Two particles \(A\) and \(B\) of masses \(m\) and \(km\),
respectively, are at rest on a smooth horizontal surface. The direction of the line passing through \(A\) and \(B\) is perpendicular to a vertical wall which is on the other side of \(B\) from \(A\). The particle \(A\) is now set in motion towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e_1\) and between \(B\) and the wall is \(e_2\). Show that there will be a second collision between \(A\) and \(B\) provided
$$
k< \frac {1+e_2(1+e_1)} {e_1}\;.
$$
Show that, if \(e_1=\frac13\), \(e_2=\frac12\) and \(k<5\), then the kinetic energy of \(A\) and \(B\) immediately after \(B\) rebounds from the wall is greater than \(mu^2/27\).
Solution: First collision:
Since the \(e = e_1\), the speed of approach is \(u\) the speed of separation will be \(e_1u\) and so \(v_B = v_A + e_1u\).
\begin{align*}
\text{COM}: && mu &= mv_A + km(v_A + e_1u) \\
\Rightarrow && v_A(1+k) &= u(1-ke_1) \\
\Rightarrow && v_A &= \frac{1-ke_1}{1+k} u \\
&& v_B &= \frac{1-ke_1}{1+k} u + e_1 u \\
&&&= \frac{1-ke_1 + e_1+ke_1}{1+k}u \\
&&&= \frac{1+e_1}{1+k}u
\end{align*}
Once the ball rebounds from the wall it will have velocity (still taking towards the wall as +ve) of \(-\frac{1+e_1}{1+k}e_2u\). There will be another collision if it is travelling faster than \(A\), ie if:
\begin{align*}
-\frac{1+e_1}{1+k}e_2u &< \frac{1-ke_1}{1+k} u \\
\Leftrightarrow && 0 &< (1-ke_1) + (1+e_1)e_2 \\
\Leftrightarrow && ke_1 &< 1 +e_2 (1+e_1) \\
\Leftrightarrow && k &< \frac{1 +e_2 (1+e_1)}{e_1} \\
\end{align*}
If \(e_1 = \frac13, e_2 = \frac12\), then \(v_A = \frac{1-\frac13k}{1+k}u = \frac{3-k}{3(1+k)}u\) and \(v_B = \frac{4}{3(1+k)}u\).
Therefore
\begin{align*}
&& \text{total k.e.} &= \underbrace{\frac12 m v_A^2}_{\text{k.e. of }A} + \underbrace{\frac12 (km) (e_2 v_B)^2}_{\text{k.e. of }B} \\
&&&= \frac12 m \frac{(3-k)^2}{9(1+k)^2}u^2 + \frac12 km \frac14 \frac{16}{9(1+k)^2}u^2 \\
&&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( (3-k)^2+4k \right) \\
&&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( 9-2k+k^2 \right) \\
&&&= \frac{mu^2}{18} \frac{9-2k+k^2}{1+2k+k^2}
\end{align*}
We wish to minimize this as a function of \(k\).
\begin{align*}
\frac{\d}{\d k} \left ( \frac{9-2k+k^2}{1+2k+k^2}\right) &= \frac{(1+k)^2(2k-2)-2(1+k)(k^2-2k+9)}{(1+k)^4} \\
&= \frac{2(k^2-1) - 2(k^2-2k+9)}{(1+k)^3} \\
&= \frac{2(2k-10)}{(1+k)^3}
\end{align*}
Therefore the minimum will be when \(k = 5\) can't be a maximum by considering \(k \to 0\).
This value is \(\frac{2}{3}\) and therefore \(\frac{mu^2}{18} \frac{2}{3} = \frac{mu^2}{27}\) is the smallest energy (which isn't quite achievable since \(k < 5\).
