Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

experiment sorry for terrible graphical representation, I did an experiment, i took 6 coins fixed 4 of them in one place by placing some real heavy objects on them , then i took a 5th coin placed it in the final position at the last , all these coins were touching each other and only the fifth one was free to move . Now i took a striker (6th coin) and collided it with this chain of coins and every time the final coin moved as if the force was transmitted through all these coins in the middle to the last one while themselves NOT(assumption) moving at all . How can this happen how can the force be transmitted to the final coin if the coins in the middle didn't moved at all the experiment works with as much as 10 fixed coins in place of 4 . By far the only explanation i can give is that the coins in the middle do move(very little ) but cant prove this theory .

share|cite|improve this question
up vote 2 down vote accepted

This is similar, but not quite identical to Newton's cradle, with the difference being the heavy objects placed on the middle coins.

To explain things, first consider the simpler case where there is no heavy object on top of the coins, and suppose the 5 nonmoving coins in "frame 1" are separated by a distance $L$.

When two objects of mass $m$ and velocities $v_1$ and $v_2=0$ undergo a perfectly elastic collision, there is no energy loss to heat, and so solving energy and momentum conservation $\frac{1}{2}m_1 v_1^2+0=\frac{1}{2} m_1 {v'_1}^2+\frac{1}{2}m_2 {v'_2}^2$ and $m_1 v_1+0=m_1 v'_1+m_2 v'_2$ for the final velocities $v'_1$ and $v'_2$ yield $$v'_1=0 \mbox{ and }v'_2=v_1.$$

As a result, when the striker hits the first coin, the striker stops and the first coin starts moving. After traveling a distance $L$, the first coin strikes the second coin; the first coin stops moving and the second coin starts moving. After traveling a distance $L$, the second coin strikes the third coin, and so on.

When the second to last coin strikes the last coin, the second to last stops moving and the final coin starts moving. Thus the striker and all the other coins except the last coin are standing still, and the final coin is moving away.

Now, if heavy objects are on top of the coins in a line, then there will be a friction that opposes movement of the coins in the line, and the energy loss due to friction, being equal to force times distance, becomes $\mu_k M g(n-1)L$, where $\mu_k$ is the kinetic friction coefficient of the coin/object interface and $M$ is the mass of the heavy object on the coins, and the factor of $n-1$ comes from the fact that there are $n-1$ spaces between the coins in the line.

However, considering the limit $L\rightarrow 0$ as you drew it in your picture, this energy loss vanishes, and yet the collision behavior still remains the same, even if the object on top of the middle coins is very heavy. Hopefully you find this explanation reasonably simple and understandable.

In reality, things are probably a bit more complicated. A more detailed analysis of Newton's cradle requires one to consider the fact that metals are nonrigid compressible materials, and thus the transfer of energy between metal spheres in contact requires an analysis of the compression waves which propagate through the system, as mentioned here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.