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Newton's cradle

  1. Jan 28, 2005 #1
    im working on a mathematical model of newton's cradle. im trying to explain mathematically why, when two balls are released, two balls pop up the other side etc.
    i said that if there are N balls in the cradle, each with mass m, and the two balls you displace have gained a velocity v by the time they hit the others, initially the momentum is 2mv, and KE is mv^2. Then i said that if n balls emerged from the other side with the same velocity, it would need to equal 2v/n for conservation of momentum. Then KE is equal to (2mv^2)/n and we see KE can only be conserved when n=2.
    I was just wondering if anyone can help me prove this result for when the velocity of the n balls emerging from the left is not equal. i.e then, the average velocity of the n balls would need to be 2v/n, but i dont know how to prove that n must equal 2. Thanks
  2. jcsd
  3. Jan 28, 2005 #2
    basically, if you let v1,v2,......vn, be the velocities of the n balls that 'pop out' after the collision, therefore v1<v2<v3......<vn. From the conservation of momentum we get:


    and from the conservation of KE we get:

    (v1)^2+(v2)^2+.......(vn)^2= 2v^2

    is there any way i can show that the only solution to this is vn=v(n-1)=2 and v1=v2=......v(n-2)=0???
  4. Jan 28, 2005 #3
    alternatively can anyone prove that, if we have N balls, and we displace 2, 2 will 'pop out' the other side in another way? any ideas welcome!! thanks
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