Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

34 of 130

Question 34 of 130

The block of mass MM moving on the frictionless horizontal surface collides with a spring of spring constant K\mathrm{K} and compresses it by length L\mathrm{L}. The maximum momentum of the block after collision is

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Answer: (A) MKL\sqrt{\mathrm{MK}} \mathrm{L}

12KL2=P22 m∴P=MKL\frac{1}{2} \mathrm{KL}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \quad \therefore \mathrm{P}=\sqrt{\mathrm{MK}} \mathrm{L}

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