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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediummultiple choice

Two masses $(m < M)$ are released from rest inside a smooth fixed hemispherical bowl. They collide at the bottommost point.

Options:(select one or more)

Answer:

A, B

Solution:

Both reach the bottom with $u = 2 g R$ , moving oppositely. Elastic formulas: $v_{m}^{'} = \frac{( m - M ) u + 2 M ( - u )}{m + M} = - \frac{( 3 M - m ) u}{m + M}$ and $v_{M}^{'} = \frac{( M - m ) ( - u ) + 2 m u}{m + M} = \frac{( 3 m - M ) u}{m + M}$ . Heights: $h_{m} = R (\frac{3 M - m}{m + M})^{2}$ and $h_{M} = R (\frac{M - 3 m}{m + M})^{2}$ . Answer: (A), (B)

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

⏱ 3mℹ️ Source: QB

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