Minimum Speed for Pendulum Bob to Complete Vertical Circle

[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 a lot 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

 

[Doc Al]

I'm having a lot 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

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.)

 

[Tyst]

Heh, sorry, 'twas at the end of the chapter on elastic, i didn't think about it!
Thanks a lot 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 transferred 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?

 

[Doc Al]

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 $0.5 M V^2$?)

 

[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.