Newton's Cradle: Explaining 2-Ball vs. 1-Ball Dynamics

[foyer]

when one ball is pulled out, released, and collides with the other four in Newton's Cradle, one ball will pop out the other side. Likewise, when two balls are pulled back, two balls will pop out the other side. Why doesn't one ball pop out, but with twice the speed?

You are to use the concepts of conservation of momentum and conservation of energy to mathematically show that only two balls may be ejected on the other side when two balls are pulled back instead of one ball with twice the speed.i know that both momentum and energy are both conserved because the collisons are elastic, but i just get stuck there

P initial=P finalany help would be appreciated

 

[Doc Al]

Compare the energy of two balls with speed v to the energy of one ball with speed 2v. Are they the same?

 

[thomate1]

Hello,

Thank you for your question. I am happy to provide a response to help you understand the dynamics of Newton's Cradle.

Firstly, let's start by defining the principles of conservation of momentum and energy. Conservation of momentum states that in a closed system, the total momentum remains constant, meaning that the initial momentum before a collision is equal to the final momentum after the collision. Similarly, conservation of energy states that in a closed system, the total energy remains constant, meaning that the initial energy before a collision is equal to the final energy after the collision.

Now, let's apply these principles to the scenario of 1-ball vs. 2-ball dynamics in Newton's Cradle. When one ball is pulled out and released, it collides with the other four balls and transfers its momentum and energy to them. This results in the first ball coming to a stop and the other four balls moving with the same speed and in the same direction. This is because the total momentum and energy before the collision is equal to the total momentum and energy after the collision.

In the case of two balls being pulled back and released, the same principles apply. However, now we have two balls colliding with the other four balls. The first ball transfers its momentum and energy to the other four balls, just like in the 1-ball scenario. But now, the second ball also transfers its momentum and energy to the other four balls, resulting in a greater total momentum and energy being transferred to the other four balls. This results in two balls being ejected on the other side, instead of one ball with twice the speed.

To mathematically show this, we can use the equations for conservation of momentum and energy. Let's assume that the initial velocity of the first ball is v and the initial velocity of the second ball is u. After the collision, the velocity of the first ball is zero and the velocity of the second ball is v. Using the conservation of momentum equation, we get:

(m1 + m2)v = m1u + m2v

Since v=0, we can simplify the equation to:

m2v = m1u + m2v

Now, using the conservation of energy equation, we get:

1/2m1v^2 + 1/2m2v^2 = 1/2m1u^2 + 1/2m2v^2

Simplifying