fisica1-youn-e-freedman-exercicios-resolvidos

8.72: a) The stuntman’s speed before the collision is v
0s
= 2gy
= 9.9 m / s.
The speed
after the collision is
v
m
= s
v0
ms
+ m
s
v
=
80.0 kg
0.100 kg
( 9.9 m/s) = 5.3 m/s.
b) Momentum is not conserved during the slide. From the work-energy theorem, the
distance x is found from
1 2
m v = µ m gx , or
2
total
k
total
x =
2
v
2µ
kg
=
2
( 5.28 m / s)
2
( 0.25)( 9.80 m / s )
2
= 5.7 m.
Note that an extra figure was needed for V in part (b) to avoid roundoff error.
8.73: Let v be the speed of the mass released at the rim just before it strikes the second
mass. Let each object have mass m.
1 2
Conservation of energy says mv = mgR;
v = 2gR
2
This is speed v1
for the collision. Let v2
be the speed of the combined object just after
the collision. Conservation of momentum applied to the collision
gives mv
1
= 2mv2
so v2
= v1
2 = gR 2
Apply conservation of energy to the motion of the combined object after the collision.
Let y3
be the final height above the bottom of the bowl.
2
( 2m ) v2
= ( m) gy3
1
2
2
2
v2
1 ⎛ gR ⎞
y
3
= = ⎜ ⎟ = R / 4
2g
2g
⎝ 2 ⎠
Mechanical energy is lost in the collision, so the final gravitational potential energy is
less than the initial gravitational potential energy.