Conservation of momentum example
Consider two model cars of mass 1.2 and 1.4 kg colliding at the speeds shown:
The total momentum before the collision is the sum of both momentums:
Momentum of blue car A:
\(p_A={m_A}v\)
\(= 1.2\times 0.5\)
\(=0.6kg\,m{s^{ - 1}}\)
Momentum of red car B:
\(p_B={m_B}v\)
\({m_B}v = 1.4 \times 0.2\)
\(= 0.28kg\,m{s^{ - 1}}\)
The red car is travelling in the negative Information to give the direction of travel, or the direction of a force, for example, a speed of 20 m s-1 to the left, or a force of 15 N to the right. so the momentum is subtracted.
Total momentum:
\(0.6 - 0.28\)
\(= 0.32kg\,m{s^{ - 1}}\)
If the two cars stick together after the collision and move as one then the velocity \({v _{AB}}\) of the two cars can be determined because the total momentum after the collision is the same (ie 0.32 kg ms-1).
The total mass \({m_{AB}}\) is now 2.6 kg.
Total momentum after collision is 0.32 kg ms-1:
\(0.32 = {m_{AB}} \times {v_{AB}}\)
\(0.32 = 2.6 \times {v_{AB}}\)
\({v_{AB}} = \frac{{0.32}}{{2.6}}\)
\(= 0.12m{s^{ - 1}}\)
The velocity is positive so this shows that the two cars move off in the positive (left to right) direction.