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Momentum calculations

  1. Oct 28, 2015 #1
    First off - English isn't my native language, so please go easy on me if my translations are wrong.

    The problem:
    A ball with the mass 20g and the speed 3.0m/s collides with another ball with the mass 80g which is standing still. The collision is a sentral, fully elastic collision. Find the speed of the two balls after the collision.
    I thought it was a question about the combined speed of the two after the collision, but appearently it's asking about both of the balls individual speeds after the collision.
    PS: I know the answere, but I'd like to understand the equations.

    The two formulas that is relevant for me (according to the book itself) is:

    I'm using 'u' as the speed after the collision and 'v' as the speed before the collision.
    First ball is ball (A) and the second is ball (B).

    u(A)= unknown
    u(B)= unknown

    Seeking u(A) and u(B) (separately, not the combined speed)

    I've tried several different setups, but the one I think I'm supposed to use is something like this:

    (1) m(A)u(A)+m(B)u(B) = m(A)v(A)+m(B)v(B)
    (2) 1/2m(A)u^2(A)+1/2m(B)u^2(B) = 1/2m(A)v^2(A)+1/2m(B)v^2(B)

    and then switch them around, divide/multiply, and so on until I'm left with the correct equations, but I'll spare you for that now as I've failed so far. I've reached a point where I see that my equation is totally off, and I've asked quite a few of the other students, but they haven't had any luck with it either.

    In advance, thanks for any tips.
  2. jcsd
  3. Oct 28, 2015 #2

    Doc Al

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    Staff: Mentor

    Since it's an elastic collision, there is no "combined speed". The balls bounce off of each other and go their own ways.

    That's what you need.

    Keep playing around with those and you should be able to solve for the final velocities. (You might want to eliminate v(B) to simplify.) It's a bit messy, but doable.

    There's another trick you can use (which is derived from those two equations): For a perfectly elastic straight-line collision, the relative velocity is reversed during the collision.
  4. Oct 28, 2015 #3


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    Staff: Mentor

    Hi Monkeybusiness, Welcome to Physics Forums.

    You've got the right idea, writing equations for the two conservation laws and then trying to solve the two equations for the two unknowns (the post-collision velocities). This is the standard approach based on the conservation laws which is always taught first. Of course, dealing with the squared speeds in the KE equation can make the process tedious and it's easy to make slips in the algebra.

    Here's a tip that you might appreciate. Something that's not always taught right away is that the conservation laws have a particularly useful consequence for perfectly elastic collisions. It always holds that the speed of separation of the two objects after the collision is equal to the closing speed before the collision. In terms of your variables:

    ##v(B) - v(A) = u(A) - u(B)##

    This can replace the kinetic energy conservation equation as one of your two equations to solve. Note that it avoids all the messy squared terms.

    EDIT: Ha! Doc Al got there before me.
  5. Oct 28, 2015 #4


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    If we start with the equation for conservation of momentum, this gives:

    m(A)u(A) + m(B)u(B) = m(A)v(A)

    Can you rearrange this to give you u(B) in terms of u(A)?

    By the way, I would use ##m## and ##M## for the small and large balls. And ##v##, ##u## and ##U## for their velocities. It makes it simpler, but we can stick with your notation of you want.
  6. Oct 29, 2015 #5
    Thanks for some (really) quick responses, and some good answers! I've actually started writing my equations with m, M, v, V, u and U now, and yes, it did make it easier. Thanks! But I'll stick to my original notations in this post to avoid any confusion by changing them mid-way.
    I found the final velocities by substituting the entire EK equation with v(B)-v(A)=u(A)-u(B), but another question popped up in my head. And I've spent quite some time trying to figure it out, but I guess my algebra skills is non-existent.
    The question is:
    How do you manage to get rid of the mass so you end up with: v(B)-v(A)=u(A)-u(B) ?
    I just want to understand the entire string of operations, and it's bugging me quite a lot that I'm failing at something that seemed so easy to begin with isn't working out as I thought it would.

    Yet again, thanks!

    Edit: When I've googled for the answers, I've seen quite a few places that people reduce and factorise the two equations until they are left with, i.e.:
    m(A)(v(A)-u(A))(v(A)+u(A)) = m(B)(u(B)-v(B))(u(B)+v(B)

    and then they divide the first equation with the last one, but I've never learned any rules that allows a mathematical operation such as that (dividing two different equations with each other).
    Last edited: Oct 29, 2015
  7. Oct 29, 2015 #6


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    In the problem you stated, the larger object was initially at rest. So, v(B) = 0. You should be able to solve the problem using the equations for momentum and energy in this case. A good exercise is to do this algebraically and derive a geneneral expression for ##u(A)## and ##u(B)## in terms of ##m(A), m(B)## and ##v(A)##.

    Have you done that?

    If ##v(B) \ne 0## you could think about changing reference frames to simplify the problem.
  8. Oct 29, 2015 #7


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    You know that you can divide though a given equation by a value (say a non-zero constant) and the equation remains balanced. Well since an equation by definition has both sides of equal value, you can divide one equation by another since the first is divided by equal values on each side.
  9. Oct 29, 2015 #8
    Oh, that does make a lot of sense. I feel silly for even asking that question now without putting up two equations with non-zero values to attempt it first.
    Thanks a lot to all of you who have helped me, I really appreciate it! And thanks for some great tips that I'll take with me into other problems.
    Case solved!

    PS: You guys are great!
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