Homework Help: Elastic Collision question

1. Mar 26, 2006

Tyst

Q: A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pundulum bob will barely swing through a complete vertical circle?

I'm having alot of trouble with this question! I just can't seem to get my head around where to start, i've implemented the conservation of momentum equation;
mv(i) + MV(i) = mv(f) + MV(f)
as
mv = m(v/2) + MV
but i really have no idea how they want the question answered, or how to go about anwering it... :uhh: anyone have any insight?
Thanks for the time

2. Mar 26, 2006

Staff: Mentor

So far, so good! Conservation of momentum does apply during the collision of bullet with pendulum. So this allows you to relate bullet speed (pre-collision) with pendulum speed (post-collision).

Now you need to know what minimum speed must the pendulum have to just make it swing up in a complete circle. Hint: What's conserved as the pendulum swings?

(By the way, the collision of bullet with pendulum is not an elastic collision.)

3. Mar 26, 2006

Tyst

Heh, sorry, 'twas at the end of the chapter on elastic, i didn't think about it!
Thanks alot for your help... would i be right, then, in thinking that the pendulum needs only enough E to reach just past the top of it's rotation, after which, gravity will act to complete the swing.
In that case, all of the kinetic E transfered to the bob will have been converted to potential E at the top of the swing...
U=mgh=K transfered=.5Mv^2
mg2l=.5M(v/2)^2
...
...
2*(mgl/M)^-2
is this along the right track?

4. Mar 26, 2006

Staff: Mentor

You're almost there. The KE of the pendulum at the bottom position (just after the collision) must be enough to get it to the top position: use energy conservation. But what's the KE of the bob in terms of the bullet's speed? (In other words, what is $0.5 M V^2$?)

5. Mar 26, 2006

Tyst

[2*(mgl/M)^1/2 not 2!!! is what i ment :P]

Aaahhh! now i get it, Thanks very much for your help Doc, really appreciated.