The block of mass M moving on the frictionless horizontal surface collides with a spring of spring constant K and compresses it by length L . The maxi - Physics MCQ AIEEE 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

At maximum compression, the kinetic energy of the block is fully converted into elastic potential energy of the spring. \frac{1}{2} M v^{2} = \frac{1}{2} K L^{2} M v^{2} = K L^{2} ⟹ v = L \frac{K}{M} Momentum P = M v : P = M (L \frac{K}{M}) = L M K Alternatively, P = 2 M (K E) = 2 M (\frac{1}{2} K L^{2}) = L M K .

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

M K L

\frac{K L ^{2}}{2 M}

zero

\frac{M L ^{2}}{K}

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