The conservation of energy and momentum is very
useful in the study of the motion of objects after an explosion. The same ideas can be
applied to firing a rifle, a radium nucleus emitting an alpha particle and a bow firing an
arrow etc.

By the law of conservation of momentum the two objects must move off in opposite directions.

Let the kinetic energy of the shell, mass m, be E and that of the cannon, mass M, be E'.

Now: E = ½ mu

By the conservation of momentum mu = -Mv and therefore:

This gives:

and so the fragment with the smaller mass has the larger kinetic energy.

A bullet with a mass one hundredth that of the gun that fires it will have one hundred times the kinetic energy of the gun.

1. A rifle of mass 3 kg fires a bullet of mass 0.025 kg at 100 ms

Kinetic energy of bullet = 0.5x0.025x100x100 = 125

Kinetic energy of rifle = [125 x 0.025]/3 = 1.04J

2. Uranium-235 is an alpha-emitter the resulting nucleus having a mass of 231 units.

If a stationary uranium -235 nucleus emits an alpha particle (mass 4 units) what is the ratio of the kinetic energy of the alpha-particle to that of the residual nucleus?

Kinetic energy of alpha/Kinetic energy of nucleus = E/E' = M/m = 231/4 = 57.75

This principle can be studied with two spring loaded trolleys (Figure 2). If they are put back to back and the spring in one is released they will explode. Both trolleys move apart with the same speed because they are of equal mass. In the second diagram trolley 2 moves off with half the speed of trolley 1 as it has twice the mass.

An interesting case is where no particle is emitted – e.g. the bow is bulled back under tension and then released with no arrow being fired. There is a danger of breaking the bow since all the energy stored in the stretched string goes into the motion of the bow. You can test this by throwing a ball and then doing the throwing action again but this time without a ball in your hand – it hurts your arm as all the stored energy must be absorbed by your arm again in bringing it to rest.

The following example demonstrates the use of relative velocity in a collision problem.

A mass m moving to the right with velocity v collides elastically with a mass M moving in the opposite direction also at velocity v (see Figure 3). If M is very much greater than m, calculate:

(a) the velocity of m immediately after the collision

(b) the change in the momentum of in, and

(c) the change in the kinetic energy of m.

(a) Consider M to be at rest; then the velocity of m relative to M is 2v. After the collision the velocity of m relative to M must be –2v, since the collision is perfectly elastic.

Therefore (since M»m) the velocity of M will be virtually unchanged by the collision, and hence the velocity of m relative to the external observer will be (-2v) ± (-v) = -3v.

(b) The momentum change is then mv - m(-3v) = -4mv.

(c) The kinetic energy change is 4mv