# Homework Help: Calculate momentum and velocity of ball after collision

1. Dec 14, 2004

### IGeekbot

HOw do I do these two problems. I know it has something to do with angles the ball trike each other, but I am so Lost.

Problem 1.
Ball A, rolling west at 3.0m/s, has a mass of 2.0 kg. Ball B is 2.0 kg,and is stationary. After colliding with ball b, ball a moves south at 2 m/s. calculate the momentum and velocity of ball B after the collision.

Problem 2.
A cue ball, moving with 7.0 Ns of momentum strikes the nine ball at rest. The nine ball moves off with 2.0Ms in the original direction of the cue ball and 2.0Ns perpendicual to that direction. What is the momentum of the cue ball after the collision.

I have no idea what to do with these.

2. Dec 14, 2004

### cyby

Are you familiar with 2d collisions and the principles of conservation of momentum?

3. Dec 14, 2004

### IGeekbot

yes, i have my book here in front of me with equations, but im not sure which one to use.

4. Dec 14, 2004

### cyby

Well, consider using conservation of momentum for two dimensions (meaning that you break them up into x and y components)...

Show us what you are trying to do in equations, and we'll work from there :D

5. Dec 14, 2004

### IGeekbot

thats just it, i dont know what to do. Our teachers idea of introducing a new concept is go through it in five minutes to move on to a lab. He justrecites the book so it is no help. I know we have to use the sin and cos of one of the veloctiies, but i am not sure which one. When i get osm calculations going here, ill show ya

6. Dec 14, 2004

### cyby

In your particular case, you don't have to. Are you familiar with unit vector notation to express vector quantities?

7. Dec 14, 2004

### cyby

Let's just look at #1 for now.

By conservation of momentum, you know that the momentum in each directional component has to be equal before and after the collision. Right?

Now, what is the original momentum:

1. Of ball A in the x (east-west) component?
2. Of ball A in the y (north-south) component?

For simplicity purposes, let's just east as +x, and north as +y.

8. Dec 14, 2004

### IGeekbot

ive probably seen them, but that is not ringing a bell. could you please explain what it is?

9. Dec 14, 2004

### cyby

The unit vector notation basically breaks down a vector quantity into its directional components (i, j, k) or (x, y, z), depending on what notation you like to use. This way, you can express a quantity say, 3 units to the east and 4 units to the north, as 3i + 4j, for instance. This quantity is the same as saying 5 units (~51 degrees north of east)... familiar?

10. Dec 14, 2004

### IGeekbot

The ball would have the original momentum of 3kg x m/s moving west. I got that, but this is where I get lost, because this is all new to me.

11. Dec 14, 2004

### cyby

Ok. Why don't we start even further back. Can you type out the conservation of momentum equations at least? :-)

12. Dec 14, 2004

### cyby

13. Dec 14, 2004

### cyby

you mean 3.0m/s * 2.0kg = 6.0 kg*m/s...

14. Dec 14, 2004

### IGeekbot

To calculate momentum, it is P=mv.
In a closed system you have Pa+Pb=P'a+P'b

Then we have the ones for elastic and inelastic equations, but i don't think those are needed here.

15. Dec 14, 2004

### IGeekbot

yeah, wrong key.

16. Dec 14, 2004

### IGeekbot

and no, we have never done vectors like that. I think i will read on that more.

17. Dec 14, 2004

### cyby

Now. Note that momentum is a vector quantity (because velocity is).

So, you can further break it down to:
Pa_x + Pb_x = P'a_x + P'b_x
Pa_y + Pb_y = P'a_y + P'b_y

Now, in your case, you know that
Pa_x = 6, and Pb_x = 0, and Pb_y = 0

You also know...
Pa'_y = 4

With units in kg*m/s

Do you see where I was getting my numbers?

18. Dec 14, 2004

### cyby

Check the beginning of your book - almost all books have a section on that stuff at the very beginning - I suppose you are on the high school/intro college level?

19. Dec 14, 2004

### IGeekbot

so I need to use the horizontal and vertical numbers? Then, when you get those, they should equal the initial momentum of the ball, which is 6 kg*m/s?

20. Dec 14, 2004

### IGeekbot

Yeah, its an intro class, and the teacher expects you to pick it up the frist time, and if you ask for help, he still won't really explain it. He just repeats what he says. And I foudn that section, and we have just used it a little bit.