fisica1-youn-e-freedman-exercicios-resolvidos

8.34: The initial momentum of the car must be the x-component of the final momentum
as the truck had no intial x-component of momentum, so
px ( mcar
+ mtruck
) v cos θ
vcar
= =
mcar
mcar
2850 kg
= (16.0 m s ) cos (90° − 24°
)
950 kg
= 19.5 m s.
2850
Similarly, v
truck
= (16.0 m s) sin 66°
= 21.9 m s.
1900
8.35: The speed of the block immediately after being struck by the bullet may be found
from either force or energy considerations. Either way, the distance s is related to the
2
v by v = µ gs. The speed of the bullet is then
speed
block
2 k
mblock
+ mbullet
v
bullet
=
2µ
kgs
m
bullet
1.205 kg
=
2(0.20)(9.80 m s
−3
5.00 × 10 kg
= 229 m s,
2
)(0.230 m)
or
2
2.3× 10 m s to two places.
8.36: a) The final speed of the bullet-block combination is
−3
12.0 × 10 kg
V =
(380 m s) = 0.758 m s.
6.012 kg
1 2
Energy is conserved after the collision, so ( m + M ) gy = ( m + M ) V , and
1 V
y =
2 g
2
=
1
2
2
(0.758 m s)
2
(9.80 m s )
2
= 0.0293 m = 2.93 cm.
b)
c)
K
1
=
1
2
mv
2
=
From part a),
1
2
K
(12.0 × 10
2
=
1
2
−3
kg)(380 m
s)
(6.012 kg)(0.758 m
2
= 866 J.
s)
2
= 1.73 J.