Problem source

The identical uniform smooth spherical marbles $A_{1},A_{2},\ldots,A_{n},$ where $n\geqslant3,$ each of mass $m,$ lie in that order in a smooth straight trough, with each marble touching the next. The marble $A_{n+1},$ which is similar to $A_{n}$ but has mass $\lambda m,$ is placed in the trough so that it touches $A_{n}.$ Another marble $A_{0},$ identical to $A_{n},$ slides along the trough with speed $u$ and hits $A_{1}.$ It is given that kinetic energy is conserved throughout. 
\begin{questionparts}
\item Show that if $\lambda<1,$ there is a possible subsequent motion in which only $A_{n}$ and $A_{n+1}$ move (and $A_{0}$ is reduced to rest), but that if $\lambda>1,$ such a motion is not possible. 
\item If $\lambda>1,$ show that a subsequent motion in which only $A_{n-1},A_{n}$ and $A_{n+1}$ move is not possible. 
\item If $\lambda>1,$ find a possible subsequent motion in which only two marbles move. 
\end{questionparts}

Solution source

Without loss of generality, let $m = u = 1$.

\begin{questionparts}
\item 
\begin{align*}
\text{COM}: && 1&=  v_n +  \lambda v_{n+1} \\
&&  &= v_n + \lambda v_{n+1}\\
\text{COE}: && \frac12   &= \frac12  v_n^2 + \frac12 \lambda  v_{n+1}^2 \\
&& 1 &= v_n^2 +\lambda v_{n+1}^2 \\
\\
\Rightarrow && v_n^2 + 2\lambda v_n v_{n+1}  + \lambda^2 v_{n+1}^2 &= v_n^2 + \lambda v_{n+1}^2 \\
&&  \lambda v_{n+1}^2 &= v_{n+1}^2 - 2 v_n v_{n+1} \\
&& \lambda v_{n+1} &= (v_{n+1} - 2v_n)  \\
&& (1-\lambda)v_{n+1} &= 2v_n
\end{align*}

Since $v_{n+1} > v_n > 0$ this is only possible if $\lambda < 1$ 

\item \begin{align*}
\text{COM}: &&  1&=  v_{n-1}+v_n+\lambda  v_{n+1} \\
&&  1&= v_{n-1} + v_n + \lambda v_{n+1} \\
\text{COE}: && \frac12  &= \frac12 v_{n-1}^2+\frac12v_n^2+\frac12\lambda  v_{n+1}^2 \\
&& 1&= v_{n-1}^2 + v_n^2 + \lambda v_{n+1}^2 \\
\\
\Rightarrow && 1 &= v_{n-1}^2 + v_n^2 + \lambda \frac{(1-v_{n-1}-v_n)^2}{\lambda^2} \\
&&&= v_{n-1}^2 + v_n^2 + \frac{(1-v_{n-1}-v_n)^2}{\lambda} \\
\Rightarrow && 1 &< v_{n-1}^2 + v_n^2 + (1-v_{n-1}-v_n)^2 \\
&&&= 2v_{n-1}^2+2v_n^2 + 1-2v_{n-1}-2v_{n-2} +2v_{n-1}v_n\\
\Rightarrow && v_{n-1}+v_n & <(v_{n-1}+v_n)^2 - v_{n-1}v_n
\end{align*}

but this cannot be true since $0 < v_{n-1}+v_n < 1$ and $v_n v_{n-1} > 0$

\item The only way this is possible is if the first and last marble are moving.

\begin{align*}
\text{COM}: && 1 &= v_0 +\lambda v_{n+1} \\
\text{COE}: && \frac12 &= \frac12 v_0^2 + \frac12 \lambda^2 v_{n+1} \\
&& 1 &= v_0^2 + \lambda v_{n+1}^2 \\
\Rightarrow && 2v_0 + \lambda v_{n+1} &= v_{n+1} \\ 
\Rightarrow && v_{n+1} &=\frac{2}{1-\lambda} v_0 \\
\Rightarrow && v_0 &= \frac{1-\lambda}{1+\lambda} \\
&& v_{n+1} &= \frac{2}{1+\lambda}  
\end{align*}

which will work since $v_0$ can travel backwards.
\end{questionparts}