Inelastic Collision of a bullet and block

[12boone]

Homework Statement

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?

Homework Equations

1/2mv^2+mgh=1/2mvf^2+mgh
mv=mvf

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 I am doing wrong.

 

[Doc Al]

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

Good!

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

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.

 

[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.

 

[Doc Al]

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

Show how you calculated that speed.

 

[12boone]

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

 

[Doc Al]

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

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

 

[12boone]

wow thank you!