26. A 75.0-kg ice skater, moving at 10.0 m/s, crashes into a stationary skater of equal mass. After the collision,
the two skaters move as a unit at 5.00 m/s. Suppose the average force a skater can experience without breaking a
bone is 4 500 N. If the impact time is 0.100 s, does a bone break?
6.26
For each skater, the impulse-momentum theorem gives
F=
Δp m ( Δv ) ( 75.0 kg )( 5.00 m s )
= 3.75 × 103 N .
=
=
Δt
Δt
0.100 s
Since F < 4500 N , there are no broken bones .
27. A railroad car of mass 2.00 x 104 kg moving at 3.00 m/s collides and couples with two coupled railroad cars,
each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the
three coupled cars after the collision? (b) How much kinetic energy is lost in the collision?
6.27
(a) If M is the mass of a single car, conservation of momentum gives
( 3 M ) v f = M ( 3.00 m s ) + ( 2 M ) (1.20 m s ) , or v f =
1.80 m s
(b) The kinetic energy lost is KElost = KEi − KE f , or
KElost =
1
1
1
2
2
2
M ( 3.00 m s ) + ( 2 M ) (1.20 m s ) − ( 3 M ) (1.80 m s )
2
2
2
With M = 2.00 × 104 kg , this yields KElost = 2.16 × 10 4 J .
28. A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of
200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.
6.28
Let us apply conservation of energy to the block from the time just after the
bullet has passed through until it reaches maximum height in order to find its
speed V just after the collision.
1 2
1
1
mv i + mgy i = mv 2f + mgy f becomes mV 2 + 0 = 0 + mgy f
2
2
2
or
V=
2gyf =
(
)
2 9.80 m s 2 ( 0.120 m ) = 1.53 m s
Now use conservation of momentum from before until just after the collision in
order to find the initial speed of the bullet, v.
(7.0 × 10
from which
−3
)
(
)
kg v + 0 = (1.5 kg )(1.53 m s ) + 7.0 × 10−3 kg ( 200 m s ) ,
v = 5.3 × 102 m s .
37. A 25.0-g object moving to the right at 20.0 cm/s overtakes and collides elastically with a 10.0-g object
moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.
6.37
Conservation of momentum gives
( 25.0 g ) v1 f + (10.0 g ) v 2 f = ( 25.0 g )( 20.0 cm s ) + (10.0 g )(15.0 cm s )
(1)
For head-on, elastic collisions, we know that v1i + v1 f = v 2i + v 2 f .
Thus, 20.0 cm s + v1 f = 15.0 cm s +v 2 f .
(2)
Solving (1) and (2) simultaneously yields
v1 f = 17.1 cm s , and v 2 f = 22.1 cm s .
38. Four railroad cars, each of mass 2.50 x 104 kg, are coupled together and coasting along horizontal tracks at
speed vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car
and gives it a big push, increasing its speed to 4.00 m/s south. The remaining three cars continue moving south,
now at 2.00 m/s. (a) Find the initial speed of the cars. (b) How much work did the actor do?
6.38
(a) The internal forces exerted by the actor do not change total momentum.
vi
m
m
m
m
2.00 m/s
m
m
4.00 m/s
m
m
From conservation of momentum
( 4m ) v i = ( 3m ) ( 2.00 m s ) + m ( 4.00 m s )
vi =
6.00 m s + 4.00 m s
= 2.50 m s
4
1
1
2
2
2
(b) Wactor = K f − Ki = ⎡( 3m ) ( 2.00 m s ) + m ( 4.00 m s ) ⎤ − ( 4m ) ( 2.50 m s )
⎣
⎦
2
2
Wactor =
( 2.50 × 10
2
4
kg
) ⎡12.0 + 16.0 − 25.0⎤( m s )
⎣
⎦
2
= 3.75 × 10 4 J
20. A rifle with a weight of 30 N fires a 5.0-g bullet with a speed of 300 m/s. (a) Find the recoil speed of the rifle.
(b) If a 700-N man holds the rifle firmly against his shoulder, find the recoil speed of man and rifle.
6.20
w
30 N
=
= 3.1 kg . We choose the direction
g 9.80 m s 2
of the bullet’s motion to be negative. Then, conservation of momentum gives
(a) The mass of the rifle is m =
(m
v
rifle rifle
+ mbullet v bullet
(
) = (m
f
v
rifle rifle
+ mbullet v bullet
)
i
)
or ( 3.1 kg ) v rifle + 5.0 × 10−3 kg ( −300 m s ) = 0 + 0 and v rifle = 0.49 m s .
(b) The mass of the man plus rifle is m =
730 N
= 74.5 kg . We use the same
9.80 m s 2
⎛ 5.0 × 10−3 kg ⎞
approach as in (a), to find v = ⎜
( 300 m s ) = 2.0 × 10−2 m s .
⎝ 74.5 kg ⎟⎠
21. A 45.0-kg girl is standing on a 150-kg plank. The plank, originally at rest, is free to slide on a frozen lake,
which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of
1.50 m/s6.21
The velocity of the girl relative to the ice, v gi , is v gi = v gp + v pi where
v gp = velocity of girl relative to plank , and
v pi = velocity of plank relative to ice . Since we are given that
v gp = 1.50 m s , this becomes v gi = 1.50 m s + v pi .
(1)
⎛ mg ⎞
(a) Conservation of momentum gives mg v gi + mp v pi = 0 , or v pi = − ⎜ ⎟ v gi .
⎝ mp ⎠
(2)
⎛
mg ⎞
Then, Equation (1) becomes ⎜ 1 +
⎟ v gi = 1.50 m s
mp ⎠
⎝
or v gi =
1.50 m s
= 1.15 m s .
⎛ 45.0 kg ⎞
1+ ⎜
⎝ 150 kg ⎟⎠
(b) Then, using (2) above,
⎛ 45.0 kg ⎞
v pi = − ⎜
(1.15 m s ) = − 0.346 m s
⎝ 150 kg ⎟⎠
or v pi = 0.346 m s d irected opposite to the girl ′s motion .