Show that for an inelastic scattering the fractional loss of kinetic energy is
$$
\frac{T-T'}{T} =
\left(\, 1 – \mathrm{e}^{2}\, \right)\,{m_{2} \over m_{1} + m_{2}}
$$
where $T$ is the initial kinetic energy and $T'$ the final kinetic energy. Hence deduce that $\mathrm{e} \leq 1$.
$$
\mbox{I know that}\quad T =
{1 \over 2}\,m_{1}u_{1}^{2}
\quad\&\quad
T' =
{1 \over 2}\,m_{1}v^{2}_{1} + {1 \over 2}\,m_{2}v^{2}_{2}
$$
But how do I manipulate an equation for
$\displaystyle{v^{2}_{1}\,\,\, \&\,\,\, v^{2}_{2}}$ to help get the RHS
equation ?.
Best Answer
Isn't this just solving the following two simultaneous equations for $v_1$ and $v_2$ in terms of $m_1$, $m_2$, $u_1$, $u_2$, and $e$: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ and $$\left(v_2-v_1\right)=e\,\left(u_2-u_1\right)\,,$$ where $u_i$ is the initial speed of the particle $i$ with mass $m_i$ and $v_i$ is its terminal speed, for $i\in\{1,2\}$? Then, with $T=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$ and $T'=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$, the rest should follow immediately. For a scattering process, $u_2=0$.