fisica1-youn-e-freedman-exercicios-resolvidos

8.7:
∆p
∆t
( 0.0450 kg)( 25.0 m s)
= −3 = 563 N. The weight of the ball is less than half a newton, so the
2.00×
10
s
weight is not significant while the ball and club are in contact.
8.8: a) The magnitude of the velocity has changed by
( 45 .0 m s) − ( − 55.0 m s) = 100.0 m s,
and so the magnitude of the change of momentum
is ( 0.145 kg) (100.0 m s) = 14.500 kg m s, to three figures. This is also the magnitude of
the impulse. b) From Eq. (8.8), the magnitude of the average applied force is
14.500 kg.m/s
=7.25×
10 3 N.
−3
2.00×
10 s
8.9: a) Considering the +x-components,
p2 = p1
+ J = (0.16 kg)(3.00 m s) + (25.0 N) × (0.05 s) = 1.73 kg ⋅ m s, and the velocity is
10.8 m s in the +x-direction. b) p 2
= 0.48 kg ⋅ m s + (–12.0 N)(0.05 s) = –0.12
kg ⋅ m s , and the velocity is +0.75 m s in the –x-direction.
8.10: a) F r t=(1.04 × 10 kg m s)
)
5 ⋅ j . b) (1.04× 10 5 kg ⋅ m s) ĵ .
5
( 1.04×
10 kg. m
s )
c) ˆj
= (1.10 m s) ˆj
.
(95,000 kg)
d) The initial velocity of the shuttle is not known; the
change in the square of the speed is not the square of the change of the speed.