# Bullet physics question

1. Oct 21, 2007

### mathcrzy

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

A gun is fired vertically into a 1.80 kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 190 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
_____m

2. Relevant equations

Vf^2-Vi^2=2ax

3. The attempt at a solution

Vf=0
Vi=190
a=9.8
x=?

0^2-190^2=2*9.8*x
x=1841.83m

2. Oct 21, 2007

### G01

The answer you have found is how high the bullet will rise in the air if it was NOT embedded in the block, i.e. if the block was never in the way. Here is a hint to get you on the right track:

Remember we are talking about how high the BLOCK moves after it is hit by the bullet. Does conservation of momentum tell you anything about the block's motion?

3. Oct 21, 2007

### mathcrzy

Conservation of momentum is going to be the same before and after the collision. so the momentum of the block upwards is going to be the same as the bullet before it hits the block.? would i then use the equation (before collision)M1V1+M2V2=M1V1+M2V2(after collision)

4. Oct 21, 2007

### G01

Exactly

Now, what would that equation tell you? Do you know where to go from there?

5. Oct 21, 2007

### mathcrzy

M1 would be the bullet .021kg
V1 is the bullets velocity 190m/s (initial) 0m/s (final)
M2 is the block 1.8kg
V2 is the block's velocity 0m/s (initial) ?m/s (final)

Then find the V2 final
.021*190+1.8*0=.021*0+1.8*?
3.99=1.8?
2.2=V2 final

Now how do i find delta x for the block?

6. Oct 21, 2007

### G01

Ok, now that you have the velocity of the block after the collision, this becomes a kinematics problem. Can you find a way to solve it with kinematics? HINT: You had this part down in you original post...