Respuesta :
by momentum conservation we have
[tex]P_i = P_f[/tex]
[tex]0 = m(v_p) + 4m(v_a)[/tex]
[tex]v_p = - 4v_a[/tex]
also here we will use the concept of energy conservation
[tex]U_i + K_i = U_f + K_f[/tex]
[tex]\frac{Ke(2e)}{r^2} + 0 = \frac{Ke(2e)}{4r^2} + \frac{1}{2}mv_p^2 + \frac{1}{2}(4m)(v_a^2)[/tex]
[tex]\frac{2Ke^2}{r^2}- \frac{Ke^2}{2r^2} = \frac{1}{2}mv_p^2 + \frac{1}{2}(4m)v_a^2[/tex]
[tex]\frac{3Ke^2}{2r^2} = \frac{1}{2}mv_p^2 + 2m(\frac{v_p}{4})^2[/tex]
[tex]\frac{3e^2}{8\pi\epsilon_0 r^2} = \frac{5}{8}mv_p^2[/tex]
[tex]v_p^2 = \frac{3e^2}{5m\pi\epsilon_0 r^2}[/tex]
[tex]v_p = \sqrt{\frac{3e^2}{5m\pi\epsilon_0 r^2}}[/tex]
unknown quantities are as following
E final speed of the proton
F final speed of the alpha particle
The speed of the proton when expressed in the given terms is gotten as;[tex]v_{p} = \sqrt{\frac{2e^{2}}{5\pi \epsilon_{0} rm} }[/tex]
What is the speed of the proton?
We are given;
Charge of proton = e
Mass of the proton = m
Charge of the alpha particle = 2e
Mass of alpha particle = 4m
The expression to find the initial electrostatic potential energy is;
U_i = (K * e * 2e)/r
where k is coulombs constant
However, final potential energy is;
U_f = (k*e*2e)/2r
Since the proton and alpha particle are momentarily at rest, it means that the initial velocity of both particles is zero.
Using law of conservation of momentum, we have;
(m_p × v_p) - (m_α × v_a) = 0
where;
m_p is mass of proton
v_p is velocity of proton
m_α is mass of alpha particle
v_α is velocity of alpha particle
We are told that mass of alpha particle is 4m and mass of proton is m. Thus;
(m × v_p) - (4m × v_α) = 0
m × v_p = 4m × v_α
v_α = v_p/4
Using law of conservation of energy, we have;
[tex]k\frac{e*2e}{r} = \frac{1}{2} mv_{p}^{2} + \frac{1}{2} 4mv_{\alpha}^{2} + k\frac{e*2e}{2r}[/tex]
Putting v_p/4 for v_α and simplifying gives us;
[tex]v_{p} = \sqrt{\frac{8}{5}\frac{ke^{2}}{rm} }[/tex]
Now k is a constant with the formula;
k = 1/(4πε₀)
Thus, putting  1/(4πε₀) for k gives us;
[tex]v_{p} = \sqrt{\frac{2e^{2}}{5\pi \epsilon_{0} rm} }[/tex]
Read more about speed of proton at; https://brainly.com/question/17238282
