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Homework Help: Can Someone help me out here, please?

  1. Apr 2, 2010 #1
    1.1. A 7 kg glider on an air track starts with an initial velocity of 32 m/s and then runs into a 2 kg glider which is initially at rest. During the collision, which glider has a larger magnitude change in momentum? Explain your answer.

    2. We observe the center of mass of a system of objects and realize that it is accelerating. Could all of the forces acting on the system be internal forces? Why or why not?

    3. You are playing pool. You hit the cue ball into the 6 ball. If there is no spin on the ball and you hit it straight on center, the cue ball stops dead, and the 6 ball rolls off with approximately the same velocity that the cue ball had before. Is the collision between the balls:

    Almost completely elastic
    Almost completely inelastic
    Roughly halfway between completely elastic and completely inelastic
    You can't tell from the information given.

    4. Explain your answer to the multiple choice question

    2. Relevant equationsK=(1/2)mv^2

    3. 1. They both experience the same magnitude change, since this is an isolated system, and these two objects exert the same force on each other (in magnitude, but opposite directions, according to Newton's 3rd law).

    2. No, if all the forces acting on it were internal, then net force would equal zero, and there would be no acceleration. Thus, there must be some external forces.

    3. Almost completely elastic

    4. In this example, the total kinetic energy would be the same after the collision, as before the collision. Before the collision, one of the balls is moving with a certain velocity, and one is at rest. After the collision, the other ball has started moving with the same velocity, while the other one is now at rest. The balls are of roughly equal mass, and thus, the initial total kinetic energy is equal to the final total kinetic energy.

    Thanks for any help!!!
  2. jcsd
  3. Apr 2, 2010 #2


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    Hi SoulInNeed! :smile:

    2,3,4: Completely correct. :smile:

    1: I'd give a slightly longer answer …​

    the force between the two bodies is not constant, so talking of "the force" is unrealistic. :wink:
  4. Apr 2, 2010 #3
    Try and refer to the force at every point in time, and refer to the duration of the interaction. That should help you round off the answer quite nicely. :)

    If it'll help jog your memory, the impulse momentum theorem:
    [tex]J=\int F dt = \Delta P[/tex] where [tex]P[/tex] is the total momentum of the system, and [tex]F[/tex] is the net external force acting on the system.
  5. Apr 2, 2010 #4
    Was I at least correct in stating that they experience the same change in magnitude (in terms of momentum)?
  6. Apr 2, 2010 #5
    OK, well I understand that the change in momentum determines impulse, and if we treat this two objects as a system, then the impulse must stay constant from the initial conditions to the final conditions, so that, even if the momentums change, the rate has to be the same so that the impulse stays the same?
  7. Apr 3, 2010 #6


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    No, "the impulse must stay constant" doesn't make sense :redface:

    the impulse (= ∫ force dt) is the total

    that's like saying "the average rainfall for 2009 was constant throughout the year" ! :wink:

    We know what you mean … and you're right … but you need a good way of saying it. :smile:
  8. Apr 3, 2010 #7
    OK, well, even if the force is inconsistent, J= p(f) - p(i) is true. The rate of change in momentum depends on the amount of force and the time interval. If I treat these two objects as a system, and the force is constant (even if it isn't, I just use the average force), then the rate of change simply depends on the time interval that both go through, and since this time interval applies to both objects, both will go through the same change in momentum. Does this sound right? lol

    BTW, thanks for the help, and the patience.
  9. Apr 3, 2010 #8


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    Hi SoulInNeed! :smile:

    (btw, have you done integration? would you for example understand what is meant by "work is the integral of force with respect to distance"?)
    "the rate of change simply depends on the time interval that both go through" … what does that mean? :confused:

    Force is a rate of change, and so only applies at an instant, not over an interval (unless of course the force is constant, which it obviously isn't here), so if you're using a principle that mentions force, you can only apply it instantaneously.

    Hint: you might find it simpler to use a frame of reference in which, long before the collision, the centre of mass is stationary, ie the gliders have equal and opposite momentum, and to consider when that might change. :wink:
  10. Apr 3, 2010 #9
    I was saying that J=F * time interval= change in momentum, and if force is constant, then it would depend on the time interval for any difference. It's a really bad way of saying it, I know, lol.

    And yeah, we've done work=force * displacement, although we didn't call it an integral.
  11. Apr 3, 2010 #10
    OK, let me start over, at the point of collision, don't the objects exert an equal but opposite in direction force on each other? And if force=rate of change in momentum, then what one object gains in momentum, the other will lose in momentum, but the rate of change is the same, right?
  12. Apr 4, 2010 #11


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    Yes, the important point is that, at any instant of time, the two rates of change (of momentum) are equal and opposite …

    and since this is true for the rate of change at any instant, it must be true for the total change (not a good way of saying it, but you can try to improve on it :wink:).

    (and work = force*displacement or change of momentum = force*time only work if the force is constant … if it isn't, then it's work = ∫ force*d(displacement) or change of momentum = ∫ force*d(time), so if force1 alwyas equals minus force2, then change of momentum1 = change of momentum2)
  13. Apr 4, 2010 #12
    This is simply a statement of the conservation of momentum, right?
  14. Apr 4, 2010 #13


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    Yes, but I'm not sure whether you're being allowed to assume conservation of momentum, or whether you're expected to start from Newton's laws, and derive conservation of momentum from them.

    If you are allowed conservation of momentum, then that immediately answers the glider question anyway. :wink:
  15. Apr 4, 2010 #14
    I'll ask the prof, and see what he says.
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