Why do 2 balls bounce in a Newton's Cradle instead of just one?

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In a Newton's Cradle, when one ball is lifted and released, it transfers its momentum and energy to the next ball, resulting in one ball bouncing out on the opposite side. When two balls are lifted and released, both balls impact the stationary balls, causing two balls to bounce out due to the conservation of momentum and kinetic energy. This phenomenon occurs because elastic collisions ensure that the total momentum and kinetic energy before and after the collision remain constant. If only one ball were to bounce out after two balls were released, it would violate the conservation laws, leading to an incorrect kinetic energy calculation. The mechanics of Newton's Cradle illustrate fundamental principles of physics that apply to various scenarios, including billiards.
Yuc4h
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Hello everyone

There is this device called Newton's Cradle

http://www.heurekashop.fi/files/magneetti/productpics/496picture2Upload.jpg

You lift one ball and let it impact with other balls, the impact is followed by other ball bouncing from the other side.

However, when you lift 2 balls, the impact is followed by 2 other balls bouncing. Why is that? I would initially guess that only one ball bounces, the answer can not be very simple since I actually asked a professor of physics about this and he didn't really know the answer.
 
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The simple answer is that this involves elastic collisions where no deformation occurs. Both kinetic energy and momentum are conserved. When you satisfy both conditions, you get 1 ball = 1 ball, 2 balls = 2 balls, etc as the only possible solution.
 
But also 2 balls = 1 ball would satisfy the conditions if the ball would get twice the momentum and it also would make more sense.
 
Yuc4h said:
But also 2 balls = 1 ball would satisfy the conditions if the ball would get twice the momentum and it also would make more sense.
How would that conserve energy? And why would it make more sense?
 
Both conditions have to be satisfied. If you have 2 balls, each of mass m, moving with velocity v before the collision, the combined momentum is 2mv and the combined kinetic energy is mv^2. After the collision, these must still be true.
If you had only one ball leaving, its momentum would have to be 2 mv. Since its mass is m, its velocity must be 2v. However, that would make the kinetic energy 2mv^2. So, this cannot be a solution.
This is a standard problem in Physics. The solution is well known and accepted. Thousands (millions?) of students have played with Newton's cradles and tried to get a different result. All have failed.
If this solution were not true, it would be impossible to play the game of pool (billiards) as we know it.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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