The initial electric potential energy of the system is:
$$U_i=k\frac{(2q)(q)}{r_i}$$
Where k is the electrostatic constant and \(r_i=0.225m\). The final kinetic energy of the system is:
$$K_f=\frac{1}{2}mv_p^2+\frac{1}{2}(4m)v_\alpha^2$$
Where \(v_p\) and \(v_\alpha\) are the final speeds of the proton and alpha particle, respectively.
By conservation of energy, we have:
$$U_i=K_f$$
$$k\frac{(2q)(q)}{r_i}=\frac{1}{2}mv_p^2+2(4m)v_\alpha^2$$
$$k\frac{(2q)(q)}{0.225m}=\frac{1}{2}mv_p^2+8mv_\alpha^2$$
$$9\times10^9\frac{Nm^2}{C^2}\frac{2(1.6\times10^{-19}C)(1.6\times10^{-19}C)}{0.225m}=\frac{1}{2}(1.67\times10^{-19}kg)v_p^2+8(1.67\times10^{-27}kg)v_\alpha^2$$
$$7.94\times10^{-18}J=1.67\times10^{-27}kg(v_p^2+8v_\alpha^2)$$
$$4.74\times10^{9}m^2s^{-2}=v_p^2+8v_\alpha^2$$
Because of momentum conservation, we have:
$$0=(2q)v_p+(4q)v_\alpha$$
$$-2v_p=4v_\alpha$$
Substituting into the previous equation:
$$4.74\times10^{9}m^2s^{-2}=v_p^2+8\left(-\frac{1}{2}v_p\right)^2$$
$$4.74\times10^{9}=v_p^2+v_p^2$$
$$4.74\times10^{9}=2v_p^2$$
$$v_p=\sqrt{\frac{4.74\times10^9}{2}}=\sqrt{2.37\times10^9}$$
$$\boxed{v_p=4.86\times10^4 m/s}$$
