A block of mass 'm' is initially lying on a wedge of mass 'M' with an angle of inclination θ, as shown in the figure. Calculate the displacement of the wedge when the block reaches bottom of the wedge.
A
mhM+m
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B
mhcotθM+m
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C
mhcosθM+m
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D
mhsinθM+m
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Solution
The correct option is BmhcotθM+m
l=hcosθ
Let displacement of wedge be x towards right: ΔxCM=m(l+x)+MxM+m 0=ml+(M+m)x x=−mlM+m=−mhcotθM+m