Laws of Motion

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 =

√3

2

v1 (along x-axis)
⇒ v2 = (

√3

2

)√6g = √45 m/s