Inelastic Collision of a bullet and block

1. Oct 19, 2008

12boone

1. The problem statement, all variables and given/known data
A bullet is fired vertically into a 1.20 kg block of wood at rest directly above it. If the bullet has a mass of 24.0g and a speed of 540 m/s , how high will the block rise after the bullet becomes embedded in it?

2. Relevant equations

1/2mv^2+mgh=1/2mvf^2+mgh
mv=mvf
3. The attempt at a solution

I know that for this i must use momentum to find the final v which would be

mV=(m+M)V'
then i use that V' to find height

1/2mV'^2=(m+M)gh

I solved this and I got an answer of 413 meters this is wrong. I do not know what im doing wrong.

2. Oct 19, 2008

Staff: Mentor

Good!

Make sure you use "m+M" on both sides. (They cancel, of course.)

If that's not the problem (and I don't think it is), show your calculations, step by step.

3. Oct 19, 2008

12boone

ok i did that and it is still wrong. i did .5(.240+1.20)(90)^2=(.240+1.20)(9.80)h

i put the left side in the calc. and got 5832=(.240+1.20)(9.80)h then divided the right numbers by the left and my answer is 413.26m which is wrong according to mastering physics.

4. Oct 19, 2008

Staff: Mentor

Show how you calculated that speed.

5. Oct 19, 2008

12boone

I multiplied .240kg(540m/s) then divided that by (.240+1.2).

6. Oct 19, 2008

Staff: Mentor

There's the problem. The bullet's mass is 24 grams = 0.024 Kg (not 0.24 Kg).

7. Oct 19, 2008

12boone

wow thank you!