F
m
v
vo
t
to
F
10000 lbm
110 ft/sec
9 sec
F
1.2×105ft lbm
sec2
mGvG
mBvB
0
or
mGvG
mBvB
Force and Motion
MOMENTUM PRINCIPLES
Rev. 0
Page 7
CP-03
Example:
The velocity of a rocket must be increased by 110 ft/sec to achieve proper orbit around
the earth. If the rocket has a mass of 5 tons and it takes 9 sec. to reach orbit, calculate
the required thrust (force) to achieve this orbit.
Solution:
Even though the initial velocity (v ) and final velocity (v) are unknown, we do know the
o
change in velocity (v-v ), which is 110 ft/sec. Therefore, using Equation 3-4 we can find
o
the solution.
Conservation of Momentum
One of the most useful properties of momentum is that it is conserved. This means that if no net
external force acts upon an object, the momentum of the object remains constant. Using
Equation 3-6, we can see that if force (F) is equal to zero, then P = 0. It is most important for
collisions, explosions, etc., where the external force is negligible, and states that the momentum
before the event (collision, explosion) equals the momentum following the event.
The conservation of momentum applies when a bullet is fired from a gun. Prior to firing the gun,
both the gun and the bullet are at rest (i.e., V and V are zero), and therefore the total
G
B
momentum is zero. This can be written as follows:
When the gun is fired, the momentum of the recoiling gun is equal and opposite to the
momentum of the bullet. That is, the momentum of the bullet (m v ) is equal to the momentum
B B
of the gun (m v ), but of opposite direction.
G G
