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 elastic collision momentum is conserved. So, conservation of momentum along x-axis yields.
`m u= mv_(1) cos theta_(1) + mv_(2) cos theta_(2)`
i.e., `u= v_(1) cos theta_(1) + v_(2) cos theta_(2)`….(1)
and along y-axis yields
`0=v_(1) sin theta_(1) - v_(2) sin theta_(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 collisions is elastic `(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) ne 0 and v_(2) ne 0`
so `cos (theta_(1) + theta_(2))= 0, "i.e.," (theta_(1) + theta_(2))= 90^(@)`