NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
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.
Options
- Av B = m A m B 2 g ℓ 1 + m A / m B
- Bv B = m A m B 4 g ℓ 1 + m A / m B
- Cv B = m A m B 2 g ℓ 1 - m A / m B
- Dv B = m A m B 4 g ℓ 1 - m A / m B
Correct answer
A. v B = m A m B 2 g ℓ 1 + m A / m B
Step-by-step solution
Linear momentum will be conserved along the horizontal axis. P → trolly + P → ball = 0 ⇒ P → trolly = - P → ball ⇒ 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 = P 2 2m B + P 2 2m A . . ..(2) from (1) & (2) P 2 = 2m A g ℓ × m B m A m A + m B ⇒ P = 2m A 2 g ℓ × m B