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The time intervals on both sides of the equation are the same because both forces
act over the same interval of time. So the equation can be simplified to:
m B v B = – m A v A
Therefore, m A v A + m B v B = 0
or p A + p B = 0
Other Examples of the Law of Conservation of Momentum
If you have ever played pool or billiards you will be familiar with the law of conservation
of momentum. When the cue ball, or white ball, hits another ball, momentum is
conserved. For example, if all the momentum is transferred from the cue ball to the
billiard ball, the cue ball will stop and the billiard ball will move.
Sample Problem 4.4.1 — Conservation of Momentum
A railway car of mass 6.0 × 10 3 kg is coasting along a track with a velocity of 5.5 m/s when suddenly a 3.0 × 10 3 kg
load of sulphur is dumped into the car. What is its new velocity?
What to think about
1. What do I know about the problem?
2. What am I trying to solve?
3. What formula applies to this situation?
4. Find the final velocity v 2 .
How to Do It
The momentum of the railway car will not change
because of the law of conservation of momentum.
Let initial mass be m 1 and initial velocity be v 1 .
The final mass of the railway car will be m 2 and the
final velocity v 2 .
Solve for the final velocity v 2 .
m 1 v 1 = m 2 v 2
(6.0 × 103 kg)(5.5 m/s) = (6.0 × 103 kg +
3.0 × 103 kg)(v2 )
33 × 103 kg•m/s = (9.0 × 103 kg)(v2 )
v2 = 33 103 kg•m/s
9.0 103 kg
184 Chapter 4 Newton’s Laws of Motion — DRAFT © Edvantage Interactive 2011 ISBN 978-0-9864778-3-6