A 1.00 10-2 kg bullet is fired horizontally into a 2.44 kg wooden block attached

Question

A 1.00 10-2 kg bullet is fired horizontally into a 2.44 kg wooden block attached to one end of a massless, horizontal spring (k = 849 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

Since the system oscillates with amplitude of 0.2m, we know that the block can compress the spring *at most* by 0.2m. This means that the block's kinetic energy when it was just hit by the bullet is equal to the potential energy of the spring when it is compressed by 0.2m. We also know that the kinetic energy of the block when it was just hit by the bullet is equal to the kinetic energy of the bullet before it hit to block. So,

0.5*(mass of bullet)*(speed of bullet)2 = 0.5*(spring constant)*(compression)2

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