# Combination of conservation laws, and older things. Need Help.

1. Dec 4, 2011

### yttuncel

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

A small wooden block, of mass M, lies in the middle of a horizontal table of length L
and height h above the floor. The coefficient of kinetic friction between this block
and the surface of the table is μ. A bullet, of mass m, is shot with a horizontal
velocity into the block. As a result, the block (with the embedded bullet) starts
moving across the table and finally lands on the floor a horizontal distance D from
the edge of the table. Use this information to find the speed of the bullet.

2. Relevant equations

ΔP=0 , f = N.μ, ΔE = Woth (1/2mV2 or mgh or f.x), x=V.t, h=1/2gt2

3. The attempt at a solution
Let Vi= speed of the bullet

Well, first i found the final velocity of the block on the table with the bullet inside, then applied energy law to the block BTWN the moments 1 to 2. (1. just off the table 2. on the ground & stopped.) Then i found t, to calculate D from projectile motion relations, and used all i found in the energy law; thus i found :

Vi= √(μ.L+(D2g)/4h)*√2*((M+m)/m)

Is that answer true, I could not find it anywhere else so I need verification please ...
Thanks! :)

2. Dec 4, 2011

### LawrenceC

Vi= √(μ.L+(D2g)/4h)*√2*((M+m)/m)

How can you add L + D2g/4h? Something is wrong.

mu*L has dimensions of length
2g*D/4h does not have dimension of length

You also have an inelastic collision here.

3. Dec 4, 2011

### yttuncel

It is not 2g, it is D^2 times g. But still the dimensions are not alike.

4. Dec 4, 2011

### BruceW

I got a slightly different answer to you yttuncel. I think you should go through it once again. You can tell it is not right because this: μ.L doesn't have the correct units.

Your answer is close, so maybe you accidentally wrote something in the wrong place while doing the calculation?

5. Dec 4, 2011

### yttuncel

By the way LawrenceC, I did not applied the energy law directly before the collision, I found the velocity after collision then applied energy law. So as energy after collision is conserved, that would be no problem I think.

BruceW, is it Vi= √(μ.g.L+(D2g)/4h)*√2*((M+m)/m) ?

I forgot putting g in Normal force :/

6. Dec 4, 2011

### BruceW

You are even closer, but still not quite right. Did you take into consideration that the block only travelled a distance L/2 along the table? (Since it was originally placed in the middle)

7. Dec 4, 2011

### LawrenceC

Then delta P = 0 is conservation of momentum. Fine.....I did not realize that's what it meant upon first looking at post.

Last edited: Dec 4, 2011
8. Dec 4, 2011

### yttuncel

Didnt even see that mentioned in the question :/ Thank you both for your comments and help :)

Edit: I think I did it wrong again. Because the final velocity on the table is not equal to the velocity just after the collision

Edit2: Ok got it. Lol. Thanks again!

Last edited: Dec 4, 2011