Explaining Newton's Cradle: Proving Conservation of Momentum & KE

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SUMMARY

This discussion focuses on the mathematical modeling of Newton's Cradle to demonstrate the conservation of momentum and kinetic energy (KE). The user establishes that when two balls are released, the initial momentum is 2mv and the KE is mv². For n balls to emerge with the same velocity, conservation of momentum dictates that n must equal 2 for KE to be conserved. The user seeks assistance in proving that if the velocities of the n balls are not equal, the only solution leads to two balls emerging with a velocity of 2, while the others remain at zero velocity.

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adamg
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im working on a mathematical model of Newton's cradle. I am 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 don't know how to prove that n must equal 2. Thanks
 
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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:

v1+v2+...vn=2v

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?
 
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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