1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculate momentum and velocity of ball after collision

  1. Dec 14, 2004 #1
    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. jcsd
  3. Dec 14, 2004 #2
    Are you familiar with 2d collisions and the principles of conservation of momentum?
  4. Dec 14, 2004 #3
    yes, i have my book here in front of me with equations, but im not sure which one to use.
  5. Dec 14, 2004 #4
    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
  6. Dec 14, 2004 #5
    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
  7. Dec 14, 2004 #6
    In your particular case, you don't have to. Are you familiar with unit vector notation to express vector quantities?
  8. Dec 14, 2004 #7
    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.
  9. Dec 14, 2004 #8
    ive probably seen them, but that is not ringing a bell. could you please explain what it is?
  10. Dec 14, 2004 #9
    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?
  11. Dec 14, 2004 #10
    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.
  12. Dec 14, 2004 #11
    Ok. Why don't we start even further back. Can you type out the conservation of momentum equations at least? :-)
  13. Dec 14, 2004 #12
  14. Dec 14, 2004 #13
    you mean 3.0m/s * 2.0kg = 6.0 kg*m/s...

  15. Dec 14, 2004 #14
    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.
  16. Dec 14, 2004 #15

    yeah, wrong key.
  17. Dec 14, 2004 #16
    and no, we have never done vectors like that. I think i will read on that more.
  18. Dec 14, 2004 #17
    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?

  19. Dec 14, 2004 #18
    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?

  20. Dec 14, 2004 #19
    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?
  21. Dec 14, 2004 #20

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Calculate momentum and velocity of ball after collision