A ball of mass `m` moving horizotally which velocity `u` hits a wedge of mass `M`. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in horizotal plane. calculate
a. the velocity of wedge `V`.
b. the velocity `(v)` at which the ball moves in vertical direction.
c. the impulse imparted by the ball on the wedge.
d. the coefficient of restitution `e=?`

Explanation

a. As no external force is acting on the system in horizontal direction the linear momentum should be constant and conserved in horizontal direction.
Let velocity of wedge after collision be `V`.
Then `"mu"=MVimpliesV=("mu")/M`………..i
b. As there is no impulse on the ball in the direction parallel to sloping side hence the veloicty of ball along the slope (tangent direction) should remain unchanged,
`implies ucostheta=vsintheta` ..........ii
c. we can find the values of `v` and `V` by impulse approach, when ball hits the wedge the impulse is generated between ball and wedge in the direction perpendicular to slopping surface. (normal direction).

For ball: `"mu"-Jsintheta=0`
`Jsintheta="mu"`
`J cos thetas =mv`
from eqn i and ii `v=ucottheta`
From eqn iii `J="mu" cosec theta,`
`Jsintheta=mVimpliesV=(Jsintheta)/M`
Which gives `V=(mu)/M`
d. The direction of restitution is given as

`(v_(2)-v_(1))_(n)-e(u_(1)-u_(2))_(n)`
` [Vsintheta-(-vcostheta)]=e[usintheta-0]`
`e=(Vsintheta+vcostheta)/(usintheta)`
`e=V/u+v/ucot theta=("mu")/M+(cottheta)cotheta`
`e=m/M+cot^(2)theta`