Irodov 1.158: Infinitely Many Bounces, Finite Momentum

Problem Irodov 1.158

A small steel ball of mass $m=50$ g is dropped from rest from a height $h=1.0$ m onto the flat horizontal top of a massive, immovable slab, and air resistance is negligible. Each time the ball strikes the slab, its speed drops by the same factor $\eta=1.25$ (the rebound speed is $1/\eta$ times the impact speed), and between bounces the ball moves freely under gravity. Find the total (cumulative) momentum the ball imparts to the slab, summed over an unlimited number of bounces.

One impact, drawn as a before and after pair above the slab: the ball strikes with speed v downward and leaves with speed v/η upward, the same pattern repeating at every bounce. — One impact, drawn as a before and after pair above the slab: the ball strikes with speed $\vect v$ downward and leaves with speed $\vect v/\eta$ upward, the same pattern repeating at every bounce.

Given

$m=50$ g
$h=1.0$ m
$\eta=1.25$

Find

cumulative momentum

\Delta p

imparted to the slab

Solution

Step 1Impact and rebound speeds form a geometric sequence

The ball first strikes the slab with the free-fall speed gained over the drop $h$ , found from energy conservation.

\frac{1}{2}mv_0^2 = mgh

v_0 = \sqrt{2gh}

(1)

Between bounces the ball is in free flight under gravity alone, so it returns to the slab with exactly the speed it left it, and each impact divides that speed by $\eta$ . So the impact speed on the $n$ -th bounce and the rebound speed leaving it are, for $n=1,2,3,\dots$ ,

v_n^{\text{in}} = \frac{v_0}{\eta^{n-1}}, \qquad v_n^{\text{out}} = \frac{v_0}{\eta^{n}}.

Step 2Momentum imparted to the slab at one impact

The ball’s momentum change at the $n$ -th impact is $m(-v_n^{\text{out}}) - m(v_n^{\text{in}})$ ; the momentum delivered to the slab is the negative of that.

\begin{align*}\Delta p_n &= m\left(v_n^{\text{in}}+v_n^{\text{out}}\right) \\ &= mv_0\left(\frac{1}{\eta^{n-1}}+\frac{1}{\eta^{n}}\right)\end{align*}

Step 3Sum the geometric series over all bounces

The cumulative momentum is the sum of $\Delta p_n$ over every bounce, split into two geometric series with common ratio $1/\eta$ .

\begin{align*}\Delta p &= \sum_{n=1}^{\infty}\Delta p_n \\ &= mv_0\left(\sum_{n=1}^{\infty}\frac{1}{\eta^{n-1}}+\sum_{n=1}^{\infty}\frac{1}{\eta^{n}}\right) \\ &= mv_0\left(\frac{1}{1-1/\eta}+\frac{1/\eta}{1-1/\eta}\right) \\ &= mv_0\left(\frac{\eta}{\eta-1}+\frac{1}{\eta-1}\right)\end{align*}

\Delta p = mv_0\,\dfrac{\eta+1}{\eta-1}

(2)

Step 4Evaluate numerically

The closed form (2) needs the impact speed first, so evaluate $v_0=\sqrt{2gh}$ from (1) with $g=9.8$ m/s $^2$ and the given drop height.

v_0 = \sqrt{2(9.8\text{ m/s}^2)(1.0\text{ m})} = 4.427\text{ m/s}

That speed, the given mass and the given $\eta=1.25$ now go into (2).

\Delta p = (0.050\text{ kg})(4.427\text{ m/s})\cdot\frac{1.25+1}{1.25-1}

\Delta p = m\sqrt{2gh}\,\dfrac{\eta+1}{\eta-1} = 2.0\ \mathrm{kg\!\cdot\!m/s}

(3)

Check

The two partial sums that build $\Delta p$ are geometric series in $1/\eta$ , which converge only for $\eta>1$ , exactly the physical requirement that each bounce loses speed and the ball eventually comes to rest on the slab; $\eta=1.25>1$ satisfies it. In the limit $\eta\to\infty$ (the ball barely rebounds at all), the formula gives $\Delta p\to mv_0$ , the momentum of the first impact alone, matching the expectation that a vanishing rebound contributes negligibly to every later bounce. The book’s own printed key states $0.2$ kg $\cdot$ m/s for this result, but evaluating the book’s own formula with its own data gives $1.99$ , i.e. $2.0$ kg $\cdot$ m/s: the printed key carries a misplaced decimal point, not a different physics.

Remark

Two more cumulative quantities sum the same way, but not with the same ratio. Each flight time is proportional to the speed leaving the slab, $t_n=2v_n^{\text{out}}/g$ , so the time series decays by the same ratio $1/\eta$ per bounce as the momentum series above. Each rise height, however, goes as the square of that speed, $h_n=(v_n^{\text{out}})^2/(2g)$ , so the distance series decays by $1/\eta^2$ per bounce, a faster ratio (for $\eta=1.25$ , $0.64$ against $0.8$ ). All three sums, momentum, time and total distance, are nonetheless finite for the same underlying reason: a geometric series converges whenever its ratio is strictly less than one, and $\eta>1$ makes both $1/\eta$ and $1/\eta^2$ so.