A horizontal rod AB of mass m and length 2l is rotating freely on a frictionless horizontal plane about a fixed vertical axis passing through A with angular velocity ω = 3v/l. The rod collides at B with a particle of same mass m moving with velocity v at the moment when the rod is perpendicular to the path of the moving particle. The coefficient of restitution is 0.5 and the particle rebounds and retraces its path with velocity 13 m/s after the collision. Find v in m/s. (Just before the impact the particle and the point B were approaching each other.)

Engineering

Physics

Collision

Momentum and Energy Conservation for System of Particles

Moment of Inertia

Question

A horizontal rod AB of mass m and length 2l is rotating freely on a frictionless horizontal plane about a fixed vertical axis passing through A with angular velocity ω = 3v/l. The rod collides at B with a particle of same mass m moving with velocity v at the moment when the rod is perpendicular to the path of the moving particle. The coefficient of restitution is 0.5 and the particle rebounds and retraces its path with velocity 13 m/s after the collision. Find v in m/s. (Just before the impact the particle and the point B were approaching each other.)

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$e = \frac{V_{seperation}}{V_{approch}}$

Let ω be the final angular velocity of rod

$\frac{7}{2} v = 13 - 2 ωℓ$ ...(i)

Conservation of angular moemtum about 'A'

$\frac{m (2 ℓ)^{2}}{3} \times \frac{3 v}{ℓ} - 2 mvℓ = 13 \times 2 ℓ + \frac{m (2 ℓ)^{2}}{3} \times ω$ ...(ii)

From eqⁿ (i) and (ii)
v = 8