1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Elastic Collisions

  1. Aug 16, 2006 #1
    Here's the problem:

    "Two steel balls are suspended on (massless) wires so that their centers align. One ball, with mass 2.30 kg, is pulled up and to the side so that it is 0.0110 m above its original position. Then it is released and strikes the other ball in an elastic collision. If the second ball has a mass of 3.10 kg, to what height does it rise above its original position?"

    I started by assuming that the change in upward momentum would equal zero, but I came out with the wrong answer. Do I need to incorporate potential and kinetic energy into this problem? If so, I'm unsure exactly how to do so.
     
  2. jcsd
  3. Aug 16, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    This question is all about energy.

    [Note: Because the masses are suspended, momentum is not conserved generally. Momentum and energy is conserved, however, at the moment of collision since there is no vertical momentum and no constraint on horizontal motion (no horizontal force from the wires). ]

    Use conservation of momentum to determine the speed of the first and second ball immediately after collision and then use energy conservation to determine how high the second ball rises.

    AM
     
  4. Aug 16, 2006 #3
    Okay, but I was told to use the conservation of energy first (the kinetic energy of the ball as soon as it is released is equal to its initial potential energy...so you can find it's initial velocity). I set its initial kinetic energy equal to the final kinetic energy of the two balls combined. Then, I solved for the final velocity of the first ball.

    Then, using the conservation of momentum, I solved the the final velocity of the first ball again. I set the equations equal to each other, but I ended up with a quadratic and the square root of a negative number.

    Any ideas?
     
  5. Aug 16, 2006 #4
    It doesn't really matter which one you do "first", as long as you have both before you start doing algebra.

    Here are the two things that you've already established:

    Kinetic energy is conserved, as is momentum at the collision:
    [tex]\frac{m_1v_1^2}2+\frac{m_2v_2^2}2=\frac{m_1v_1'^2}2+\frac{m_2v_2'^2}2[/tex]
    [tex]m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'[/tex]

    You already know how to solve for [tex]v_1[/tex], and you know that [itex]v_2 = 0[/itex]. Solve for [itex]v_{1}'[/itex] and [itex]v_{2}'[/itex]. Make sure you keep your algebra straight, because otherwise you will become extremely frustrated with wrong answers.

    From there, you know that energy is conserved in the second ball. So, after solving for and obtaining the value of [itex]v_{2}'[/itex], you can use conservation of energy to find the final height of the second ball.
     
  6. Aug 16, 2006 #5
    What you're saying is that you would set its KE equal to its PE?

    Did you actually work out the problem? Because I couldn't get an answer by doing the algebra. I did it several times.
     
  7. Aug 16, 2006 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The total energy of the system is the potential energy of the first ball at the moment of its release. This total energy never changes. So:

    [tex]E_{sys} = m_1gh_1[/tex]

    Immediately prior to the collision:

    [tex]E_{sys} = \frac{1}{2}m_1v_{1i}^2[/tex]

    After collision by cons. of momentum:

    [tex]m_1v_{1f} + m_2v_{2f} = m_1v_{1i} = m_1\sqrt{2gh_1}[/tex]

    (1) [tex]v_{1f} = \sqrt{2gh_1} - m_2v_{2f}/m_1[/tex]

    and by cons.of energy:

    [tex]\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 = E_{sys} = m_1gh_1[/tex]

    [tex]\frac{1}{2}m_1v_{1f}^2 = m_1gh_1 - \frac{1}{2}m_2v_{2f}^2 [/tex]

    (2) [tex]v_{1f} = \sqrt{2(m_1gh_1 - \frac{1}{2}m_2v_{2f}^2)/m_1}[/tex]

    Subtracting 1 from 2 will allow you to solve for v_{2f}:

    [tex]\sqrt{2gh_1} - m_2v_{2f}/m_1 = \sqrt{2(m_1gh_1 - \frac{1}{2}m_2v_{2f}^2)/m_1}[/tex]

    [tex]2gh_1 + m_2^2v_{2f}^2/m_1^2 - 2\sqrt{2gh_1}m_2v_{2f}/m_1 = 2gh_1 - m_2v_{2f}^2/m_1[/tex]


    [tex]\left(\frac{m_2^2}{m_1^2} + \frac{m_2}{m_1}\right)v_{2f} - \left(\frac{2\sqrt{2gh_1}m_2}{m_1}\right) = 0[/tex]

    [tex]v_{2f} = \left(\frac{2\sqrt{2gh_1}m_2}{m_1}\right) / \left(\frac{m_2^2}{m_1^2} + \frac{m_2}{m_1}\right)[/tex]

    AM
     
    Last edited: Aug 16, 2006
  8. Aug 17, 2006 #7
    After the collision, the second ball has a certain kinetic energy. This energy has to be converted into potential energy. When all of the KE has been converted to PE, then the ball reaches its maximum height. You can solve for the height by setting KE = PE.

    After you find [itex]v_{2f}[/itex], use the following expression (KE = PE) to find the height of the second ball.
    [tex]\frac{1}{2}m_2v_{2f}^2 = m_2gh[/tex], where h is the height.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Elastic Collisions
  1. Elastic Collisions? (Replies: 3)

  2. Elastic Collisions (Replies: 5)

  3. Elastic collision (Replies: 6)

Loading...