Particle `1` experiences a perfectly collision with a staionary particle `2`. Determine their mass ratio , if the particles move perpendicularly to each other.

Explanation

Momentum conservation
`x - ax is : m_(1) u + m_(2) xx 0 = m_(1) v_(1) cos theta + m_(2) v_(2) cos (90 - theta)` (i)
`y - ax is : 0 = m_(1) v_(1) sin theta - m_(2) v_(2) sin (90 - theta)` (ii)
`K_(i) = K_(f)`
`(1)/(2) m_(1)u^(2) = (1)/(2) m_(1) v_(1)^(2) + (1)/(2) m_(2) v_(2)^(2)` (iii)
Solving , `(m_(1))/(m_(2)) = 1`
The calculations will be simple if we do not break momentum in two directions.
We know , `p = mv , K = (1)/(2) mv^(2) , K = (p^(2))/(2m)`
Resultant momentum after the collision should be along the `x`-axis.
By the momentum conservation :
`p = sqrt(p_(1)^(2) + p_(2)^(2)) rArr p^(2) = p_(1)^(2) + p_(2)^(2)` (i)
Since the collision is elastic , `K_(i) = K_(f)`
`(p^(2))/(2m_(1)) = (p_(1)^(2))/(2m_(1)) + (p_(2)^(2))/(2 m_(2))` (iii)
`(p_(1)^(2) + p_(2)^(2))/(2m_(1)) = (p_(1)^(2))/(2 m_(1)) + (p_(2)^(2))/(2 m_(2)) rArr (m_(1))/(m_(2)) = 1`