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A wedge Q of mass M is placed on a horizontal frictionless surface AB and a block P of mass m is released on its frictionless slope. As P slides by a length L on this slope of inclination θ, Q would slide by a distance of

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a

(mM)L cosθ

b

mL(M+m)

c

(M+m)(mL cos θ)

d

(mL cos θ)(m+M)

answer is D.

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Detailed Solution

Let Q slides by a distance r towards left, then net horizontal displacement of P w.r.t. ground= L cos θ-xNow apply m1x1 = m2x2 ⇒Mx = m(L cosθ-x)⇒x = mL cosθM+m

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