/sci/ - Science & Math

[No.2532563]

"From a large distance away, a particle of mass m1 and positive charge q1 is fired at speed v in the positive x direction straight toward a second particle, originally stationary but free to move, with mass m2 and positive charge q2. Both particles are constrained to move only along the x axis. Find the distance of closest approach."

How do I solve this, /sci/?

[No.2532712]

bump

[No.2532775]

Can somebody help me out?
Please?
Will bump with moar troll sci

[Anonymous No.2532791]

I'm gonna take a guess and say by deriving something.

[Anonymous No.2532816]

>>2532563
There are no numbers given though.

[No.2532817]

>>2532791
but wat?

[No.2532845]

>>2532816
I'm supposed to answer it algebraically in terms of the given variables

[had to submit assignment already b4 I missed the deadline, but I would still like to know for future reference]

[Anonymous No.2532865]

You're going to have to integrate their positions over time, which would be a differential equations problem.

Or possibly, if you take the initial kinetic energy of the first particle, you can convert that into a potential energy of closest approach... but you'd also have to take into account their shared kinetic energy at that point, so maybe that wouldn't be easier.

[Anonymous No.2532867]

Bro, I don't know much about electrostatics. But if you want someone to answer you might want to post some humorous trollphysics pics. These are just bad.

[Anonymous No.2532883]

I think the best solution for you would be accepting a career in management with Mcdonalds. Problem solved

[Anonymous No.2532912]

>>2532775
this one is funny because it says lightspeed is slower than 99% lightspeed

[Anonymous No.2532961]

>>2532845

Method 1) If you assume mass two doenst move (m2>>m1). This is usually how this problem is given.

Find the force of of attraction on mass one, and the force of replusions. Set them equal to eachother, solve for the distance at which this happens.

Method 2) Just set the kinetic energy of mass 1 equal to the potential energy between the two masses and solve for r. Technically this doesnt give the right answer, but is a good enouigh approximation for what you need.

Real Method) If you assume both masses move, then you need to use Lagragaian mechnaics, and you will get a set of coupled differntail equations. You then need to solve for the position fucntions of both particles.

[Anonymous No.2533133]

You have the kinetic energy of the particles, and you have the potential energy. The closest possible approach will be when the potential energy is equal to total energy.

T_0 = 0.5mv^2 + 0.5MV^2
U_0 = -Qqk/r_0
E_0 = 0.5mv^2 + 0.5MV^2 - Qqk/r_0 = E_f = U_f
T_f = 0

r_c = Qqk/U_f
= q1q2k/(0.5m1v1^2 - q1q2k/r0)
= 1/(m1v1^2/2q1q2k - 1/r0)

[Anonymous No.2533147]

>>2532961
Gwen hierarchy:
Best Gwen
Ben 10

Shit Gwen
Everything else.

[No.2533198]

>>2533133
so, if the particles start infinitely far apart, then the closest they get would be

2k*q1*q2
------------
m1*v1^2

right?

[Anonymous No.2533237]

>>2533198
That looks right, in the sense that the minimum distance should be proportional to the charges, but inversely proportional to the velocity and mass.

Although I think I swapped signs on the potential energy out of habit.