Aproton 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 particles
b final separation of the particles
c initial speed of the proton
d initial speed of the alpha particle
e final speed of the proton
f final speed of the alpha particle
g mass of the proton
h mass of the alpha particle
i charge of the proton
j charge of the alpha particle

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

okay i have so many questions. but the big one here is: um wt..f is this mess? ?