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Newton's cradle height problem

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data
    An executive toy consists of three suspended steel balls of masses M, n and m arranged in order with their centres in a horizontal line. The ball of mass M is drawn aside in their common plane until its centre has been raised by h and is then released. if M ≠ m and all collisions are elastic, how must n be chosen so that the ball of mass m rises to the gratest possible height? What is that height? (Neglect multiple collisions)

    2. Relevant equations
    velocity after a perfectly elastic collision
    v2' = (m2 - m1)v2/(m1+m2) + 2m1v1/(m1+m2)

    3. The attempt at a solution
    After releasing M, its vellocity immediately before the first collision is
    [itex]V = \sqrt{2gh}[/itex]

    Then, M collides with n, and n's velocity immediately after collision is
    [itex]v = \frac{2M}{M+n}\sqrt{2gh}[/itex]

    Analogously, m's velocity immediately after n colides
    [itex]u = \frac{2n}{n+m}\frac{2M}{M+n}\sqrt{2gh}[/itex]

    m shall rises H
    [itex]mu^{2}/2 = mgH[/itex]

    [itex]H = 16h\frac{M^{2}n^{2}}{(M+n)^{2}(m+n)^{2}} [/itex]

    How am I supposed to maximize H from it? What should be done to solve it?
    Last edited: Aug 11, 2014
  2. jcsd
  3. Aug 11, 2014 #2


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    Looks good so far. You have found H as a function of n. Are you familiar with using calculus to find the maximum of a function?
  4. Aug 12, 2014 #3
    But if you derivate with respect to n, you get
    2n³ + (M+m-1)n² - (M+m)n - Mm = 0
    probably, three roots, one of them might get H to max. The problem is to find these roots.
    By the way, this problem is from a test which you should take 8 minutes each question: there must be a better way to solve it.
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4
    I've found a solution

    What sorcery did it do to get n = √(Mm)??
  6. Aug 14, 2014 #5


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    Note that the equation [itex]H = 16h\frac{M^{2}n^{2}}{(M+n)^{2}(m+n)^{2}} [/itex] can be written as $$H = 16hM^2\left(\frac{n}{(M+n)(m+n)}\right)^2$$

    H will be maximum when the expression ##\frac{n}{(M+n)(m+n)}## is a maximum. So, you just need to maximize the function $$f(n) = \frac{n}{(M+n)(m+n)}$$
  7. Sep 27, 2014 #6
    Oh, thanks, TSny, I didn't see the reply. I got it
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