Question

A wedge of mass M rests on a smooth horizontal floor A block of mass "m" is kept on it with a horizontal velocity v The height reached by the block after breaking off from the wedge is (wrt initial velocity level) (Friction is absent)​

Answer 1

Answer:

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Answer 2

SOLUTION :

As the relative sliding between M and m comes to a halt at the maximum height, both M and m  have the same velocity. Linear momentum is conserved in the horizontal direction beacause no external force operates horizontally.

We have,

$(p_{x})_{f}$ = $(p_{x}) _{i}$

where ,

$(p_{x}) _{i}$ = $Mv_i$ and $(p_{x}) _{f}$ = $(M+m)v_{o}i$

Then we have ,  $(M+m)v_{o}$ = $Mv$

using W.E Theorem , $W_{ext}$ + $W_{int}$ = ΔK          → (1)

where , $W_{ext}$ = -Mgh , $W_{int}$ = $W_{contact}$     =  0

And, ΔK = [$\frac{1}{2}$ (M+m) $v_{o} ^{2}$ - $\frac{Mv^{2} }{2}$]

This gives ,

-Mgh = $\frac{1}{2}$(M+m) $v_{o} ^{2}$ - $\frac{Mv^{2} }{2}$    → (2)

substituting v from Eq. (1) in Eq. (2) , we have ,

h = $\frac{v^{2}}{2(1+\frac{M}{m} )g}$