A bullet of mass 0.06 kg hits a 5.0 kg block with an initial speed of 225 m/s. T

Question

A bullet of mass 0.06 kg hits a 5.0 kg block with an initial speed of 225 m/s. The block is connected to a spring. The friction between the block and the table is negligible. Upon impact the bullet bounces back from the box with a speed of 75 m/s.
As a result of the collision the spring compress to a maximum of 0.2 m. Find the spring constant.

This is from a practice test that I am working on. I thought that you used the equation F=kx but that doesn't seem to be getting me the right answer.
The given answer is 1620 N/m and I don't know how they got it.

Explanation / Answer

The mass of bullet, m1 = 0.06 kg The initial velocity of bullet, u1 = 225 m/s The final velocity of the bullet, v1 = 75 m/s The mass of block, m2 = 5 kg The initial velocity of the block, u2 = 0 The final velocity of the block = v2 Since there is some collision is done, the momentum is conserved So, m1u1 + m2u2 = m1v1 + m2v2 From the above the final velocity of the block is given by the formula v2 = ( m1u1 + m2u2 - m1v1)/m2 = 1.8 m/s So the velocity of the block after collision is, v2 = 1.8 m/s We have from the law of conservation of energy (1/2)kx^2 = (1/2)m2v2^2 From the above spring constant, k = m2v2^2/x^2                           = 5(1.8)^2/(0.2)^2                                    = 405 N/m

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