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For the figure shown, block of mass m is...

For the figure shown, block of mass `m` is released from the rest. Find the distance of the wedge from initial position, when block `m` arrives at the bottom of the wedge. All surfaces are frictionless.

Text Solution

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As there is net external force in the `x` direction, thus the momentum of system in `x` direction `(P_("sys"))_("x")` is conserved.
`"Mv"_(1)="mv"^(2)`,Initially`(P_("sys"))_("x")`
`therefore` displacement of centre of mass in x-direction=0
i.e. `"MX"_(1)="mx"_(2)` ..........(i)
Let the displacement of wedge `M` be `x` backwards
`therefore` displacement of block `=hcotalpha-x`
Using equation(i)
`m(hcotalpha-x)=Mx`
`x=(m)/(m+M)hcotalpha`
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