- #1

XxBollWeevilx

- 78

- 0

**[SOLVED] Angular momentum and collisions**

## Homework Statement

A wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length [tex]\ell[/tex] and of negligible mass. The rod is pivoted at the other end. A bullet of mass m and traveling parallel to the horizontal surface and normal to the rod with speed [tex]\vec{v}[/tex] hits the block and gets embedded in it. What fraction of the original kinetic energy is lost in the collision?

## Homework Equations

Fraction = (K[tex]_{f}[/tex]-K[tex]_{0}[/tex]) / K[tex]_{0}[/tex]

v[tex]_{f}[/tex] = (mv / M+m)

## The Attempt at a Solution

The main thing that confused me here is the way the collision works and what the energy is before and after the collision. I said that before the energy before would simply be (1/2)mv^2, the kinetic energy of the bullet. Then after, the kinetic would be (1/2)(M+m)v^2 where the masses are combined and v is the final speed. I got the equation for final velocity above from my collision formulas for inelastic collisions, but each time I try to work it out, I cannot get the correct answer, which is M / (M+m). Also, I am confused by how I find the fraction of the original energy lost...is my fraction formula above correct, or is it simply final energy over initial energy? I'm just having trouple conceptualizing it. Thanks so much.