Newton's cradle height problem

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Homework Statement


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)


Homework Equations


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


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?
 
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Answers and Replies

  • #2
TSny
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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?
 
  • #3
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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.
 
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  • #4
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I've found a solution
pendulonewton.png


What sorcery did it do to get n = √(Mm)??
 
  • #5
TSny
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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 $$f(n) = \frac{n}{(M+n)(m+n)}$$
 
  • #6
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Oh, thanks, TSny, I didn't see the reply. I got it
 

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