Newton's Cradle same diameter, different mass

In summary, the mass of the balls does not affect the equations for kinetic energy and momentum/impulse as long as all the balls have the same mass and air resistance can be ignored. This is because both kinetic energy and momentum are proportional to the mass, so it can be cancelled or divided out of equations. The difference in impact between the metal and marble balls is due to the difference in their mass, but they both need the same amount of impulse to reach the same height.
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lostagain
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Homework Statement
I have two cradles, one with 5 metal balls 2.54cm in diameter and 66g and one with 5 marble balls 2.54cm in diameter and 21g. The 1st metal ball was released with no applied force from 10cm which kicked the 5th ball out to 10cm's. I assumed since P and KE were both different for the metal and the marble balls, that the marble ball, when tested from the same drop height would not bounce as far. It did, 5th ball went out 10cm too.
How can the mass not affect the 5th ball more with the metal ball?
Relevant Equations
Ball Type Weight Grams
Metal 66
Marble 21

V = SQRT(D * (acceleration due to gravity or 9.8 m/s^2))2
D = .1 in meters

P=MV
KE = =(MV^2) / 2
Ball drops cm P (Momentum)
Metal 10cm 92.4
Marble 10cm 29.4

Ball drops cm KE
Metal KE 10cm 64.68
Marble KE 10cm 20.58
 
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  • #2
As long as all balls in a cradle have the same mass (which they do here), and as long as they're dense enough to allow us to ignore air resistance (this too is OK here), the mass doesn't matter.

Why? Because both KE and momentum/impulse are proportional to the ball mass M, so for any equation involving M we can remove M by either cancelling (when it appears in the numerator and denominator of a fraction) or dividing the whole equation by M.

Looked at another way: the metal ball scenario does affect the 5th ball more: it imparts approximately three times (66/21) the impulse to that ball compared to what is imparted to the fifth ball in the marble-ball cradle. But it needs three times the impulse to make the ball reach the same height, because the ball is three times as heavy.
 
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FAQ: Newton's Cradle same diameter, different mass

How does the mass of the balls affect the motion of a Newton's Cradle?

The mass of the balls in a Newton's Cradle affects the motion by determining the amount of kinetic energy transferred between the balls during a collision. Heavier balls will transfer more energy and therefore result in a more pronounced swinging motion.

Why do the balls in a Newton's Cradle have to be the same diameter?

The balls in a Newton's Cradle must have the same diameter in order to ensure that the collisions between them are perfectly elastic. This means that no energy is lost during the collisions, resulting in a continuous motion of the balls.

Can the Newton's Cradle still work if the balls have different masses but the same diameter?

Yes, the Newton's Cradle can still work if the balls have different masses but the same diameter. However, the motion may not be as smooth and continuous as when the balls have the same mass.

How does the height of the drop affect the motion of a Newton's Cradle with balls of different masses?

The height of the drop affects the motion of a Newton's Cradle with balls of different masses by determining the amount of potential energy that is converted into kinetic energy during the collision. A higher drop height will result in a greater amount of energy being transferred, resulting in a more pronounced swinging motion.

What is the scientific explanation for the movement of a Newton's Cradle with balls of different masses?

The movement of a Newton's Cradle with balls of different masses can be explained by the laws of conservation of energy and momentum. When one ball is lifted and released, it gains potential energy. As it falls and collides with the other balls, this potential energy is converted into kinetic energy, causing the other balls to move. The collisions between the balls also follow the law of conservation of momentum, where the total momentum of the system remains constant.

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