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Question

A block of mass m is pushed towards a movable wedge of mass nm and height h, with a velocity u. All surfaces are smooth. The minimum value of u for which the block reaches the top of the wedge is -

A
2gh
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B
2ngh
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C
2gh(1+1n)
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D
2gh(11n)
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Solution

The correct option is C 2gh(1+1n)
When block just reaches the top, it is at rest wrt wedge, so that u is minimum.
Let common velocity be v

From conservation of linear momentum,
mu=(m+nm)v
v=u1+n
From energy conservation
12mu2=mgh+12(m+nm)v2
u=2gh(1+1n)

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