Chapter 9, Problem 114GP

(III) A particle of mass m_A traveling with speed v_A collides elastically head-on with a stationary particle of smaller mass m_B. (a) Show that the speed of m_B after the collision is

${v^{\prime }}_B=\frac{2v_{\text{A}}}{1+m_{\text{B}}/m_{\text{A}}}$

(b) Consider now a third particle of mass m_C at rest between m_A and m_B so that m_A first collides head on with m_C and then m_C collides head on with m_B. Both collisions are elastic. Show that in this case.

${v^{\prime }}_B=4v_{\text{A}}\frac{m_{\text{C}}{m}_{\text{A}}}{\left(m_{\text{C}}+m_{\text{A}}\right)\left(m_{\text{B}}+m_{\text{C}}\right)}.$

(c) From the result of part (b), show that for maximum ${v^{\prime }}_{\text{B}}$, $m_{\text{C}}=\sqrt{m_{\text{A}}{m}_{\text{B}}}$. (d) Assume m_B = 2.0 kg, m_A = 18.0 kg and v_A = 2.0 m/s. Use a spreadsheet to calculate and graph the values of ${v^{\prime }}_{\text{B}}$ from $m_{\text{C}}=0.0\text{ kg}$ to $m_{\text{C}}=50.0\text{ kg}$ in steps of 1.0 kg. For what value of m_C is the value of ${v^{\prime }}_{\text{B}}$ maximum? Does your numerical result agree with your result in part (c)?