Newton's Cradle

question physics

Question

In a Newton’s cradle, the leftmost ball collides with the stationary balls with a velocity $u$. Is it possible for the two rightmost balls to then move to the right together at a speed $\frac{1}{2}u$ after the collision? All of the balls have the same mass, $m$.


Hints

Consider the instants before and and after the collision. What quantities are conserved? What equations can you make from this?


Answer

The two quantities conserved in the collision are momentum, and energy.

At the instant before the collision, the leftmost ball has velocity $u$. Immediately after, the two rightmost balls have velocity $v=\frac{u}{2}$. Considering momentum: $$ mu = m(\frac{u}{2}) + m(\frac{u}{2}) = mu $$ Momentum is therefore conserved in this situation.

Considering energy: $$ \frac{1}{2} m u^2 = \frac{1}{2} m (\frac{u}{2})^2 + \frac{1}{2} m (\frac{u}{2})^2 = \frac{1}{4} m u^2 $$ There is a contradiction here, as the initial energy and final energy are not equal. There are two possibilities here:

  • some energy is lost in the collision - equal to $ \frac{1}{4} m u^2 $
  • if the collision is elastic, then this is impossible

In real life, the collision is almost elastic, and you'll find that the number of balls released will always equal the number of balls that move off from the cradle, conserving both energy and momentum.