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Two things colliding at a imaginary moving surface

  1. Dec 8, 2006 #1
    Suppose you imagined a vertically aligned surface and with respect to this imagined (massless) surface were two objects shooting towards each other.

    Suppose these objects both had mass [itex]m[/itex], and each their kinetic energies (with respect to each other) would be [itex].5mv^2[/itex].

    Suppose they are moving along the same line perpendicular to the surface.

    [itex]v[/itex] is the sum of the velocities with respect to the surface, call these velocities [itex]a[/itex] and [itex]b[/itex].

    Is there any significance to the quantity [itex]m(a+b)^2-ma^2-mb^2[/itex] or [itex]mab[/itex]??

    In physics, does the product of two different velocities along the same dimension have any meaning? Does this have anything to do with the left over kinetic energy with respect to the imagined surface?
    Last edited: Dec 8, 2006
  2. jcsd
  3. Dec 8, 2006 #2


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    I can be of no help with any of the math stuff. It seems to me, intuitively, that any massless vertical barrier would of necessity have no actual matter involved. It should therefore have no effect upon the colliding objects. The interaction should take place as if the barrier doesn't exist.
  4. Dec 8, 2006 #3
    Yes, that is what I am implying.
  5. Dec 12, 2006 #4
    wrong assumption

    basically your first assumption is only wrong "each have kinetic energies (with respect to each other) 0.5 mv^2" is wrong if one is moving with velocity v(with respect to fixed frame) towards other which is also moving with velocity v(with respect to fixed frame) their velocity(with respect to each other is "2v" and not v. how does your -sign evolves as mass and velocity square(of course positive) is always +ve.your equation simply points to kinetic energy of any arbitrary body moving with velocity v(may be different from v). your equation would be completely wrong if it yields a -ve answer as kinetic energy can never be -ve.
  6. Dec 12, 2006 #5

    Doc Al

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    Staff: Mentor

    kmarinas86 has stipulated that "v" is their relative velocity (with respect to each other), not their velocity with respect to some other frame.
  7. Dec 13, 2006 #6
    kmarinas has said that KE with respect to each other is 0.5mv^2. But KE surely depends upon velocity with respect to your frame.(HERE BOTH BODIES ARE MOVING)
  8. Dec 13, 2006 #7

    Doc Al

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    Yes, both bodies are moving with respect to something. So what? To say that the KE with respect to each other is 0.5mv^2 means that their relative velocity is v. (With respect to itself, each body is at rest.)

    That means: As measured in the frame of body a, body b has speed v and KE = 0.5mv^2; And, as measured in the frame of body b, body a has speed v and KE = 0.5mv^2. So what?
  9. Dec 15, 2006 #8
    but both are moving in same direction hence with respect to frame of body a, b is moving with velocity 2v and not v.for better visualization sit on body a
  10. Dec 15, 2006 #9


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    No, Doc Al's point is that the original post referred to "kinetic energies (with respect to each other)", not with respect to some third frame of reference.

    I might also point out that it said "two objects shooting towards each other", not "in the same direction", but that's really irrelevant to the point.
  11. Dec 15, 2006 #10

    Doc Al

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    I don't know where you got the impression that each body was moving with speed v with respect to a third frame, which would make their relative speed 2v. The OP was clear that they move with speed v with respect to each other:

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