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Collision w/ zero Energy Loss not making sense

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data

    Block m1 has mass 10kg and 10m/s
    Block m2 has mass 5kg and 0m/s

    Energy loss is 0%

    2. Relevant equations

    Conservation of momentum
    Conservation of energy
    Quadratic Equation

    3. The attempt at a solution

    (mv)1i + (mv)2i = (mv)1f + (mv)2f

    10(10) + 5(0) = 10(v)1f + 5(v)2f

    [itex] v_{2f} = 20 - 2(v_{1f})[/itex] Then substituting v2f into the following...


    [itex]E_{i} = E_{f}[/itex]

    [itex](\frac{1}{2}mv^{2})_{1i} + (\frac{1}{2}mv^{2})_{2i} = (\frac{1}{2}mv^{2})_{1f} + (\frac{1}{2}mv^{2})_{2f}[/itex]

    [itex] 1000 = 10v^{2}_{1f} + 5v^{2}_{2f} [/itex]

    [itex] 1000 = 10v^{2}_{1f} + 5(20 - 2v_{1f})^{2} [/itex]

    [itex] 0 = 30v^{2}_{1f} - 400v_{1f} + 1000[/itex]

    by quadratic formula:

    [itex] v_{1f} = 10m/s [/itex] or 3.33m/s

    Then [itex] v_{2f} = 20 - 2(3.33m/s) = 13.333333[/itex]


    [itex] v_{2f} = 20 - 2(10m/s) = 0m/s[/itex]

    So, the two possible sets of final velocities are:

    Possibility 1) velocity of [itex] m_{1f} = 10m/s [/itex] and [itex] m_{2f} = 0m/s[/itex]


    Possibility 2) velocity of [itex] m_{1f} = 3.333m/s [/itex] and [itex] m_{2f} = 13.333m/s[/itex]

    Neither case makes sense to me. In the first case, how can something traveling at 10m/s hit a stationary object and continue traveling at 10m/s while the stationary object remains stationary? Maybe I have it mixed up or something? Maybe the object traveling at 10m/s actually becomes stationary while the stationary object is propelled to 10m/s?

    In the second case, how can an object traveling at 10m/s collide with an object and cause the object to travel faster than it? Is that even possible?

  2. jcsd
  3. Sep 18, 2015 #2


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    There's an interesting toy called Newton's Cradle which illustrates the phenomenon you are having trouble grasping:

    The energy from the moving ball on one end is transmitted thru the other balls, which remain stationary. The ball at the opposite end, which is initially at rest, is flung away from the rest of the balls. All of these balls in the cradle have identical masses, so the velocities would remain the same. Any ball which had a smaller mass would be flung away at a higher velocity, in order to conserve momentum.
  4. Sep 18, 2015 #3


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    In the first case, it's clear that both Kinetic Energy and Momentum are conserved. It's just that the collision does not happen. the final conditions are unchanged from the initial conditions.

    In the second case: That is exactly what we expect for an elastic collision.
  5. Sep 18, 2015 #4
    I remember seeing that contraption as a kid on an office desk somewhere and thought it was amazing. Even more amazing at this time, now that I have the power of science to begin understanding what's going on there. Thank you.

    I was trying to imagine a bowling ball colliding with a pool ball and then visual if the pool ball would actually bounce off the bowling ball at a faster speed than the bowling ball was originally travelling. You are saying it is absolutely possible. So if I go to the bowling alley and pick a 10lb bowling ball and roll it into a pool ball, will I see the pool ball roll away faster than the bowling ball was originally rolling? If it should happen and does not, I'd imagine it would be due to the downward rotation of the bowling ball opposing the forward rolling of the pool ball at the time of impact (like friction)? And the friction of the floor and air?

    And hypothetically, if I go into zero gravity with no air resistance and I propel a bowling ball at some speed directly toward a stationary (relative to the bowling ball) pool ball, you are saying I can bet with confidence that the pool ball will bounce forward at a speed greater than the original speed of the bowling ball?

    Thank you
  6. Sep 18, 2015 #5


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    What happens to the Pins when the bowling ball strikes them?
  7. Sep 18, 2015 #6
    The pins definitely do seem to crash away at a faster velocity than the ball travels down the lane. So this is in fact the case? If so, it can be stated that a object in space, when colliding with a less massive object, will cause the less massive object to always bounce away at a faster velocity than the heavier ball was originally travelling (if both objects move in a linear path and spinning/rotation is not occuring). Correct?
  8. Sep 18, 2015 #7


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    Right .

    For a head on, elastic collision, the light object willm move away at close to twice the speed, is the mass difference is great enough.

    Look into "center of mass".
  9. Sep 18, 2015 #8
    Will do. Thanks.
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