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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium2 minQB

PhysicsMediumsingle choice

Passage / Comprehension

A particle (P) of mass m is released from rest at point A on the curved surface of a wedge (Q) of mass M, free to move on a smooth horizontal surface. There is no information regarding the nature of the surface of the wedge. The speed of P w.r.t. Q when it reaches B is $v_{2}$ and the speed of Q at this instant is $v_{1}$ .

Consider the system comprising of P and Q. Linear momentum can not be conserved in vertical direction because, there is non zero vertical net force of gravity and normal from ground. The conservation of linear momentum in horizontal direction for this system gives

Options:

Answer:

Solution:

P's ground-frame horizontal velocity is $v_{1} - v_{2} cos θ$ (wedge velocity plus the horizontal component of P's velocity relative to Q).

$M v_{1} + m (v_{1} - v_{2} cos θ) = 0$

Ans. (C)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 2mℹ️ Source: QB

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