Question

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the speed of the proton by the time the distance between theproton and the alpha particle doubles. Both particles arepositively charged. The charge and the mass of the proton are,respectively, e and m. The e charge and the mass of the alphaparticle are, respectively, 2e and 4m.Find the speed of the proton (vf)p by the time the distancebetween the particles doubles.Express your answer in terms of some or all of the quantities,e, m, r, and ?0.Which of the following quantities are unknown?A initial separation of the particlesB final separation of the particlesC initial speed of the protonD initial speed of the alpha particleE final speed of the protonF final speed of the alpha particleG mass of the protonH mass of the alpha particleI charge of the protonJ charge of the alpha particle

Answer 1

by momentum conservation we have

$P_i = P_f$

$0 = m(v_p) + 4m(v_a)$

$v_p = - 4v_a$

also here we will use the concept of energy conservation

$U_i + K_i = U_f + K_f$

$\frac{Ke(2e)}{r^2} + 0 = \frac{Ke(2e)}{4r^2} + \frac{1}{2}mv_p^2 + \frac{1}{2}(4m)(v_a^2)$

$\frac{2Ke^2}{r^2}- \frac{Ke^2}{2r^2} = \frac{1}{2}mv_p^2 + \frac{1}{2}(4m)v_a^2$

$\frac{3Ke^2}{2r^2} = \frac{1}{2}mv_p^2 + 2m(\frac{v_p}{4})^2$

$\frac{3e^2}{8\pi\epsilon_0 r^2} = \frac{5}{8}mv_p^2$

$v_p^2 = \frac{3e^2}{5m\pi\epsilon_0 r^2}$

$v_p = \sqrt{\frac{3e^2}{5m\pi\epsilon_0 r^2}}$

unknown quantities are as following

E final speed of the proton

F final speed of the alpha particle