Problem source

\begin{questionparts}
\item
In an experiment, a particle $A$ of mass $m$ is at rest on a smooth horizontal table. A particle $B$ of mass $bm$, where $b >1$, is projected along the table directly towards $A$ with speed $u$. The collision is perfectly elastic. 
Find an expression for the speed of $A$ after the collision in terms of $b$ and $u$, and show that, irrespective of the relative masses of the particles,  $A$ cannot be made to move at twice the initial speed of $B$. 
\item 
In a second experiment, a particle $B_1$ is projected along the table directly towards $A$ with speed $u$. This time,  particles $B_2$, $B_3$, $\ldots\,$, $B_n$ are at rest in order on the line between $B_1$ and $A$. The mass of $B_i$ ($i=1$, $2$, $\ldots\,$, $n$) is $\lambda^{n+1-i}m$, where $\lambda>1$. All collisions are perfectly elastic. Show that, by choosing $n$ sufficiently large, there is no upper limit on the speed at which $A$ can be made to move. In the case $\lambda=4$, determine the least value of $n$ for which $A$ moves at more than $20u$. You may use the approximation $\log_{10}2 \approx 0.30103$.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$bm$};
        \def\massb{$m$};
        \def\velocityua{$u$};
        \def\velocityub{$0$};
        \def\velocityva{$v_B$};
        \def\velocityvb{$v_A$};


        \draw (0,0) circle (1);
        \draw (2.5,0) circle (1);

        \draw (8,0) circle (1);
        \draw (10.5,0) circle (1);
        
        
        \node at (1.25, \h) {before collision};
        \node at (9.25, \h) {after collision};

        \draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);

            \node at (0,0) {\massa};
            \draw[->, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
            \node[above, red] at (0, \arrowheight) {\velocityua};
            \node at (2.5,0) {\massb};
            \draw[->, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight);  \node[above, red] at (2.5, \arrowheight) {\velocityub};
            \node at (8,0) {\massa};
            \draw[->, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);  
            \node[red,above] at (8, \arrowheight) {\velocityva};
            \node at (10.5,0 ) {\massb};
            \draw[->, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);  
            \node[red,above] at (10.5, \arrowheight) {\velocityvb};
    \end{tikzpicture}
\end{center}

Since the collision is perfectly elastic, the speed of approach and separation are equal, ie $v_B = v_A - u$

\begin{align*}
\text{COM}: && bmu &= bm(v_A - u) + mv_A \\
\Rightarrow && (b+1)v_A &= 2bu \\
\Rightarrow && v_A &= \frac{2b}{b+1} u
\end{align*}

Since $0 < \frac{b}{b+1} < 1$, the largest $0 < v_A < 2u$

\item After the first collision with each $B_i$ we will have $\displaystyle  v_{i+1} = \frac{2\lambda}{\lambda + 1}v_i$, ie $\displaystyle v_{i+1} = \left (\frac{2\lambda}{\lambda + 1} \right)^i u$ and so $\displaystyle v_A = \left (\frac{2\lambda}{\lambda + 1} \right)^n u$ which can be arbitrarily large. Suppose $\lambda = 4$, then

\begin{align*}
&& 20u &< v_A \\
&&&= \left (\frac{8}{5} \right)^n u \\
\Rightarrow && \log_{10} 20 < n \log_{10}(16/10) \\
&& \log_{10} 2 + 1 < n 4\log_{10} 2 - n \\
\Rightarrow && n &> \frac{ \log_{10} 2 + 1}{ 4\log_{10} 2 - 1} \\
&&&\approx \frac{0.30103+1}{4 \times 0.30103 -1}\\
&&&= \frac{1.30103}{0.20412} \\
&&&>6
\end{align*}

So $n =7$ is the smallest possible
\end{questionparts}