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Linear Momentum and Collisions
CHAPTER OUTLINE 9.1 Linear Momentum and Its Conservation 9.2 Impulse and Momentum 9.3 Collisions in One Dimension 9.4 Two-Dimensional Collisions 9.5 The Center of Mass 9.6 Motion of a System of Particles 9.7 Deformable Systems 9.8 Rocket Propulsion
ANSWERS TO QUESTIONS
*Q9.1 (a) No. Impulse,
F ∆ t , depends on the force and the time for which it is applied.
(b) No. Work depends on the force and on the distance over which it acts.
*Q9.2 The momentum magnitude is proportional to the speed and the kinetic energy is proportional to the speed squared.
(i) The speed of the constant-mass object becomes 4 times larger and the kinetic energy 16 times larger. Answer (a).
(ii) The speed and the momentum become two times larger. Answer (d).
*Q9.3 (i) answer (c). For example, if one particle has 5 times larger mass, it will have 5 times smaller speed and 5 times smaller kinetic energy.
(ii) answer (d). Momentum is a vector.
*Q9.4 (i) Equal net work inputs imply equal kinetic energies. Answer (c).
(ii) Imagine one particle has four times more mass. For equal kinetic energy it must have half the speed. Then this more massive particle has 4(12) = 2 times more momentum. Answer (a).
Q9.5 (a) It does not carry force, for if it did, it could accelerate itself.
(b) It cannot deliver more kinetic energy than it possesses. This would violate the law of energy conservation.
(c) It can deliver more momentum in a collision than it possesses in its flight, by bouncing from the object it strikes.
*Q9.6 Mutual gravitation brings the ball and the Earth together. As the ball moves downward, the Earth moves upward, although with an acceleration on the order of 10 25 times smaller than that of the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is conserved. Answer (d).
Q9.7 (a) Linear momentum is conserved since there are no external forces acting on the system. The fragments go off in different directions and their vector momenta add to zero.
(b) Kinetic energy is not conserved because the chemical potential energy initially in the explosive is converted into kinetic energy of the pieces of the bomb.
209
Q9.8 Momentum conservation is not violated if we choose as our system the planet along with you. When you receive an impulse forward, the Earth receives the same size impulse backwards. The resulting acceleration of the Earth due to this impulse is much smaller than your acceleration forward, but the planet’s backward momentum is equal in magnitude to your forward momentum.
*Q9.9 (i) During the short time the collision lasts, the total system momentum is constant. Whatever momentum one loses the other gains. Answer (c).
(ii) When the car overtakes the manure spreader, the faster-moving one loses more energy than the slower one gains. Answer (a).
Q9.10 The rifle has a much lower speed than the bullet and much less kinetic energy. Also, the butt distributes the recoil force over an area much larger than that of the bullet.
*Q9.11 (i) answer (a). The ball gives more rightward momentum to the block when the ball reverses its momentum.
(ii) answer (b). In case (a) there is no temperature increase because the collision is elastic.
Q9.12 His impact speed is determined by the acceleration of gravity and the distance of fall, in v^2 f^ = vi^2^ − 2 g (^) ( 0 − yi ). The force exerted by the pad depends also on the unknown stiffness of the pad.
Q9.13 The sheet stretches and pulls the two students toward each other. These effects are larger for a faster-moving egg. The time over which the egg stops is extended, more for a faster missile, so that the force stopping it is never too large.
*Q9.14 Think about how much the vector momentum of the Frisbee changes in a horizontal plane. This will be the same in magnitude as your momentum change. Since you start from rest, this quantity directly controls your final speed. Thus f is largest and d is smallest. In between them, b is larger than c and c is larger than g and g is larger than a. Also a is equal to e, because the ice can exert a normal force to prevent you from recoiling straight down when you throw the Frisbee up. The assembled answer is f > b > c > g > a = e > d.
Q9.15 As one fi nger slides towards the center, the normal force exerted by the sliding finger on the ruler increases. At some point, this normal force will increase enough so that static friction between the sliding finger and the ruler will stop their relative motion. At this moment the other finger starts sliding along the ruler towards the center. This process repeats until the fingers meet at the center of the ruler. Next step: Try a rod with a nonuniform mass distribution. Next step: Wear a piece of sandpaper as a ring on one finger to change its coefficient of friction.
*Q9.16 (a) No: mechanical energy turns into internal energy in the coupling process.
(b) No: the Earth feeds momentum into the boxcar during the downhill rolling process.
(c) Yes: total energy is constant as it turns from gravitational into kinetic.
(d) Yes: If the boxcar starts moving north the Earth, very slowly, starts moving south.
(e) No: internal energy appears.
(f) Yes: Only forces internal to the two-car system act.
210 Chapter 9
212 Chapter 9
(b) original chemical energy in girl’s body = total final kinetic energy Uchemical = (12)(65 kg)(2.9 m s) 2 + (12)(40 kg)(4.71 m s) 2 = 717 J
(c) System momentum is conserved with the value zero. The net forces on the two siblings are of equal magnitude in opposite directions. Their impulses add to zero. Their final momenta are of equal magnitude in opposite directions, to add as vectors to zero.
P9.3 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with speed given by
v^2 f^ − vi^2^ = 2 a x ( f − xi ): 0 − v^2 i = 2 ( −9 80. m s 2 )( 0 250. m)
vi = 2 20. m s
Total momentum of the system of the Earth and me is conserved as I push the planet down and myself up:
0 = (^) (5 98. × (^10 24) ) −( ) + (85 0. )( 2 20. ) ~
kg v kg m s v
e e^110 − (^23) m s
*P9.4 (a) For the system of two blocks ∆ p = 0 ,
or p (^) i = pf Therefore, 0 = M vm + ( 3 M )( 2 00. m s)
Solving gives vm = −6 00. m s (motion toward the left).
(b)
kx^2^ = M v^2 M + ( 3 M ) v 32 M = 8 40. J
(c) The original energy is in the spring. A force had to be exerted over a distance to com- press the spring, transferring energy into it by work. The cord exerts force, but over no distance.
(d) System momentum is conserved with the value zero. The forces on the two blocks are of equal magnitude in opposite directions. Their impulses add to zero. The final momenta of the two blocks are of equal magnitude in opposite directions.
P9.5 (a) The momentum is p = mv , so v = (^) mp and the kinetic energy is
K m m p m
p m
2 2 2 v
(b) K = m
v^2 implies v = 2 K m
, so p m m
K
m mK = v = =
FIG. P9.
Linear Momentum and Collisions 213
Section 9.2 Impulse and Momentum
*P9.6 From the impulse-momentum theorem, (^) F (∆ t ) = ∆ p = m v (^) f − mvi , the average force required to hold onto the child is
F
m t
= (^ f^ − i )
( )
= (^ )^ (^ − ) −
v v ∆
kg mi h 1
. s
mm s mi h
N
⎝⎜^
= − ×
In trying to hang onto the child, he would have to exert a force of 6.44 kN (over 1 400 lb) toward the back of the car, to slow down the child’s forward motion. He is not strong enough to exert so large a force. If he were belted in and his arms were firmly tied around the child, the child would exert this size force on him toward the front of the car. A person cannot safely exert or feel a force of this magnitude and a safety device should be used.
P9.7 (a) I = (^) ∫ Fdt =area under curve
I = (^) ( × − )( ) = ⋅
1 50. 10 3 s 18 000 N 13 5. N s
(b) F =
×
N s 1.50 10 s kN
(c) From the graph, we see that F max = 18 0 kN.
P9.8 The impact speed is given by 1 2 1
2 m v = mgy 1. The rebound speed is given by mgy (^) 2 m 2
= v. The impulse the floor imparts to the ball is the change in the ball’s momentum, m m m m gh gh
v (^) 2 v 1 (^) v (^) 2 v 1 (^2 2 )
up down up up
− = ( + )
kg m s 2 m m up kg m s upward⋅
P9.9 ∆ ∆
p^ ^ = F
t p (^) y m v (^) fy viy m v cos 60 0. ° mv cos 60 .. sin. sin. sin
∆ p x = m (− v − v ) = − mv ..
= − ( )( )( ) = − ⋅
kg m s kg m ss kg m s s
F avg p N t
= ∆ x = −^ ⋅^ = − ∆
P9.10 Assume the initial direction of the ball in the − x direction.
(a) Impulse,
I = ∆ p = p (^) f − p (^) i = ( 0 060 0. )( 40 0. ) i ˆ^ − ( 0 060 0. )( 50 .0 0 ) −( ˆ i (^) ) =^ 5 40. ˆ i N s⋅
(b) Work (^) = K (^) f − Ki = 1 ( ) (⎡⎣ ) − ( )⎤⎦ = − 2
0 060 0. 40 0. 2 50 0. 2 27 0. JJ
FIG. P9.
FIG. P9.
Linear Momentum and Collisions 215
*P9.14 After 3 s of pouring, the bucket contains (3s)(0.25 Ls) = 0.75 liter of water, with mass 0.75 L(1 kg1 L) = 0.75 kg, and feeling gravitational force 0.75 kg(9.8 m s^2 ) = 7.35 N. The scale through the bucket must exert 7.35 N upward on this stationary water to support its weight. The scale must exert another 7.35 N to support the 0.75-kg bucket itself. Water is entering the bucket with speed given by mgytop = (12) mvimpact^2 vimpact = (2 gytop )^1 ^2 = [2(9.8 ms 2 )2.6 m]^1 ^2 = 7.14 m s downward The scale exerts an extra upward force to stop the downward motion of this additional water, as described by mvimpact + Fextra t = mvf The rate of change of momentum is the force itself: ( dm dt ) vimpact + Fextra = 0 Fextra = −( dm dt ) vimpact = −(0.25 kgs)(−7.14 ms) = +1.78 N Altogether the scale must exert 7.35 N + 7.35 N + 1.78 N = 16.5 N
Section 9.3 Collisions in One Dimension
P9.15 Momentum is conserved for the bullet-block system
10 0 10 5 01 0 600 301
(.^ ×^3 ) = (.^ )(. )
− (^) kg kg m s
m s
v v
P9.16 (a) m v (^) 1 i + 3 m v (^) 2 i = 4 mvf where m = 2 50. × 10 4 kg
v (^) f =
. m s
(b) K f − K i = ( m ) f − ⎡ m i + ( m ) i
⎣⎢^
v^2 v 1^23 v 22 (^) ( 2 .5 50 × (^10 4) ) (12 5. − 8 00. −6 00. ) = −3 75. × 104 J
Ki = Kf + ∆ Eint ∆ Eint = +37.5 kJ
P9.17 (a) The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor 4 3 2 00 4 00 6 00 4
m (^) i m m
i
( ) = ( )( ) + ( )
= +
v
v
m s m s m s .. (^00). 4
m s (^) = 2 50m s
(b) W actor (^) = K (^) f − K (^) i = 1 ⎡⎣( m )( m s ) + m ( m s) 2
3 2 00. 2 4 00. 2 ⎤⎤⎦ − 1 ( )( ) 2
4 m 2 50. m s^2
W actor
kg = m s ( × ) ( + − )( ) =
... 3 37 5. kJ
(c) The event considered here is the time reversal of the perfectly inelastic collision in the previous problem. The same momentum conservation equation describes both processes.
FIG. P9.
216 Chapter 9
P9.18 Energy is conserved for the bob-Earth system between bottom and top of swing. At the top the stiff rod is in compression and the bob nearly at rest.
K (^) i + U (^) i = K (^) f + Uf : 1 2
M v^2 b + 0 = 0 + Mg 2
vb^2^ = g 4 so vb = 2 g Momentum of the bob-bullet system is conserved in the collision:
m v = m v + M ( g )
2 v =
4 M
m g
P9.19 First we find v 1 , the speed of m 1 at B before collision. 1 2 2 9 80 5 00 9 90
1 1 2 1
1
m v m gh
v
= (. )(. ) =. m s
Now we use the text’s analysis of one-dimensional elastic collisions to find v 1 f , the speed of m 1 at B just after collision.
v 1^1 2 v 1 2
1
f 9 90^ 3 30
m m m m
= − (. ) m s = −. m s
Now the 5-kg block bounces back up to its highest point after collision according to
m gh 1 m 1 2
max =^ (^ −3 30. )^ h max
( − ) ( )
m s^2 m s 2 m
*P9.20 (a) We assume that energy is conserved in the fall of the basketball and the tennis ball. Each reaches its lowest point with a speed given by K U K U
mgy m
g g
i b
b
( + ) =^ ( + )
release bottom 0 1 2
v^20
v == 2 gyi = 2 9 8(. m s (^2) )( 1 20. m ) = 4 85. m s
FIG. P9.
continued on next page
v / FIG. P9.
218 Chapter 9
P9.22 We assume equal firing speeds v and equal forces F required for the two bullets to push wood fi bers apart. These equal forces act backward on the two bullets.
For the first, K (^) i + ∆ E (^) mech= Kf^1 2 ( 7 00.^ ×^10 −^3 kg)^ v^^2 −^ F (8 00.^ ×^10 −^2 m) =^0
For the second, pi = pf (7 00. × 10 −^3 kg) v = (1 014. kg) v (^) f
v
v f =^
( 7 00^ ×^10 −) 1 014
.^3
Again, K (^) i + ∆ E (^) mech = Kf : 1 2
(.^ × −^3 kg)^ v^^2 −^ Fd^ =^ (1 014^. kg) v^2 f
Substituting for v (^) f ,
1 2
3 2 7 00^103
( × ) −^ =^ (^ )^
− × −
kg v Fd kg vv 1 014
2 .
⎝⎜^
Fd = (^) ( × ) − (^1) − ( × −) 2
3 2 3 2
..^2 . v v
Substituting for v ,
Fd = F ( × ) −
⎛ ×
⎝⎜^
8 00 10 − 1 7 00^10 −
. 2.^3
m d = 7 94. cm
P9.23 (a) From the text’s analysis of a one-dimensional elastic collision with an originally stationary target, the x -component of the neutron’s velocity changes from vi to v 1 f = (1 − 12) vi 13 = − 11 vi 13. The x -component of the target nucleus velocity is v 2 f = 2 vi 13. The neutron started with kinetic energy (12) m 1 vi^2 The target nucleus ends up with kinetic energy (12) (12 m 1 )( 2 vi 13) 2 Then the fraction transferred is 12 1 2 12 1 2
m m
i i
v v = = Because the collision is elastic, the other 71.6% of the
original energy stays with the neutron. The carbon is functioning as a moderator in the reactor, slowing down neutrons to make them more likely to produce reactions in the fuel.
(b) (^) K (^) n = ( 0 716. ) (^) ( 1 6. × 10 − 13 J) = 1 15. × 10 −^13 J
K (^) C = ( 0 284. ) (^) ( 1 6. × 10 −^13 J (^) ) = 4 54. × 10 −^14 J
Linear Momentum and Collisions 219
P9.24 (a) Using conservation of momentum, ( ∑ p^ ) before = ( ∑ p )after , gives
( 4 0. kg )( 5 0. m s ) + ( 10 kg )( 3 0. m s ) + (3 0. kg) −( 4 .. 0 m s ) = (^) [( 4 0. + 10 +3 0.)kg] v
Therefore, v = +2 24. m s, or 2 24. m s toward the right
(b) No. For example, if the 10-kg and 3.0-kg mass were to stick together first, they would move with a speed given by solving (^13 kg^ ) v 1^ = (^10 kg^ )(3 0.^ m s^ ) + (3 0.^ kg^ ) −( 4 0.^ m s), or^ v 1 = +1 38^.^ m s
Then when this 13 kg combined mass collides with the 4.0 kg mass, we have
( 17 kg ) v = ( 13 kg )(1 38. m s ) + ( 4 0. kg )(5 0. m s), and v = +2 24. m s
just as in part (a). Coupling order makes no difference to the final velocity.
P9.25 During impact, momentum of the clay-block system is conserved:
m v (^) 1 = ( m (^) 1 + m 2 ) v 2
During sliding, the change in kinetic energy of the clay-block-surface system is equal to the increase in internal energy: 1 2 1 2
1 2 2 2 1 2
2 2
(^ m^ + m^ ) =^ f df^ =^ ( m^ + m^ ) gd
( ) =
v
v
μ
. kg 0 0 650 0 112. (^) (. kg (^) )( 9 80. m s (^2) )(7 50. m)
v 22 = 95 6. m 2 s^2 v 2 = 9 77. m s
(^ 12 0.^ ×^10 −^3 kg^ ) v 1^ = (^ 0 112.^ kg^ )(9 77^. m s) v 1 =^ 91 2.^ m s
Section 9.4 Two-Dimensional Collisions
*P9.26 (a) Over a very short time interval, outside forces have no time to impart significant impulse—thus the interaction is a collision. The opponent grabs the fullback and does not let go, so the two players move together at the end of their interaction—thus the collision is completely inelastic.
FIG. P9.
continued on next page
Linear Momentum and Collisions 221
P9.28 We use conservation of momentum for the system of two vehicles for both northward and eastward components, to fi nd the original speed of car number 2. For the eastward direction: M (13 0. m s) = 2 MV (^) f cos 55 0.° For the northward direction: M v 2 i = 2 MVf sin 55 0.° Divide the northward equation by the eastward equation to find: v 2 i = (13 0. m s ) tan 55 0. °= 18 6. m s =41 5. mi h
Thus, the driver of the north bound car was untruthful. His original speed was more than 35 mih.
P9.29 p^ xf = pxi m m m
v (^) O v Y m s
cos. cos. .
= ( ) 0 799. v (^) O + 0 602. v Y = 5 00. m s (1) p p m m
yf = yi v (^) O sin 37 0. ° − v Ysin 53 0. °= 0
0 602. v (^) O = 0 799. v Y (2)
Solving (1) and (2) simultaneously,
v O = 3 99. m s and v Y = 3 01. m s
P9.30 p xf = pxi : m v O cos θ+ m v Ycos ( 90 0°. −θ) = m vi
v (^) O cos θ + v (^) Ysinθ= vi (1)
p yf = pyi : m v O sin θ− mv Ysin ( 90 0. °−θ) = 0
v (^) O sin θ = v Ycosθ (2)
From equation (2),
v (^) O = v Y⎛⎝ ⎞⎠ cos sin
θ θ (3) Substituting into equation (1),
v (^) Y cos v (^) Y v sin
sin
(^2) θ θ
⎛ θ ⎝⎜^
so v (^) Y ( cos 2 θ +sin 2 θ (^) ) = vi sinθ, and v (^) Y = vi sin θ
Then, from equation (3), v (^) O = vi cos θ.
We did not need to write down an equation expressing conservation of mechanical energy. In the prob- lem situation, the requirement of perpendicular final velocities is equivalent to the condition of elasticity.
FIG. P9.
FIG. P9.
FIG. P9.
222 Chapter 9
P9.31 m 1 (^) v^ 1 (^) i + m (^) 2 v ^ (^) 2 i = (^) ( m (^) 1 + m 2 ) v f : 3 00 5 00. (. ) ˆ i^ − 6 00. ˆ j^ =5 00. v^
v = i − j ( 3 00.^ ˆ^ 1 20.^ ˆ^ )m s
P9.32 x -component of momentum for the system of the two objects: p (^) 1 ix + p (^) 2 ix = p 1 (^) fx + p 2 fx : − m vi + 3 m vi = 0 + 3 m v 2 x
y -component of momentum of the system: 0 + 0 = − m v 1 (^) y + 3 mv 2 y
by conservation of energy of the system: + 1 + = + ( + )
m vi^2^ m v^2 i m v 12 (^) y 3 mv 22 (^) x v 22 y
we have v (^) 2 2 v x 3 = i also v 1 (^) y = 3 v 2 y
So the energy equation becomes
4 9 4 3
2 v v (^) 22 v (^) v i y i = + + y
8 3
2 2
v (^) i v 2 = y
or v (^) 2 2 v y 3 = i
(a) The object of mass m has fi nal speed v (^) 1 y = 3 v (^) 2 y = 2 vi and the object of mass 3 m moves at
v (^) 22 v (^) 22 v^ v
x y 9
v (^) 22 v (^) 22 2 v x y (^) 3 i
(b)^ θ =^
tan −^1 2
v v
y x
θ =
tan −^1 2 = .° 3
v 35 3 v
i i
P9.33 m 0 = 17 0. × 10 −^27 kg v i = 0 (the parent nucleus)
m 1 = 5 00. × 10 −^27 kg (^) v (^) 1 = 6 00. × 106 ˆ j m s
m 2 = 8 40. × 10 −^27 kg v (^) 2 = 4 00. × 106 i ˆ m s
(a) m (^) 1 v 1 (^) + m (^) 2 v^2 (^) + m 3 v 3 = 0 where m (^) 3 = m (^) 0 − m (^) 1 − m 2 = 3 60. × 10 −^27 kg
(^ 5 00.^ ×^10 −^27 ) ( 6 00.^ ×^10 6 ˆ^ j^ ) +^^ (8 40.^ ×^10 −^27 ) 4 00.^ ×^100 3 60^10
9 33 10 8
6 27 3
3 6
. ˆ^.
i v
v i
( ) +^ ( × ) =
= − × −
333 × 10 6
( ˆ j^ )m s
FIG. P9.
continued on next page
224 Chapter 9
P9.36 Let the x axis start at the Earth’s center and point toward the Moon.
x m x^ m x CM m m
kg 0 = +
1 1 2 2 ×^ +^ ×
1 2
5 98. 1024 7 36. 1022 kg m kg
8 24
( × ) ×
= 4.67^ ×^106 m from the Earth’s center
The center of mass is within the Earth, which has radius 6 37. × 10 6 m. It is 1.7 Mm below the point on the Earth’s surface where the Moon is straight overhead.
P9.37 Let A 1 represent the area of the bottom row of squares, A 2 the middle square, and A 3 the top pair. A A A A M M M M M A
M
A
1 2 3 1 2 3 1 1 A 1 = 300 cm 2 , (^) A 2 = 100 cm 2 , (^) A 3 = 200 cm 2 , A = 600 cm 2
M M
A
A
M
M
M M A
A
1 1
2 2
cm cm
2 2 000 600 6 200 (^3600) 3
cm cm cm cm
2 2 2
M M
M M
A
A
= ⎛⎝ ⎞⎠ = 22 M = M
x x M x M x M M
M M
CM
cm cm
1 1 2 2 3 3 15 0.^ (^12 ) +5 00.^ ((^16 ) +^ (^ )
=
=
(^13)
(^12)
cm
cm cm
CM
CM
M
M
x
y M )) + ( ) + ( )( ) =
=
M M
M
y
cm cm cm
CM 1 13 3.^ cm
P9.38 (a) Represent the height of a particle of mass dm within the object as y. Its contribution to the
gravitational energy of the object-Earth system is ( dm gy ). The total gravitational energy is
U (^) g = (^) ∫ gy dm = g (^) ∫ y dm all mass
. For the center of mass we have y M CM =^ ∫ y dm
, so
U (^) g = gMy CM
(b) The volume of the ramp is 1 2
(3 6. m )(15 7. m )(64 8. m ) = 1 83. × 10 3 m 3. Its mass is
ρ V = ( 3 800 kg m (^3) ) ( 1 83. × 10 3 m (^3) ) = 6 96. × 106 kg. As shown in the chapter, its center
of mass is above its base by one-third of its height, y CM = m = m
15 7. 5 23.. Then
U (^) g = Mgy CM= 6 96. × 10 6 kg (^) ( 9 8. m s (^2) ) 5 23. m= 3 57. × 1088 J
FIG. P9.
Linear Momentum and Collisions 225
P9.39 This object can be made by wrapping tape around a light stiff uniform rod.
(a) M = (^) ∫ λ dx = ⎡⎣ + x ⎤⎦ dx 0
0 300
0
0 30 50 0 20 0
.. ..
m g m g m^2
0 0 m ∫
M = ⎡⎣ 50 0 x +10 0 x^2 ⎤⎦ 0 = 15 9 0 300
... . g m g m 2 g m
(b) x
xdm
M M xdx M CM all mass x
m = = = g
∫ ∫
0
0 300 λ
.
. mm g m 2
m ∫ ⎡⎣ +20 0^2 ⎤⎦ 0
0 300 .
. x dx
x x x CM
2 g g m g m = ⎡ + ⎣⎢^
2 3 0
0 300 .
. mm = 0 153. m
P9.40 Take the origin at the center of curvature. We have L = r
2 π ,
r
L
π
. An incremental bit of the rod at angle θ from the x axis has mass given by dm rd
M
θ L = , dm Mr L
= d θ where we have used the definition of radian measure. Now
y M
y dm M
r Mr L CM d^ r all mass
= (^) ∫ = (^) ∫ = = °
45
135 sin θ θ θ
22
45
135
2 45
L
d
L L
sin
cos
θ θ
π θ
°
°
°
°
∫
2 2
L L
π π
The top of the bar is above the origin by r
L
π , so the center of mass is below the middle of the
bar by 2^ L^^4 22 L^^2 1 2 2 L 0 063 5 L π π π π
⎠⎟^
Section 9.6 Motion of a System of Particles
P9.41 (a) ^
v
v (^) v v
i
CM
kg m
( )
∑ m M
m m M
i i (^) 1 1 2 2
2 00. (^) ( 2 00. ˆ^ ss −3 00. ˆ j^ m s (^) ) + ( 3 00. kg ) (^) ((1 00. ˆ i^ m s +6 00. ˆ j m s) 5 00. kg v i j CM =^ (1 40^.^ ˆ^ +2 40.^ ˆ) m s
(b) p ^ = M v (^) CM = ( 5 00. kg ) (^) (1 40. i ˆ^ +2 40. ˆ j (^) )^ m s= ( 7 00. i ˆ^ ++12 0. ˆ j ) (^) kg m s⋅
FIG. P9.
x
y
Linear Momentum and Collisions 227
(b) Before, v i CM
kg m s kg
( 0 200. )( 1 50. ) ˆ^ + ( 0 300. ) −0 400. m s kg
( ) ˆ .
i 0 500 v i CM = (^ 0 360.^ m s)ˆ Afterwards, the center of mass must move at the same velocity, because the momentum of the system is conserved.
Section 9.7 Deformable Systems
*P9.45 (a) Yes. The only horizontal force on the vehicle is the frictional force exerted by the floor, so it gives the vehicle all of its final momentum, (6 kg)(3 ˆ i m s) = 18.0 ˆ i kg⋅m s
(b) No. The friction force exerted by the floor on each stationary bit of caterpillar tread acts over no distance, so it does zero work.
(c) Yes, we could say that the final momentum of the cart came from the floor or from the planet through the floor, because the floor imparts impulse.
(d) No. The floor does no work. The final kinetic energy came from the original gravitational energy of the elevated load , in amount (12)(6 kg)(3 m s)^2 = 27.0 J.
(e) Yes. The acceleration is caused by the static friction force exerted by the floor that pre- vents the caterpillar tracks from slipping backward.
*P9.46 (a) Yes. The floor exerts a force, larger than the person’s weight over time as he is taking off.
(b) No. The work by the floor on the person is zero because the force exerted by the floor acts over zero distance.
(c) He leaves the floor with a speed given by (12) mv^2 = mgyf v = [2(9.8 ms^2 )0.15 m] 1 ^2 = 1.71 ms, so his momentum immediately after he leaves the floor is (60 kg)(1.71 m s up) = 103 kg⋅m s up
(d) Yes. You could say that it came from the planet, that gained momentum 103 kg⋅ms down, but it came through the force exerted by the floor over a time interval on the per- son, so it came through the floor or from the floor through direct contact.
(e) (12)(60 kg)(1.71 ms)^2 = 88.2 J
(f ) No. The energy came from chemical energy in the person’s leg muscles. The floor did no work on the person.
228 Chapter 9
*P9.47 (a) When the cart hits the bumper it immediately stops, and the hanging particle keeps moving with its original speed vi. The particle swings up as a pendulum on a fixed pivot, keeping constant energy. Measure elevations from the pivot: (12) mvi^2 + mg (− L ) = 0 + mg (− L cos θ) Then vi =^ [2 gL (1^ −^ cos^ θ)]^1 ^2
(b) vi = [2 gL (1 − cos θ)] 1 ^2 = [2(9.8 ms 2 )(1.2 m)(1 − cos 35°)]^1 ^2 = 2.06 ms
(c) Yes. The bumper must provide the horizontal force to the left to slow down the swing of the particle to the right, to reverse its rightward motion, and to make it speed up to the left. When the particle passes its straight-down position moving to the left, the bumper stops exerting force. It is at this moment that the cart-particle system momentarily has zero horizontal accelera- tion for its center of mass.
*P9.48 Depending on the length of the cord and the time interval ∆ t for which the force is applied, the sphere may have moved very little when the force is removed, or we may have x 1 and x 2 nearly equal, or the sphere may have swung back, or it may have swung back and forth several times. Our solution applies equally to all of these cases.
(a) The applied force is constant, so the center of mass of the glider-sphere system moves with constant acceleration. It starts, we define, from x = 0 and moves to ( x 1 + x 2 )2. Let v 1 and v 2 represent the horizontal components of velocity of glider and sphere at the moment the force stops. Then the velocity of the center of mass is vCM = ( v 1 + v 2 )2 and because the acceleration is constant we have ( x 1 + x 2 ) 2 = [( v 1 + v 2 )2]∆ t 2 ∆ t = 2( x 1 + x 2 )( v 1 + v 2 ) The impulse-momentum theorem for the glider-sphere system is F ∆ t = mv 1 + mv 2 F 2( x 1 + x 2 )( v 1 + v 2 ) = m ( v 1 + v 2 ) F 2( x 1 + x 2 ) m = ( v 1 + v 2 ) 2 F 2( x 1 + x 2 ) 4 m = ( v 1 + v 2 ) 2 4 = vCM^2 Then vCM = [ F ( x 1 + x 2 ) 2 m ] 1 ^2
(b) The applied force does work that becomes, after the force is removed, kinetic energy of the constant-velocity center-of-mass motion plus kinetic energy of the vibration of the glider and sphere relative to their center of mass. The applied force acts only on the glider, so the work-energy theorem for the pushing process is Fx 1 = (12)(2 m ) vCM^2 + Evib Substitution gives Fx 1 = (12)(2 m ) F ( x 1 + x 2 ) 2 m + Evib = Fx 1 2 + Fx 2 2 + Evib Then E (^) vib = Fx 1 2 − Fx 2 2 When the cord makes its largest angle with the vertical, the vibrational motion is turning around. No kinetic energy is associated with the vibration at this moment, but only gravita- tional energy: mgL (1 − cos θ) = F ( x 1 − x 2 ) 2 Solving gives θ = cos−^1 [1 − F ( x 1 − x 2 ) 2 mgL ]
P9.49 A picture one second later differs by showing five extra kilograms of sand moving on the belt.
(a) ∆ ∆
p t
x (^) = (^ 5 00^ )(^ 0 750 ) = 1 00
kg m s s
N
(b) The only horizontal force on the sand is belt friction, so from p^ xi +^ f^ ∆ t^^ = pxf this is (^) f p t = ∆ x = ∆
3 75. N
(c) The belt is in equilibrium: ∑^ Fx^ =^ max :^ +^ F ext^ −^ f =^0 and^ F ext =^ 3 75. N
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