# How is the rate of decay of a Newton's Cradle affected by the number of spheres in the system?

How does the number of spheres in a Newton's Cradle affect its rate of decay? (assuming only 1 sphere is released initially)

From experiment, I observed that the rate of decay of a 5 sphere Newton's Cradle is larger than that of a 4 sphere, which in turn is larger than that of a 3 sphere and which is larger than a 2 sphere decay rate.

However, how do we explain this? My initial reasoning was that the collisions in a Newton's Cradle are inelastic, thus energy is lost even though momentum is conserved, but for a N-sphere Newton's Cradle, regardless of N, the number of collisions will still be the same, thus this cannot be correct.

My next thought was that energy would be dissipated through the strings connected to the spheres, thus with more spheres there would be more energy dissipation per oscillation, but I am unsure if this is correct, and if so, how do we calculate this energy dissipated?

Thus, how can we calculate the energy lost for every oscillation of a N-sphere Newton's Cradle? (neglecting the effects of air resistance)

• Are you able to make any measurements of the rate of decay? Experimental evidence would be very useful for confirming any theory given in answer to your question. – sammy gerbil Sep 17 '18 at 10:31

Every ball passes energy and momentum by inelastic collision. If the fraction of energy lost is $\delta$ a first crude estimate of the energy in the system after n collisions would be $(1-\delta)^n$, which amounts to $1-n\delta$ if $\delta$ is small enough. So roughly speaking the decay rate should be proportional to the number of balls.