A small block of mass `M` moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from `60^(@)` to `30^(@)` at point `B`. The block is many at rest at `A`. Assume that collisions between the block id the incline are totally inelastic.
The speed of the block at point `C`, immediately before it leaves the second incline

Options

A. `sqrt(60) m//s`
B. `sqrt(45) m//s`
C. `sqrt(30) m//s`
D. `sqrt(15) m//s`

Answer

B

Explanation

Between A and B, height fallen by the block is
`h_1 = sqrt(3)tan 60^@ = 3m`
Hence, speed of the block just before striking the second incline is:
`v_2 = sqrt(2gh_1) = sqrt(2 xx 10 xx 3) = sqrt(60) m//s`
In perfectly inelastic collision, component of `v_1` perpendicular to BC will become zero, while component of `v_1` parallel to BC will remain unchanged.
`v_2` = component of `v_1` along BC, `" " = v_1 cos 30^@u = (sqrt(60)) ((sqrt(3))/2) = (sqrt(45)) m//s`