What Is the Amplitude of the Block's Motion After a Bullet Is Fired Into It?

[longhorn991]

1. A 4kg block is suspended vertically from a spring with k = 500N/m. A 0.05kg bullet is fired into the block from directly below with a velocity of 150 m/s and becomes embedded in the block. Find the amplitude of the resulting simple harmonic motion.

Homework Equations

V cm = (Bullet mass)(Bullet velocity)/(Bullet mass + block mass)
(1/2)mv^2 + (1/2)Kx^2 = MgX

The Attempt at a Solution

This seems like a pretty straightforward problem, when I solve for the velocity after the the inelastic collision of the bullet and the block, I get 1.8519 m/s, which gives an initial kinetic energy of 6.9448 J. However, there is no solution to the above equation for any positive kinetic energy, so I must be setting up the work theorem equation wrong. Can anyone point out what mistake I am making?

 

[gneill]

When the combined mass is at the top of its range (velocity reaches zero), then all the initial kinetic energy will be converted to gravitational PE and spring PE. So if KE is your initial KE and h₁ is the height above the starting position, and dealing with the magnitudes of the energies,

KE = Mgh₁ + (1/2)k(h₁)²

On the way back down, the mass will pass through the same location with the same kinetic energy (only this time the velocity will be directed downwards). Write another equation similar to the above that will find h₂, the distance below this initial level where the spring will now hold the KE and the additional gravitational PE. The amplitude of the oscillations will be h₁ + h₂.