# Conservation of Momentum problem

1. Jan 13, 2009

### toasted

1. The problem statement, all variables and given/known data

A gun is fired vertically into a 1.40kg block of wood at rest directly above it. If the bullet has a mass of 21.0g and a speed of 310m/s, how high will the block rise into the air after the bullet becomes imbedded into it?

2. Relevant equations
m1v1+m2v2= m1v11+ m2v12

3. The attempt at a solution
So I first converted 21.0 grams into .021kg because I know you use kg for this equation.
After substituing in a few things this is as far as I have gotten:

m1v1+m2v2= m1v11+ m2v12
(.021)(0)+(1.40)(0)=(.021)(310)+(1.40)(V)
V= -4.65m/s as the final velocity of the block

so from here I have a few questions first did I do this part correctly and from here how would I proceed in order to determine how high the block will go?

2. Jan 13, 2009

### LowlyPion

Not sure where your - sign is from, but for positive y up, 4.6 m/s looks about right.
(Your equations look like there are typos.)

Now you know your mv for the combined block/bullet and that Kinetic energy = ½mv² , and that energy will eventually be consumed by gravity with mgh which is the height you are looking for.

3. Jan 13, 2009

### toasted

Thank you ever so much! That makes so much sense!