Problem from Newtonian Mechanics by A. P. French

Jan 2020


This is my first post I'm sorry if I'm not following some rule. I've spent hours on this problem. I proved it in a spreadsheet but they didn't even have spreadsheets when this book was written. I can get the formula for each collission but can't get them to collapse into any easy formula based on n. And I have no idea where e comes from. Theres a quarter page of approximations used throughout the book but none of them contain e.


Forum Staff
Apr 2008
On the dance floor, baby!
I didn't see the two expansions that are needed! This will help. When \(\displaystyle x = \alpha n << 1\), then \(\displaystyle e^{-x} \approx 1 - x\). For the second \(\displaystyle x = \alpha n \to \infty\) then \(\displaystyle e^{-x} \approx 0\). (Note: There may be an asymptotic expression for when x is infinite, but I couldn't find one.)

Jan 2020
9-13 A block of mass $m$, initially at rest on a frictionless surface, is bombarded by a succession of particles each of mass $\delta$m ($\ll$m) and initial speed $v_0$ in the positive x direction. The collisions are perfectly elastic and each particle bounces back in the negative x direction. Show that the speed acquired by the block after the *n*th particle has struck it is given very nearly by v = $v_0(1 - e^{-\alpha n})$, where $\alpha = 2\delta m/m$. Consider the validity of this result for $\alpha n \ll 1$ as well as $\alpha n\to\infty$

This is my work so far:
If $u_1$ and $u_2$ are the two initial velocities and $v_1$ and $v_2$ are the two final velocities, then $u_1$ = $v_0$ always and $u_2$ = 0 initially and is $v_2$ of the previous collision thereafter.

By elasticity: $u_1$ - $u_2$ = $v_2$ - $v_1$ $\Rightarrow$ $v_1$ = $v_2$ + $u_2$ - $v_0$

By conservation of momentum: $$\delta m v_0 + m u_2 = \delta m v_1 + m v_2$$

By combination:
$$\delta m v_0 + m u_2 =\delta m (v_2 + u_2 - v_0) + m v_2\Rightarrow v_2 = {2\delta v_0 + (1 - \delta)u_2 \over (1 + \delta)}$$

I can't get this to collapse to any equation based on n, let alone one approximated by anything involving $e$. I can prove it with a spreadsheet but this book was written in the early 70s.

Obviously, as $\alpha n \to \infty$, $v = v_0$ so the approximation given holds there. I can prove by rote calcuation it works well for the first few collisions given small $\delta$ even with early 1970s technology. Is that all one can do at the introductory mechanics level?

If $m_n$ is velocity of object $m$ after $n^{th}$ impact, $\beta = {2\delta v_0\over(1+\delta)}$, $\gamma = {(1-\delta)\over(1+\delta)}$, then $v_{n+1} = \beta + \gamma v_n,$

$m_0 = 0 $

$m_1 = \beta$

$m_2 = \beta + \beta\gamma$

$m_3 = \beta + \beta\gamma + \beta\gamma^2$

$m_4 = \beta + \beta\gamma + \beta\gamma^2 + \beta\gamma^3 $

$m_n = \beta\sum_{i=0}^{n-1}\gamma^i$

But by the approximation,
$(1+x) \approx e^x when x \ll 1$ therefore, since $\delta \ll 1,\gamma \approx \frac{e^{-\delta}}{e^\delta} = e^{-2\delta} \Rightarrow \gamma^i \approx e^{-2\delta i} $


$m_n \approx \frac{2\delta v_0}{e^\delta}\sum_{i=0}^{n-1}e^{-2\delta i}$

So, the question is why does $\frac{2\delta}{e^\delta}\sum_{i=0}^{n-1}e^{-2\delta i} \approx 1 - e^{-2 \delta n}$
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Jan 2020
By the aid of this guy here: approximations of finite series of $e^x$

So, the question is why does $\frac{2\delta}{e^\delta}\sum_{i=0}^{n-1}e^{-2\delta i} \approx 1 - e^{-2 \delta n}$

$$\sum_{i=0}^N r^i = \frac{1-r^{N+1}}{1-r}$$
with $r = e^{-2\delta}$ and $N=n-1$. Note that
$$\frac{2\delta}{e^\delta} \sum_{i=0}^{n-1} (e^{-2\delta})^i = \frac{2\delta}{e^\delta} \frac{1-(e^{-2\delta})^n}{1-e^{-2\delta}}$$

Since $e^x \approx (1 + x)$ when $\lvert x\rvert \ll 1$, $(1 - e^{-2\delta}) \approx 1 - (1 - 2\delta) = 2\delta$ and $e^\delta \approx 1$ so $$\frac{2\delta}{e^\delta}\sum_{i=0}^{n-1}e^{-2\delta i} \approx 1 - e^{-2 \delta n}$$