A ball of mass `m` collides with a stationary wedge of mass `M`, perpendicular to its inclined face, inclined at an angle as shown in the figure. If the coefficient of restitution between the wedge and ball is `e`, calculate the ratio of modulus of velocity of the ball immediately after and before collision. Also calculate the velocity of wedge just after collision.

Answer

`(musnalpha(e+1)/(M+sin^(2)alpha))`

Explanation

Let velocity of wedge after colision be `v_(2)` and that of ball be `v_(1)` . We can show that `v_(1)ltu` , because otherwise final `KE` would be greater than initial `KE` which is not possible `v_(1)` will be perpendicular to wedge surface.
From Newoton's experimental law:
` e=(v_(2)sinalpha-(-v_(1)))/uimpliesv_(2)=(eu-v_(1))/(sinalpha)`.....i

Conservastion of linear mometum along `x`-axis.
`musinalpha=-mv_(1)sinalpha+Mv_(2)` ...........ii
From i and ii `v_(1)/u=(Me-msin^(2)alpha)/(M+sin^(2)alpha)`
Put the value of `v_(1)` in eqn i to get `v_(2)`
`v_(2)=1/(sinalpha)[eu-((Me-msin^(2)alpha)u)/(M+msin^(2)alpha)]=(musinalpha(e+1))/(M+msin^(2)alpha)`