# Inelastic collision kinetic energy lost?

inelastic collision....kinetic energy lost?

1. Homework Statement
a bullet with a mass of 6 g is fired through a 1.25kg block of wood on a frictionless surface. initial speed of the bullet is 896 m/s and speed of bullet after it passes through it is 435 m/s. whats the final velocity? how much KE is lost?

2. Homework Equations
ke lost=ke initial-ke final
ke lost=1/2mvf-1/2mvi

i got 11.2 m/s for the final velocity of the wood

3. The Attempt at a Solution
KEi=1/2(.006)(896^2) + 0
KEf=1/2(.006)(435^2) + 1/2(1.25)(11.225^2)

i got 2408.228 - 488.925

thats not negative..where did i go wrong??

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PhanthomJay
Homework Helper
Gold Member
Re: inelastic collision....kinetic energy lost?

Don't forget to square the velocity when calculating KE. But first try using conservation of momentum, and show your work, please.

Re: inelastic collision....kinetic energy lost?

edited

PhanthomJay
Homework Helper
Gold Member
Re: inelastic collision....kinetic energy lost?

1. Homework Statement
a bullet with a mass of 6 g is fired through a 1.25kg block of wood on a frictionless surface. initial speed of the bullet is 896 m/s and speed of bullet after it passes through it is 435 m/s. whats the final velocity? how much KE is lost?

2. Homework Equations
ke lost=ke initial-ke final
ke lost=1/2mvf-1/2mvi

i got 11.2 m/s for the final velocity of the wood
how did you arrive at this number?
3. The Attempt at a Solution
KEi=1/2(.006)(896^2) + 0
KEf=1/2(.006)(435^2) + 1/2(1.25)(11.225^2)

i got 2408.228 - 488.925

thats not negative..where did i go wrong??
The change in energy (final minus initial) is negative....if you use (initial minus final), you get your signs reversed. Don't do it that way, the minus signs are bad enough without compounding them.