A ball of mass `m=1kg` falling vertically with a velocity `v_0=2m//s` strikes a wedge of mass `M=2kg` kept on a smooth, horizontal surface as shown in figure. If impulse between ball and wedge during collision is J. Then make two equations which relate J with velocity components of wedge and ball. Also find impulse on wedges from ground during impact.

Answer

`v_(1) = (1)/(sqrt3) ms^(-1), v_(2) = (2)/(sqrt3) ms^(-1)`

Explanation

Let `v =` velocity of the ball after collision along the normal
`J =` impulse on ball
`= v - (-2 cos 30^(@)) = v + sqrt3`
Impulse on wedge `J sin 30^(@) = mv_(1) = 2v_(1)`
`rArr v = 4v_(1) - sqrt3` …..(i)
`e = - ((v_(2) - v_(1))/(u_(2) - u_(1))) rArr (1)/(2) = ((v + (v_(1))/(2)))/(2 cos 30^(@)`
`rArr v = (sqrt3)/(2) - (v_(1))/(2)` ....(ii)
Solving we get `v_(1) - (1)/(sqrt3) m//s`
For th eball velocity along incline remains constant.
`:. v' = 2 sin 30^(@) = 1 m//s`
`:.` Final velocity of ball `= sqrt(1^(2) + ((1)/(sqrt3))^(2)) = - (2)/(sqrt3) m//s`