Help!!!

[Lorax]

63. A pool shark is forced to do a tricky shot in order to win a game. He needs to sink the target ball in the corner pocket at an angle of 30 degrees away from the collision location.

The Cue ball needs to bounce off of the other ball at an angle of 315 degrees with a velocity of of 0.75 m/s. If he gives a cue ball a velocity of 1.00 m/s @ ) degrees what is the velocity of the target ball after the collision?


--I'm just having a hard time visualizing this one, and if someone could help start me off that would be swell.

[vsage]

thinking of 315 degrees as -45 degrees might help some. It's just a simple matter of resolving the momentums in the x and y directions. Let's take the path the cueball takes to the site of collision to be our 0 degrees reference point. Before the collision, it's moving at 1m/s at 0 degrees from the reference angle. After the collision, the cueball is moving at -45 degrees at a velocity of 0.75 m/s. You should have all the information you need to determine the velocity of the other ball given its direction is 30 degrees from the reference angle. Tell me if you'd like to expound or put into equations what I just said.

[Lorax]

Okay so what I've done so far...

I did the M-V-P for the cue ball before collision

M: x
V: 1 m/s
P: 1x

Then I did the after collision mvp for both cue ball and target ball.

Cue: M:x V:0.75 P: 0.75x
Target: M: x V:? P:?

What should I do next?
I'm stuck again, but I'll continue to work on it. Hopefully you can help me out soon. =p

[Lorax]

Anyone? =(

[vsage]

(Didn't see the private message, so absent-minded I am) You have to work out the momentums in the x and y directions separately. That's the only way you can resolve this question. Tell me if you need help setting them up but I don't want to tell you more than you'd like to know :)

[Lorax]

I do I'm completely stumped on this question. If you could maybe write out what your telling me in formula form, that might even help more. Thanks.