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A wedge of mass M = 4m lies on a frictio...

A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by:

A

`(2 v^(2))/(5g)`

B

`(v^(2))/(g)`

C

`(2 v^(2))/(7g)`

D

`(v^(2))/(2g)`

Text Solution

Verified by Experts

The correct Answer is:
A
By conservation of L momentum
`mv_(0) = 5 mv`
`v' = (mv_(0))/(5m) = (v_(0))/(5)`
Conservation of M.E
`(1)/(2) MV_(0)^(2) = (1)/(2) 5m (v')^(2) + mgH, H = (2v_(0)^(2))/(5g)`
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