Elastic collisions and snooker
Although you would not play snooker using a protractor and a calculator to work out your
shots the study of momentum is crucial in what actually happens!
1.
Collisions in a straight line
We will first consider the more simple case where
the two balls collide along a line joining their centres. We have to assume that the two balls
are not only exactly the same mass (m) but that the collision between them is perfectly
elastic - remember that this means that no kinetic energy is lost in a collision. First think of
just one stationary red being hit head on by the white.
After the collision the white ball stops dead and
the red ball moves off with the velocity that the white ball had before the collision. (This is
only exactly true if the two balls are sliding and not rolling!)
We can prove this as
follows:
Momentum before collision = momentum after collision mu
A =
mv
A + mv
BBut because the collision is elastic kinetic energy
is also conserved and so:
½ mu
A2 = ½
mv
A2 + ½ mv
B2The masses can be
cancelled giving:-
u
A = v
A +v
B and
u
A2 = v
A2 + v
B2
Using a little algebra to combine these two equations will show you that:
vA = 0 and VB = uA
In fact it can be proved that when two equal masses collide elastically their velocities are swapped
over. If mass A had a velocity u before collision and mass B a velocity v, then after collision
A moves with velocity v and B with velocity u.
Example problem
A ball of mass 0.5 kg travelling from left to right at 1.5 ms-1 collides with a ball of equal mass travelling in the same direction at 1.0 ms-1. Prove that they swap velocities after impact.
Momentum before collision = 0.5x1.5 + 0.5x1.0 = 1.25 Ns
Momentum after impact = 1.25 Ns
Kinetic energy before impact = kinetic energy after impact = 0.8125 J
Therefore:
vA + vB = 1.25/0.5 = 2.5 ms-1
vA2 + vB2 = 0.8125x4 = 3.25 = vA2 + (2.5 - vA)2 = vA2 + 6.25 - 5vA + vA2 = 3.25
and so 2vA2 - 5vA + 3 = 0 and when solved this gives vA = 1ms-1 or 1.5 ms-1 but since ball A cannot move through ball B the velocity of ball A must be 1 ms-1.
They have swapped velocities!
2. Oblique or angled
collisions.
Both balls move off at an angle to the original direction of the white.
The interesting thing is that the angle between the red and the white balls is exactly 90o, in
theory. The big problem here is spin - if you put some "side" (spin) on the white the angle
will be different.
Assuming no spin and that the collision is elastic:
½
muA2 = ½ mvA2 + ½
mvB2
using a vector diagram for the momenta of the
two balls before and after the collision we see that:
muA = mvA +
mvB and since from the equation for kinetic energy conservation
uA2 =
vA2 + vB2
The vector triangle for velocity
must be right angled (Pythagoras). Remember that velocity and therefore momentum is a
vector but kinetic energy is a scalar and so we can simply add the kinetic energies of the
balls after impact without allowing for their different directions of
motion.
schoolphysics oblique collisions animation
To see an animation of an oblique collision click on the animation link.
Elastic collisions between particles of unequal
mass
1. Collisions in a straight line
If a body A with a mass m
collides with a second body B with a mass M then the kinetic energy lost by A
is:
(a) maximum when m = M
(b) zero when M = infinity
2.
Oblique collisions
This type of collision can often be seen in cloud chamber and
bubble chamber photographs and is very useful in comparing the masses of the colliding
particles.
It can be proved that if a mass m collides with a mass M then
(a) if m << M
the angle between their paths after collision is > 90
o(b) if m = M the angle
between their paths after collision is 90
o(c) if m > M the angle between their
paths after collision is < 90
o
