The simple pendulum A of mass m A and length l is suspended from the trolley B of mass m B . If the system is released from rest at θ = 0 , determine the velocity v B of the trolley. Friction is negligible.

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Aicharya · Accepted Answer

Linear momentum will be conserved along the horizontal axis.

{\vec{P}}_{trolly} + {\vec{P}}_{ball} = 0 \Rightarrow {\vec{P}}_{trolly} = - {\vec{P}}_{ball} \Rightarrow P_{trolly} = P_{ball} = P ...(1)

so linear momentum of trolly at any instant will be equal to linear momentum of the ball but in opposite direction.

From energy conservation Loss of PE of ball = gain of KE (trolly + ball)

m_{A} g ℓ = \frac{P^{2}}{{2m}_{B}} + \frac{P^{2}}{{2m}_{A}} . . ..(2)

from (1) & (2) P^{2} = \frac{{2m}_{A} g ℓ \times m_{B} m_{A}}{m_{A} + m_{B}} \Rightarrow P = \sqrt{\frac{{2m}_{A}^{2} g ℓ \times m_{B}}{m_{A} + m_{B}}}

P = m_{A} \sqrt{\frac{{2m}_{B} g ℓ}{m_{A} + m_{B}}} \Rightarrow m_{A} v_{A} = m_{A} \sqrt{\frac{{2m}_{B} g ℓ}{m_{A} + m_{B}}} \Rightarrow v_{A} = \sqrt{\frac{{2m}_{B} g ℓ}{m_{A} + m_{B}}}

& m_{B} v_{B} = m_{A} \sqrt{\frac{2 g ℓ}{1 + m_{A} / m_{B}}} \Rightarrow v_{B} = \frac{m_{A}}{m_{B}} \sqrt{\frac{2 g ℓ}{1 + m_{A} / m_{B}}}

The simple pendulum A of mass m A and length l is suspended from the trolley B of mass m B . If the system…

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