Asymmetrical Newton's cradle

  • Thread starter d4rr3n
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  • #1
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In a newtons cradle the ball at one end of the ball train moves with similar velocity to the other minus the gradual slowing due to loss of energy in both.

What would be the result if I made a newtons cradle in which the size/mass of the balls gradually decrease?
My instinct tells me that the small ball will move with several times grater velocity then the large ball but not sure why. Also assuming that assertion in correct is it necessary to have a great number of balls of different sizes, what is the minimum number that would result in the same velocity difference?
 

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  • #2
.Scott
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If the large ball was lifted and released, when it struct the others, all would tend to spread out.
If the small one was lifted and release, it would bounce off, and again, the balls would tend to spread.
In general, the motion would not become periodic as it is with the equal-mass cradle. But you could probably find mass ratios for the balls that would create interesting periodic patterns.
 
  • #3
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If the large ball was lifted and released would not its kinetic energy be transmitted through the decreasing size balls causing the last to accelerate away from the others with a grater velocity then the large ball that initiated the process?
 
  • #4
.Scott
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If the large ball was lifted and released would not its kinetic energy be transmitted through the decreasing size balls causing the last to accelerate away from the others with a grater velocity then the large ball that initiated the process?
Let's say that the masses were 4, 2, 1.
4 hits 2 with velocity 1.
-- The average velocity is reduce by the mass increase by (4/6) to 0.667, but they will still recede from each other at v=1.
-- So 4 moves forward at 0.667-0.333 = 0.333.
-- 2 moves forward at 0.667 + 0.667 = 1.333.
2 immediately hits 1 with velocity 1.333
-- The average velocity is reduced by the mass increase by (2/3) to 0.8889, but they will still recede from each other at v=1.333.
-- So 2 moves forward at 0.8889-0.4444 = 0.4444. This is still faster that the 4 ball, so there is no further collision.
-- 1 moves away at 0.88889+0.88889 = 1.77778.

So immediately after 4 strikes 1 and 2 all move forward: 4 at 0.333; 2 at 0.444; and 1 at 1.778.
 
  • #5
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In a newtons cradle when one ball strikes one end only the last moves forward, the force is merely transmitted through the others...but you are saying they will all move forward? In the example you gave I would expect 4 to strike 2 and 1 to rapidly accelerate away from 2, I would expect 2 to hardly move but merely act as a path of the kinetic energy.
 

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