A 0.0252-kg bullet is fired horizontally into a 3.05-kg wooden block attached to

Question

A 0.0252-kg bullet is fired horizontally into a 3.05-kg wooden block attached to one end of a massless, horizontal spring (k = 873 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 within 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 0.262 m. What is the speed of the bullet?

Explanation / Answer

apply the law of conservation of energy and the law of conservation momentum, and solve for speed of buulet.

v = (x/m)*((m+M)*k)1/2 = (0.262/0.0252)*((0.0252+3.05)*873)1/2 = 538.7 m/s

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