Prove that in case of oblique elastic collision of two particles of equal mass if one is at rest, the recoiling particles always move off at right angles to each other.

Explanation

In elactic collision momentum is conserved.
So, conservation of momentum along x - axis yields.
`mu = mv_(1) cos theta_(1) + mv_(2) cos theta_(2)`
`i.e., u = v_(1)costheta_(1)+v_(2)costheta_(2)" " ....(1)`
and along y - axis yields
`0=v_(1)sin theta_(1)-v_(2)sintheta_(2) " " ....(2)`
Squaring and adding Eqns. (1) and (2), we get
`u^(2)=v_(1)^(2)+v_(2)^(2)+2v_(1)v_(2)cos(theta_(1)+theta_(2))....(3)`
As the collision is elactic `(1)/(2)m u^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)`
`i.e., u^(2)=v_(1)^(2)+v_(2)^(2)....(4) "Put" " "u^(2)=v_(1)^(2)+v_(2)^(2)` in Eqn (3).
we get `2v_(1)v_(2)cos(theta_(1) + theta_(2)) = 0`
As it is given that `v_(1)!=0 and v_(2)!=0`
so `cos(theta_(1) + theta_(2)) = 0, i.e., (theta_(1) + theta_(2)) = 90^(@)`