A bullet of mass 1.9×10−3 kg embeds itself in a wooden block with mass 0.987 kg, which then compresses a spring (k = 120 N/m) by a distance 3.5×10−2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.42. What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

Question

A bullet of mass 1.9×10−3 kg embeds itself in a wooden block with mass 0.987 kg, which then compresses a spring (k = 120 N/m) by a distance 3.5×10−2 m before coming to rest. The coefficient of kinetic friction between the block and table is 0.42. What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block?

(delta K) / K = ?

drwls · Answer

(1) Use the compression of the spring to deduce the velocity of the block with embedded bullet, immediately after impact. Assume all of the kinetic energy of the block/bullet combination is converted to spring potential energy PLUS frictional work as the block drags over the table.

(2) Divide the initial KE of the bullet by the combined bullet/block KE that you get from part 1.

Show your work if you wish further assistance

kkj · Answer

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