Question

Two oppositely and equally charged particles of equivalent mass 'm' a distance 'd' apart. find the time it takes for the two particles to collide.

Answer 1

U(x) = qV(x)
V(x) = kq/x
U = kq^2/x
KE = 1/2mv^2
v(x) = sqrt( 2kq^2/mx )
dx/dt = v(x)
1/v(x) dx = dt
t = Int[0,1/2d] 1/v(x) dx
t = sqrt( mx/8kq^2 )
Pretty sure I did it incorrectly, was wondering what other people thought.

Answer 2

eh, have no Idea what this is because I'm not very far in physics, but I can provide a medal. hope you don't mind

Answer 3

suppose both are intially at rest ..and moving with velocity v (both moving with equal velocity)
opposite charGe attract with force F
F=(q1q2/d^2)9*10^9
F=Ma
a=q1q2/d^2M
v=u+at
v+0+(q1q2/d^2M)t
t=vd^2M/Q1Q2

Answer 4

So how come energy doesn't derive the same answer you got using forces?

Answer 5

The kinetic energy is equal to the DIFFERENCE between the potential energy at a given point and the potential energy at the starting point.

Answer 6

@jordanlovato
It went wrong here :

v(x) = sqrt( 2kq^2/mx )

Because you have forgotten the electric potential energy at the starting point, namely: kq²/d

Answer 7

This question is quite difficult because you need the value of a generalised integral.

Answer 8

Another issue is that as they meet the velocity of approach becomes infinite! You will be integrating a singularity.

Answer 9

Okay, second attempt at this..

U(x) = qV(0) - qV(x)
V(0) = kq^2/d
V(x) = kq^2/d-2x -> both particles have moved x distance closer to each other

U(x) = kq^2(2x/d(d-2x))
U(x) = KE(x)
kq^2(2x/d(d-2x)) = 1/2mv(x)^2

v(x) = sqrt(2kq^2/m) * sqrt( 2x/d(d-2x) )
v(x) = dx/dt
t = Int[0,1/2d] 1/v(x) dx
t = sqrt(m/2kq^2) * Int [0,1/2d] sqrt(d(d-2x)/2x)dx

Am I getting any closer?

Answer 10

Keep x as the distance that will vary from d to 0.
v² will be proportional to 1/x - 1/d
To keep v positive , choose v = –dx/dt

Answer 11

In the end, I think this integral is needed:

http://www.wolframalpha.com/input/?i=integrate+%281%2Fx-1%29^%28-1%2F2%29+dx+from+x%3D0+to+1

Answer 12

Cool, thank you sir!