A ball of mass m makes head-on elastic collisions with…

A ball of mass $m$ makes head-on elastic collisions with bodies of mass $15m$ (each at rest before the collision). After how many collisions has its energy dropped by a factor of 100?

Explanation of Method

Energy loss over repeated elastic collisions.

Step 1: Energy kept per collision

\[ \frac{E'}{E} = \left(\frac{m - 15m}{m + 15m}\right)^2 = \left(\frac78\right)^2 = \frac{49}{64} \approx 0.766. \]

Step 2: Number of collisions

\[ \left(\frac{49}{64}\right)^n \le \frac{1}{100}, \]

\[ n \ge \frac{\log 100}{\log(64/49)} = \frac{2}{0.1160} \approx 17.2. \]

\[n = 18 \ \text{collisions}\]

After 17 collisions 1.07% of the energy is left, just above 1%.

$E_n/E_0$ $n$ $0$ $5$ $10$ $15$ $18$ $20$ $0.5$ $1$ $1\%$ $\left(\tfrac{49}{64}\right)^n$

The ball keeps 49/64 of its energy at each collision; after 17 collisions 1.07% remains, after 18 only 0.82%.

Result

\[\boxed{\,n = 18 \ \text{collisions}\,}\]

Remark

About the original solution. The method is right, but the logarithm was evaluated wrongly: $\log_{10}(49/64) = -0.116$, so $n \approx 17.2$, not 76.6. Seventy-seven collisions would reduce the energy to about $10^{-9}$ of its initial value.