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 initially at rest at A. Assume that collisions between the block and 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

Between `A` and `B` height fallen by the block is
`h_(1)=sqrt(3)tan60^(@)=3m`
Hence speed of the block just before striking the second incline is
`v_(2)=sqrt(2gh_(1))=sqrt(2xx10xx3)=sqrt(60)m//s`

In perfectly inelastilc collision component of `v_(1)` perpendicular to `BC` will become zero, while component of `v_(1)` parallel of `BC` will remain unchanged.
`v_(2)="component of " v_(1)` along `BC`
`=v_(1)=cos30^(@)=(sqrt(60))((sqrt(3))/2)=(sqrt(45))=m//s`