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A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle θ₀. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ₁. Then M is given by

Asked in JEE Main 12th Jan 1st Shift 2019 · Velocity exchange and mass ratios

Answer: (2) m((θ₀+θ₁)/(θ₀-θ₁))

Step-by-step solution

For small angles the speed at the bottom is proportional to the angle: v₀∝θ₀ before, v₁∝θ₁ after.

Elastic collision with M at rest: v₁=(m-M)/(m+M)v₀. The bob bounces back, so v₁ is reversed and M>m: (θ₁)/(θ₀)=(M-m)/(M+m).

Cross-multiplying: M(θ₀-θ₁)=m(θ₀+θ₁), so M=m((θ₀+θ₁)/(θ₀-θ₁)), larger than m as a rebound requires.

(The book's key prints the inverted ratio, which would make M<m and could not send the bob back.)

Why the other options are wrong

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