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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.

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I'm adding the homework tag here not to suggest that it's actually a homework question, but because it's one of those "educational value" problems where the point is to understand the method, not just to get the answer. (See meta.physics.stackexchange.com/q/714 for example) –  David Z Jun 20 '11 at 17:54

3 Answers 3

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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. :-)

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yeah, that's basically what I did for myself... I just didn't go into this much detail in my answer because of the whole "educational value" thing ;-) –  David Z Jun 20 '11 at 19:03
    
@Manu I didn't. I intentionally avoided giving a full solution to the problem for the moment. But it's just a question of playing with possible relationships between $y_B$ and $x_A$, as David said in his answer. –  mmc Jun 20 '11 at 19:27

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.

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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.

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I'm a little bit confused here, and perhaps I'm missing something. I thought the equation $x = x_{0} + v_{0}t + \frac{1}{2}at^{2}$ applied only to uniform acceleration. That's not the case here is it? –  EuYu Jun 20 '11 at 18:44
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Also, we're not given that the force is a power law $F=kd^Q$: it could be any function of $d$. –  Ted Bunn Jun 20 '11 at 19:00
    
@Yuqing --> You are right. My last two equations assumed constant acceleration, which is incorrect. Thank you. Post edited. –  Vintage Jun 20 '11 at 20:24
    
@ Ted ---> You are correct. However, the OP says that force is dependent on d, which means SOME non-zero power of d must be in the equation somewhere, even if that power is 1 (d^1 = d). I did over-simplify (even though I was trying to be general) by not allowing such things as ln(d); but the question is moot, since whatever the function of d is, it drops out of the equation when you solve for ratios of acceleration. Ay/Ax = y/x, no matter what the force function is. –  Vintage Jun 20 '11 at 20:31

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