Further motion concepts – WJECExamples of momentum

Moving objects have momentum, and forces cause it to change. The total momentum in an explosion or collision is conserved and stays the same. Equations of motion apply to uniformly accelerated motion.

Part ofPhysics (Single Science)Forces, space and radioactivity

Examples of momentum

Look at this example.

A red and a blue snooker ball both have a mass of 160 g. They collide with speeds and directions as shown in this diagram.

A red snooker ball of mass 160 g moves at 0.28 m/s and collides head on with a blue ball moving at -0.12 m/s. After the collision the blue ball has a speed of 0.18 m/s.

The following steps show you how to calculate the velocity of the red ball after the collision, and use kinetic energy to determine whether the collision was elastic or not.

\(\text{total momentum before} = {\text{total momentum after}}\)

\({\text{m}}_{\text{red}}{\text{u}_{\text{red}}} + {\text{m}}_{\text{blue}}{\text{u}_{\text{blue}}} = {\text{m}}_{\text{red}}{\text{v}_{\text{red}}} + {\text{m}}_{\text{blue}}{\text{v}_{\text{blue}}}\)

\(0.16 \times 0.28 + 0.16 \times ( - 0.12) = 0.16 \times {\text{v}}_{\text{red}} + 0.16 \times 0.18\)

Rearranging, ie right to left.

\(0.16 \times{\text{v}}_{\text{red}} = (0.0448 - 0.0192) - 0.0288\)

\(= 0.0256 - 0.0288\)

\(= \frac{{ - 0.0032}}{{0.16}}\)

\({\text{v}}_{\text{red}}~=~- 0.02{\text{ ms}}^{-1}\)

To determine if the collision is elastic or not, you must work out the kinetic energy before and after the collision.

Remember the equation for calculating kinetic energy in general.

\(\text{E}_{\text{k}}=\frac{1}{2}{\text{mv}^{2}}\)

This example deals with two objects. The kinetic energy before the collision involves their initial speed, so the equation becomes

\(\text{E}_{\text{k}}=\frac{1}{2}{\text{m}}_{\text{red}}{{\text{(u}}_{\text{red}}{\text{)}}}^{2}+\frac{1}{2}{\text{m}}_{\text{blue}}{{\text{(u}}_{\text{blue}}{\text{)}}}^{2}\)

\(= 0.5 \times 0.16 \times {(0.28)^2} + 0.5 \times 0.16 \times {(0.12)^2}\)

\(= 0.00627 + 0.00115\)

\(= 0.00742{\text{ J}}\)

Total kinetic energy after collision

\(\frac{1}{2}{\text{m}}_{\text{red}}{{\text{(v}}_{\text{red}}{\text{)}}}^{2}+\frac{1}{2}{\text{m}}_{\text{blue}}{{\text{(v}}_{\text{blue}}{\text{)}}}^{2}\)

\(=0.5\times0.16\times(0.02)^2+0.5\times0.16\times(0.18)^2\)

\(=0.000032+0.00259\)

\(=0.00262{\text{ J}}\)

Total kinetic energy lost

\(=0.00742-0.00262\)

\(=0.0048{\text{ J}}\)

48 mJ of energy is lost so the collision is inelastic.

Question

If two bumper cars collide head-on in a fairground and both cars come to a stop due to the collision, kinetic energy is obviously not conserved. Is momentum conserved even though both cars stop?

The total momentum of two objects before a collision is the same as their total momentum after the collision - provided there are no external forces.