A 32.0 kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k= 32.0 N/m. A 2.00E-2 kg bullet travelling with a speed of 5.500E+2 m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.

asked by Sam

8 years ago

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1 answer

m1v1 = (m1+m2)vf
1/2 (m1 + m2) vf^2 = 1/2 kx^2
solve for x

answered by Chanz

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Question

A 32.0 kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k= 32.0 N/m. A 2.00E-2 kg bullet travelling with a speed of 5.500E+2 m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.

1 answer

m1v1 = (m1+m2)vf
1/2 (m1 + m2) vf^2 = 1/2 kx^2
solve for x

Ask a New Question or answer this question.

Chanz · Accepted Answer

m1v1 = (m1+m2)vf 1/2 (m1 + m2) vf^2 = 1/2 kx^2 solve for x