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A small block of mass M moves on a fric...

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.
If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the blocks at point `B`, immediately after it strikes the second incline is

A

`sqrt(30)m//s`

B

`sqrt(15)m//s`

C

0

D

`-sqrt(15)m//s`

Text Solution

Verified by Experts

The correct Answer is:
C
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:]
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