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I Momentum, Energy, and Elastic Collisions

  1. Oct 27, 2017 #1
    Hello guys.
    So this is less about about the physics problem and more about the concepts behind it. I don't need help with any solution but I am framing my questions around the example given.

    So there are two steel balls with equal mass that are hanging from a point by a string. Both strings are the same length. One ball is lifted so that one string makes an angle to the other string. The ball is let go and swings downward along a circular path and hits the other steel ball elastically.

    So these are questions/confusions.

    When speaking of the momentum of the masses, assuming that the system is made up of both steel balls combined, the momentum is conserved. Kinetic energy is also conserved. This means that the momentum of ball one (the ball that was lifted) is equal to the momentums of both steel balls after the collision.

    The kinetic energy of ball one right before the collision is equal to the kinetic energy of both balls after the collision.

    When looking up solutions for this homework problem, I read that the velocity after the collision of ball one (the ball that was picked up) is zero. So that all momentum of ball one before the collision is transfered to ball two after the collision.

    But the solution doesn't explain, and I can't figure out through my math, how they arrive at this conclusion.

    Couldn't it also be that ball one hits ball two and both have velocities? I would think that in doing this experiment, ball one wouldn't just "stop" and ball two starting moving....

  2. jcsd
  3. Oct 27, 2017 #2


    Staff: Mentor

    If that happened, what would happen to the KE?
  4. Oct 27, 2017 #3


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    No. Let's say that B1 has a velocity of 10 m/s when it strikes B2 and that both balls have a mass of 1 kg (rather heavy, but this is just to make the math easier). The momentum of the system just prior to the collision is ## p = mv = 10 kg \frac{m}{s}## and the kinetic energy is: ##k_{b1} = \frac{1}{2}mv^2= 50 j##.

    Just after the collision B1 has a velocity of zero and B2 has a velocity of 10 m/s. Thus both momentum and energy are conserved.

    Let's investigate what will happen if some other velocity is chosen. Let's say that both B1 and B2 have a velocity of 5 m/s after the collision. The momentum is now 5 for B1 and 5 for B2, so the total momentum of the system is 10, which conserves momentum. However, what about the kinetic energy? The new kinetic energy is now: ##k_{b1} + |k_{b2}= \frac{25}{2} + \frac{25}{2} = 25 j## .

    As you can see, we lost 25 joules. If you plug in other numbers for the velocity of B1 and B2, you'll find that the only numbers which work are 0 for B1 and 10 for B2. It is impossible to conserve both momentum and energy in a 2 body elastic collision, where one body starts at rest, unless the impacting body ends up at rest and the other body takes up all of the momentum and kinetic energy of the impacting body.

    Note that this is only true for elastic collisions. Non-elastic collisions do not conserve kinetic energy, though they do conserve momentum.
  5. Sep 10, 2018 #4


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    Homework Helper

    Alternate method:

    Directly calculate the collision impulse using formula Δp=2μΔv (elastic collision) where μ is the reduced mass [m1*m2/(m1+m2)] of the colliding bodies and Δv their relative velocity. In this case μ=m^2/(2m)=m/2 and Δv=v (initial velocity of ball 1). Hence Δp=2 x m/2 x v=mv.

    After collision:

    Ball 1: p - Δp = mv - mv = 0
    Ball 2: 0 + Δp = mv.
  6. Sep 12, 2018 #5


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    Science Advisor

    Momentum is conserved during what interaction? You have an external force from the strings, surely?
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