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
A
`sqrt(120) m//s`
B
`sqrt(105) m//s`
C
`sqrt(90)m//s`
D
`sqrt(75) m//s`
Text Solution
Verified by Experts
The correct Answer is:
D
Height fallen by the block from `B` to `C` `h_(2)=3sqrt(3) tan30^(@)=3m` Let `v_(3)` be the speed of the block, at point `C` just before it leaves the second incline, then `v_(3)=sqrt(v_(2)^(2)+2gh_(2))=sqrt(45+2xx10xx3)=sqrt(105) m//s`
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