Page 106 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 8] MOMENTUM 91

CONSERVATION OF LINEAR MOMENTUM
According to the law of conservation of linear momentum, when the vector sum of the external forces that act
on a system of bodies equals zero, the total linear momentum of the system remains constant no matter what
momentum changes occur within the system.
Although interactions within the system may change the distribution of the total momentum among the
various bodies in the system, the total momentum does not change. Such interactions can give rise to two general
classes of events: explosions, in which an original single body flies apart into separate bodies, and collisions, in
which two or more bodies collide and either stick together or move apart, in each case with a redistribution of
the original total momentum. Collisions will be examined in the next section.

SOLVED PROBLEM 8.7
A rocket explodes in midair. How does this affect (a) its total momentum and (b) its total kinetic energy?

(a) The total momentum remains the same because no external forces acted on the rocket.
(b) The total kinetic energy increases because the rocket fragments received additional KE from the explosion.
(Of course, the total energy of the rocket, including the original chemical potential energy of its fuel, is constant.)

SOLVED PROBLEM 8.8
An airplane’s velocity is doubled. (a) What happens to its momentum? Is the law of conservation of
momentum obeyed? (b) What happens to its kinetic energy? Is the law of conservation of energy obeyed?
(a) The airplane’s momentum mv also doubles. Momentum is conserved because the increase in the airplane’s
velocity is accompanied by the backward motion of air through the action of its engines, so the total momentum
of airplane plus air remains the same when their opposite directions are taken into account.
1
2
(b) The airplane’s kinetic energy mv increases fourfold. Energy is conserved because the additional KE comes
2
from chemical potential energy released in the airplane’s engines.
SOLVED PROBLEM 8.9
A 15-g (0.015-kg) bullet is fired from a 5-kg rifle at a muzzle velocity of 600 m/s. Find the recoil velocity
of the rifle.

From conservation of momentum, m r v r = m b v b , and so

m b 0.015 kg
v r = (v b ) = (600 m/s) = 1.8 m/s
m r 5kg
SOLVED PROBLEM 8.10

An astronaut is in space at rest relative to an orbiting spacecraft. His total weight is 300 lb, and he throws
away a 1-lb wrench at a velocity of 15 ft/s relative to the spacecraft. How fast does he move off in the
opposite direction?
From conservation of momentum,
m a v a = m w v w

w a w w
(v a ) = (v w )
g g

w w 1lb
v a = (v w ) = (15 ft/s) = 0.05 ft/s
w a 300 lb
We notice that the g’s have canceled, so it was not necessary to find the mass values first; the ratio of two masses is
always the same as the ratio of the corresponding weights.