A 0.0315-kg bullet is fired horizontally into a 4.83-kg wooden block attached to

Question

A 0.0315-kg bullet is fired horizontally into a 4.83-kg wooden block attached to one end of a massless, horizontal spring (k = 806 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 0.171 m. What is the speed of the bullet?

Explanation / Answer

using energy conservation,

intial kinetic energy = (4.83 +0.0315)v^2 /2

initial spring PE = 0

final KE = 0

final spring PE = kx^2 /2 = 806 x 0.171^2 /2

(4.83 +0.0315)v^2 /2 + 0 = 0 + 806 x 0.171^2 /2

v = 2.20 m/s

now using momentum conservation on block - bullet,

initial momentum = final momentum

0.0315u + 4.83 x 0 = (4.83 +0.0315) x 2.20

u = 339.81 m/s

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