# Center of mass for two small particles

1. Aug 27, 2014

### myko

1. Two small particles of mass m1 and mass m2 attract each other with a force that varies with the inverse cube of their separation. At time t0 , m1 has velocity v directed towards m2 , which is at rest a distance d away. At time t1 , the particles collide.

How far does m1 travel in the time interval (t0 and t1 )? Express your answer in terms of some or all of the variables m1, m2, t1, t0, v, and d

3. The attempt at a solution. I think it can be solved by noticing that since there are no external forces acting on the system, the center of mass will not move. So finding it, would give the final position of the particles at t1. Taking position of m1 at t0 as reference point:
$$r_{cm}=m2*d/(m1+m2)$$

But this is wrong. Could someone point me where the mistake is?

2. Aug 27, 2014

### Staff: Mentor

But the centre of mass IS moving! One of the bodies is moving, the other is stationary. At the moment they contact, our interest in the situation ends.

3. Aug 27, 2014

### myko

Ya. I see it. Thankyou. So then, would it be correct to proceed this way?
Sincé the center of mass is moving with velocity v and there is no external forcé acting, it will keep moving same velocity. So it will travel distance $$d_{cm}=v(t1-t0)$$ from it's initial position.
So at t1 it will be at $$r_{cm}=m2*d/(m1+m2)+v(t1-t0)$$.
At this momento particles collide, so it will be the position of m1. Is this correct?

4. Aug 27, 2014

### Staff: Mentor

EDITED
I think that's right, the CoM will not change its velocity while the pair interact.

Not velocity v.

5. Jan 9, 2016

### michel_s

Correct answer is $$\frac{m_2\cdot d}{m_1+m_2}+v_{cm}(t_1-t_0)$$ with $$v_{cm}=\frac{m_1\cdot v}{m_1+m_2}$$.