Clickety-clack

Classical Mechanics Level 3

One single ball at the end of a Newton's cradle is pulled back by a very small angle and released. Every time the ball hits the four other hanging balls, a single ball lifts off the other side with 0.99 of the energy of the incoming ball (as in the first 10 seconds of this video ). Each impact also makes a click. If T T is the time it takes between the first and second clicks, and T 25 T_{25} is the time it takes between the 25th and 26th clicks, what is T / T 25 T/T_{25} ?

Details and assumptions

  • The acceleration of gravity is − 9.8 m/s 2 -9.8~\mbox{m/s}^2 .
  • Make the approximation that the bottom balls remain stationary and that no time passes between the impact of one ball and the lift off of the opposite side ball.


The answer is 1.

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David Mattingly by David Mattingly

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10 solutions

Meet Udeshi
Dec 30, 2013

The time period of a pendulum for small angles is independent of the maximum velocity. Different velocity means it will go to a different max angle, but it will take exactly the same time.

The balls of the cradle are oscillating like a pendulum so their kinematics will be the same too. The only thing changing after one "click" is the velocity so the next "click" will occur after the same time interval as before.

Thus T / T 25 = 1 T/T_{25}=1

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Well done Meet Udeshi!!

Balaji Dodda - 7 years, 5 months ago

oh cool solution man! thanks bro

amgalan amgaa - 7 years, 4 months ago

Let's take some numerical values: m = 1 kg, length of the string = 1 m and assume the initial energy of the ball to be = 1 J. These are just some numbers just to get some feeling for the calculation. With these numbers we can get some estimate for the maximum angle of oscillation after the first collision. The angle turns out to be approximately 26 degrees. 26 degrees cannot be considered as small angle, so small angle approximation cannot be used. There should be more assumptions stated for the question. Explicitly they should state: 'use small angle approximation to answer the question".

Sashi Kumar Krishnan - 7 years, 5 months ago

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Hey it is already mentioned that the first ball is pulled back by a very small angle and released. Your assumption of 1J is wrong.

Balaji Dodda - 7 years, 5 months ago

Ever read NCERT? Small amplitudes means small in radians .. We can approximate sinx as x well between the range of 0 to 50 degrees.. i.e. 0 to ""1.05"" radians..

Toshi Parmar - 7 years, 4 months ago
Rui-Xian Siew
Dec 19, 2014

The period of oscillation is only dependent of the length of pendulum. Without the change of length of pendulum strings, the time taken is constant and same ad infinitum. Therefore the answer is 1.

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For small angles we have T = 2 π l g T = 2\pi\sqrt{\frac{l}{g}} and hence , being the difference in radiants between the angles minimum, the time between the clicks is approximately the same , so that T / T 25 = 1 T/T_{25} = 1 .

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Nahom Yemane
Jan 5, 2014

Time period of a pendulum: 2 π ( L g ) 2\pi \sqrt(\frac{L}{g})

So time period is totally independnt of m a s s mass or v e l o c i t y velocity hence independent of K . E K.E

So T = T 2 = T 3 . . . = T 25 T=T_2=T_3...=T_{25}

So T T 25 = 1 \frac{T}{T_{25}}=\boxed{1}

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Toshi Parmar
Jan 18, 2014

For small angular amplitudes, the time period only depends on length of string and acc. due to gravity. And thus, dissipation of energy only results in smaller and smaller amplitudes BUT has no effect on time period.

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Pietro Pelliconi
Jan 6, 2014

because the angle is small, and remains small, T(n) is costant. so T/T(25)=1

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Su Ku
Jan 5, 2014

The ball is released from a very small angle, so we can assume the motion to be SHM. Although energy is less of the ball at the other end of the cradle to begin with, that will not change the time period of SHM of this ball, it will only change the amplitude. So time between first and second click is same as time between 25th and 26th click.

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Venkatesh G
Jan 5, 2014

For small oscillations the time period is independent of amplitude. So, the ratio of between two clicks remains the same i.e 1.

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William Song
Jan 1, 2014

The balls are essentially pendulums under going periodic motion. The period of a pendulum only depends on the length of the string and the acceleration of gravity, since both of those remain constant the time T must be the same between "clicks."

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Arjun Chaudhuri
Dec 30, 2013

Time period of a simple pendulum =2pi \times(\sqrt{l/g})

0 Helpful 0 Interesting 0 Brilliant 0 Confused

That is only true when the amplitude of oscillation is small. When the amplitude of oscillation is large,that is, when the angle of oscillation is not small the period formula is not correct.

Sashi Kumar Krishnan - 7 years, 5 months ago

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