Let V1 be the speed of the block at point B just before it strikes the second incline. That is, Mg (3) =
1
2
Mv12 ⇒ v1 = √6g (m/s)
Now, vx = v1cos30° = (
√3
2
)v1 vy = v1sin30° = −
v1
2
After the inelastic collision with the second incline, we get vy′ = 0 and vx′ = vx (centre of mass) Therefore, after collision, the speed of the block is v2 =