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The block of mass `M` moving on the frictionless horizontal surface collides with the spring constant `k` and compresses it by length `L` . The maximum momention of the block after collision isA. `(ML^(2))/(K)`B. zeroC. `(KL^(2))/(2M)`D. `sqrt(MKL)` |
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Answer» Correct Answer - D When block of mass `M` collides with the spring, its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy, `(1)/(2) Mv^(2) = (1)/(2) KL^(2) : v = sqrt((K)/(M)) L` Where `v` is the velocity of block by which it collides with spring. So, its maximum momentum, `P = Mv = M sqrt((K)/(M)) L = sqrt(MK) L` After collision, the block will rebound with same linear momentum. |
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