# Bullet Velocity

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.

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Doc Al
Mentor
I was thinking mgh = 1.2mv^2 but that wasn't it.
That's part of what you need. Conservation of energy applies after the collision.

What applies during the collision?

tiny-tim
Homework Helper
AI was thinking mgh = 1.2mv^2 but that wasn't it.
You're nearly there. But you haven't used M, have you?

Try again … Conservation of Momentum.

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..

Doc Al
Mentor
Looks good.

tiny-tim
Homework Helper
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..
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! Doc Al
Mentor
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.
Huh??? (Mass cancels out.)

tiny-tim
Homework Helper
Huh??? (Mass cancels out.)
No - the inital KE is in the bullet (and the final KE is zero).

The PE involves the block also.

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

The PE involves the block also.
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.

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