A block Q of mass M is placed on a horizontal frctionless surface AB and a body P of mass m is released on its frictionless slope. As P slides by a length L on this slope of inclination θ, the block Q would slide by a distance
A
mMLcosθ
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B
mM+mL
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C
M+mmLcosθ
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D
mLcosθm+M
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Solution
The correct option is DmLcosθm+M As there is no external force along X direction. Here, the centre of mass of the system remains unchanged in X direction.
When the mass m moves a distance Lcosθ on M, let the mass (m+M) moves a distance x in the backward direction. ∴(M+m)x−mLcosθ=0 ∴x=(mLcosθ)/(m+M)
Hence, the correct answer is option (d).