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Head on collision

  1. Jun 11, 2006 #1
    Hi everyone,

    I've got a problem with solving this one.

    Two spheres X and Y (no mentioning about them being identical in mass but drawn in equal sizes) are about to collide head on. Their initial velocities are +5ms^-1 and -5ms^-1 initially, moving towards each other with X being towards east. Assuming completely elastic and head-on collision, what happens to the spheres after collision?

    The answer I was given was that X comes to a rest while Y moves off in the reverse direction, that is, east with 10ms^-1.

    Why is this so? I don't really understand why is it that Y would move off and not the other way around where X moves off towards west with 10ms^-1? Why is all momentum transfered to Y and not X? I apologise that I don't have any workings with me when I should have but I'm confused with the principles here. Also, does the relative speed of approach only applicable to elestic collisions?

    Thanks again to all for the constant help.
  2. jcsd
  3. Jun 11, 2006 #2

    Andrew Mason

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    The answer you were given is obviously wrong, if the masses are the same. Conservation of momentum requires that the centre of mass does not change. The centre of mass is at rest in the lab frame. They both recoil at 5 m/sec in opposite directions.

  4. Jun 11, 2006 #3
    Yes, as AM said the answer is definitely wrong .
    Imagine this, there is nothing special given about mass X or Y . That is the names can be used interchangeably(except for the direction of velocity), this implies that whatever causes an effect on both objects simultaneously (like a collission) should show symmetrical effects. So the answer is clearly wrong ( As you have observed, why isn't it the other way round ?).
    The only way to distinguish further between X and Y is to give their masses and they are not the same too.
  5. Jun 11, 2006 #4
    Thanks guys. For the formula on the relative speed of separation, can it be only used on perfectly elastic collisions and on objects of equal masses?

    Anyway the interesting part is that the other option (its a MCQ) is both speheres come to rest after collision.

    Do you have any websites that explains that COM does require the CG of the system to be the same? Couldn't really find some good ones.

    Many thanks!
    Last edited: Jun 11, 2006
  6. Jun 11, 2006 #5

    Andrew Mason

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    The formula can be used only on perfectly elastic collisions but the masses need not be the same.

    What is CG? If it is the centre of gravity, then you are asking whether the intertial mass is identical to gravitational mass. Try googling "Eotvos experiment".

  7. Jun 12, 2006 #6
    The masses will definitely not come to rest, if the collission is elastic (why ?).
    If you could give us all the options in the MCQ , we could direct you to the right answer, perhaps it involves eliminating all other options, since the masses are not given .
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