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Question

# A particle of mass $m_{1}$ with an initial kinetic energy ${T}_{1}$ makes an elastic collision with a particle of mass $m_{2}$ initially at rest. $m_{1}$ is deflected from its original direction with a kinetic energy $T_{1}^{\prime}$ through an angle ${\phi}_{1}$. Letting $\alpha=m_{2} / m_{1} \text { and } \gamma=\cos \phi_{1}$, show that the fractional kinetic energy lost by $m_{1}, \Delta T_{1} / T_{1}=\left(T_{1}-T_{1}^{\prime}\right) / T_{1}$, is given by $\frac{\Delta T_{1}}{T_{1}}=\frac{2}{1+\alpha}-\frac{2 \gamma}{(1+\alpha)^{2}}(\gamma+\sqrt{\alpha^{2}+\gamma^{2}-1})$

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Basically, what we are trying to calculate in this problem is fractional kinetic energy lost by $m_1$

\begin{align*} \frac{T_1 - T_1'}{T_1} = \frac{\Delta T_1}{T_1} &= 1 - \underbrace{\frac{T_1' }{T_1}}_{\equiv r} = 1 -r , \tag{1} \end{align*}

where $T_1 = \dfrac{1}{2} m_1 v_1^2$ is initial kinetic energy of mass $m_1$, $T_1' = \dfrac{1}{2} m_1 v_1^{2 \prime}$ final kinetic energy of mass $m_1$ and $r= \dfrac{ T_1' }{T_1} = \dfrac{v_1^{2 \prime}}{v_1^2}$ ration of kinetic energies of scattered particle to incident particle energy. Because this is similar problem that is in textbook (figure 7.6.1.), we can make use of figures in textbook; if we look at figure 7.6.2., we can recognize following relation

\begin{align*} \overline{v}_1' \overline{v}_1' = (v_1' - v_{ \text{cm}}) \cdot (v_1' - v_{ \text{cm}}), \tag{2} \end{align*}

where $\overline{v}_1'$ is scattering speed of particle of mass $m_1$ in the center of mass system, $v_1'$ is scattering speed of particle of mass $m_1$ in lab system and $v_{ \text{cm}}$ is speed of center of mass. Then we have from equation (2)

\begin{align*} \overline{v}^{2 \prime} &= v_1^{2 \prime} + v_{\text{cm}}^2 - 2 v_1' v_{\text{cm}} \cos \phi_1 \\ v_1^{2 \prime} &= \overline{v}_1^{2 \prime} - v_{\text{cm}}^2 + 2 v_1' v_{\text{cm}} \gamma, \tag{3} \end{align*}

where $\gamma = \cos \phi_1$ ($\phi_1$ is scattering angle of mass $m_1$). With the use of expression (3) and definition of r we have

\begin{align*} r &= \frac{ v_1^{2 \prime} }{v_1^2} \\ r &= \frac{ \overline{v}_1^{2 \prime} }{ v_1^2 } - \frac{ v_{\text{cm}}^2 }{ v_1^2 } + \frac{ 2 v_1' v_{\text{cm}} \gamma }{v_1^2 }. \tag{4} \end{align*}

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