Bullet being shot into a block

Question

A 0.0150kg bullet is shot into a 4.0kg block which is at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 3.8cm. The spring constant is 2500 N/m.

What percentage of mechanical energy was lost in this experiment?

Answer 1

The first part of this problem is an inelastic collision momentum problem to find the velocity after collision But you are missing two pieces of information. So, lets tackle the spring problem first to find the velcoity after collision.

using the conservation of energy: K_before = U_{sp after}

1/2 (4.015) v² = 1/2 (2500) (0.038)² this v is v after collision

v = 0.94 m/s

Now, we do our mommentum problem:

(0.0150)(v_before) = 4.015 (0.948)

v_before = 254 m/s

Now find the ratio or percent of energy left and subtract from 100% to get energy lost.

1/2 (2500)*(0.038)² / 1/2 (0.015)(254)² = 1.805 / 484 = 0.37% so 99.63% lost

This seems like a strange answer but I cannot see any computational errors. The process is sound.