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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 $v_{1}$ .

Now, let the surface of wedge be smooth so that there is no loss of energy. For the system of P and Q, the conservation of mechanical energy yields.

Options:

Answer:

Solution:

$v_{P}^{2} = (v_{1} - v_{2} cos θ)^{2} + (v_{2} sin θ)^{2} = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos θ$

$\frac{1}{2} M v_{1}^{2} + \frac{1}{2} m (v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos θ) = m g h$

Ans. (A)

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

⏱ 2mℹ️ Source: QB

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