Two particles of mass 2m and 3m are d meters apart and move toward each other due to the gravitational force on each other. What is the distance of the position where they collide to the starting position of the mass 2m?

I have no idea about how to solve it. I tried Google but found nothing helpful.
I have basic idea on gravitation and stuff. If you can tell me some keywords to search in Google or YouTube, I will look it up myself. If I need to learn a new chapter, please tell me it's name. I will learn it and then try this.
Otherwise I have no idea about how to start. Please help.

This is actually a lot easier than it looks. Consider the two bodies as a single system, with a center of mass. What happens to the center of mass as the two bodies move towards each other, given that there are no external forces acting on the system?

Well the center of mass then should be closer to the mass 3m... And it is 3d/5 meters away from the mass 2m. Hey! Whatever the value of d is, the center of mass will always be at this setting. That means they will meet at the Point of center of mass.
So the answer must be 3d/5 meters from the mass 2m.
Am I right?

So the answer must be 3d/5 meters from the mass 2m.
Am I right?

Yes sir!

Thanks sir!

I never thought it was that easy.
However if they asked me to calculate their velocity before they collided, I would probably have to use gpe to ke idea and law of conservation of momentum. Right?

If they however asked me to calculate the time taken for them to collide, or their position or speed at a certain time, then it would get really difficult right? If so then such questions won't be in exam.

If there is any shortcut for calculating velocity in terms of position or time, please tell me. Because I must prepare very well for physics Olympiad. If however they include very complex calculations, I will put them off for later. So please tell me.
Thanks .

To calculate velocity as a function of position, start with:

vdv = adx

Integrate:

\frac {v^2} 2 = \int a dx

Acceleration 'a' comes from Newtons Law of Gravity. For body 1:

a_1 = \frac {GM_2}{(x_2-x_1)^2}

Which gives:

\frac {v_1^2} 2 = \int \frac {GM_2}{(x_2-x_1)^2} dx_1\ =\ \frac {GM_2}{(x_2-x_1)}+ C

where C = the constant of integration. In this case since v=0 at the start (when x2-x1 = d), we get:

\frac {v_1^2} 2\ =\ \frac {GM_2}{(x_2-x_1)}\ - \ \frac {GM_2}d

If the particles are of zero physical size that at the moment of collision x1=x2, which means according to classical physics at the moment of collision the velocity is infinite.

For the final formula, you wrote v squared over 2 equals gm2/(x2-x1) - GM2/d... You mean gm2/(x2-x1) EQUALS gm2/d right?

No. The GM_2/d term is the constant of integration. It's required to yield v = 0 when x2-x1 = d (when the particles are at their initial positions).

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