Prove that in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central impact.
Assume particle 1 travels in the x direction with speed v1(before) and strikes particle 2, which is at rest. After the collision, particle 1 makes an angle theta1 with the x axis, and particle 2 makes angle theta2. We need to show that theta1 + theta2 = pi/2.
Conservation of linear momentum
mv1(before) = SUM(mvi(after))
x component: mv1(before) = mv1(after) cos(theta1) + mv2(after) cos(theta2)
y component: 0 = mv2(after) sin(theta2) - mv1(after) sin(theta1)
Conservation of kinetic energy
.5mv1(before)^2 = .5mv1(after)^2 + .5mv2(after)^2
The Attempt at a Solution
I'm afraid I really don't know what to do with this. I notice that if theta1 + theta2 form a right angle, then by the Pythagorean theorem,
v1(after)^2 + v2(after)^2 = [v1(after) sin(theta1) + v2(after) sin(theta2)]^2
= v1(after)^2 sin^2(theta1) + (2)(v1(after))(v2(after)) sin(theta1) sin(theta2) + v2(after)^2 sin^2(theta2)
But this doesn't really look very useful. Hint please?