# Conservation of momentum - Collision of pool balls

1. Apr 8, 2005

### crazedbeat

Hi,

I am having trouble recalling my concepts in momentum. I just remember its conserved

Plus, I am horrible with directions, so need help understanding this problem, as well as the underlying concepts.

Two pool balls are rolling along a pool table. The orange pool ball has a velocity of 15 cm/s west and the green ball a velocity of 15 cm/s north 30° east. What are the velocities of the balls after the collision?

What I am doing is as follows:

15*m + 15*m = m(15 + 15)
which doesn't take me anywhere

Do I need to resolve into x and y components?

So any help on this question would be greatly appreciated. I will try and return the help by helping others. (i.e. pay it forward!)

Last edited: Apr 8, 2005
2. Apr 8, 2005

### Nylex

Yeah, you need to resolve the velocities into components.

3. Apr 8, 2005

### crazedbeat

Okay, so if I resolve them, what exactly does North 30 degrees east mean? Is it north or east or north east? If its north east, whats the 30 degrees about?

How do I draw the diagram? The way I am drawing it right now, they never appear to collide?

4. Apr 8, 2005

### whozum

The oragne ball will be a vector in the negative x direction magnitude 15cm. The green ball has 15cm north+30 degrees east. I would interpret this to be 30 degrees east of north, which would be 60 degrees from the x axis. At the point of collision, the two vectors will be nose-to-nose, so you can start from that point and draw the vectors from there. You can break down the green ball's vectors by making a triangle and using sin and cosin to find each component.

Then conserve momentum in each direction.

5. Apr 8, 2005

### Quantum Cat

remember the i component is (magnitude)cos(angle from + x axis) and the j component is (magnitude)sin(angle from + x axis) then all you have to do is add them, and convert them back into angles and magnitudes (tan^-1 j/i is angle from + x axis, magnitude is sqrt (i^2 + j^2))

Or you can just draw a vector diagram and work it out with sin and cos rules.

6. Apr 9, 2005

### crazedbeat

So then would this be correct?

Momentum is conserved

x direction:
$$mv_{1} + mv_{2} = 2mv_{x}$$
$$10 + 10cos(60) = 2v_{x}$$
$$15 = 2v_{x}$$
$$7.5 = v_{x}$$

y direction:
It remains the same $$10sin(60)$$ for $$m_{2}$$

I just want to know if I did this correctly :) After this, I would simply have to find the direction and composite velocity for the green ball.