A 1.15 10^-2 kg bullet is fired horizontally into a 2.49 kg wooden block attache

Question

A 1.15 10^-2 kg bullet is fired horizontally into a 2.49 kg wooden block attached to one end of a massless, horizontal spring (k = 841 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt with in it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 m. What is the speed of the bullet?

Explanation / Answer

First, from the conservation of energy we can find the speed of the block/bullet combo

For the Spring, PE = .5kA2    For the mass, KE = .5mv2

So,

(841)(.2)2 = (2.49 + 1.15 X 10-2)(v2)

v = 3.67 m/s

Then, from the conservation of momentum, we can find the initial speed of the bullet

The initial momentum is only that of the bullet.

(1.15 X 10-2)vb = (2.49 + 1.15 X 10-2)(3.67)

vb = 612 m/s

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