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A block of mass M with a smooth (frictionless) circular cut of radius R rests on a frictionless horizontal table. The block's right edge is initially at x = 0. A small particle of mass m is released from rest at the top of the circular track and slides down. When the particle finally leaves the block, let its position be x and speed be v. Which of the following statements are correct? (Multiple correct)

1. The horizontal displacement of the centre of mass of block M is -mR/(M + m)
2. The position of the point mass is x = -sqrt(2)*mR/(M + m)
3. The speed of the point mass m is v = sqrt(2gR/(1 + m/M))
4. The speed of the block M is V = (m/M)*sqrt(2gR)

Correct answer: The speed of the point mass m is v = sqrt(2gR/(1 + m/M))

Solution

Momentum conservation gives MV = mv, and energy conservation mgR = (1/2)mv² + (1/2)MV²; eliminating V yields v² = 2gR/(1 + m/M), so option (C) is correct (this is the classic JEE 2017 answer where A, B, D as printed are not all correct).

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