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Working out the velocity after a collision

  1. Jan 5, 2005 #1
    Hi guys, first time poster whose not sure if he's in the right forum, but here goes.

    I have a question that reads:

    A body, A, of mas 4kg moves with a velocity of 2ms-1 and collides head-on with another body, B, of mass 3kg moving in the opposite direction at 5 ms-1. After the collision the bodies move off together with the velocity v. Calculate v.

    This is probably basic to most, but I've just returned to college. I'm not looking for the answers as such, but rather if someone could point me to the right equation. The ones I've tried using are:

    m1 u1 + m2 v2 = m1 v1 + m2 v2


    mv = mv1+mv2


    mv1 = (M = m)v2 (I think this is wrong because its for an inelastic collision)

    Thanks for reading!

  2. jcsd
  3. Jan 5, 2005 #2
    I think that

    m1v1 +m2v2=(m1+m2)V

    should work since the balls stick together their mass combines and they have the same velocity.
  4. Jan 5, 2005 #3
    What makes you think the collision is elastic?

  5. Jan 5, 2005 #4
    Hey guys, thanks for the quick replies!

    Justin, until I re-read the question I didn't realise that it says 'the bodies move off together' so I guess it could be inelastic.

    Thanks for the tip Arlamos. This is the solution I come up with.

    m1• u1 + m2• u2 = m1• v1 + m2• v2
    4 • 2 + 3 • 5 = 4 • 6 + 3 • v2
    8 + 15 = 24 + 3v2
    23 = 24 + 3v2
    -23 -24
    -1 = 3v2
    v2 = - 1/3

    I had a similar problem to this a few weeks ago and the way I worked it out looked similar. I'm going to college tomorrow (first day back) so I can check with my friends and I'll come back and tell you my findings. I can't hang stay online now, I'm in the UK and it's getting a bit late ;)

    Thanks again for your feedback, it's been most useful!

  6. Jan 5, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that v1 = v2 = v. The collision is perfectly inelastic.
    Better to write it as Arlamos suggested:
    m1*u1 + m2*u2 = (m1 + m2)*v
    Realize that the initial velocities are in opposite directions; so if one is positive, the other must be negative:
    4*2 + 3*(-5) = (4 + 3)*v
  7. Jan 6, 2005 #6
    Thanks for replying Doc Al. I had a discussion with one of my friends who pointed out that it needed to be negative. I've had a quick sketch on my pad and have come up with;

    m1*u1 + m2*u2 = (m1+m2)*v
    4*2 + 3*(-5) = (4+3)*v
    8 + -15 = 7v
    -7 = 7v
    /7 /7

    v = -1 ms-1
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