# Conservation of angular and linear momentum

1. Nov 5, 2013

### Nikitin

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

http://home.phys.ntnu.no/brukdef/undervisning/tfy4145/ovinger/Ov10.pdf
Look at the picture in "oppgave 1".

Suppose you have an incoming mass which hits the very thin rod straight on in a completely inelastic collision. the incoming mass is $m$, the rod has a mass of $M$, and the little mass hits the rod at a length $l$ from the top.

According to the text, the linear momentum right before and right after are NOT preserved, while the angular momentum is.

3. The attempt at a solution

I calculated the total linear momentum before and after, and indeed I got:

$$p_0 = m v$$
and
$$p_1 = \frac{m v + M L/2l}{MvL^2 /3ml^2 +1}$$

so the two momentums are seemingly unpreserved. Why is this so? I realize it's a translational motion going over to a rotational one, but what does this have to do with linear momentum not being preserved??

Last edited: Nov 5, 2013
2. Nov 5, 2013

### voko

You forgot to mention that the rod is hinged at its top. What is the implication of that?

3. Nov 5, 2013

### Nikitin

That the motion after the collision is rotational?

4. Nov 5, 2013

### voko

That too. But on a more fundamental level, conservation of momentum works only when no external forces act on the system. Is that the case with the hinged rod?

5. Nov 5, 2013

### Nikitin

ahh, of course. How stupid of me. thanks for the help :)