Constrained particles under distance dependent force This question is from the 1975 Canadian Association of Physicists Exam. No solutions are posted and I am quite lost on how to proceed with it.
A particle is constrained to move along the x-axis of a Cartesian co-ordinate system and an identical particle is constrained along the y-axis. Show that if the particles are originally at rest and attract each other according to any law which depends only on the distance between them, then they will reach the origin simultaneously.
 A: The operative word is "attract," which means that the force exerted on each particle points toward the other particle. Of course, not all of that force actually affects the motion of the particle, since each particle is constrained to move only along one axis. So a good first step would be to find the component of force on each particle that acts in the direction that particle is free to move in. Once you do that, try playing around with the equations to relate the motions of the two particles to each other.
Keep in mind that you don't know how the force between the two particles depends on the distance between them, so you'll need to write it as an unknown function, $f(r)$ for example.
A: If we call the particles $A$ and $B$, we can write the force that $B$ exerts upon $A$ as
$\mathbf{F}_{AB} = F\left(\|\mathbf{r}_{AB}\|\right)\frac{\mathbf{r}_{AB}}{\|\mathbf{r}_{AB}\|}$
where $\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A$ is the displacement vector between $B$ and $A$.
As only of the x coordinate of $A$ and the y coordinate of $B$ take nonzero values, we can simplify the expression of the force to
$\mathbf{F}_{AB} = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(-x_A, y_B)}{\sqrt{x_A^2 + y_B^2}}$
But this expression is not the total force acting over $A$, because $A$ is constrained to move over the x axis. This means that we need to keep only the projection of $\mathbf{F}_{AB}$ over the x axis to get the total force over $A$
$\mathbf{F}_A = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(-x_A, 0)}{\sqrt{x_A^2 + y_B^2}}$
By an identical process, we can arrive to
$\mathbf{F}_B = F\left(\sqrt{x_A^2 + y_B^2}\right) \frac{(0, -y_B)}{\sqrt{x_A^2 + y_B^2}}$
Using this expressions for the forces, together with Newton's 2nd Law, we get the following system of differential equations for $x_A$ and $y_B$ (the only nonzero coordinates of the particles)
$\frac{d^2x_A}{dt^2} = \frac{1}{m} F\left(\sqrt{x_A^2 + y_B^2}\right)\frac{-x_A}{\sqrt{x_A^2 + y_B^2}}$
$\frac{d^2y_B}{dt^2} = \frac{1}{m} F\left(\sqrt{x_A^2 + y_B^2}\right)\frac{-y_B}{\sqrt{x_A^2 + y_B^2}}$
The two differential equations are almost identical, with only one difference. Lets rewrite the equations in a way that emphasizes this difference
$G(x_A, y_B) = \frac{F\left(\sqrt{x_A^2 + y_B^2}\right)}{m\sqrt{x_A^2 + y_B^2}}$
$\frac{d^2x_A}{dt^2} = -x_A G(x_A, y_B)$
$\frac{d^2y_B}{dt^2} = -y_B G(x_A, y_B)$
I think that this is enough (maybe more than enough?) of a starting point. It's essentially the first part of David's answer translated to equations. :-)
A: Let the distance (d) related force between the two particles be (K)(d^Q), where K is a constant and Q is any power (e.g., for gravity K = MxMy times the gravitational constant, and Q = (-2)).
x, Vx, Ax, and Mx is the positon, velocity, acceleration, and mass, respectively, of the particle constrained to the x-axis.  X0 = position at t=0.
y, Vy, Ay, and My is the positon, velocity, acceleration, and mass, respectively, of the particle constrained to the y-axis. Y0 = position at t=0
d = SQRT(x^2 + y^2).
Assuming the x and y particles start out somewhere on the positive x and y axes, the force on the particle constrained to the x-axis, which results in motion is:
f(x) = -(K)(d^Q)cos(tan^-1(y/x).
Assuming the x and y particles start out somewhere on the positive x and y axes, the  force on the particle constrained to the y-axis, which results in motion is:
f(y) = -(K)(d^Q)sin(tan^-1(y/x).
Mx = My = M.  f = MA
f(y)/f(x) = MyAy/MxAx = Ay/Ax.
f(y)/f(x) = [-(K)(d^Q)sin(tan^-1(y/x)] / [-(K)(d^Q)cos(tan^-1(y/x)]
f(y)/f(x) = tan(tan^-1(y/x)) = y/x.
Therefore:  Ay/Ax = y/x.  Ay = Ax(y/x).

EDIT:  The below is incorrect (see comments section against this post)
x = (Ax)(t^2)/2 + X0.  When x=0 ---> t = SQRT(-2(X0)(Ax))
y = (Ax)(y/x)(t^2)/2 + Y0.  When y=0 ---> t = SQRT(-2(Y0)(Ax)(y/x))

You wanted to know how to get started.  Armed with the fact that Ay/Ax = y/x, I think you can take from there. 
