# Bullet Velocity

1. Mar 8, 2008

### am08

A bullet of mass m= 3.10×10-2kg is fired along an incline and imbeds itself quickly into a block of wood of mass M= 1.35kg. The block and bullet then slide up the incline, assumed frictionless, and rise a height H= 1.25m before stopping. Calculate the speed of the bullet just before it hits the wood.

What equation do I need?

I was thinking mgh = 1.2mv^2 but that wasn't it.

2. Mar 8, 2008

### Staff: Mentor

That's part of what you need. Conservation of energy applies after the collision.

What applies during the collision?

3. Mar 8, 2008

### tiny-tim

You're nearly there. But you haven't used M, have you?

Try again …

4. Mar 8, 2008

### physixguru

Conservation of Momentum.

5. Mar 8, 2008

### am08

So i used v2 = squareroot (2gy) and found the final velocity to be 5 m/s

then i used v1 = (mass bullet + mass block / mass bullet) * v2

is this right? I'm using conservation of momentum..

6. Mar 8, 2008

Looks good.

7. Mar 8, 2008

### tiny-tim

I'm not convinced.

Don't forget - the "final velocity" is zero - you're looking for the initial velocity (the velocity of the bullet just before everything happens).

Your "v = √(2gy)" is wrong - the mass that should be on the left is not the same as on the mass on the right, and so you must put them both in.

(btw, if you type alt-v, it prints √ for you)

Try the √2gy equation (conservation of energy) again!

8. Mar 8, 2008

### Staff: Mentor

Huh??? (Mass cancels out.)

9. Mar 8, 2008

### tiny-tim

No - the inital KE is in the bullet (and the final KE is zero).

The PE involves the block also.

10. Mar 8, 2008

### Staff: Mentor

Energy is conserved after the collision:

$$1/2 (M + m)v_2^2 = (M + m)gy$$

Thus:
$$v_2 = \sqrt{2gy}$$
is perfectly correct. (Note that v_2 is the speed of block + bullet immediately after the collision.

11. Mar 8, 2008

### Oerg

woohooo tiny-tim managed to force the answer out of Doc AI!