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Homework Help: Two balls hanging (lol)

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Two small balls are suspended side by side from two strings of length L so that they just touch when in their lowest positions. Their masses are m and 2m, repectively. If the left ball (of mass m) is pulled aside and released from a height h, it will swing down and collide with the right ball (of mass 2m) at the lowest point. Both balls again swing down, and they collide once more at the lowest point. How high will each swing after this second collision? Assume both collisions are elastic.

    2. Relevant equations

    So, for now, I'm just using:

    mv1 +mv2 = mv1' + mv2'
    Ek = 0.5mv^2
    Ep = mgh

    3. The attempt at a solution

    So I know v2 = 0, but I'm not sure if v1 would also be 0.
    For ball#1 (mass m), Ek = Ep => 0.5v1'^2 = mgh
    Sort of stuck as with what to do next...
    Any help is appreciated.
  2. jcsd
  3. Nov 13, 2009 #2
    For the first collision:

    Your mass m1 will reach its maximum velocity at the point of impact (why?) so yes, that also means that its kinetic energy will be a maximum there and its potential energy will be zero (again, why?). From these conditions you can calculate the initial velocity of the first mass m. We also know that, since it's an elastic collision energy will be conserved.

    That should be enough to get you started.
  4. Nov 14, 2009 #3
    So I basically tried what you were saying...but I can't seem to find an expression for v1 to find h. I was wondering if you could help me get started with finding it. THanks
  5. Nov 14, 2009 #4
    Ok from:

    Conservation of momentum:
    mv1 + mv2 = mv1' + mv2'

    and Conservation of energy:
    0.5mv1^2 + 0.5mv1'^2 = 0.5mv2^2 + 0.5mv2'^2

    ...I got:

    (1)... m(v1+v1') = 2m(v2'-v2)
    (2)... 0.5m(v1+v1')(v1-v1') = m(v2'+v2)(v2'-v2)

    Since v1' and v2 are 0 and m's can cancel:

    (3)... v1 = 2(v2')
    (4)... 0.5(v1)^2 = (v2')^2

    However, substituting (3) into (4) results into an expression that makes no sense:

    0.5[4(v2')^2] = (v2')^2 --> 2(v2')^2 = (v2')^2 ?????????????????

    It just dosen't make sense =\
    Can someone please point out where I went wrong?
  6. Nov 15, 2009 #5
    I'm sorry, I made a typo here :redface:

    It should be:

    The maximum energy is the maximum potential energy (where the first mass is at its highest point) so

    [tex]E_{max} = E_p = m_1gh[/tex]

    at the bottom of its swing, this potential energy will be zero, but its kinetic energy will be a maximum and from conservation of energy we have

    [tex]E_p = E_k[/tex]
    [tex]m_1gh = \frac{1}{2}m_1v_1^2[/tex]

    Now, we have the initial conditions (I use u for initial velocity and v for final, all those primes just confuse me)

    [tex]u_1 = 0[/tex] (the initial velocity of the first mass before it's released)

    [tex]v_1 = \sqrt{2gh}[/tex] (the final velocity of the first mass at impact)

    [tex]v_2 = 0[/tex] (the final velocity of the second mass when it reaches the top of its swing)

    [tex]u_2 = ?[/tex] (what we're looking for)

    From conservation of momentum

    [tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
    [tex]m_1u_1 + 2m_1u_2 = m_1v_1 + 2m_1v_2[/tex]
    [tex]u_1 + 2u_2 = v_1 + 2v_2[/tex]
    [tex]0 + 2u_2 = \sqrt{2gh} + 0[/tex]
    [tex]u_2 =\frac{1}{2} \sqrt{2gh}[/tex]

    The energy of the second mass is at a max shortly after the collision and we can calculate its kinetic energy

    [tex]E_k = \frac{1}{2}u_2^2[/tex]

    This should get you going towards finding the height the second mass reaches after the impact.

    I've given you quite a lot here and I hope it helps. I'm not going to be around for the next couple of days, so please contact one of the homework helpers if you're still stuck (have a look under the "Staff" tab at the top of the PF page).

    Good luck.
  7. Nov 15, 2009 #6
    Alright...one question.
    I'm not sure if this is a mistake on your part of if I'm missing something but..
    Why does Ek = 0.5(u2)^2?...Shouldn't it be Ek = 0.5m(u2)^2?
  8. Nov 15, 2009 #7


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    Homework Helper

    Yes, it should be 0.5m(u2)^2.
  9. Nov 15, 2009 #8
    I'm confused. Shouldn't u2 be 0, since the 2nd ball is initially hanging?
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