Chapter 28 - The Electric Potential - Exercises and Problems - Page 837: 74

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We know that the repulsive force between the proton and the alpha particle will make them stop before colliding with each other. We assume that the system [the proton+the alpha particle] is perfectly isolated, so the energy and the momentum are conserved. $$E_i=E_f$$ $$U_{i}+K_{i,p}+K_{i,\alpha}=U_{f}+K_{f,p}+K_{f,\alpha}$$ Initially, the distance between them was far enough to make the electric potential energy zero. $r_i\rightarrow\infty$, hence $U_i=0$. Thus, $$0+K_{i,p}+K_{i,\alpha}=U_{f}+K_{f,p}+K_{f,\alpha}$$ $$\frac{1}{2}m_pv_i^2+\frac{1}{2}m_\alpha v_i^2=\dfrac{k_e(e)(2e)}{r_f}+\frac{1}{2}m_pv_{fp}^2+\frac{1}{2}m_\alpha v_{f,\alpha}^2 $$ The initial speeds of both particles are the same. $$\frac{v_i^2}{2}(m_p+m_\alpha) =\dfrac{2k_ee^2}{r_f}+\frac{1}{2}m_pv_{fp}^2+\frac{1}{2}m_\alpha v_{f,\alpha}^2 \tag 1$$ Now if we assumed that the final speeds of the two particles at the closest approach are zero, then the conservation of momentum principle will be viloated. Now let's see what conservation of momentum gives us: $$p_i=p_f$$ $$m_pv_{i }-m_\alpha v_{i }=m_pv_{fp}+m_\alpha v_{f,\alpha}$$ The negative sign of the speed of the alpha particle is due to the direction since both particles are moving toward each other which means one of them is moving to the right and the other is moving toward the left. $$(m_p -m_\alpha )v_{i }=m_pv_{fp}+m_\alpha v_{f,\alpha}$$ Now if we assumed that the two particles stopped at the closest approach, then the mass of the two particles must be the same which is not the case, so that violates this law. So we can assume that the repulsive force will be able to stop the lightest particle from moving instantaneously while the other will continue moving in the same direction at a lower speed. The alpha particle is 4 times the mass of the proton. So, $v_{fp}=0$. So that, $$(m_p -m_\alpha )v_{i }=m_p(0)-m_\alpha v_{f,\alpha}$$ Thus, $$v_{f,\alpha}=\dfrac{-(m_p -m_\alpha )v_{i }}{m_\alpha}\tag 2$$ Plug (2) into (1), and recall that $v_{fp}=0$ $$\frac{v_i^2}{2}(m_p+m_\alpha) =\dfrac{2k_ee^2}{r_f}+0+\frac{1}{2}m_\alpha \left[\dfrac{-(m_p -m_\alpha )v_{i }}{m_\alpha}\right]^2 $$ $$ v_i^2 (m_p+m_\alpha) =\dfrac{4k_ee^2}{r_f}+ \dfrac{ (m_p -m_\alpha )^2v_{i }^2}{m_\alpha } $$ $$ \dfrac{4k_ee^2}{r_f}= v_i^2 (m_p+m_\alpha) - \dfrac{ (m_p -m_\alpha )^2v_{i }^2}{m_\alpha } $$ $$ \dfrac{4k_ee^2}{r_f}= \dfrac{ v_i^2\left[ m_\alpha(m_p+m_\alpha) -(m_p -m_\alpha )^2 \right]}{m_\alpha } $$ $$ \dfrac{4k_ee^2}{r_f}= \dfrac{ v_i^2\left[ m_\alpha m_p+m_\alpha^2-m_p^2 +2m_pm_\alpha-m_\alpha^2 \right]}{m_\alpha } $$ $$ \dfrac{4k_ee^2}{r_f}= \dfrac{ v_i^2\left[ 3m_\alpha m_p -m_p^2 \right]}{m_\alpha } $$ Therefore, $$r_f=\dfrac{4k_em_\alpha e^2}{ v_i^2\left[ 3m_\alpha m_p -m_p^2 \right]}$$ Plug the known; $$r_f=\dfrac{4(9\times 10^9)(4\times 1.67\times 10^{-27}) (1.6\times 10^{-19})^2}{ (0.01\times 3\times 10^8)^2\left[ 12 -1 \right]( 1.67\times 10^{-27})^2}$$ $$r_f=\color{red}{\bf 2.23\times10^{-14}}\;\rm m$$