Bullet, Block and Spring

Question:

A 20 gram bullet is fired horizontally t 1200 m/s into a 300 gram block which rests on a smooth surface. The back of the block rests against a spring which is in a relaxed state. The spring has stiffness k=200N/m. Determine the distance the block moves in compressing the spring after the bullet hits the block.

Answer:

Since the collision between the bullet and the block is inelastic, the sum of kinetic and potential energy is not conserved. Some of it ends up as heat in the block and the bullet. Momentum will be conserved though. At the instant the bullet hits the block it momentum is transfered to the bullet block combination. This momentum will establish the velocity of the bullet block combination over that first tiny delta x, before the spring is compressed significantly.

Initially all the momentum is in the bullet at .02kg times 1200m/s = 24 kgm/s. As soon as the bullet becomes one with the block, the mass is .320 kg so the initial block velocity is 24 kgm/s / .320 kg = 75 m/s. The kinetic energy of the block-bullet combination is 1/2*.32*75² = 900 Joules. The spring will compress until this kinetic energy is transfered into potential energy of the spring. The potential energy of a spring is given by pe = 1/2*k*x² where x is the amount of compression. So pe=1/2*200*x²=900 Joules when the block stops compressing the spring. This gives us x²=1800/200 = 9, so x=3m.

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