A block of mass M = 5.30 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 5.30 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 6400 N/m. A bullet of mass m = 9.90 g and velocity of magnitude 500 m/s strikes and is embedded in the block (Fig. 15-38). Assuming the compression of the spring negligible until the bullet is embedded, determine the amplitude of the resulting simple harmonic motion.

Explanation / Answer

(9.90/1000)*500 = (9.90/1000+5.3)*v >>>> v = 0.93222 m/s 0.5 k A^2 = 0.5 m v^2 >>> A = (m v^2/k)^0.5 = ((9.90/1000+5.3)*0.93222*0.93222/6400)^0.5 = 0.0269 m/s

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