9.1. Model: Model the car and the baseball as particles. Solve: (a) The momentum p mv = = ( )( ) 1500 kg 10 m/s = × 15 10 4 . kg m/s . (b) The momentum p mv = = ( )( ) 0.2 kg 40 m/s = 8.0 kg m/s .
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9.1. Model: Model the car and the baseball as particles.Solve: (a) The momentum p mv= = ( )( )1500 kg 10 m/s = ×1 5 104. kg m/s.(b) The momentum p mv= = ( )( )0.2 kg 40 m/s = 8.0 kg m/s .
9.2. Model: Model the bicycle and its rider as a particle. Also model the car as a particle.Solve: From the definition of momentum,
p pcar bicycle= ⇒ = ⇒ =m v m v vm
mvcar car bicycle bicycle bicycle
car
bicycle car =
( ) =1500 kg
100 kg5.0 m/s m/s75 0.
Assess: This is a very high speed (≈168 mph). This problem shows the importance of mass in comparing twomomenta.
9.3. Visualize: Please refer to Figure Ex9.3.Solve: The impulse is defined in Equation 9.6 as
J F t dtx xt= ( ) =∫
i
ft area under the Fx(t) curve between ti and tf
⇒ ( )( ) ⇒ =6 0 8 150012. N s = ms Nmax maxF F
9.4. Model: Model the rubber ball and the steel ball as particles, and their interaction as a collision.Solve: When objects collide, Newton’s third law tells us that the forces between the objects are equal andopposite. The time of contact is the same for each object. Therefore, the impulse on each will be the same.
9.5. Model: The particle is subjected to an impulsive force.Visualize: Please refer to Figure Ex9.5.Solve: Using Equation 9.5, the impulse is the area under the curve. From 0 s to 2 ms the impulse is
Fdt = −( ) ×( ) = −−∫ 12
32 10500 N s 0.5 N s
From 2 ms to 8 ms the impulse is
Fdt = +( ) −( ) = +∫ 12 2000 N 8 ms 2 ms 6.0 N s
From 8 ms to 10 ms the impulse is
Fdt = −( ) −( ) = −∫ 12 500 N 10 ms 8 ms 0.5 N s
Thus, from 0 s to 10 ms the impulse is − + −( )0 5 6 0 0 5. . . N s = 5.0 N s.
9.6. Model: Model the object as a particle and the interaction as a collision.Visualize: Please refer to Figure Ex9.6.Solve: The object is initially moving to the right (positive momentum) and ends up moving to the left (negativemomentum). Using the impulse-momentum theorem p p Jx x xf i= + ,
− = + +2 6 kg m/s kg m/s Jx ⇒ = − = −Jx 8 kg m/s 8 N s
Since J F tx = avg∆ , we have
F tavg 8 N s∆ = − ⇒ = − = −Favg
8 N s
10 ms800 N
The force is rF = ( )800 N, left .
9.7. Model: Model the object as a particle and the interaction with the force as a collision.Visualize: Please refer to Figure Ex9.7.Solve: (a) Using the equations
pfx = pix + Jx J F t dtx x
t
t
= ∫ ( )i
f
= area under force curve
⇒ = ( ) +v xf 1.0 m/s2.0 kg
1 (area under the force curve)
= ( ) + ( ) =1.0 m/s2.0 kg
1.0 N s m/s1
1 5.
(b) Likewise,
v xf 1.0 m/s2.0 kg
= ( ) +
1 (area under the force curve)
= ( ) +
−( ) =1.0 m/s2.0 kg
1.0 N s m/s1
0 5.
Assess: For an object with positive velocity, a negative impulse slows an object, whereas a positive impulseincreases an object’s speed. The opposite is true for an object with negative velocity.
9.8. Model: Use the particle model for the sled, the model of kinetic friction, and the impulse-momentumtheorem.Visualize:
Note that the force of kinetic friction fk imparts a negative impulse to the sled.Solve: Using ∆p Jx x= , we have
p p F t dt f dt f tx x x
t
t
t
t
f i k k
i
f
i
f
− = = − = −∫∫ ( ) ∆ ⇒ − = − = −mv mv n t mg tx xf i k kµ µ∆ ∆
We have used the model of kinetic friction fk = µkn, where µk is the coefficient of kinetic friction and n is thenormal (contact) force by the surface. The force of kinetic friction is independent of time and was therefore takenout of the impulse integral. Thus,
∆tg
v vx x= −( )1
µki f =
( )( ) −( )1
0 25. 9.8 m/s8.0 m/s 5.0 m/s
2 = 1.22 s
9.9. Model: Use the particle model for the falling object and the impulse-momentum theorem.Visualize:
Note that the object is acted on by the gravitational force, whose magnitude is mg.Solve: Using the impulse-momentum theorem,
p p J F t dty y y y
t
t
f i
i
f
− = = ∫ ( ) ⇒ − = −mv mv mg ty yf i ∆
⇒ =−
∆tv v
gy yi f =
− − −( ) =5 5
0 50.
. m/s 10.4 m/s
9.8 m/s s2
Assess: Since F mgy = − is independent of time, we have taken it out of the impulse integral.
9.10. Model: Model the tennis ball as a particle, and its interaction with the wall as a collision.Visualize:
The force increases to Fmax during the first two ms, stays at Fmax for two ms, and then decreases to zero during thelast two ms. The graph shows that Fx is positive, so the force acts to the right.Solve: Using the impulse-momentum theorem p p Jx x xf i= + ,
0.06 kg m/s 0.06 kg 32 m/s6 ms
( )( ) = ( ) −( ) + ( )∫320
F t dtx
The impulse is
F t dx F F F F
F
x ( ) = = ( ) + ( ) + ( ) = ( )
⇒ = ( )( ) + ( )( ) =
∫0
1
20 002 0 002
1
20 002 0 004
32
0 004960
6 ms
max max max max
max
area under force curve s s s s
0.06 kg m/s 0.06 kg 32 m/s
s N
. . . .
.
9.11. Model: Model the glider cart as a particle, and its interaction with the spring as a collision.Visualize:
Solve: Using the impulse-momentum theorem p p Fdtx xf i− = ∫ ,
0.6 kg 3 m/s 0.6 kg 3 m/s area under force curve 36 N( )( ) − ( ) −( ) = = ( )( )12 ∆t ⇒∆t = 0.20 s
9.12. Model: Choose skydiver + glider to be the system.Visualize:
Note that there are no external forces along the x-direction, implying conservation of momentum along the x-direction.Solve: The momentum conservation equation pfx = pix is
680 kg 60kg 60 kg 680 kg m/sG D−( )( ) + ( )( ) = ( )( )v vx x
30
Immediately after release, the skydiver’s horizontal velocity is still vxD m/s( ) = 30 . Thus
620 kg 60 kg 30 m/s 680 kg 30 m/sG( )( ) + ( )( ) = ( )( )vx
⇒ ( ) =vxG m/s30
Assess: The skydiver’s motion in the vertical direction has no influence on the glider’s horizontal motion.
9.13. Model: Choose car + gravel to be the system.Visualize:
Solve: There are no external forces on the car + gravel system, so the horizontal momentum is conserved. Thismeans pfx = pix. Hence,
10,000 kg 4,000 kg 10,000 kg 2.0 m/s 4,000 kg 0.0 m/sf+( ) = ( )( ) + ( )( )v x ⇒ =v xf m/s1 43.
9.14. Model: Choose car + rainwater to be the system.Visualize:
There are no external horizontal forces on the car + water system, so the horizontal momentum is conserved.Solve: Conservation of momentum is pfx = pix. Hence,
m m m mcar water car water20 m/s 22 m/s 0 m/s+( )( ) = ( )( ) + ( )( )⇒ +( )( ) = ( )( )5,000 kg 20 m/s 5,000 kg 22 m/swaterm ⇒ =mwater 500 kg
9.15. Model: We will define our system to be archer+arrow. The force of the archer (A) on the arrow (a) isequal to the force of the arrow on the archer. These are internal forces within the system. The archer is standing on
frictionless ice, and the normal force by ice on the system balances the weight force. Thus r rFext = 0 on the system,
and momentum is conserved.Visualize:
The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment pfx
must also be zero.Solve: We have MAvA + mava = 0 kg m/s. Therefore,
vm v
ma a
AA
0.1 kg 100 m/s
50 kg0.20 m/s= − =
−( )( ) = −
The archer’s recoil speed is 0.20 m/s.Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrowhas forward momentum, the archer will have backward momentum.
9.16. Model: We will define our system to be Bob+rock. Bob’s (B) force on the rock (R) is equal to the rock’sforce on Bob. These are internal forces within the system. Bob is standing on frictionless ice, and the normal force
by ice on the system balances the weight. r rFext = 0 on the system, and thus momentum is conserved.
Visualize:
The initial momentum pix of the system is zero because Bob and the rock are at rest. Thus pfx = 0 kg m/s.Solve: We have mBvB + mRvR = 0 kg m/s. Hence,
vm
mvB
R
BR
0.5 kg
75 kg30 m/s 0.20 m/s= − = −
( ) = −
Bob’s recoil speed is 0.20 m/s.Assess: Since the rock has forward momentum, Bob’s momentum is backward. This makes the total momentum zero.
9.17. Model: We will define our system to be Dan+skateboard.Visualize:
The system has nonzero initial momentum pix. As Dan (D) jumps backward off the gliding skateboard (S), theskateboard will move forward in such a way that the final total momentum of the system pfx is equal to pix. This
conservation of momentum occurs because r rFext = 0 on the system.
Solve: We have m v m v m m vx x xS f S D f D S D i( ) + ( ) = +( ) . Hence,
5.0 kg 8.0 m/s 50 kg 5.0 kg 50 kg 4.0 m/sf D( )( ) + ( )( ) = +( )( )v x ⇒ ( ) =v xf D
m/s3 6.
9.18. Model: We will define our system to be bird+bug. This is the case of an inelastic collision because thebird and bug move together after the collision. Horizontal momentum is conserved because there are no externalforces acting on the system during the collision.Visualize:
Solve: The conservation of momentum equation pfx = pix is
m m v m v m vx x x1 2 1 1 2 2+( ) = ( ) + ( )f i i ⇒ +( ) = ( )( ) + ( ) −( )300 g 10 g 300 g 6.0 m/s 10 g m/sfv x 30 ⇒ =v xf m/s4 84.
Assess: We left masses in grams, rather than convert to kilograms, because the mass units cancel out from bothsides of the equation. Note that (vix)2 is negative.
9.19. Model: The two cars are not an isolated system because of external frictional forces. But during thecollision friction is not going to be significant. Within the impulse approximation, the momentum of theCadillac+Volkswagen system will be conserved in the collision.Visualize:
Solve: The momentum conservation equation pfx = pix is
m m v m v m vx x xC VW f C i C VW i VW+( ) = ( ) + ( )
⇒ 0 kg mph 2000 kg 1.0 mph 1000 kg i VW= ( )( ) + ( )( )v x ⇒ ( ) = −v xi VW
mph2 0.
You need a speed of 2.0 mph.
9.20. Model: Because of external friction and drag forces, the car and the blob of sticky clay are not exactly anisolated system. But during the collision, friction and drag are not going to be significant. The momentum of thesystem will be conserved in the collision, within the impulse approximation.Visualize:
Solve: The conservation of momentum equation pfx = pix is
m m v m v m vx x xC B f B i B C i C
+( )( ) = ( ) + ( )
⇒ = ( )( ) + ( ) −( ) ⇒ =0 kg m/s 10 kg 1500 kg 2.0 m/s ( ) 300 m/si B i Bv vx x
Assess: This speed of the blob is around 600 mph which is very large. However, we must point out that a verylarge speed is expected in order to stop a car with only 10 kg of clay.
9.21. Model: This problem deals with the conservation of momentum in two dimensions in an inelastic collision.Visualize:
Solve: The conservation of momentum equation r rp pbefore after= is
m v m v m m vx x x1 1 2 2 1 2i i f( ) + ( ) = +( ) m v m v m m vy y y1 1 2 2 1 2i i f( ) + ( ) = +( )Substituting in the given values,
.02 kg 3.0 m/s kg m/s .02 kg .03 kg f( )( ) + = +( )0 v cosθ
0 kg m/s .03 kg 2.0 m/s .02 kg .03 kg f+ ( )( ) = +( )v sinθ
⇒ =vf 1.2 m/scosθ vf 1.2 m/ssinθ =
⇒ = ( ) + ( ) =vf 1.2 m/s 1.2 m/s 1.7 m/s2 2
θ = = ( ) = °− −tan tan1 1 1 45v
vy
x
The ball of clay moves 45° north of east at 1.7 m/s.
9.22. Model: We assume that the momentum is conserved in the collision.Visualize: Please refer to Figure Ex9.22.Solve: The conservation of momentum equation yields
p p p px x x xf f i i( ) + ( ) = ( ) + ( )1 2 1 2⇒ ( ) + = −p xf kg m/s 2 kg m/s 4 kg m/s
10 ⇒ ( ) = −p xf 2 kg m/s
1
p p p py y y yf f i i( ) + ( ) = ( ) + ( )1 2 1 2
⇒ ( ) − = +p yf 1 kg m/s 2 kg m/s 1 kg m/s1
⇒ ( ) =p yf 4 kg m/s1
Thus, the final momentum of particle 1 is − +( )2 4ˆ ˆi j kg m/s .
9.23. Model: The moon is treated as a particle.Solve: A particle moving in a circular orbit of radius r with velocity v has an angular momentum L = mvr. Theperiod of the moon is T = 27.3 days = 2.36 × 106 s. The radius of the orbit r = 3.84 × 108 m. The mass of the moonm = 7.36 × 1022 kg. Using v r T= 2π , we can write
L mT
r= 2 2π = ××
×( )7 36 102
2 36 103 84 1022
68 2
.. sec
. kg mπ = ×2 89 1034. kg
m
s
2
9.24. Model: Model the puck as a particle.Visualize:
Solve: The angular momentum is
L mvr= = 3.0 kg m /s2 = (0.2 kg) (0.5 m)v ⇒ v = 30 m/s
The force that keeps the puck in circular motion is the tension rT in the string. Thus,
Tmv
r= = ( )( )2 2
0.2 kg 30 m/s
0.5 m = 360 N
Assess: A tension of 360 N in the string whose farthest end has a puck moving at approximately 60 mph isreasonable.
9.25. Model: Model the ball as a particle. We will also use constant-acceleration kinematic equations.
Solve: (a) The momentum just after throwing is
p p mvx0 0 030 30 30= ° = ° = ( )( ) ° =cos cos cos0.050 kg 25 m/s 1.083 kg m/s
p p mvy0 0 030 30 30= ° = ° = ( )( ) ° =sin sin sin0.050 kg 25 m/s 0.625 kg m/s
The momentum at the top is
p mv mvx x x1 1 0 1 083= = = . kg m / s p mvy y1 1 0= = kg m / s
Just before hitting the ground, the momentum is
p mv mvx x x2 2 0= = = 1.083 kg m/s p mvy y2 2= = − = −mv y0 0.625 kg m/s
(b) px is constant because no forces act on the ball in the x-direction. Mathematically,
Fdp
dtpx
xx= = ⇒ =0 N constant
(c) The change in the y-component of the momentum during the ball’s flight is
∆py = –0.625 kg m/s − 0.625 kg m/s = –1.250 kg m/s
From kinematics, the time to reach the top is obtained as follows:
v v a t ty y y1 0 1 0= + −( ) ⇒ =−
=−( )−
=tv
ay
y1
0 1 27612.5 m/s
9.8 m/ss2 .
The time of flight is thus ∆t t= = ( ) =2 21 1.276 s 2.552 s . Multiplying this time by –mg, the y-component of theweight, yields –1.25 kg m/s. This follows from the impulse-momentum theorem:
∆ ∆p mg ty = −
9.26. Model: Model the rocket as a particle, and use the impulse-momentum theorem. The only force acting onthe rocket is due to its own thrust.Visualize: Please refer to Figure P9.26.Solve: (a) The impulse is
J F dtx x= ∫ = area of the Fx(t) graph between t = 0 s and t = 30 s = ( )( )12 1000 N 30 s = 15,000 N s
(b) From the impulse-momentum theorem, pfx = pix + 15,000 Ns. That is, the momentum or velocity increases as longas Jx increases. When Jx increases no more, the speed will be a maximum. This happens at t = 30 s. At this time,
mvfx = mvix + 15,000 N s ⇒ ( ) = ( )( ) +425 kg 425 kg 75 m/s 15,000 N sfv x ⇒ =v xf 110.3 m/s
9.27. Model: Let the system be ball + racket. During the collision of the ball and racket, momentum isconserved because all external interactions are insignificantly small.Visualize:
Solve: (a) The conservation of momentum equation pfx = pix is
m v m v m v m vx x x xR f R B f B R i R B i B( ) + ( ) = ( ) + ( )1.0 kg 0.06 kg 40 m/s 1.0 kg 10 m/s 0.06 kg m/sf R
( )( ) + ( )( ) = ( )( ) + ( ) −( )v x 20 ⇒ ( ) =v xf R m/s6 4.
(b) The impulse on the ball is calculated from p p Jx B x B xf i( ) = ( ) + as follows:
0.06 kg m/s 0.06 kg m/s( )( ) = ( ) −( ) +40 20 Jx ⇒ = = =∫J Fdt F tx 3.6 N s avg∆
⇒ Favg
3.6 Ns
10 ms360 N= =
Let us now compare this force with the ball’s weight w m gB B20.06 kg m/s 0.588 N= = ( )( ) =9 8. . Thus, Favg = 612 wB.
Assess: This is a significant force and is reasonable because the impulse due to this force changes the direction aswell as the speed of the ball from approximately 45 mph to 90 mph.
9.28. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. Wewill also use constant-acceleration kinematic equations.Visualize:
Solve: To find the ball’s velocity just before and after it hits the floor:
v v a y y v
v v a y y v v
y y y y
y y y y y
12
02
1 0 1
32
22
3 2 22
2
2 0 2 9 8 0 6 261
2 0 2 9 8 1 5 0 5
= + −( ) = + −( ) −( ) ⇒ = −
= + −( ) ⇒ = + −( ) −( ) ⇒ =
m /s m/s 2.0 m m/s
m /s m/s m m
2 2 2
2 2 2
. .
. . .422422 m/s
The force exerted by the floor on the ball can be found from the impulse-momentum theorem:
mv mv Fdt mvy y y2 1 1= + = +∫ area under the force curve
⇒ ( )( ) = −( )( ) + ×( )−0.2 kg m / s 0.2 kg m / s s5 422 6 261 5 1012
3. . maxF
⇒ =Fmax 935 N
Assess: A force of 935 N exerted by the floor is reasonable.
9.29. Model: Model the rubber ball as a particle that is subjected to an impulsive force when it comes in contactwith the floor. We will also use constant-acceleration kinematic equations and the impulse-momentum theorem.Visualize:
Solve: (a) To find the magnitude and direction of the impulse that the floor exerts on the ball, we use the impulse-momentum theorem:
p p Jy y yf i= + ⇒ = − = −( )J p p m v vy y y y y2 1 2 1
Let us now find v1y and v2y by using kinematics. For the falling and rebounding ball,
v v a y y v
v v a y y v v
y y y y
y y y y y
12
02
1 0 1
32
22
3 2 22
2
2 0 2 9 8 0 5 940
2 0 2 9 8 0 4 850
= + −( ) = + −( ) −( ) ⇒ = −
= + −( ) ⇒ = + −( ) −( ) ⇒ =
m /s m/s m 1.8 m m/s
m /s m/s 1.2 m m m
2 2 2
2 2 2
. .
. . //s
Going back to the impulse-momentum equation, we find
Jy = ( ) − −( )[ ]0.040 kg m/s m/s4 850 5 940. . = 0.432 N s
The impulse is 0.432 N s upward.(b) As the ball compresses, the force of contact increases and the ball slows to vy = 0 m/s. Then in decompressionthe ball is accelerated upward. To a good approximation, the force due to the floor as a function of time is shown inthe figure.(c) For a rubber ball, ∆t is likely in the range 5 to 10 ms. For 10 ms,
J F t Fy = = ⇒ ≈avg avg0.432 N s 4 N∆ 0
The force is ≈8 N0 if ∆t = 5 ms. Altogether, 40 to 80 N is a reasonable estimate.
9.30. Model: Model the cart as a particle rolling down a frictionless ramp. The cart is subjected to an impulsiveforce when it comes in contact with a rubber block at the bottom of the ramp. We will use the impulse-momentumtheorem and the constant-acceleration kinematic equations.Visualize:
Solve: From the free-body diagram on the cart, Newton’s second law “before the collision” is
F w max x( ) = =∑ sinθ ⇒ = = ° =amg
mg
gx
sinsin
θ30
2
Using this acceleration, we can find the cart’s speed just before its contact with the rubber block:
v v a x xx x x12
02
1 02= + −( ) = + −0 2 1 0 012 m /s m m2 2 ( )( . )g ⇒ =v x1 3 130. m/s
Now we can use the impulse-momentum theorem to obtain the velocity just after the collision:
mv mv F dt mvx x x x2 1 1= + = +∫ area under the force graph
⇒ ( ) = ( )( ) − ( ) ×( )−0 5 3 130212. . kg 0.5 kg m/s 200 N 26.7 10 s3v x ⇒ = −v x2 2 21. m/s
Note that the given force graph is positive, but in this coordinate system the impulse of the force is to the left up the
slope. That is the reason to put a minus sign while evaluating the F dtx∫ integral.
We can once again use a kinematic equation to find how far the cart will roll back up the ramp:
v v a x xx x x32
22
3 22= + −( ) ⇒ ( ) = −( ) + −( ) −( )0 2 21 22 2 1
2 3 2 m/s m/s. g x x ⇒ −( ) =x x3 2 0.50 m
9.31. Model: Use the particle model for the ball of clay (C) and the 1.0 kg block (B). The two objects are asystem and it is a case of a perfectly inelastic collision. Since no significant external forces act on the system in thex-direction during the collision, momentum is conserved along the x-direction.Visualize:
Solve: (a) The conservation of momentum equation pfx = pix is
1.0 kg 0.02 kg 0.02 kg m/s 1.0 kg m/sf+( ) = ( )( ) + ( )( )v x 30 0 ⇒ =v xf m/s0 588.
The impulse of the ball of clay on the block is calculated as follows:
p p Jx x xf B i B B( ) = ( ) + ( ) ⇒ ( ) = ( ) − ( )J m v m vx x xB B f B B i B= ( )( ) −1.0 kg N sfv x 0 = 0.588 N s
(b) The impulse of the block on the ball of clay is calculated as follows:
p p Jx x xf C i C C( ) = ( ) + ( ) ⇒ ( ) = − ( )J m v m vx x xC C f C i C= ( )( ) − ( )( )0.02 kg m/s 0.02 kg m/s0 588 30. = − 0.588 N s
(c) Yes, (Jx)B = − (Jx)C.Assess: During the collision, the ball of clay and the block exert equal and opposite forces on each other for thesame time. Impulse is therefore also equal in magnitude but opposite in direction.
9.32. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfectlyinelastic collisions. Momentum is conserved in the collisions of this problem.Visualize:
Solve: In the collision between the three-car train and the single car:
mv m v mvx x x1 + ( ) =3 42 3 ⇒ + =v v vx x x1 2 33 4 ⇒ ( ) + ( ) =4 0 3 2 0 4 3. . m/s m/s v x ⇒ =v x3 2 5. m/s
In the collision between the four-car train and the stationary car:
4 5m v mv m vx x x( ) + = ( )3 4 5 ⇒ + =4 0 53 5v vx x m/s ⇒ = = ( )( ) =vv
xx
534
50 8 2 5 2 0. . . m/s m/s
9.33. Model: Model the earth (E) and the asteroid (A) as particles. Earth+asteroid is our system. Since the twostick together during the collision, this is a case of a perfectly inelastic collision. Momentum is conserved in thecollision since no significant external force acts on the system.Visualize:
Solve: (a) The conservation of momentum equation pfx = pix is
m v m v m m vx x xA i A E i E A E f( ) + ( ) = +( )
⇒ ×( ) ×( ) + = × + ×( )1.0 10 kg 4 10 m s kg m / s 1.0 10 kg 5.98 10 kg13 4 13 24f0 v x ⇒ = × −v xf m / s6 7 10 8.
(b) The speed of the earth going around the sun is
vr
TE
m
s m / s= =
×( )×
= ×2 2 1 50 10
3 15 103 0 10
11
74π π .
..
Hence, v vxf E = × = ×− −2 10 2 1012 10%.Assess: The earth’s recoil speed is insignificant compared to its orbital speed because of its large mass.
9.34. Model: Model the skaters as particles. The two skaters, one traveling north (N) and the other travelingwest (W), are a system. Since the two skaters hold together after the “collision,” this is a case of a perfectlyinelastic collision in two dimensions. Momentum is conserved since no significant external force in the x-y planeacts on the system in the “collision.”Visualize:
Solve: (a) The x-component of the conservation of momentum is
m m v m v m vx x xN W f N i N W i W+( ) = ( ) + ( ) ⇒ +( ) = + ( ) −( )75 kg 60 kg kg m / s 60 kg m / sfv x 0 3 5. ⇒ = −v xf m / s1 556.
The y-component of the conservation of momentum is
m m v m v m vy y yN W f N i N W i W+( ) = ( ) + ( ) ⇒ +( ) = ( )( ) +75 kg 60 kg 75 kg m / s kg m / sfv y 2 5 0.
⇒ =v yf m / s1 389. ⇒ = ( ) + ( ) =v v vx yf f f m / s2 2
2 085.
The time to glide to the edge of the rink is
radius of the rink m
2.085 m / sfv= 25
= 12.0 s
(b) The location is θ = = °−tan .1 41 8v vy xf f north of west.
Assess: A time of 12.0 s in covering a distance of 25 m at a speed of ≈2 m/s is reasonable.
9.35. Model: This problem deals with a case that is the opposite of a collision. The two ice skaters, heavier andlighter, will be modeled as particles. The skaters (or particles) move apart after pushing off against each other.During the “explosion,” the total momentum of the system is conserved.Visualize:
Solve: The initial momentum is zero. Thus the conservation of momentum equation pfx = pix is
m v m vx xH f H L f L kg m / s( ) + ( ) = 0 ⇒ ( )( ) + ( )( ) =75 kg 50 kg kg m / sf H f L
v vx x 0
Using the observation that the heavier skater takes 20 s to cover a distance of 30 m, we find v xf H30 m 20 s( ) = =
1 5. m / s . Thus,
75 kg m / s 50 kg kg m / sf L( )( ) + ( )( ) =1 5 0. v x ⇒ ( ) = −v xf L
m / s2 25.
Thus, the time for the lighter skater to reach the edge is
30 m 30 m
2.25 m / s13.3 s
f Lv x( ) = =
Assess: Conservation of momentum leads to a higher speed for the lighter skater, and hence a shorter time toreach the edge of the ice rink.
9.36. Model: This problem deals with a case that is the opposite of a collision. Our system is comprised ofthree coconut pieces that are modeled as particles. During the blow up or “explosion,” the total momentum of thesystem is conserved in the x-direction and the y-direction.Visualize:
Solve: The initial momentum is zero. From pfx = pix, we get
+ ( ) + ( ) =m v m vx1 1 3 30f f kg m / scosθ ⇒ ( ) =
− ( )=
− −( ) =vm v
m
m
mx
ff m / s
2 m / s
3
1 1
3
2010cosθ
From pfx = pix, we get
+ ( ) + ( ) =m v m vy2 2 3 30f f kg m / ssinθ ⇒ ( ) =
− ( )=
− −( ) =vm v
m
m
m
y
f
f m / s
2 m / s
3
2 2
3
2010sinθ
⇒ ( ) = ( ) + ( ) =vf 10 m / s 10 m / s m / s3
2 2 14 1. θ = ( ) = °−tan 1 1 45
9.37. Model: The billiard balls will be modeled as particles. The two balls, m1 (moving east) and m2 (movingwest), together are our system. This is an isolated system because any frictional force during the brief collisionperiod is going to be insignificant. Within the impulse approximation, the momentum of our system will beconserved in the collision.Visualize:
Note that m1 = m2 = m.Solve: The equation pfx = pix yields:
m v m v m v m vx x x x1 1 2 2 1 1 2 2f f i i( ) + ( ) = ( ) + ( ) ⇒ ( ) + = ( ) + ( )m v m v m vx x1 1 1 1 2 20f i i kg m / scosθ
⇒ ( ) = ( ) + ( )v v vx xf i i1 1 2cosθ = − =2 0 1 0 1 0. . . m / s m / s m / s
The equation pfy = piy yields:
+ ( ) + ( ) =m v m vy y1 1 2 20f f kg m / ssinθ ⇒ ( ) = −( ) = −v v yf f m / s
1 21 41sin .θ
⇒ ( ) = ( ) + −( ) =vf m / s m / s m / s1
2 21 0 1 41 1 73. . .
θ =
= °−tan .1 54 71.41 m / s
1.0 m / s
The angle is below +x axis.
9.38. Model: This is a two part problem. First, we have an inelastic collision between the wood block and thebullet. The bullet and the wood block are an isolated system. Since any external force acting during the collision isnot going to be significant, the momentum of the system will be conserved. The second part involves the dynamicsof the block+bullet sliding on the wood table. We treat the block and the bullet as particles.Visualize:
Solve: The equation pfx = pix gives
m m v m v m vx x xB W f i B W i W+( ) = ( ) + ( )B
⇒ +( ) = ( )( ) + ( )( )0.01 kg 10 kg 0.01 kg 10.0 kg m / sf i Bv vx x 0 ⇒ = ( )v vx xf i B
1
1001From the model of kinetic friction,
f n m m g m m axk k k B W B W= − = − +( ) = +( )µ µ ⇒ = −a gx µk
Using the kinematic equation v v a x xx x x12
22
1 02= + −( ) ,
v v g x xx x12
22
1 02= − −( )µk ⇒ = −0 221 m s2 2
f kv g xx µ ⇒
( ) =1
10012
22
1v g xxi B kµ
⇒ ( ) = = ( )( )( )v g xxi B k2 m / s 0.05 m1001 2 1001 2 0 2 9 81µ . . = 443 m/s
Assess: The bullet’s speed is reasonable (≈900 mph).
9.39. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B).Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since nosignificant external force acts on the system. The second part involves the dynamics of the Fred+Brutus systemsliding on the ground.Visualize:
Note that the collision is head-on and therefore one-dimensional.Solve: The equation pfx = pix is
m m v m v m vx x xF B f F i F B i B+( ) = ( ) + ( ) ⇒ +( ) = ( ) −( ) + ( )( )60 kg 120 kg 60 kg m / s 120 kg m / sfv x 6 0 4 0. .
⇒ =v xf m / s0 667.
The positive value indicates that the motion is in the direction of Brutus.The model of kinetic friction yields:
f n m m g m m a a gx xk k k F B F B k= − = − +( ) = +( ) ⇒ = −µ µ µ
Using the kinematic equation v v a x xx x x12
02
1 02= + −( ) , we get
v v g x v x
x x
x x x12
02
12
1
21 1
2 0 2 0 30 9 8
0 0 667 5 88 7 57
= − ⇒ = − ( )( )⇒ = ( ) − ( ) ⇒ =
µk2 2
f2
2 2 2
m s m / s
m s m / s m / s cm
. .
. . .
They slide 7.57 cm in the direction Brutus was running.Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. Astopping distance of 7.57 cm is reasonable.
9.40. Model: Model the package and the rocket as particles. This is a two-part problem. First we have aninelastic collision between the rocket (R) and the package (P). During the collision, momentum is conserved sinceno significant external force acts on the rocket and the package. However, as soon as the package+rocket systemleaves the cliff they become a projectile motion problem.Visualize:
Solve: The minimum velocity after collision that the package+rocket must have to reach the explorer is v0x, whichcan be found as follows:
y y v t t a t ty y1 0 0 1 012 1 0
2= + −( ) + −( ) ⇒ − = + + −( )200 m . m m m / s20 0 9 812 1
2t ⇒ =t1 6.389 s
With this time, we can now find v0x using x x v t t a t tx x1 0 0 1 012 1 0
2= + −( ) + −( ) . We obtain
30 m 0 m 6.389 s m= + ( ) +v x0 0 ⇒ = =v vx x0 4 696. m / s f
We now use the momentum conservation equation pfx = pix which can be written
m m v m v m vx x xR P f R i R P i P+( ) = ( ) + ( )
⇒ +( )( ) = ( )( ) + ( )( )1.0 kg 5.0 kg m / s 1.0 kg 5.0 kg m / si R4 696 0. v x ⇒ ( ) =v xi R
m / s28 2.
9.41. Model: This is a two-part problem. First, we have an explosion that creates two particles. The momentumof the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equationsand the model of kinetic friction to find the displacement of the lighter fragment.Visualize:
Solve: The initial momentum is zero. Using momentum conservation pfx = pix during the explosion,
m v m v m v m vx x x xH H L L H H L L1 1 0 0( ) + ( ) = ( ) + ( ) ⇒ ( ) + ( ) =7 kg m / sH L
m v m vx x1 1 0 ⇒ v vx x117 1( ) = −( )( )H L
Because mH slides to x2H = −8.2 m before stopping, we havef n w m g m ak k H k H k H H H= = = =µ µ µ ⇒ =a gH kµ
Using kinematics,
v v a x xx x2
2
1
22( ) = ( ) + −( )
H H H 2H 1H ⇒ = ( ) ( ) + − −( )0 2 017
2
1
2 m s 8.2 m m2 2
L kv gx µ ⇒ ( ) = −v x1 88 74L k m / s. µ
How far does mL slide? Using the information obtained above in the following kinematic equation,
v v a x xx x2
2
1
22( ) = ( ) + −( )
L L L 2L 1L ⇒ = ( ) −0 88 74 22 m s2 2k k 2Lµ µ. gx ⇒ =x2L 402 m
Assess: Note that aH is positive, but aL is negative, and both are equal in magnitude to µkg. Also, x2H is negativebut x2L is positive.
9.42. Model: We will model the two fragments of the rocket after the explosion as particles. We assume theexplosion separates the two parts in a vertical manner. This is a three-part problem. In the first part, we will usekinematic equations to find the vertical position where the rocket breaks into two pieces. In the second part, we willapply conservation of momentum to the system (that is, the two fragments) in the explosion. In the third part, wewill again use kinematic equations to find the velocity of the heavier fragment just after the explosion.Visualize:
Solve: The rocket accelerates for 2.0 s from rest, so
v v a t ty y y1 0 1 0 0 10 20= + −( ) = + ( ) −( ) =m / s m / s 2 s 0 s m / s2
y y v t t a t ty y1 0 0 1 012 1 0
2 12
20 0 10= + −( ) + −( ) = + + ( )( ) =m m m / s 2 s 20 m2
At the explosion the equation pfy = piy is
m v m v m m vy y yL L H H L H2 2 1( ) + ( ) = +( ) ⇒ ( )( ) + ( )( ) = ( )( )500 kg 1000 kg 1500 kg 20 m / sL H
v vy y2 2
To find (v2y)H we must first find (v2y)L, the velocity after the explosion of the upper section. Using kinematics,
v v y yy y3
2
2
22 9 8( ) = ( ) + −( ) −( )
L L
23L 2L m / s. ⇒ ( ) = ( ) −( ) =v y2 2 9 8 99 98
L
2 m / s 530 m 20 m m / s. .
Now, going back to the momentum conservation equation we get
500 kg m / s 1000 kg 1500 kg m / sH
( )( ) + ( )( ) = ( )( )99 98 202. v y ⇒ ( ) = −v y2 20 0H
m / s.
The negative sign indicates downward motion.
9.43. Model: Let the system be bullet + target. No external horizontal forces act on this system, so thehorizontal momentum is conserved. Model the bullet and the target as particles. Since the target is much moremassive than the bullet, it is reasonable to assume that the target undergoes no significant motion during the briefinterval in which the bullet passes through.Visualize:
Solve: (a) By assuming that the target has negligible motion during the interval in which the bullet passesthrough, the time is that needed to slow from 1200 m/s to 900 m/s in a distance of 30 cm. We’ll use kinematics tofirst find the acceleration, then the time.
( ) ( )
( ) ( ) ( (
( ..
( ) ( )
( ) ( )
v v a x
av v
x
v v a t
tv v
a
x x x
xx x
x x x
x x
x
1 0
1 0 6
1 0
1 0
2
2
900 1200
2 0 301 05 10
B2
B2
B2
B2 2 2
2
B B
B B
m / s) m / s)
m) m / s
= + ∆
⇒ = −∆
= − = − ×
= + ∆
⇒ ∆ = − = 900900 1 00
1 05 102 86 10 2866
4 m / s 2 m / s
m / s s s2
−− ×
= × =−
.. µ
The average force on the bullet is Favg = m|ax| = 26,200 N.(b) Now we can use the conservation of momentum equation p1x = p0x to find
m v m v m v m v m v
vm v m v
m
x x x x x
xx x
T T B B T T B B B B
TB B B B
T
kg
350 kg m / s) m / s) m / s
( ) ( ) ( ) ( ) ( )
( )( ) ( ) .
( ( .
1 1 0 0 0
10 1
0
0 0251200 900 0 0214
+ = + = +
⇒ = − = −( ) =
9.44. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First,we have a collision between the bullet and the first block (A). Momentum is conserved since no external force actson the system (bullet+block A). The second part of the problem involves a perfectly inelastic collision between thebullet and block B. Momentum is again conserved for this system (bullet+block B).Visualize:
Solve: For the first collision the equation pfx = pix is
m v m v m v m vx x x xL L A A L L A A1 1 0 0( ) + ( ) = ( ) + ( )
⇒ ( )( ) + ( )( ) = ( )( ) +0.01 kg 0.500 kg m / s 0.01 kg m / s kg m / sL
v x1 6 400 0 ⇒ ( ) =v x1 100L
m / s
The bullet emerges from the first block at 100 m/s. For the second collision the equation pfx = pix is
m m v m vx xL B L L+( ) = ( )2 1 ⇒ +( ) = ( )( )0.01 kg 0.5 kg 0.01 kg m / sv x2 100 ⇒ =v x2 1 96. m / s
9.45. Model: Model Brian (B) along with his wooden skis as a particle. The “collision” between Brian andAshley lasts for a short time, and during this time no significant external forces act on the Brian+Ashley system.Within the impulse approximation, we can then assume momentum conservation for our system. After finding thevelocity of the system immediately after the collision, we will apply constant-acceleration kinematic equations andthe model of kinetic friction to find the final speed at the bottom of the slope.Visualize:
Solve: Brian skiing down for 100 m:
v v a x xx x x1
2
0
22( ) = ( ) + −( )B B 1B 0B = + −( )0 22 m s 100 m 0 m2 ax ⇒ ( ) = ( )v ax x1 B
200 m
To obtain ax, we apply Newton’s second law to Brian in the x and y directions as follows:
F w f m ax xon B B k B( ) = − =∑ sinθ F n w
yon B B N( ) = − =∑ cosθ 0 ⇒ =n wcosθ
From the model of kinetic friction, f n wk k k B= =µ µ θcos . The x-equation thus becomes
w w m axB k B Bsin cosθ µ θ− =
⇒ = −( )a gx sin cosθ µ θk = ( ) ° − ( ) °[ ]9 8 20 0 06 20. sin . cos m / s2 = 2 799. m / s2
Using this value of ax, v x1 2 799 23 66( ) = ( )( ) =B
2200 m m / s m / s. . . In the collision with Ashley the conservation
of momentum equation pfx = pix is
m m v m vx xB A B B+( ) = ( )2 1 ⇒ =
+( ) =
+( )v
m
m mvx x2 1
B
B AB
80 kg
80 kg 50 kg23.66 m / s = 14 56. m / s
Brian+Ashley skiing down the slope:
v v a x xx x x32
22
3 22= + −( ) = ( ) + ( )( )14.56 m/ s 2 2.799 100 m2 m / s 2 ⇒ =v x3 27 8. m / s
That is, Brian+Ashley arrive at the bottom of the slope with a speed of 27.8 m/s. Note that we have used the samevalue of ax in the first and the last parts of this problem. This is because ax is independent of mass.Assess: A speed of approximately 60 mph on a ski-slope of 200 m length and 20° slope is reasonable.
9.46. Model: Model the spy plane (P) and the rocket (R) as particles. The plane and the rocket undergo aperfectly inelastic collision. During the brief collision time, momentum for the plane+rocket system will beconserved because no significant external forces act on this system during the collision. After finding the velocityof the system immediately after the collision, we will apply projectile equations to find where the system will hitthe ground.Visualize:
Solve: For the collision we have pfx = pix and pfy = piy. This means the magnitude of the final momentum is
p p p p px y x yf f f i i= ( ) + ( ) = ( ) + ( )2 2 2 2⇒ +( ) = ( )[ ] + ( )[ ]m m v m v m vx yR P f R i R P i P
2 2
⇒ =+( ) ( )( )[ ] + ( )( )[ ]vf 1280 kg 575 kg
1280 kg m / s 575 kg m / s1
725 4502 2
= 519 4. m / s
θ =
= ( )( )( )( )
= °− −tan tan .1 1 15 6
p
py
x
f
f
575 kg 450 m / s
1280 kg 725 m / s north of east
We call this direction the x′ axis in the x-y plane. Let us now look at the falling plane+rocket system in the x′-zplane where z is the vertical axis perpendicular to the x-y plane. We have
z z x t0 1 0 02700 0 0 0= = ′ = = m m m s
v vz x0 00 519 4= =′ m / s m / s.
Using the kinematic equation z z v t t a t tz z1 0 0 1 012 1 0
2= + −( ) + −( ) , we can find the time to fall:
0 270 0 9 812 1
2m 0 m m m / s2= + + −( ). t ⇒ t1 = 23.47 s
In this time t1, the wreckage travels horizontally to
′ = ′ + −( )′x x v t tx1 0 0 1 0 = + ( )( )0 m 519.4 m / s 23.47 s = 12,192 m = 12.2 km
The enmeshed plane+rocket system lands 12.2 km from the collision point at an angle of 15.6° north of east.
9.47. Model: Model the bullet and the vehicle as particles, and use the impulse-momentum theorem to find theimpulse provided to the vehicle by bullet(s). Because the final speed of the vehicle is small compared to the bulletspeed, and the mass of the bullet is so much smaller than the mass of the vehicle, we will assume that each bulletexerts the same impulse on the vehicle.Visualize:
Solve: For one bullet, the impulse-momentum theorem ∆px = Jx allows us to find that the impulse exerted on thebullet by the vehicle’s sail is
( ) ( ) ( . ( ( .J m vx B B B kg) m / s) m / s) kg m / s= ∆ = − −( ) = −0 020 200 400 12 0
From Newton’s third law, the impulse that the bullet exerts on the vehicle is (Jx)V = –(Jx)B = +12.0 kg m/s. A seconduse of the impulse momentum theorem allows us to find the vehicle’s increase in velocity due to one bullet:
( )( ) .
.∆ = = =vJ
mx
VV
V
kg m / s
100 kg m / s
12 00 120
This increase in velocity is independent of the vehicle’s speed, so as long as the impulse per bullet stays essentiallyconstant (which it does because the bullet speeds are so much larger than the vehicle speed), the number of impactsneeded to increase the vehicle speed to 12 m/s is
Nbullets
m / s
0.12 m / s per bullet bullets= =12
100
To reach this speed in 20 s requires the bullets to be fired at a rate of 5 bullets per second.
9.48. Model: Choose the system to be cannon + ball. There are no significant external horizontal forces duringthe brief interval in which the cannon fires, so within the impulse approximation the horizontal momentum isconserved. We’ll ignore the very small mass loss of the exploding gunpowder.Visualize:
Solve: The statement that the ball travels at 200 m/s relative to the cannon can be written (v1x)B = (v1x)C + 200 m/s.That is, the ball’s speed is 200 m/s more than the cannon’s speed. The initial momentum is zero, so theconservation of momentum equation p1x = p0x is
p m v m v m v m v
vm
m m
x x x x x
x
1 1 1 1 1
1
200 0
20010
200 3 9
= + = + +( ) =
⇒ = −+
= − = −
C C B B C C B C
CB
C B
m / s kg m / s
m / s kg
510 kg m / s m / s
( ) ( ) ( ) ( )
( ) ( ) ( ) .
That is, the cannon recoils to the left (negative sign) at 3.9 m/s. Thus the cannonball’s speed relative to the groundis (v1x)B = –3.9 m/s + 200 m/s = 196 m/s.
9.49. Model: This is an isolated system, so momentum is conserved in the explosion. Momentum is a vectorquantity, so the direction of the initial velocity vector
rv1 establishes the direction of the momentum vector. The
final momentum vector, after the explosion, must still point in the +x-direction. The two known pieces continue tomove along this line and have no y-components of momentum. The missing third piece cannot have a y-componentof momentum if momentum is to be conserved, so it must move along the x-axis – either straight forward orstraight backward. We can use conservation laws to find out.Visualize:
Solve: From the conservation of mass, the mass of piece 3 is
m m m m3 1 2= − − = ×total57.0 10 kg
To conserve momentum along the x-axis, we require
p m v p p p p m v m v pi total i f 1f 2f 3f 1f 2f 3f=[ ] = = + + = + +[ ]1 2
⇒ = − − = + ×p m v m v m v3f total i 1f 2f kg m / s1 2131 02 10.
Because p3f > 0, the third piece moves in the +x-direction, that is, straight forward. Because we know the mass m3,we can find the velocity of the third piece as follows:
vp
m3f3f
5
kg m / s
7.0 10 kg m / s= = ×
×= ×
3
1371 02 10
1 46 10.
.
9.50. Model: Model the proton (P) and the gold atom (G) as particles. The two constitute our system, andmomentum is conserved in the collision between the proton and the gold atom.Visualize:
Solve: The conservation of momentum equation pfx = pix is
m v m v m v m vx x x xG f G P f P P i P G i G( ) + ( ) = ( ) + ( )
⇒ ( )( ) + ( ) − × ×( ) = ( ) ×( ) +197 u 1 u m / s 1 u m / s u m / sf Gv x 0 90 5 0 10 5 0 10 07 7. . .
⇒ ( ) = ×v xf G m / s4 82 105.
9.51. Model: Model the proton (P) and the target nucleus (T) as particles. The proton and the target nucleusmake our system and in the collision between them momentum is conserved. This is due to the impulseapproximation because the collision lasts a very short time and the external forces acting on the system during thistime are not significant.Visualize:
Solve: The conservation of momentum equation pfx = pix is
m v m v m v m vx fx x xT f T P P T i T P i P( ) + ( ) = ( ) + ( )
⇒ ×( ) + ( ) − × ×( ) = + ( ) ×( )mT m / s 1 u m / s u m / s 1 u m / s3 12 10 0 75 2 5 10 0 2 5 105 6 6. . . .
⇒ mT = 14 u
Assess: This is the mass of the nucleus of a nitrogen atom.
9.52. Model: This problem deals with a case that is the opposite of a collision. It is a case of an “explosion” inwhich a 214Po nucleus (P) decays into an alpha-particle (A) and a daughter nucleus (N). During the “explosion” ordecay, the total momentum of the system is conserved.Visualize:
Solve: Conservation of mass requires the daughter nucleus to have mass mN = 214 u – 4 u = 210 u. Theconservation of momentum equation pfx = pix is
m v m v m m vx xN f N A f A N A P( ) + ( ) = +( ) ⇒ ( )( ) + ( ) − ×( ) =210 u 4 u m s u m / sf Nv x 1 92 10 07. /
⇒ ( ) = ×v xf N m / s3 66 105.
9.53. Model: The neutron’s decay is an “explosion” of the neutron into several pieces. The neutron is anisolated system, so its momentum should be conserved. The observed decay products, the electron and proton,move in opposite directions.Visualize:
Solve: (a) The initial momentum is pix = 0 kg m/s. The final momentum p m v m vxf e e p p= + is
2 73 10 1 67 1023 22. .× − ×− − kg m / s kg m / s = − × −1 40 10 22. kg m / s
No, momentum does not seem to be conserved.(b) and (c) If the neutrino is needed to conserve momentum, then p p pe p+ + =neutrino kg m / s0 . This requires
p p pe pneutrino kg m / s= − +( ) = + × −1 40 10 22.
The neutrino must “carry away” 1.40 × 10 − 22 kg m/s of momentum in the same direction as the electron.
9.54. Model: Model the two balls of clay as particles. Our system comprises these two balls. Momentum isconserved in the perfectly inelastic collision.Visualize:
Solve: The x-component of the final momentum is
pfx = pix = ( ) + ( )m v m vx x1 1 2 2i i
= ( )( ) − ( )( ) °0.020 kg m / s 0.030 kg 1.0 m / s2 0 30. cos = 0.0140 kg m / s
The y-component of the final momentum is
pfy = piy = ( ) + ( )m v m vy y1 1 2 2i i
= ( )( ) − ( )( ) °0.02 kg m / s 0.03 kg m / s0 1 0 30. sin = −0.0150 kg m / s
⇒ = ( ) + −( ) =pf 4 kg m / s 0.015 kg m / s 0.0205 kg m / s0 012 2
.
Since p m m vf f m / s= +( ) =1 2 0 0205. kg , the final speed is
vf
0.0205 kg m / s
0.02 0.03 kg0.410 m / s=
+( )=
and the direction is
θ = = = °− −tan tan.
..1 1 0 015
0 01447 0
p
p
y
x
f
f
south of east
9.55. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3(moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum ofthe system is conserved.Visualize:
Solve: The three initial momenta arer rp m v j ji1 i1 0.02 kg 2 m / s 0.04 kg m / s= = ( )( ) =1
ˆ ˆ
r rp m v i ii2 i2 0.03 kg m / s 0.09 kg m / s= = ( ) −( ) = −2 3 ˆ ˆ
r rp m v i ji3 i3 0.04 kg m / s m / s= = ( ) ( ) ° − ( ) °[ ]3 4 45 4 45cos ˆ sin ˆ = −( )0 113 0 113. ˆ . ˆi j kg m / s
Since r r r rp p p p pf i i1 i2 i3= = + + , we have
m m m v i j1 2 3 0 023 0 073+ +( ) = −( )rf kg m / s. ˆ . ˆ ⇒ r
v i jf 6 m / s= −( )0 25 0 811. ˆ . ˆ
⇒ = ( ) + −( ) =vf m / s m / s m / s0 256 0 811 0 8502 2. . .
θ = = = °− −tan tan.
..1 1 0 811
0 25672 5
v
v
y
x
f
f
below +x
9.56. Model: Model the truck (T) and the two cars (C and C′) as particles. The three forming our system sticktogether during their collision, which is perfectly inelastic. Since no significant external forces act on the systemduring the brief collision time, the momentum of the system is conserved.Visualize:
Solve: The three momenta arer rp m v i iiT T iT 2100 kg 2 m / s 4200 kg m / s= = ( )( ) =ˆ ˆ
r rp m v j jiC C iC 1200 kg 5 m / s 6000 kg m / s= = ( )( ) =ˆ ˆ
r rp m v i iiC C iC 1500 kg 10 m / s 15,000 kg m / s′ ′ ′= = ( )( ) =ˆ ˆ
r r r r rp p p p pf i iT iC iC= = + + ′ = +( )19,200 6000 kg m / sˆ ˆi j
⇒ = + +( ) = ( ) + ( )′p m m m vf T C C f 19 200 kg m / s 6000 kg m / s,2 2
⇒ =vf m / s4 19. θ = = = °− −tan tan,
.1 1 6000
19 20017 4
p
py
x
above +x
Assess: A speed of 4.19 m/s for the entangled three vehicles is reasonable since the individual speeds of the carsand the truck before entanglement were of the same order of magnitude.
9.57. Model: The 14C atom undergoes an “explosion” and decays into a nucleus, an electron, and a neutrino.Momentum is conserved in the process of “explosion” or decay.Visualize:
Solve: The conservation of momentum equation r rp pf i= = 0 kg m/s is
r r rp p pe n N N+ + = 0 ⇒
r r rp p pN e n= − +( ) = − −m v m ve e n n
r r
= − ×( ) ×( ) − ×( )− −9 11 10 31. ˆ ˆ kg 5 10 m / s 8.0 10 kg m / s7 24i j = − × + ×( )− −45.55 10 8.0 10 kg m / s24 24ˆ ˆi j
⇒ = = ×( ) + ×( )− −p m vN N N kg m / s45 55 10 8 0 1024 2 24 2. .
⇒ 2 34 10 4 62 1026 23. .×( ) = ×− − kg kg m / sNv ⇒ = ×vN m / s1 97 103.
9.58. Model: The angular momentum of a particle in circular motion is conserved if there is no net tangentialforce on the particle. In the present case, the block of dry ice is treated as a particle. In the absence of external
forces, including that of friction, angular momentum rL is conserved.
Visualize:
Solve: We have L Lf i= . This means
mv r m v r0
02 f i= ⇒ = = ( ) =v vf i 2.0 m / s 4.0 m / s2 2
Assess: An increase in speed to 4.0 m/s (from 2.0 m/s) is reasonable as half the dry ice disappears.
9.59. Model: Because there’s no friction or other tangential forces, the angular momentum of the block + rodsystem is conserved.Visualize:
Solve: The rod is massless, so the angular momentum is entirely that of a mass in circular motion. The conservationof angular momentum equation L Lf i= is
mv r mv rf f i i= ⇒ ( ) = ( )r r r rf f f i i iω ω ⇒ ω ωfi
fi
30 cm
100 cm50 rpm=
=
( )r
r
2 2
= 4.5 rpm
Assess: An angular speed of 4.5 rpm is reasonable, since the angular speed varies inversely as the square of theobject’s distance from the rotation axis.
9.60. Model: The heavier person (H) and the lighter person (L) are our system. The board they are standing onis essentially massless. The angular momentum of the system is constant because no external tangential force actson the system.Visualize: Please refer to Figure P9.60.Solve: The initial angular momentum is zero. The conservation of angular momentum equation L Lf i= = 0 kg m2/s is
L LH L2 kg m s+ = 0 / ⇒ L LH L= −
That is, the angular momenta of the two people must be equal but opposite to keep the total angular momentumzero. The angular momentum of a particle in circular motion is L = mrvt, where vt is the tangential component of
rv .
The lighter person’s jump causes him to move counterclockwise around the fulcrum. Thus, according to the signsthat we established for circular motion in Chapter 7, vL = –1.5 m/s. Consequently,
L L m r v m r v
vm r
m rv
t t
t t
H L H H H L L L
HL L
H HL
kg)(2.4 m)
(75 kg)(1.6 m) m/ s) m / s
= − ⇒ = −
⇒ = − = − − = +
( ) ( )
( ) ( )(
( . .50
1 5 1 5
The positive value for vt indicates that the heavier person moves clockwise around the fulcrum, or verticallyupward. So just after the lighter person jumps, the heavier person is moving upward at 1.5 m/s.Assess: You might think the heavier person would fall downward. But in order to jump, the lighter person mustkick his side of the board downward. That lifts the heavier person. The fact that the speeds are equal is aconsequence of the fact that the initial balance requires mHrH = mLrL.
9.61. Model: Model the puck and the 200 g weights as particles. The puck is in circular motion and the forcesacting on the puck are its weight downward, the radial tension in the string, and a normal force upward. There is notangential force acting on the puck, and thus the angular momentum of the puck is conserved.Visualize:
Solve: (a) For the puck to move in a circle, the force causing the centripetal acceleration is provided by the two200 g weights. Thus,
0.40 kg 9.8 m / s 0.010 kg0.020 m
2puck
i
i
i( )( ) = = ( ) ( )m
v
r
v2 2
⇒ =vi m / s2 80.
(b) The conservation of angular momentum equation L Lf i= implies
m v r m v rpuck f f puck i i= ⇒ v r v rf f i i22.8 m / s 0.20 m m / s= = ( )( ) = 0 56.
Now for the new weight to cause circular motion
m v
rpuck f
f
20.20 kg 9.8 m / s2
= ( )( ) ⇒ = =rm v v
fpuck f
2f
21.96 kg m / s 19.6 m / s
2 2
Substituting this expression into the conservation of angular momentum equation, we have
vv
ff
22
19.6 m / s m / s
2
0 56
= . ⇒ =vf3 3 m / s3 10 976. ⇒ vf m / s= 2 22.
Now we can obtain rf as
rv
ff
2 m / s
2.22 m / s
19.6 m / s0.252 m 25.2 cm= = ( ) = =
2 2
219 6.
9.62. (a) A 100 g ball traveling to the left at 30 m/s is batted back to the right at 40 m/s. The force curve for theforce of the bat on the ball can be modeled as a triangle with a maximum force of 1400 N. How long is the ball incontact with the bat?(b)
(c) The solution is ∆t = 0.010 s = 10 ms.
9.63. (a) A 200 g ball of clay traveling to the right overtakes and collides with a 400 g ball of clay traveling tothe right at 3.0 m/s. The balls stick and move forward at 4.0 m/s. What was the speed of the 200 g ball of clay?(b)
(c) The solution is v xi m / s( ) =2
6 0. .
9.64. (a) A 2000 kg auto traveling east at 5.0 m/s suffers a head-on collision with a small 1000 kg auto travelingwest at 4.0 m/s. They lock bumpers and stick together after the collision. What will be the speed and direction ofthe combined wreckage after the collision?(b)
(c) The solution is v xf m / s= 2 0. along +x direction.
9.65. (a) A 150 g spring-loaded toy is sliding across a frictionless floor at 1.0 m/s. It suddenly explodes into twopieces. One piece, which has twice the mass of the second piece, continues to slide in the forward direction at 7.5m/s. What is the speed and direction of the second piece?(b)
(c) The solution is (vfx)1 = –12 m/s. The minus sign tells us that the second piece moves backward at 12 m/s.
9.66. Model: This is a three-part problem. In the first part, the shell, treated as a particle, is launched as aprojectile and reaches its highest point. We will use constant-acceleration kinematic equations for this part. Theshell, which is our system, then explodes at the highest point. During this brief explosion time, momentum isconserved. In the third part, we will again use the kinematic equations to find the horizontal distance between thelanding of the lighter fragment and the origin.Visualize:
Solve: The initial velocity is
v vx0 125 55 71 7= = ( ) ° =cos cos .θ m / s m / s
v vy0 55 102 4= = ( ) ° =sin sin .θ 125 m / s m / s
At the highest point, v y1 0= m/s and v x1 71 7= . m / s . The conservation of momentum equation p px xf i= is
m v m v m m vx x xL L H L H1 1 1( ) + ( ) = +( )H
The heavier particle falls straight down, so (v1x)H = 0 m/s. Thus,
15 kg kg m s 15 kg 60 kg m / s m / sL L
( )( ) + = +( )( ) ⇒ ( ) =v vx x1 10 71 7 358.
That is, the velocity of the smaller fragment immediately after the explosion is 358 m/s and this velocity is in thehorizontal x-direction. Note that (v1y)L = 0 m/s. To find x2, we will first find the displacement x1 − x0 and then x2 − x1.For x1 − x0,
v v a t t t ty y y1 0 1 0 1 10 9 8= + −( ) ⇒ = ( ) + −( ) −( ) ⇒ = m s 102.4 m / s m / s 0 s 10.45 s2.
x x v t t a t t x xx x1 0 0 1 012 1 0
2
1 0 0 749= + −( ) + −( ) ⇒ − = ( )( ) + =71.7 m / s 10.45 s m m
For x2 − x1:
x x v t t a t t x xx L x2 1 1 2 112 2 1
2
2 1 0= + ( ) −( ) + −( ) ⇒ − = ( )( ) +358 m / s 10.45 s m = 3741 m
⇒ = −( ) + −( ) = +x x x x x2 2 1 1 0 3741 m 749 m = 4490 m
Assess: Note that the time of ascent to the highest point is equal to the time of descent to the ground, that is, t1 – t0
= t2 – t1.
9.67. Model: The cart+man (C+M) is our system. It is an isolated system, and momentum is conserved.Visualize:
Solve: The conservation of momentum equation pfx = pix is
m v m v m v m vx x x xM f M C f C M i M C i C( ) + ( ) = ( ) + ( )Note that (vf)M and (vfx)C are the final velocities of the man and the cart relative to the ground. What is given in thisproblem is the velocity of the man relative to the moving cart. The man’s velocity relative to the ground is
v vx xf M f C m / s( ) = ( ) − 10
With this form for (vfx)M, we rewrite the momentum conservation equation as
m v m v m mx xM f C C f C M C m / s 5 m / s 5 m / s( ) −[ ] + ( ) = ( ) + ( )10
⇒ ( ) ( ) −[ ] + ( )( ) = +( )( )70 kg m / s 1000 kg 1000 kg 70 kg 5 m / sf C f Cv vx x10
⇒ ( ) +[ ] = ( )( ) + ( )( )v xf C1000 kg 70 kg 1070 kg m / s 70 kg 10 m / s5 ⇒ ( ) =v xf C
m / s5 65.
9.68. Model: Model Ann and cart as particles. The initial momentum is pi = 0 kg m/s in a coordinate systemattached to the ground. As Ann begins running to the right, the cart will have to recoil to the left to conserve momentum.Visualize:
Solve: The difficulty with this problem is that we are given Ann’s velocity of 5.0 m/s relative to the cart. If thecart is also moving with velocity vcart, then Ann’s velocity relative to the ground is not 5.0 m/s. Using the Galileantransformation equation for velocity, Ann’s velocity relative to the ground is
v vx xf Ann f cart m/s( ) = ( ) + 5 0.
Now, the momentum conservation equation p px xi f= is
0 kg m / s Ann f Ann cart f cart= ( ) + ( )m v m vx x ⇒ = ( ) ( ) +[ ] + ( )( )0 5 kg m / s 50 kg 0 m / s 500 kgf cart f cart
v vx x.
⇒ ( ) = −v xf cart m/s0 45.
Using the recoil velocity (vfx)cart relative to the ground, we find Ann’s velocity relative to the ground to be
(vfx)Ann = 5.00 m/s − 0.45 m/s = 4.55 m/s
The distance Ann runs relative to the ground is ∆x = (vfx)Ann∆t, where ∆t is the time it takes to reach the end of thecart. Relative to the cart, which is 15 m long, Ann’s velocity is 5 m/s. Thus ∆t = ( ) ( ) =15 m 5 m / s 3.00 s. Herdistance over the ground during this interval is
∆ ∆x v tx= ( )f Ann= ( )( )4 55. m/s 3.00 s = 13.6 m
9.69. Model: The projectile+wood ball are our system. In the collision, momentum is conserved.Visualize:
Solve: The momentum conservation equation p px xf i= is
m m v m v m vx x xP B f P i P B i B+( ) = ( ) + ( ) ⇒ +( ) = ( )( ) +1.0 kg 20 kg 1.0 kg kg m/sf i P
v vx x 0
⇒ ( ) =v vx xi P f21
We therefore need to determine vfx. Newton’s second law for circular motion is
T w T m m gm m v
rx− = − +( ) =
+( )P B
P B f2
Using Tmax = 400 N, this equation gives
400 N . 1.0 kg 20 kg m/s1.0 kg 20 kg
2.0 mf− +( )( ) =
+( )9 8
2v x ⇒ ( ) =v xf m/smax
.4 30
Going back to the momentum conservation equation,
v vx P xi f( ) = 21 = ( )( ) =21 .4 30 m/s 90.3 m/s
That is, the largest speed this projectile can have without causing the cable to break is 90.3 m/s.
9.70. Model: This is an “explosion” problem and momentum is conserved. The two-stage rocket is our system.Visualize:
Solve: Relative to the ground, the conservation of momentum equation p px xf i= is
m v m v m m vx x x1 1 2 2 1 2 1f f( ) + ( ) = +( )⇒ ( ) + ( ) = ( )( )3 m 4 m m/s2 f f 2v m vx x1 2 2
1200 ⇒ ( ) + ( ) =3 48001 2
v vx xf f kg m/s
The fact that the first stage is pushed backward at 35 m/s relative to the second can be written
v vx xf f m/s( ) = − + ( )1 235
Substituting this form of (vfx)1 in the conservation of momentum equation,
3 35 42 2
− + ( )[ ] + ( ) = m/s 800 kg m/sf fv vx x ⇒ ( ) =v xf m/s2
1226
9.71. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved.Visualize: The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf)B = (vf)R +139,000 m/s.
Solve: Consider the firing of one bullet when the rocket has mass M and velocity vi. The conservation of momentumequation pf = pi is
( ( ( ,
( ,
M v v Mv
v v vM
− + + =
⇒ ∆ = − = −
5 5 139 000
5139 000
kg) kg) m / s)
kg m / s)
f f i
f i
The rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M willnot change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed perbullet is ∆v = –347.5 m/s = –1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000km/h to 15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate ∆v for each bulletand reduce M by 5 kg after each shot. The loss of mass causes ∆v to increase slightly for each bullet. An eight-stepcalculation then finds that 8 bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and9 would slow the rocket far too much.
9.72. Visualize:
Solve: Ladies and gentlemen of the jury, how far would the chair slide if it was struck with a bullet from myclient’s gun? We know the bullet’s velocity as it leaves the gun is 450 m/s. The bullet travels only a small distanceto the chair, so we will neglect any speed loss due to air resistance. The bullet and chair can be considered anisolated system during the brief interval of the collision. The bullet embedded itself in the chair, so this was aperfectly inelastic collision. Momentum conservation allows us to calculate the velocity of the chair immediatelyafter the collision as follows:
p px xi f= ⇒ ( ) = +( )m v m m vB i B B C f ⇒ =( )+
= ( )( ) =vm v
m mfB i B
B C
450 m/s 0.010 kg
20.01 kg m/s0.225
This is the velocity immediately after the collision when the chair starts to slide but before it covers any distance.For the purpose of the problem in dynamics, call this the initial velocity v0. The free-body diagram of the chairshows three forces. Newton’s second law applied to the chair (with the embedded bullet) is
a aF
m
f
m
n
mxx= =
( )= − = −net
tot
k
tot
k
tot
µ a
F
m
n m g
myy= =
( )= −
0 m/s2 net
tot
tot
tot
where we’ve used the friction model in the x-equation. The y-equation yields n m g= tot , and the x-equation yieldsa g= − = −µk
2 m/s1 96. . We know the coefficient of kinetic friction because it is a wood chair sliding on a wood floor.Finally, we have to determine the stopping distance of the chair. The motion of the chair ends with v1 = 0 m/s aftersliding a distance ∆x, so
v v a x12
020 2= = + m /s2 2 ∆ ⇒ = − = − ( )
−( ) = =∆xv
a02 2
2 2
0.225 m/s
1.96 m/s0.013 m 1.3 cm
2
If the bullet lost any speed in the air before hitting the chair, the sliding distance would be even less. So you can seethat the most the chair could slide if it had been struck by a bullet from my client’s gun, would be 1.3 cm. But inactuality, the chair slid 3 cm, more than twice as far. The murder weapon, ladies and gentlemen, was a much morepowerful gun than the one possessed by my client. I rest my case.
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