A block with mass M attached to a horizontal spring with force constant K is mov

Question

A block with mass M attached to a horizontal spring with force constant K is moving with simple harmonic motion having amplitude A. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it. What should be the value of the putty mass m so that the amplitude after the collision is one-half the original amplitude? For this value of m , what fraction of the original mechanical energy is converted into heat?Express your answer in terms of the variables M, k , and A.

Explanation / Answer

(0.5kA^2 = 0.5Mv_0^2) (Mv_0 = (m+M)v) (0.5(m+M)v^2 = 0.5k(A/2)^2) ((m+M)v^2 = Mv_0^2/4) We get, by subsituting for (v): (4M = m+M) (m = 3M) EDIT: Added energy lost Total energy at max amplitudes, loss = (rac{3}{4})th of the total energy lost in collision

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