A block of mass M = 5.80 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 5.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5450 N/m. A bullet of mass m = 9.20 g and velocity vector v of magnitude 690 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine the amplitude of the resulting simple harmonic motion. Number 0.35685 Units m The number of significant digits is set to 3; the tolerance is +/-2%

Explanation / Answer

initial momentum of the bullet = Pi = 9.2e-3*690 = 6.348 kg m /s

after embeded the final momentum Pf = (m+M)*V

Pf = Pi

V = 6.348/(9.2e-3+5.8) = 1.093 m/s

after collision KE of the bullet+block PE stored in spring

0.5*(M+m)*V^2 = 0.5*K*A^2

A = 0.03568 m

A = 3.57 cm

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