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## Homework Statement

A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block, as in Figure P9.67. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find

(a) the speed at which the bullet emerges from the block and

(b) the mechanical energy converted into internal energy in the collision.

## Homework Equations

conservation of momentum

KE=1/2 mv

^{2}

## The Attempt at a Solution

I solved part a, getting an answer of 100m/s for the speed at which the bullet emerges from the block. But I'm stuck at part b.

I interpreted the question as: finding the mechanical energy tat was converted = change in mechanical energy = change in kinetic energy of the bullet = 1/2 m(v

_{f}-v

_{i}) = (1/2)(5/1000)(100-400)

^{2}= 225J

The correct answer should be 374J.

Pls help, thanks! :)