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## Homework Statement

A 4.0kg block is suspended from a spring with force constant of 500N/m.

A 50g bullet is fired into the block from directly below with a speed of 150m/s and is imbedded in the block.

Find the amplitude of the resulting simple harmonic motion.

## Homework Equations

F=-kx

E_p=mgh

E_k=(mv^2)/2

E_pspring=(kx^2)/2

m1v1=m2v2

## The Attempt at a Solution

Using conservation of momentum:

0.05(150)=(4+0.05)v2

v2=1.852 m/s

I am lost on how the decide where the equilibrium position is. Is it the resting position with the block considered the equilibrium position, or is it without the block, or is it with the block and the bullet?

I said that the kinetic energy

0.5*(4.05)(1.852)^2 is equal to 0.5k(x)^2 and completely ignored mgh(potential energy). Then i assumed x is the amplitude. This seems incorrect of course.

Any help would be appreciated, thanks!