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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A particle of mass m1\mathrm{m}_1 collides head-on with another stationary particle of mass m2(m2>m1)\mathrm{m}_2\left(\mathrm{m}_2 > \mathrm{m}_1\right). The collision is perfectly inelastic. The fraction of kinetic energy which is converted into heat in this collision is
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Correct option: A
m2/(m1+m2)\mathrm{m}_2 /\left(\mathrm{m}_1+\mathrm{m}_2\right)
Option A (m2/(m1+m2)\mathrm{m}_2 /\left(\mathrm{m}_1+\mathrm{m}_2\right)) is correct because it represents the fraction of kinetic energy converted into heat during a perfectly inelastic collision, derived from conservation of momentum and energy loss calculations.

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