fisica1-youn-e-freedman-exercicios-resolvidos

8.66: The total mass of the car is changing, but the speed of the sand as it leaves the car
is the same as the speed of the car, so there is no change in the velocity of either the car
or the sand (the sand acquires a downward velocity after it leaves the car, and is stopped
on the tracks after it leaves the car). Another way of regarding the situation is that vex
in
Equations (8.37), (8.38) and (8.39) is zero, and the car does not accelerate. In any event,
the speed of the car remains constant at 15.0 m/s. In Exercise 8.24, the rain is given as
falling vertically, so its velocity relative to the car as it hits the car is not zero.
8.67: a) The ratio of the kinetic energy of the Nash to that of the Packard is
2
m
v
2
mPvP
=
2
( 840 kg)( 9 m s)
2
( 1620 kg)( 5 m s)
mPvP
= 1.68. b) The ratio of the momentum of the Nash to that of the
mv
(840 kg)(9 m/s)
Packard is = = 0.933, therefore the Packard has the greater magnitude
(1620 kg)(5 m/s)
of momentum. c) The force necessary to stop an object with momentum P in time t is
F = − P/
t. Since the Packard has the greater momentum, it will require the greater force
to stop it. The ratio is the same since the time is the same, therefore F / = 0.933. d)
F P
By the work-kinetic energy theorem, F = ∆ k
d
. Therefore, since the Nash has the greater
kinetic energy, it will require the greater force to stop it in a given distance. Since the
distance is the same, the ratio of the forces is the same as that of the kinetic energies,
F
/ F P
= 1.68.
8.68: The recoil force is the momentum delivered to each bullet times the rate at which
the bullets are fired,
F =
3
⎛1000 bullets/min ⎞
36.4 N.
ave
(7.45× 10
− kg) (293 m/s) ⎜
⎟ =
⎝ 60 s/ min ⎠