A ball of mass `m` hits a wedge of mass `M` vertically with speed `u`, which is placed, on a smooth horizontal surface. Find the maximum compression in the spring, if the collision is perfectly elastic and no friction any where. Spring constant of spring is `K`.

Answer

`sqrt(M/K)`

Explanation

Let after collision, the ball has velocity `v_(1)` and wedge `v_(2)` as shown.

Aplying conservation of momentum in horizontal direction
`mv_(1)costheta=Mv_(2)` ……….i
From Newton's experimental law:
`e=1=(v_(2)sin45^(@)-[-v_(1)cos(45+theta)])/(ucos45^(@))`
`implies u=v_(2)+v_(1)costheta-v_(1)theta` ...........ii
Velocity component of the ball along the surfasce of wedge will remain the same as during collision there is no force on the ball in this direction So
`usin45^(@)=v_(1)cos(45-theta)impliesu=v_(1)costheta+v_(1)sintheta`...iii
From i and ii and iii
`v_(2)=(2"mu")/(m+2M)`
To find maximum compression in the sprig, after collision apply conservation of energy for the system of wedge and spring.
`1/2 Mv_(2)^(2)=1/2Kx^(2)`
`implies x=sqrt(M/x)v_(2)implies x((2"mu")/(m+2M))sqrt(M/K)`