# Momentum with an angle

1. Apr 10, 2007

### fishert16

Two balls of masses Ma=40g and Mb=60g are suspended as shown in the picture in your text. The lighter ball is pulled away to a 60 degree angle with the verticle and released.
a) What is the velocity of the lighter ball before imapct?
b) What is the veloctity of each ball after the elastic collision?
c) What will be the maximum height of each ball after the elastic collision.

Relevant equations
Im not sure how to incorporate the angle into the equations i know.
I have the equations MaVa=MaV'aCosThetaa' + MbV'bCosThetab'

The attempt at a solution

I have not attempted to solve this the prime in the equation is throwing me off. Also I could easily solve this problem without the angle so I do not know how to incorporate this into the equations I know.

2. Apr 10, 2007

### e(ho0n3

I don't understand your equation. In any case, forget about the equation for now and think about how you would find the velocity of the lighter ball.

3. Apr 10, 2007

### fishert16

Well wouldnt you need to find the angle and include the acceleration of gravity

4. Apr 10, 2007

### hage567

You've been given the angle. If an object is raised above the ground, what can you say about the energy it has at that point? How would you find it? Yes, acceleration due to gravity will be in there.

5. Apr 10, 2007

### fishert16

the PE is the only energy it has before released correct. I dont know how this will help me find velocity. Im really looking for the correct formula to use here.

6. Apr 10, 2007

### hage567

Think convservation of energy. If it returns to the original height what can you then say about the kinetic energy it has at the point just before it hits the other ball? What's the formula for kinetic energy?

This is getting you the formula you need. Sometimes you need to work them out yourself. Where you given the length of the string in your question/diagram?

7. Apr 10, 2007

### fishert16

yes the string is 30cm

8. Apr 10, 2007

### fishert16

the only formula for KE is KEa+KEb = KE'a + KE'b is this correct

9. Apr 10, 2007

### hage567

Yes that will probably be useful later, but that is not what I'm talking about for finding the initial velocity of the lighter ball. If it is raised with respect to its original position, it will have a potential energy of mgh, right? So when it is released it will come back down and all of that potential energy must be conserved and become kinetic energy (so 0.5*m*v^2). So, you must find a way to express h in terms of your angle and length of string. Then you can find v from equating your potential and kinetic energy terms.