# Impulse and Momentum of a bullet

1. Jul 30, 2009

### Cheddar

1. The problem statement, all variables and given/known data
A 2.5 g bullet, traveling at a speed of 425 m/s, strikes the wooden block (mass = 215 g) of a ballistic pendulum. (A) Find the speed of the bullet/block combination immediately after the collision. (B) How high does the combination rise above its initial position?

2. Relevant equations
initial velocity = (mass1 + mass2 / mass1) * final velocity
final velocity = square root of (2 * gravity * final height)

3. The attempt at a solution
is the 425 m/s the initial or final velocity?

2. Jul 30, 2009

### cepheid

Staff Emeritus

Well, is it the velocity before or after the collision?

3. Jul 30, 2009

### Cheddar

It is the velocity before the collision. If it is the velocity when it strikes the block then it is the final velocity in its own equation. Final velocity before impact and final velocity of the bullet/block combination are two differenct variables in two different equations.

4. Jul 30, 2009

### cepheid

Staff Emeritus
Initial and final are relative to the collision. The 425 m/s is the initial velocity of the bullet (before the collision). That's it.

The final velocity of the bullet would be the velocity of the bullet after the collision (which would be the same as the velocity of the block after the collision, since they are now combined.

EDIT: "before and after" may have been better words to use than initial and final (which suggests "at the start", and "at the end"), the reason being that we are not talking about a gradual process that starts and ends (like acceleration), but rather a sudden event (like a collision). But now I'm talking about semantics, not physics. I am taking initial = before and final = after in this context.

Also, saying that the 425 m/s is the final velocity of the bullet "in its own equation" makes no sense, because the only relevant equation here is the one that arises from conservation of momentum.

Last edited: Jul 30, 2009
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