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(1−1n)
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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)v′2 u=√2gh(1+1n)