A particle of mass m experienced an elastic collision with a stationary particle of mass 2m . What fraction of the kinetic energy does the striking particle lose if recoils at the right angles to its original motion direction.

A particle of mass m experienced an elastic collision with a stationar

1.	A particle of mass m experienced an elastic collision with a stationary particle of mass 2m . What fraction of the kinetic energy does the striking particle lose if recoils at the right angles to its original motion direction.
Answer» Solution :Since the total momentum before the collision is along the `x`-axis , the total momentum after the collision should be along the `x`-axis, for this particle of mass `2M` should make at angle so that the net momentum along the `y`-axis is ZERO. Fractional loss of `K.E.` of striking particle is `((1)/(2) m U^(2) - (1)/(2) mv_(1)^(2))/((1)/(2) m u^(2)) = 1 - ((v_(1))/(u))^(2)` By the momentum conservation: Along the `x`-axis: `mu = 2m v_(2) cos theta rArr v_(2) cos theta = (u)/(2)`(i) Along the `y`-axis: `0 = mv_(1) - 2mv_(2) sin theta rArr v_(2) sin theta = (v_(1))/(2)`(ii) ELIMINATING `theta` from (i) and (ii) , we get `(v_(2) cos theta)^(2) + (v_(2) sin theta)^(2) = (u^(2) + v_(1)^(2))/(4)` `v_(2)^(2) = (u^(2) + v_(1)^(2))/(4)` Since the collision is elastic `K_(i) = K_(f)` `(1)/(2) m u^(2) = (1)/(2) mv_(1)^(2) + (1)/(2) xx 2m v_(2)^(2)` `u^(2) = v_(1)^(2) + 2v_(2)^(2) = v_(1)^(2) + 2((u^(2) + v_(1)^(2))/(4)) rArr ((v_(1))/(u))^(2) = (1)/(3)` Fractional loss `= 1 - ((v_(1))/(u))^(2) = 1 - ((1)/(3)) = (2)/(3)`