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 `B` immediately after it strikes the second incline is

Options

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

Answer

B

Explanation

1. Between A and B, height fallen by block
`h_(1)=sqrt3tan60^(@)=3m`
`therefore` Speed of block just before striking the second incline,
`v_(1)=sqrt(2gh_(1))=sqrt(2xx10xx3)" "sqrt(60)m//s^(-1)`
In perfectly inelastic collision, component of `v_(1)` perpendicular to BC become zero, while component of `v_(1)` parallel to BC will remain unchanged.
`therefore`Speed of block `B` immediately after it strikes the second incline is

`v_(2)="component of "v_(1)` along BC
`=v_(1)cos30^(@)=(sqrt(60))((sqrt3)/(2))=(sqrt(45))ms^(-1)`
`therefore` correct option(B).
2. Height fallen by the block from `B` to `C`
`h_(2)=3sqrt3tan30^(@)=3m`
Let `v_(3)` be the speed of block, at point `C`, just before it leaves the second incline then:]