A particle (a mud pallet, say) of mass `m` strikes a smooth stationary wedge of mass `M` with as velocity `v_(0)` at an angle `theta` with horizontal. If the collision is perfectly inelastic, find the
a. velocity of the wedge just after the collision.
b. Chane in `KE` of the system `(M+m)` in collision.

Explanation

a. Let the system `(M+m)` moves as a single mass with a velocity `v`.
conserving the momentum of the system in horizontal, we have
`mv_(0)costheta=(M+m)v`

This gives `v=(mv_(0)costheta)/(M+m)`
b. the change in `KE` of the system is
`/_\K=1/2(M+m)v^(2)-1/2mv_(0)^(2)`
where `v=(mv_(0)costheta)/(M+m)` ltbr This gives `/_\K=((M+m))/2((mv_(0)costheta)/(M+m)^(2)-1/2mv_(0)^(2)`
`=-(mv_(0)^(2))/2(1-m/(M+m)cos^(2)theta)`
`=-((M+msin^(2)theta)mv_(0)^(2))/(2(M+m))`