1. A 0.0340 kg bullet traveling at 120. m/s embeds itself in a 1.24 kg wooden bl

Question

1.     A 0.0340 kg bullet traveling at 120. m/s embeds itself in a 1.24 kg wooden block which is at rest on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k = 99.0 N/m). Calculate how far the block-bullet compresses the spring.

Please explain with great detail.

A 0.0340 kg bullet traveling at 120. m/s embeds itself in a 1.24 kg wooden block which is at rest on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k = 99.0 N/m). Calculate how far the block-bullet compresses the spring. Please explain with great detail.

Explanation / Answer

First find the speed of the block immediately after being hit, using conservation of momentum:

(mv)1 = (mv)2

inital momentum of bullet + initial momentum of blck = final momentum of bullet + block

(0.034 kg)(120 m/s) + (1.24 kg)(0 m/s) = (1.24+0.034 kg)(v)

v = 3.2 m/s ............. velocity of bullet + block

Now use conservation of energy to find how far the spring is compressed from the moment the block-bullet mass hits the spring, till when it stops and the spring is fully compressed:

KE1 + PE1 = KE2 + PE2

KE = (1/2)mv^2

PE = (1/2)kx^21 ? spring potential energy. Since the block doesn't change height, the gravitational potential energy term is zero for this problem.

(1/2)(1.24+0.034 kg)(3.2 m/s)^2 + (1/2)(19.6 N/m)(0 m)^2 = (1/2)(1.24+0.034 kg)(0 m/s)^2 + (1/2)(99 N/m)(x)^2

x = 0.36 m or 36 cm ........compression

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