fisica1-youn-e-freedman-exercicios-resolvidos

m1v1
+ m2v2
8.28: a) From v + m v = m v + m v = ( m + m ) v,
v = . Taking positive
m
1 1 2 2 1 2 1 2
m1
+ m2
velocities to the right, v
1
= −3.00.m s and v = 1.20 m
2
s , so v = −1.60 m s .
b)
1
2
∆K = (0.500 kg + 0.250 kg)( −1.60m s)
2
1
2 1
− (0.500 kg)( −3.00 m s) − (0.250 kg)(1.20m s)
2
2
= −1.47 J.
2
8.29: For the truck, M = 6320 kg, and V = 10 m s, for the car, m = 1050 kg and
v = −15m s (the negative sign indicates a westbound direction).
a) Conservation of momentum requires ( M + m)
v′
= MV + mv , or
′ v
(6320 kg)(10 m s) + (1050 kg)( −15m s)
=
= 6.4 m s eastbound.
(6320 kg + 1050 kg)
− mv − (1050 kg)( −15 m s)
b) V = =
= 2.5 m s.
M 6320 kg
c) ∆KE = −281 kJ for part (a) and ∆KE = −138 kJ for part (b).
8.30: Take north to be the x-direction and east to be the y-direction (these choices are
arbitrary). Then, the final momentum is the same as the intial momentum (for a
sufficiently muddy field), and the velocity components are
v
x
(110 kg)(8.8 m s)
=
= 5.0 m s
(195 kg)
(85 kg)(7.2 m s)
vy
=
= 3.1m s.
(195 kg)
2
2
The magnitude of the velocity is then (5.0 m s) + (3.1m s) = 5.9 m s,
at an angle or
.1
arctan ( ) = 32°
3
east of north.
5.0