fisica1-youn-e-freedman-exercicios-resolvidos

8.81: a) Apply conservation of energy to the motion of the package from point 1 as it
leaves the chute to point 2 just before it lands in the cart. Take y = 0 at point 2, so y 1 =
4.00 m. Only gravity does work, so
K + U
1
2
v
1
mv
2
1
1
= K
2
1
+ U
+ mgy =
1
2
2
2
= v1
+ 2gy
1
=
2
mv
2
2
9.35 m/s
b) In the collision between the package and the cart momentum is conserved in the
horizontal direction. (But not in the vertical direction, due to the vertical force the floor
exerts on the cart.) Take +x to be to the right. Let A be the package and B be the cart.
P x is constant gives
v + m v = ( m + M
ma
A1 x B B1x
A B
) v2x
v
B1x
= −5.00 m/s
ο
v
A1x
= (3.00 m/s) cos 37.0 ( The horizontal velocity of the package is constant during
its free-fall.)
Solving for v 2 x
gives v2x
= −3.29 m/s. The cart is moving to the left at 3.29 m/s after
the package lands in it.
8.82: Even though one of the masses is not known, the analysis of Section (8.4) leading
to Eq. (8.26) is still valid, and v = 0.200 m/s + 0.050 m/s 0.250 m/s. b) The mass
red
=
m
red
may be found from either energy or momentum considerations. From momentum
conservation,
(0.040 kg)(0.200 m/s − 0.050 m/s)
m
red
=
= 0.024 kg.
(0.250 m/s)
As a check, note that
K
K
1
2
=
=
so K
1
1
2
1
2
(0.040 kg)(0.200 m/s)
(0.040 kg)(0.050 m/s)
2
2
2
= 8.0 × 10
1
2
−4
J, and
(0.024 kg)(0.250 m/s)
= K , as it must for a perfectly elastic collision.
+
2
= 8.0×
10
−4
J,