G m 2. pd1 = 4.40 kg ⋅ xˆ s G m pd2 = − 4.40 kg ⋅ yˆ s G m m pg = (9.00 kg) −1.30 yˆ = −11.7 kg ⋅ yˆ s s m m m G G G G p total = pd1 + pd2 + pg = 4.40 kg ⋅ xˆ − 4.40 kg ⋅ yˆ − 11.7 kg ⋅ yˆ = 4.40 kg ⋅ m s xˆ − 16.1 kg ⋅ m s yˆ s s s

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3. Determine the total momentum of the dog and cat. Let north be in the y-direction. G G G m m m m G G pd + p c = md v d + mc v c = p total = (20.0 kg) 2.50 yˆ + (5.00 kg) 3.00 xˆ = 15.0 kg ⋅ xˆ + 50.0 kg ⋅ yˆ s s s s Set the total momentum of the dog and cat equal to that of the owner. G G G p 0 = m0 v 0 = p total G 15.0 kg ⋅ ms xˆ + 50.0 kg ⋅ ms yˆ p m m v 0 = total = xˆ + 0.714 yˆ = 0.214 70.0 kg s s m0 0.714 = 73.3° 0.214

θ = tan −1

2

2

m m v0 = 0.214 + 0.714 = 0.745 m s s s

4. (a) The carts must have equal and opposite momenta for the total momentum of the system to be zero. G G G G p1 − p 2 = 0 = m1 v1 − m2 v 2

(

)

(0.45 kg) 1.1 ms mv = 0.76 m s v2 = 1 1 = m2 0.65 kg

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

(c) Ksystem =

2

1 1 1 m 1 m m1v12 + m2 v2 2 = (0.45 kg) 1.1 + (0.65 kg) 0.76 = 0.46 J 2 2 2 s 2 s

5. Determine the speed of the baseball before it hits the ground. m p 0.780 kg ⋅ s m = 5.20 v= = m 0.150 kg s Recall that v 2 = v02 − 2 g ( y − y0 ).

(

)

2

5.20 ms − (0)2 v 2 − v02 = = 1.38 m y0 − y = h = 2g 2 9.81 m2

(

s

)

G G G 6. (a) ∆p = p f − pi G G = m( v f − v i ) = (0.220 kg) 2.0 = (0.220 kg) 4.5 m = 0.99 kg ⋅ yˆ s

m m yˆ − −2.5 yˆ s s m yˆ s

∆p = 0.99 kg ⋅ m s m m − 2.5 = − 0.11 kg ⋅ m s (b) pf − pi = m(vf − vi ) = (0.220 kg) 2.0 s s

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(c) The quantity found in part (a). 7. p1 = ptotal x = ptotal cos θ = (17.6 kg ⋅ m/s)( cos 66.5°) = 7.02 kg ⋅ m/s p 7.02 kg ⋅ m/s m1 = 1 = = 2.51 kg v1 2.80 m/s p2 = ptotal y = ptotal sin θ = (17.6 kg ⋅ m/s)( sin 66.5°) = 16.14 kg ⋅ m/s p 16.14 kg ⋅ m/s = 5.21 kg m2 = 2 = v2 3.10 m/s −3 8. I = Fav ∆t = (1350 N)(6.20 × 10 s) = 8.37 kg ⋅ m/s

9. To estimate the magnitude of the force, determine the magnitude of the average force. m ∆p m(vf − vi ) (0.045 kg) 67 s − 0 = = = 3.0 kN Fav = ∆t ∆t 0.0010 s

(

10. Fav = ∆t =

11.

∆p ∆t

(

∆p (0.50 kg) 3.2 = Fav 230 N

m s

−0

)=

)

7.0 ms

I = ∆p = m∆ v m=

−9 kg ⋅ ms I = ∆v −23 m − 4.5 s

m s

= 0.3 kg

12. (a) Recall that v = 2 gh . Just before the marble hits the floor, its speed is vi = − 2 gh1 (negative downward motion). Just after it hits the floor, its speed is vf = 2 gh2 . So, G G I = ∆p G = m ∆v = m[ 2 gh2 yˆ − (− 2 gh1 yˆ )] = m 2 g ( h2 + h1 )yˆ m = (0.0150 kg) 2 9.81 ( 0.640m + 1.44m)yˆ s2 = (0.133 kg ⋅ m s ) yˆ

The impulse is 0.133 kg ⋅ m/s in the positive y-direction. (b) Because the impulse is proportional to the sum of the square roots of the initial and final heights, the impulse would have been greater than that found in part (a). 13. The impulse is equal to the change in the y-component of the momentum of the ball. m I = ∆p y = m∆v y = m[v0 cos 65° − (−v0 cos 65°)] = (0.60 kg) 5.4 (2 cos 65°) = 2.7 kg ⋅ m s s

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14. (a) Let the initial motion of the ball be along the positive x-axis and the final along the positive y-axis. G G m m m m G G I = ∆p = m( v f − vi ) = (0.14 kg) 18 yˆ − (−36) xˆ = 5.04 kg ⋅ xˆ + 2.52 kg ⋅ yˆ s s s s 2.52 θ = tan −1 = 27° 5.04 2

2

m m I = 5.04 kg ⋅ + 2.52 kg ⋅ = 5.6 kg ⋅ m s s s

G (b) Since ∆p is directly proportional to the ball’s mass, it would double in magnitude. There would be no change in the direction. G (c) If the ∆p of the ball is unchanged, then the impulse is unchanged. G G G G G G 15. I = ∆p = p f − pi = m( v f − vi ) = (0.75 kg)[(5.2 m/s)xˆ + (3.7 m/s)yˆ − (8.8 m/s)xˆ − (−2.3 m/s)yˆ ] = (−2.7 kg ⋅ m/s)xˆ + (4.5 kg ⋅ m/s)yˆ Iy 4.5 kg ⋅ m/s (a) θ = tan −1 = tan −1 = −59° + 180° = 121° from the x -axis −2.7 kg ⋅ m/s Ix

(b) I = (−2.7 kg ⋅ m/s)2 + (4.5 kg ⋅ m/s)2 = 5.2 kg ⋅ m/s 16. The sum of the canoes’ momenta must be zero. The set up is similar to Example 9-3. p1x + p2 x = 0 = m1v1x + m2 v2 x m2 =

(

−m1v1x −(340 kg) − 0.52 = v2 x 0.44 m

m s

)=

400 kg

s

17. This problem is similar to Problem 16, so we may use the above result for m2.

(

−m1v1x −(45 kg) −0.62 m2 = = v2 x 0.89 m

m s

)=

31 kg

s

18. Assume the bee’s motion is in the negative direction. pbee + ps = 0 = mbee vbee + ms vs vs =

(

− mbee vbee −(0.175 g) −1.41 = 4.75 g ms

cm s

)=

51.9 mm s

19. The sum of the pieces’ momenta must be zero. Assume the motion is one-dimensional. p1 + p2 = 0 = m1v1 + m2 v2 m1v1 = −m2 v2 m1 −v2 = m2 v1 2

2

m1 −v2 v22 = = 2 v1 m2 v1 172

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Now, use the fact that K 2 = 2 K1. K2 =

1 1 m2 v22 = 2 K1 = 2 m1v12 = m1v12 2 2 2

m1 v22 1 m1 = = m2 2v12 2 m2 1 m 1= 1 2 m2 m1 = 2 m2 The piece with the smaller kinetic energy has the larger mass.

20. The motion is one-dimensional. Assume the satellite’s motion is in the negative x-direction. pa + ps = 0 = ma va + ms vs −ms vs ma The initial distance is va =

x0 = va t =

(

)

−(1100 kg) −0.13 ms −ms vs t= (7.5 s) = 11 m ma (97 kg )

21. (a) vL = speed of lumberjack relative to the shore vL,log = speed of lumberjack relative to the log vlog = speed of the log relative to the shore

Use conservation of momentum. mL vL + mlog vlog = 0 mL (vL,log + vlog ) + mlog vlog = 0 mL vL,log + mL vlog + mlog vlog = 0 mL vL,log = (− mL − mlog )vlog

vlog =

mL vL,log −mL − mlog

=

(

)

(85 kg) 2.7 ms m = − 0.494 −85 kg − 380 kg s

vL = vL,log + vlog = 2.7

m m + −0.494 = 2.2 m s s s

(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 vL + vlog = vL − vlog = the speed of the lumberjack relative to the shore and vlog ∝

1 , which decreases as mlog increases. mlog

(

)

(85 kg) 2.7 ms m (c) vlog = = = − 0.429 −mL − mlog −85 kg − 450 kg s m m vL = vL,log + vlog = 2.7 + − 0.429 = 2.3 m s s s mL vL,log

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22. Choose the motion of the two pieces at right angles to one another to be along the positive x- and y-axes. Use conservation of momentum. p1 + p2 + p3 = 0 G = mvxˆ + mvyˆ + mv3 v3 = −v(xˆ + yˆ ) v3 = (−v)2 + (−v)2 =

2v

−v = 225° −v

θ = tan −1

θ = 225° if the two pieces with speed v lie along the postive x - and y -axes. 23. The collision is completely inelastic because the two carts stick together. Assume the motion is one-dimensional. The initial momentum is equal to the final momentum. pi = mv + m(0) = 2mvf = pf mv v vf = = 2m 2 The final kinetic energy is 2

Kf =

1 1 2 v (2m)vf 2 = m = mv 2 4 2

24. From Example 9-6, m2 v2 = tan θ m1v1

(

)

m m v tan θ (950 kg) 20.0 s tan 40.0° v2 = 1 1 = = 12 m s m2 1300 kg and (950 kg) 20.0 ms m1v1 vf = = = 11 m s (m1 + m2 ) cos θ (950 kg + 1300 kg) cos 40.0°

(

)

25. Let the motion of player 1 be in the positive x-direction and the motion of player 2 be at an angle of 120° measured counterclockwise from the positive x-axis. G G The initial momenta for the players are pli = mvxˆ and p 2i = mv(cos θ xˆ + sin θ yˆ ). The final momentum for the G G G two-player system is pf = 2mv f . Using conservation of momentum, mvxˆ + mv cos θ xˆ + mv sin θ yˆ = 2mv f . 1 G v f = v[(1 + cos θ )xˆ + sin θ yˆ ] 2 m 1 = 5.75 [(1 + cos125°)xˆ + sin125°yˆ ] 2 s m = 2.875 [(1 − 0.5736)xˆ + 0.8192yˆ ] s = (1.23 m s) xˆ + (2.36 m s ) yˆ

26. (a) The final kinetic energy of the car and truck together is less than the sum of their initial kinetic energies. Some of the kinetic energy is converted to sound and some to heat. Some of the energy creates the permanent deformations in the materials of the car and truck.

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2

2

1 1 1 m 1 m mc vc 2 + mt vt 2 = (1200 kg) 2.5 + (2600 kg) 6.2 = 54 kJ 2 2 2 s 2 s

(b) Ki =

2

Kf =

1 1 m (mc + mt )vf 2 = (1200 kg + 2600 kg) 5.0 = 48 kJ 2 2 s

27. (a) Use momentum conservation. Let the subscripts b and B denote the bullet and the block, respectively. mb vbi + mB vBi = mb vbf + mB vBf mb vbi + 0 = mb vbf + mB vBf m v − mB vBf vbf = b bi mb =

(

)

(

(0.0040 kg) 650 ms − (0.095 kg) 23 ms 0.004 kg

)

= 1.0 × 102 m s

(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 1 1 mb vbi 2 = (0.0040 kg) 650 ms = 850 J 2 2 2 1 2 1 1 1 K f = mb vbf 2 + mB vBF2 = (0.0040 kg) 103.75 ms + (0.095 kg) 23 ms = 47 J 2 2 2 2

(c) Ki =

(

)

(

)

28. (a) No (b) Use momentum conservation to determine the speed of the puddy-block system just after the collision. mb vb + mp vp = (mb + mp )vf mp = v mb + mp p mb + mp Use vf to determine K f and equate Kf with the gravitational potential energy above the original position of the block. vf =

mb (0) + mp vp

mp 1 (mb + mp ) mb + mp 2

2

vp 2 = (mb + mp ) gh mp h= mb + mp

2

vp 2 2g

2 m 2 5.60 s 0.0700 kg = 0.470 kg + 0.0700 kg 2 9.81 m2 s = 2.69 cm

(

175

(

)

)

Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

29. Use momentum conservation to determine the speed of the puddy-block system just after the collision. mb (0) + mp vp = (mb + mp )vf mp vf = v mp + mb p Use vf to determine Kf and equate Kf with the spring potential energy. mp 1 (mp + mb ) mp + mb 2

∆x =

mp 2 vp 2 k (mp + mb )

30. (a) Ki =

=

2

1 vp 2 = k ∆x 2 2

(

(

)

2

(0.0500 kg) 2 2.30 ms = 3.71 cm N (0.0500 kg + 0.430 kg) 20.0 m

)

1 1 1 m1v 2 + m2 v 2 = (m1 + m2 )v 2 2 2 2 2

Kf =

1 1 v (m1 + m2 ) = (m1 + m2 )v 2 2 32 4

2 1 K f 32 (m1 + m2 )v 1 = = 1 ( m + m )v 2 16 Ki 2 2 1

(b) Use momentum conservation. v m1v + m2 (−v) = (m1 + m2 ) 4 1 m1 − m2 = (m1 + m2 ) 4 4m1 − 4m2 = m1 + m2 3m1 = 5m2 m1 5 = 3 m2 31. m1 = the mass of the truck m2 = the mass of the car v0 = the initial speed of the truck Use conservation of momentum to find an equation for the final speed of the truck. m1v0 = m1v1f + m2 v2f m1v1f = m1v0 − m2 v2f m v1f = v0 − 2 v2f m1 There is one equation and two unknowns. Use conservation of energy to find a second equation relating v1f and v2f. 1 1 1 m1v02 = m1v1f 2 + m2 v2f 2 2 2 2 m1v1f 2 = m1v02 − m2 v2f 2 Substitute for v1f and solve for v2f.

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2

m m1 v0 − 2 v2f m 1 2 m m1v02 − 2m2 v0 v2f + 2 v2f 2 m1 m −2v0 v2f + 2 v2f 2 m1 m1 + m2 2 v2f m1

= m1v0 2 − m2 v2f 2 = m1v0 2 − m2 v2f 2 = −v2f 2 = 2v0 v2f

2m1 v2f = v0 m1 + m2 Substitute for v2f in the equation for v1f. m1 − m2 m2 2m1 2m2 v0 = 1 − v0 = v0 m1 m1 + m2 m1 + m2 m1 + m2 Using the given information, m1 = 1620 kg, v1f = v0 −

m2 = 722 kg, and v0 = 14.5 m/s, the final speeds of the truck and car are: vtruck = 5.56 m/s and vcar = 20.1 m/s .

32. This problem is analogous to Problem 31. The hammer takes the place of the truck, and the nail takes the place of the car. Therefore, the kinetic energy acquired is given by 2

K 2f =

2

2 2m1 1 1 1 2(0.550 kg) m 2 m2 v2f 2 = m2 4.5 v0 = (0.012 kg) = 0.47 J 2 2 2 s 0.550 kg + 0.010 kg m1 + m2

33. From Example 9-7, Ki = 0.197 J. 1 1 m1v1f 2 + m2 v2f 2 = Ki 2 2 Solve for v1f. Kf =

2

2(0.197 J) 0.160 kg m − 1.03 = 1.31 m s 0.130 kg 0.130 kg s Set the final y-component of momentum equal to zero to determine θ . 0 = p1 y − p2 y = m1v1f sin θ − m2 v2f v1f =

2 Ki m2 v2f 2 = − m1 m1

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sin θ =

Physics: An Introduction

m2 v2f m1v1f

m2 v2f m1v1f m2 v2f = sin −1 m (2 K − m v 2 ) i 2 2f 1

θ = sin −1

m (0.160 kg) 1.03 s = sin −1 2 (0.130 kg) 2(0.197 J) − (0.160 kg) 1.03 ms = 74.8°

(

)

(

)

34. For the neutron, 1 mv 2 v Kf f = 2 = f 2 1 Ki mvi vi 2

2

.

Recall that m − m2 v1f = 1 v0 , so m1 + m2 m−M vf = vi . m+M Kf m − M = Ki m + M

2

2

K f 1.009u − 5.49 × 10−4 u = (a) = 0.998 Ki 1.009u + 5.49 × 10−4 u 2

(b)

K f 1.009u − 1.007u −7 = = 9.84 × 10 Ki 1.009u + 1.007u

(c)

K f 1.009u − 207.2u = = 0.9807 Ki 1.009u + 207.2u

2

35. (a) Let subscript 1 refer to the elephant and subscript 2 refer to the ball. Use momentum conservation. m1v1 + m2 v2 = m1v1f + m2 v2f Use conservation of kinetic energy. 1 1 1 1 m1v12 + m2 v22 = m1v1f 2 + m2 v2f 2 . 2 2 2 2 Rearranging the first equation gives m1 (v1 − v1f ) = 1. m2 (v2f − v2 )

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Rearranging the second equation gives m1 (v12 − v1f 2 ) 2

2

m2 (v2f − v2 )

=1=

m1 (v1 − v1f )(v1 + v1f ) . m2 (v2f − v2 )(v2f + v2 )

Comparing these two equations implies that (v1 − v1f ) = 1, or v2f = v1 + v1f − v2 . (v2f − v2 ) Substitute for v2f in the first equation and solve for v1f . m1v1 + m2 v2 = m1v1f + m2 (v1 + v1f − v2 ) (m1 − m2 )v1 + 2m2 v2 = (m1 + m2 )v1f m − m2 2m2 v1f = 1 v1 + v2 m1 + m2 m1 + m2 Since v2f = v1 + v1f − v2 , v1f = v2f + v2 − v1.

Substitute for v1f in the first equation and solve for v2f . m1v1 + m2 v2 = m1 (v2f + v2 − v1 ) + m2 v2f 2m1v1 + (m2 − m1 )v2 = (m1 + m2 )v2f 2m1 m2 − m1 v2f = v1 + v2 m1 + m2 m1 + m2 2(5400 kg) m 0.150 kg − 5400 kg m = −4.30 8.11 + s 5400 kg + 0.150 kg s 5400 kg + 0.150 kg = −17 m/s

(b) Kinetic energy has been transferred from the elephant to the ball. 36. Place the x-axis along the Earth-Moon center-to-center line with the origin at the center of the Earth. Σmx X cm = M mE xE + mm xm = mE + mm mE (0) + mm xm = mE + mm mm = xm mE + mm 7.35 × 1022 kg = (3.85 × 108 m) 5.98 × 1024 kg + 7.35 × 1022 kg = 4.67 × 106 m

The mean radius of the Earth is about 6.37 × 106 m. Therefore, the center of mass is 6.37 ×106 m − 4.67 ×106 m = 1.70 ×106 m below the surface of the Earth .

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Physics: An Introduction

37. Place the origin at the center of the basket. Then, the two cartons of cereal are at xc = (−0.75 m)/2 = −0.375 m (assuming the cartons are left of center). Calculate the x-coordinate of the center of mass and set it equal to zero. Σmx 2mc xc + mm xm = =0 X cm = 2mc + mm M Solve for xm. 2mc xc + mm xm = 0 2mc xc mm 2(0.55 kg)(−0.375 m) =− 1.8 kg = 0.23 m The half gallon of milk should be placed 0.23 m from the center of the basket, opposite the cartons of cereal. xm = −

38. Calculate the x-coordinate of the center of mass. Assume that the mass of each brick is m and that the mass of each brick is distributed uniformly. 5L L 1 11L 11 Σmx m1 x1 + m2 x2 + m3 x3 m 2 + L + 4 X cm = = = = L = 3m 3 4 12 M m1 + m2 + m3

(

)

39. Place the origin at the center of the box with the plane of the missing top perpendicular to the positive z-axis. Due to symmetry, Xcm = Ycm = 0.

(

)

L Σmz mz1 + mz2 + mz3 + mz4 + mzbottom m 0 + 0 + 0 + 0 − 2 L = = =− M 5m 5m 10 The center of mass is L/10 units below the center of the box.

Z cm =

mx2 + mx3 + mx4 (The subscript refers to the quadrant.) 3m By symmetry, x2 = x3 = − x4 .

40. X cm =

x 1 X cm = ( x2 + x2 − x2 ) = 2 3 3 x2 = 3 X cm = 3(−1.2 in.) = −3.6 in. my + my3 + my4 Ycm = 2 3m By symmetry, y2 = − y3 = − y4 . y 1 Ycm = ( y2 − y2 − y2 ) = − 2 3 3 y2 = −3Ycm = −3(−1.2 in.) = 3.6 in. ( x2 , y2 ) = (−3.6 in., 3.6 in.)

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41. Due to symmetry, Xcm = 0. Calculate Ycm. Σmy Ycm = M my1 + my2 + ms ys = 2m + ms 2my1 + ms ys ( y1 = y2 ) = 2m + ms 2(16u )(0.143 nm) sin 30° + (32u )(0) = 2(16u ) + 32u = 3.6 × 10−11 m ( X cm , Ycm ) = (0, 3.6 × 10−11 m)

42. (a) Calculate Xcm. Mx1 + Mx2 + Mx3 1 = (0 + 0.50 m + 1.5 m) = 0.67 m 3M 3 Calculate Ycm. X cm =

Ycm =

My1 + My2 + My3 1 = (0.50 m + 0 + 0) = 0.17 m 3M 3

( X cm , Ycm ) = (0.67 m, 0.17 m)

(b) The location of the center of mass would not be affected. The mass drops out of the equations. 43. We define the following subscripts: f = floor nf = not on floor L = length of rope Find the equation of motion of the top of the rope. m 2.00 m = 2.20 s ytop = 0.910 t, 0 < t < s 0.910 ms Set the origin at the floor. ytop ynf = and yf = 0 2 m y + mf yf Ycm = nf nf mnf + mf Substitute. Ycm

( =

0.604 kg 2.00 m

)y

ytop top 2

0.604 kg

+0 =

2

1 1 1 m 2 2 2 ytop = 0.910 t = (0.207 m s )t 2.00 m 2 2(2.00 m) s

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Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

y (m) 1.0 0.8 0.6 0.4 0.2 O

0.5

1.0

1.5

2.0

t (s)

)

(

2 mnf + mf v t v + m (0) nf nf f 0.910 ms L mnf vnf + mf vf vnf 2 = = Vcm = t= t = (0.414 m s 2 )t 2.00 m mnf + mf mnf + mf L

(

Vcm (m/s) 1.0 0.8 0.6 0.4 0.2 O

0.5

1.0

1.5

2.0

t (s)

44. We define the following subscripts: f = floor nf = not on floor L = length of rope Find the equation of motion of the top of the rope. m 2.00 m = 2.20 s ytop = 2 m − 0.910 t, 0 < t < s 0.910 ms Set the origin at the floor. ytop ynf = and yf = 0 2 m y + mf yf Ycm = nf nf mtotal Substitute. y 0.604 kg +0 ytop top 2.00 m 2 Ycm = 0.604 kg 1 1 2 = ytop 2.00 m 2

(

)

1 2 m − 0.910 2(2.00 m) 1 = 4 m 2 − 3.64 2(2.00 m)

=

m t s m2 s

(

2

m2 t + 0.828 2 t 2 s

)

Ycm = 1.00 m − ( 0.910 m s ) t + 0.207 m s 2 t 2

182

)

Physics: An Introduction

Chapter 9: Linear Momentum and Collisions

y (m) 1.0 0.8 0.6 0.4 0.2 O

0.5

Vcm =

1.0

1.5

mnf vnf + mf vf mnf + mf

(

t (s)

2.0

v

)

(mnf + mf ) 1 + Lnf t vnf + mf (0) = mnf + mf v = 1 + nf t vnf L vnf 2 = vnf + t L

(

m m − 0.910 s = − 0.910 + s 2.00 m

)

2

t

Vcm = − 0.910 m s + (0.414 m s 2 )t Vcm (m/s) 0

0.5

1.0

t (s)

1.5

-0.2 -0.4 -0.6 -0.8 -1.0

45. (a) Before the string breaks, the reading on the scale is the total weight of m Mg = (1.20 kg + 0.150 kg) 9.81 = 13.2 N . s2 (b) After the string breaks, the reading is 13.2 N. Because the ball is moving with constant speed, the center of mass of the system undergoes no net acceleration. Therefore, the reading will not change. 46. (a) Taking up to be positive, calculate the net external force acting on the cooking pot of water and the egg. Fnet,ext = Fs − mp g − me g Now, determine the acceleration of the center of mass. Acm =

mp (0) + me M

( ) = − −g 2

me me + mp

(

g 9.81 0.0460 kg = − 2 2 0.0460 kg + 2.90 kg

183

m s2

)=

−0.0766 m s 2

Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

(b) Recall that Fnet,ext = MAcm , then me g . 2 m g m Fs = mp g + e = g mp + e 2 2 Fs − mp g − me g = −

m 0.0460 kg = 28.7 N = 9.81 2 2.90 kg + 2 s

(c) After the egg comes to rest on the bottom of the pot, the reading is the total weight of m Mg = (2.90 kg + 0.0460 kg) 9.81 = 28.9 N s2 47. (a) From Example 9-9, the velocity of the center of mass before the collision is m v + m2 v2 Vcm = 1 1 m1 + m2 m1v1 + m2 (0) = m1 + m2 m1 = v1 m1 + m2 0.750 kg m = 0.455 s 0.750 kg + 0.275 kg = 0.333 m s

(b) Use momentum conservation to find the speed of the carts after the collision. m1v1 = m1vf + m2 vf = (m1 + m2 )vf m1 vf = m1 + m2

(c) Ki = = = = =

v1 = Vcm = 0.333 m s

1 1 m1v12 + m2 v22 2 2 1 2 1 m1v1 + m2 (0)2 2 2 1 2 m1v1 2 2 1 m (0.750 kg) 0.455 2 s 0.0776 J 2

Kf =

1 1 m (m1 + m2 )vf 2 = (0.750 kg + 0.275 kg) 0.333 = 0.0568 J 2 2 s

48. (a) Before the string is cut, the force of gravity is countered by the force of the spring. 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 .

(b) Since Fnet,ext = MAcm = 0, Acm = 0 .

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Chapter 9: Linear Momentum and Collisions

49. The net force must be zero for the helicopter to hover. ∆m Fnet = 0 = thrust − mg = v − mg ∆t ∆m mg = = v ∆t

(

(5500 kg) 9.81 60.0

m s

m s2

)=

900 kg s

∆m 50. Ffriction = 3.4 N = thrust = v ∆t kg ∆m 3.4 N 3.4 N = = = 0.309 m v s ∆t 11 s

Convert ∆m/∆t from kg/s to rocks/min. 0.309 kg 60 s 1 rock = 37 rocks min s min 0.50 kg 51. Use conservation of momentum. pi = 0 = pf = (mp + ms )v + 2mb vb m + v 0 = (mp + ms )v + 2m b 18.0 s m 0 = (mp + ms )v + 2mb 18.0 + 2mb v s m −2mb 18.0 = v(mp + ms + 2mb ) s v=

(

)

(

)

−2mb 18.0 ms −2(0.850 kg) 18.0 ms m = = − 0.503 mp + ms + 2mb 57.0 kg+2.10 kg + 2(0.850 kg) s

The person will recoil with a speed of 0.503 m/s. 52. Use conservation of momentum for each time the person throws a brick. First brick pi = 0 = pf = (mp + ms + mb )v + mb vb m 0 = (mp + ms + mb )v + mb 18.0 + v s m 0 = (mp + ms + mb )v + mb 18.0 + mb v s −mb 18.0 ms −(0.850 kg) 18.0 ms m = = −0.252 v= mp + ms + 2mb 57.0 kg + 2.10 kg + 2(0.850 kg) s

(

)

(

)

Second brick pi = (mp + ms + mb )v = (mp + ms )vp + mb vb = pf m (mp + ms + mb )v = (mp + ms )vp + mb 18.0 + vp s

185

Chapter 9: Linear Momentum and Collisions

vp = =

(

(mp + ms + mb )v − mb 18 mp + ms + mb

m s

Physics: An Introduction

) (

(57.0 kg+2.10 kg + 0.850 kg) −0.252

m s

) − (0.850 kg) (18 ms )

57.0 kg + 2.10 kg + 0.850 kg m = − 0.507 s The person will recoil with a speed of 0.507 m/s.

53. (a) 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. ∆m v Scale reading = (mb + ms ) g + ∆t m kg m = (0.540 kg + 0.750 kg) 9.81 + 0.0560 3.20 2 s s s = 12.8 N m (b) W = (mb + ms ) g = (0.540 kg + 0.750 kg) 9.81 = 12.7 N s2 2

54. (a) thrust =

∆m kg m v = (mass per unit length)(speed)v = 0.13 1.4 = 0.25 N ∆t m s

(b) The scale reads more than 2.5 N. It reads the weight of the rope on the scale and the thrust due to the falling rope. 2

(c) Scale reading =

∆m kg m m v + mg = 0.13 1.4 + (0.25 kg) 9.81 2 = 2.7 N ∆t m s s

55. Place the origin at Xcm. me xe + mt xt me + mt mx xe = − t t me (72.5 kg)(555 ft) =− 5.97 × 1024 kg

X cm = 0 =

= − 6.74 × 10−21 ft

The earth moves only 6.74 × 10−21 ft . 56. (a) Use conservation of momentum. 1 v 1 mv + m(0) = m + mv2 2 3 2 2 1 mv = mv2 3 2 4 v2 = v 3 186

Physics: An Introduction

Chapter 9: Linear Momentum and Collisions

1 2 mv 2 2 2 1 v 1 1 4 Kf = m + m v 2 3 2 2 3 1 4 1 = mv 2 + mv 2 = mv 2 18 9 2 Ki = K f , therefore, the collision is elastic.

(b) Ki =

57. Use conservation of momentum to determine the horizontal speed of the bullet and block. m = the mass of the bullet M = the mass of the block mv + M (0) = (m + M )vf m vf = v m+M Recall that d y = h = (1/ 2) gt 2 for a mass that is initially stationary. Find the horizontal distance. m x = vf t = m+M

0.0100 kg m 2(0.750 m) 2h = = 2.16 m 725 v g 0.0100 kg 1.30 kg s + 9.81 m2 s

58. (a) Since egg 2 is farther from the center of mass than egg 1, the location of the center of mass will change more if egg 2 is removed. mx1 3.0 cm = 0− = − 0.25 cm 12m 12 my 3.5 cm = − 0.29 cm Ycm = Ycm,prev − 1 = 0 − 12m 12 (Xcm, Ycm) = (–0.25 cm, –0.29 cm)

(b) X cm = X cm,prev −

mx2 15 cm = 0− = −1.3 cm 12m 12 my 3.5 cm = − 0.29 cm Ycm = Ycm,prev − 2 = 0 − 12m 12 (Xcm, Ycm) = (–1.3 cm, –0.29 cm)

(c) X cm = X cm,prev −

59. Use the thrust equation to estimate the force.

(

kg

)

( )

)

m 1000 3 (1 m 2 )(31 in.) 0.0254 in. ∆m m m v= thrust = 10 = 0.24 N ∆t s (9 h) 3600 hs

(

60. (a) The change in momentum per second is the weight of the apple. ∆p = F = mg = 3.0 N ∆t (b) ∆p = F ∆t = W ∆t = Wtf = (3.0 N)(1.5 s) = 4.5 kg ⋅ m s

187

Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

61. Place the origin at the center of the wheel and the lead weight on the positive x-axis. mlw xlw + mwh X cm,prev mlw + mwh m 0.0502 kg −2 X cm,prev = − lw xlw = − (25.0 cm) = −3.54 × 10 cm mwh 35.5 kg Before the lead weight was added, the center of mass was 0.354 mm from the center of the wheel. X cm = 0 =

62. Since there are no external forces acting on the system in the x-direction, X cm of the system will not change. After the system comes to rest, the ball, hoop, and system have the same X cm . Therefore, the x-coordinate of the ball will be X cm . X cm =

Mxh + 2Mxb 1 1 R R = ( xh + 2 xb ) = 0 + 2 R − = 3M 3 3 4 2

63. (a) When the canoeist walks toward the shore, the canoe will move away from the shore according to conservation of momentum. Therefore, her distance from the shore is greater than 2.5 m. (b) Place the origin at the center of the canoe before the canoeist walks toward the shore. X cm = 0 and will not change since there is no external force with an x-component acting on the system. Assume the canoeist walks in the positive x-direction. After the canoeist walks to the end of the canoe, the distance between the canoeist and the canoe’s center of mass is xM − xm = 1.5 m, where xM and xm represent the x-component of the center of mass for the canoeist and the canoe, respectively. mx + MxM X cm = 0 = m m+M 0 = mxm + MxM = mxm + M ( xm + 1.5 m) = (m + M ) xm + (1.5 m) M M xm = −(1.5 m) m+M 63 kg = −(1.5 m) = −1.1 m 22 kg + 63 kg xM = −1.1 m + 1.5 m = 0.4 m The distance between the system’s center of mass and the shore is

3.0 m + 2.5 m = 4.0 m. So, the canoeist is 2

4.0 m – 0.4 m = 3.6 m from shore. 64. Place the origin at the center of the canoe before the canoeist walks toward the shore. X cm = 0 and will not change since there is no external force with an x-component acting on the system. Assume the canoeist walks in the positive x-direction. After the canoeist walks to the end of the canoe, the distance between the canoeists and the canoe’s center of mass is xM − xm = 1.5 m, where xM and xm represent the x-component of the center of mass for the canoeist and the canoe, respectively. mx + MxM X cm = 0 = m m+M 0 = mxm + MxM = m( xM − 1.5 m) + MxM MxM m=− xM − 1.5 m

188

Physics: An Introduction

Chapter 9: Linear Momentum and Collisions

The distance between the system’s center of mass and the shore is (3.0 m)/2 + 2.5 m = 4.0 m. So, xM = 4.0 m − 3.4 m = 0.6 m. (63 kg)(0.6 m) m=− 0.6 m − 1.5 m = 42 kg m 65. (a) Scale reading = W = (m1 + m2 ) g = (0.150 kg + 1.20 kg) 9.81 = 13.2 N s2

(b) In the absence of the liquid, the ball would fall with an acceleration equal to g. The liquid is retarding the motion of g 3 the ball with a force of m1 g − = m1 g . So, the scale reading is 4 4 3 m 3 3 m1 g + m2 g = m1 + m2 g = (0.150 kg) + 1.20 kg 9.81 = 12.9 N 4 4 4 s2 66. Use conservation of momentum. 0 = mp vp + mh vh mp = −

(

)

(1.1 kg) 6.2 ms cos13° mh vh =− = 27 kg vp −0.25 m s

67. Due to symmetry, Ycm = 0. X cm =

2(1.0u )(9.6 × 10−11m) cos 52.25° + (16u )(0) = 6.5 × 10−12 m 2(1.0u ) + 16u

( X cm , Ycm ) = (6.5 × 10−12 m, 0) kg 68. m = 0.125 (0.500 m) = 0.0625 kg m ∆m F = mg + v ∆t m = mg + v v L m kg m m = (0.0625 kg) 9.81 + 0.125 1.33 1.33 2 m s s s = 0.834 N

( )

mv0 + mr 2 69. Vcm = v0 = m + mr 3 2 1 (m + mr ) = m + mr 3 2 2 1 2 − mr = 1 − m 3 2 3 mr = 2m

v0 2

189

Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

70. (a) mv0 = (2m)vf vf =

(b)

v0 2

1 2 1 mv0 = (2m)vf2 2 2 v vf = 0 2

71. Assume gravity is the only force acting on the rocket after it is launched. After rising for 2.5 s its speed is v = 44.2 m/s − (9.81 m/s 2 )(2.50 s) = 19.7 m/s.

(a) Since the initial momentum is upward, each piece must have a momentum with a vertical component equal to half the initial momentum. 1 m p1 y = p2 y = (mv) = vf sin 45° 2 2 v 19.7 m/s vf = = = 27.9 m/s sin 45° sin 45° (b) Before the explosion Vcm = (19.7 m/s)yˆ . Since the momentum of the system is the same after the explosion, and the total mass has not changed, Vcm = (19.7 m/s)yˆ after the explosion too. (c) The only force acting on the system before and after the explosion is gravity. Therefore, Acm = (−9.81 m/s 2 )yˆ . G 72. (a) (11, 000 kg ⋅ m/s)xˆ + (−370 kg ⋅ m/s)yˆ + p 2 = (15, 000 kg ⋅ m/s)xˆ + (2100 kg ⋅ m/s)yˆ G p 2 = (4000 kg ⋅ m/s)xˆ + (2470 kg ⋅ m/s)yˆ

(b) No

Momentum depends only on mass and velocity. It is independent of position.

73. (a) Use momentum conservation. m1v1 + m2 v2 = m1vf + m2 vf (0.84 kg)(0) + (0.42 kg)(0.68 m/s) = (0.84 kg + 0.42 kg)vf vf = 0.23 m/s

(b) The energy stored in the spring bumper is equal to the loss of kinetic energy at that time. 1 1 ∆K = K f − Ki = (0.84 kg + 0.42 kg)(0.227 m/s)2 − (0.42 kg)(0.68 m/s)2 = −0.065 J 2 2 Energy stored in the bumper is 0.065 J. (c) Since this is a one-dimensional, head-on elastic collision, we can use the results of Problem 74. 0.84 kg − 0.42 kg 2(0.42 kg) v1,f = (0) + (0.68 m/s) = 0.45 m/s 0.84 kg + 0.42 kg 0.84 kg + 0.42 kg 2(0.84 kg) 0.42 kg − 0.84 kg v2,f = (0.68 m/s) = −0.23 m/s (0) + 0.84 kg 0.42 kg + 0.84 kg + 0.42 kg 190

Physics: An Introduction

Chapter 9: Linear Momentum and Collisions

74. Use momentum conservation. m1v1 + m2 v2 = m1v1f + m2 v2f Use conservation of kinetic energy. 1 1 1 1 m1v12 + m2 v22 = m1v1f 2 + m2 v2f 2 . 2 2 2 2 Rearranging the first equation gives m1 (v1 − v1f ) = 1. m2 (v2f − v2 ) Rearranging the second equation gives m1 (v12 − v1f 2 ) 2

2

=1=

m1 (v1 − v1f )(v1 + v1f ) . m2 (v2f − v2 )(v2f + v2 )

m2 (v2f − v2 ) Comparing these two equations implies that v1 + v1f = 1, or v2f = v1 + v1f − v2 . v2f + v2

Substitute for v2f in the first equation and solve for v1f . m1v1 + m2 v2 = m1v1f + m2 (v1 + v1f − v2 ) (m1 − m2 )v1 + 2m2 v2 = (m1 + m2 )v1f m − m2 2m2 v1f = 1 v1 + v2 m1 + m2 m1 + m2 Since v2f = v1 + v1f − v2 , v1f = v2f + v2 − v1. Substitute for v1f in the first equation and solve for v2f . m1v1 + m2 v2 = m1 (v2f + v2 − v1 ) + m2 v2f 2m1v1 + (m2 − m1 )v2 = (m1 + m2 )v2f 2m1 m2 − m1 v2f = v1 + v2 m1 + m2 m1 + m2 2m1 m1 − m2 m2 − m1 2m2 75. v2f − v1f = − v1i + − v2i m1 + m2 m1 + m2 m1 + m2 m1 + m2 m + m2 − m1 − m2 = 1 v1i + v2i m1 + m2 m1 + m2 = v1i − v2i

76. In each case, the potential energy of the spring is converted into the kinetic energy of the cart(s). So, the kinetic energy of the single cart is equal to the sum of the kinetic energies of the two carts. 1 2 1 1 mv = mv12 + mv22 2 2 2 1 1 1 Because of momentum conservation, v1 = −v2 . So, we have mv 2 = mvf2 + m(−vf )2 = mvf2 . 2 2 2 The final speed of each cart is vf =

2 v . 2

191

Chapter 9: Linear Momentum and Collisions

Physics: An Introduction

77. Assume v0 is in the positive x-direction. Use conservation of momentum. pxi = p xf mv0 = mv1x + mv2 x (I) v0 = v1 cos θ1 + v2 cos θ 2 p yi = p yf 0 = v1 sin θ1 + v2 sin θ 2 (II) Use conservation of kinetic energy. 1 1 1 mv02 = mv12 + mv22 2 2 2 v02 = v12 + v2 2 (III) Square (I) and subtract from (III). v02 = v12 + v2 2 −(v02 = v12 cos 2 θ1 + v2 2 cos 2 θ 2 + 2v1v2 cos θ1 cos θ 2 ) 0 = v12 (1 − cos 2 θ1 ) + v22 (1 − cos 2 θ 2 ) − 2v1v2 cos θ1 cos θ 2 = v12 sin 2 θ1 + v22 sin 2 θ 2 − 2v1v2 cos θ1 cos θ 2

(IV)

From (II), v sin θ1 v2 = − 1 and v12 sin 2 θ1 = v22 sin 2 θ 2 . sin θ 2 Substituting these results into (IV) gives v sin θ1 0 = 2v12 sin 2 θ1 − 2v1 1 cos θ1 cos θ 2 sin θ 2 cos θ1 cos θ 2 sin θ1 sin 2 θ1 = − sin θ 2 0 = cos θ1 cos θ 2 + sin θ1 sin θ 2 = cos(θ1 − θ 2 ) So,

θ1 − θ 2 = cos −1 0 = 90°. 78. Fnet, ext = (mc + ms ) Acm = mc ac + ms as ac =

Fnet, ext − ms as mc

=

(

40.0 N − (9.50 kg) 2.32 21.0 kg

m s2

)=

0.855 m s 2

79. Place the origin at the position of impact, and assume that the combined objects move away from the origin in the negative x-direction. Use conservation of momentum. pix = pfx v mv cos θ1 + mv cos θ 2 = (2m) 3 2 cos θ1 + cos θ 2 = 3 piy = pfy mv sin θ1 + mv sin θ 2 = 0 sin θ1 = − sin θ 2 = sin(−θ 2 ) θ1 = −θ 2 192

Physics: An Introduction

Chapter 9: Linear Momentum and Collisions

Substitute this result into the previous result. 2 cos θ1 + cos(−θ1 ) = 3 2 2 cos θ1 = 3 1 cos θ1 = 3 1 θ1 = cos−1 3 = 70.5° So, the initial angle was 2(70.5°) = 141°. 80. From Problem 74, m − m2 2m2 v1f = 1 v1 + v2 . + m m 1 m1 + m2 2 Choose the positive direction to be up. v1f = vm v1 = −v m1 = m v2 = v m2 = M m−M 2M vm = ( −v ) + m+M m+M Recall that v = 2 gh .

M − m + 2M v = m+M

3M − m v = v m+M

3M − m 2 ghm = 2 gh m+M 2

3M − m hm = h m+M

81. (a) If the rope’s center of mass moves upward with constant acceleration, then the velocity of the rope’s center of mass must be increasing linearly with time, since it is upward. We define the following subscripts: t = table nt = not on table M vt v + m (0) v2 t m v + mt vt Vcm = nt nt = L = t L M M Vcm is upward, and since both v and L are constant, Vcm is proportional to t. Hence Acm is upward and

(

constant.

Acm =

)

v2 , the slope of a graph of Vcm versus t. L

(b) The rope being lowered has downward momentum. Its downward momentum is decreasing as more and more of its mass comes to rest. Therefore, there must be a net upward force acting on the rope, resulting in an upward acceleration of the rope’s center of mass.

193

Chapter 9: Linear Momentum and Collisions

(

) (

Physics: An Introduction

)

M + ML vt v + ML vt (0) v2 mnt vnt + mt vt = = = v+ L M M

t Even though v is negative if the rope is moving downward, the equation for Vcm is linear with positive slope. Therefore, Acm , which is the slope of a velocity versus time graph, is positive and constant, having the same

(c) Vcm

magnitude as in part (a),

v2 . L

194

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