A block of mass ‘m’ moving on a frictionless surface at speed ‘v’ collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle ‘θ’ with its initial direction and has speed ‘ v 1 ’ . The speed of t

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A block of mass ‘m’ moving on a frictionless surface at speed ‘v’ collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle ‘θ’ with its initial direction and has speed ‘ v 1 ’ . The speed of the second block after collision is

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Correct answer: B. v 2 - v 1 2

Applying law of conservation of kinetic energy, KE (before collision) = KE (after collision) 1 2 m v 2 + 1 2 m 0 2 = 1 2 m v 1 2 + 1 2 m v 2 2 ⇒ v 2 = v 1 2 + v 2 2 ⇒ v 2 = v 2 - v 1 2 Thus, the velocity of second block after collision is v 2 - v 1 2 .

A block of mass ‘m’ moving on a frictionless surface at speed ‘v’ collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle ‘θ’ with its initial direction and has speed ‘ v 1 ’ . The speed of the second block after collision is

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