Start Free Mock LoginAlready registered? Login here

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 (from the two rims). They collide at the bottommost point.

Options:(select one or more)

Answer:

A, B

Solution:

Both arrive at the bottom with $v = 2 g R$ in opposite directions. Elastic: $v_{m}^{'} = \frac{( m - M ) v - 2 M v}{m + M} = - \frac{( 3 M - m ) v}{m + M}$ , so $h_{m} = \frac{v _{m}^{'2}}{2 g} = R (\frac{3 M - m}{m + M})^{2}$ - (A) true.

$v_{M}^{'} = \frac{( M - m ) ( - v ) + 2 m v}{m + M} = \frac{( 3 m - M ) v}{m + M}, so h_{M} = R (\frac{M - 3 m}{m + M})^{2} - (B) true.$

Answer: (A), (B)

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

⏱ 3mℹ️ Source: QB

Doubts & Discussion

Loading discussions...