Question

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=?`

Answer 1

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`