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Momentum and Velocity

  1. Sep 16, 2007 #1
    Why is it that when a small object traveling at velocity v, when colliding with a much larger object traveling at velocity w, ends up with a final velocity of 2w + v? What happens to the velocity of the large object?

    I have seen the math behind this, but conceptually, I don't see how this works. I figured the small object would bounce off with a relative velocity of -v, thus making the actual velocity w-v. What is the flaw in this reasoning?
  2. jcsd
  3. Sep 16, 2007 #2


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    It has to do with conservation of energy. The velocity changes from +v to -v. So the change of velocity is 2v, not v.
    Try writing out the kinetic energies of the objects before and after collision, and you'll see that it works out.
  4. Sep 16, 2007 #3


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    Are you sure about that? Possibly you are confusing velocity (a vector quantity) with speed (the magnitude of the velocity, so always >= 0).

    Or you are confusing relative velocities and absolute velocities? (see below)

    The relative velocity (relative to the large object) before the bounce is v-w, not v. For a very small object, and an elastic collision, the relative velocity afterwards is close to w-v and the velocity of the large object doesn't change much.

    As compuchip said, for an elastic collision, write the equations for conservation of momentum and conservation of energy, and solve them. If you assume m1 << m2 for the two masses, you can get a simpler approximate solution.
  5. Sep 16, 2007 #4
    well the first ball ends up with a velocity of 2w-v not 2w+v so if the velocity of the small ball is much larger than the larger ball it wud bounce back and the velcity of large ball will remain w

    equation : final velocity of first ball=(m-M)v/m+M +2Mw/m+M

    since M is much larger consider m=0
    Last edited: Sep 16, 2007
  6. Sep 16, 2007 #5
    This is exactly how I thought it would be as well. However, I was doing a problem about a bullet traveling to the right at velocity v1 towards a block of wood attached to a pendulum. The pendulum was at the lowest point of the swing when the two elastically collided (so essentially, the block was traveling to the left at velocity v2). It asked what the speed of the bullet would be after the collision.

    It said the bullet would end up with a velocity v1 + v2 relative to the block, so its actual speed would be v1 + 2v2.

    This diagram also shows a similar idea: http://mysite.du.edu/~jcalvert/phys/assist1.gif
  7. Sep 16, 2007 #6


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    Well, how about showing the calculations that give that? The picture in that website can't be exactly correct. You show the larger object continuing with the same speed as before. That can't happpen. In order to have conservation of momentum and conservation of energy, the larger mass must slow down if the smaller speeds up.
  8. Sep 16, 2007 #7

    Doc Al

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    The picture is correct (ok, not to infinite precision, but close enough to get the answer). One object is much bigger than the other: Imagine a ping pong ball colliding with a bowling ball. For all practical purposes, the speed of the bowling ball doesn't change. (Of course it must change a little to conserve momentum.)

    Transforming to a frame in which the heavy object is at rest, as illustrated in the diagram, is the quick and easy way to solve this kind of problem. In that frame, the lighter object just bounces back with the same speed that it started. Then transform back to the original frame and you're done.
  9. Sep 16, 2007 #8
    I don't understand - if its going at speed v -w and bounces back at speed v -w (relative to the big object at rest), where does the term v+ 2w come from (assuming the big object has speed w)?

    Edit: Hm, for some reason, I think the v + 2w answer comes up only when you use the word SPEED exclusively, and not velocity. So if the bullet at speed v hit the block going to the left at speed w, then the bullet would elastically bounce off at speed v + 2w. If everything was in terms of velocity, then it would be -v + 2w. Is this correct reasoning?
    Last edited: Sep 16, 2007
  10. Sep 16, 2007 #9

    Doc Al

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    Let w & v stand for the speeds (magnitude only). Say the big object goes to the right at +w and the small object goes to the left, thus -v. The relative velocity before the collision is -v-w; after the collision, it's v+w (its sign is reversed). Thus the final velocity of the small object is w + v + w = v + 2w.
  11. Sep 16, 2007 #10
    Ahhh Doc Al, thank you very much. I just edited my post right before you posted that, and you confirmed my prediction.
    I appreciate everyone's help a lot. This problem has been bugging me so much. It's nice to have some closure. =)

    So does this mean if w and v were velocities, and not speeds, the answer would be -v + 2w?
  12. Sep 17, 2007 #11

    Doc Al

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    That's right.
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