# Centre of mass frame

I was just looking at the equation: $$v2-v1=-e(u2-u1).$$ This equation is to describe the collision between two masses, where $$v$$ is the final velocity and $$u$$ is the initial velocity, $$e$$ is the coefficient of restitution.

The collision is the most inelastic when $$e=0$$, however, it is not true that all motion stops in an inelastic collision.

Case one: Two objects of the same mass and same speed in opposite directions, when they collide they will stick together and stop- total momentum was zero before the collision, so should also be conserved after.

Case two: Two objects of the same mass. One mass is stationary and the other one is moving, they collide and stick together.For the momentum to be conserved, after they join, they move together with the half the initial velocity of the moving mass.

I was told that when $$e=0$$, all motion stops only in the centre of mass frame, because the total momentum is always zero in the centre of mass frame.

Please could someone explain why the velocity and the momentum always zero in the centre of mass frame?

• That's how the COM frame is defined, the frame where total momentum is zero Nov 19, 2021 at 6:05
• The law of collision talks about relative motion (and hence the subtraction in the formula). It says nothing about absolute motion (or apparent motion from a far away observer). Feb 16 at 20:09

If $$\sum F_{ext}=0$$, then only centre of mass frame holds momentum conservation.

Note that internal forces which act in collision, do not play role in $$F_{ext}$$. Hope it helps

Speaking honestly, i haven’t seen that equation (i haven’t completed a course on CM that might be the reason either). But I will try to explain what I learned and what I understood from your equation.

What I had seen that is : $$m_1v_1+m_2v_2=m_1u_1+m_2u_2$$

The equation actually tell that when an object push another object then 1st object's initial velocity will be second object's initial velocity after collision. Sometimes 1st object stops after collision and sometimes not that depend on force exerted and mass. As we know, F=ma... (I was thinking to write differential form but i am considering u don’t know calculus)

Using calculus, we write that $$F=dp/dt$$. In simple term, we say that force is changes in momentum in order (with respect) to time.

When force through the whole body is 0 than momentum is conserved.

Now moving to ur equation u wrote that $$v_2-v_1=e(u_2-u_1)$$

You know what! I didn’t know meaning of restitution. After using google translate i understood little bit..

Your given only deals with collision of objects that's what i know. Cause $$e$$ is representing a rate of changes in velocity after and before collision. Now i am going to assume first object's initial velocity is second object's final velocity so, $$v_2+eu_1=eu_2+v_1$$ From my last statement, $$v_2(1+e)=eu_2+v_1$$

If $$e$$ is 0 then first object's final velocity will be second object's final velocity $$v_2=v_1$$ instead of talking about second object we should talk about first object since $$e$$ is 0 which means there's no collision so my last equation is nonsensical. But i wrote in my last paragraph that first object's initial velocity is second object's final velocity so we rewrite our equation $$u_1=v_2$$ hence $$u_1=v_1$$. So an object at rest remains rest until exerted any force to the object. And if an object is moving with velocity $$\alpha$$ than it is keep moving with velocity $$\alpha$$ until exerted any force that's Newtons first law and maybe they wanted to say something just like this by "all motion stops".

I am happy to learn something new... Thanks for that.

The velocity (V) of the center of mass (only) is zero in the center of mass system because that system is defined as moving with the center of mass. The x component (X) for the center of mass of a system of objects is defined by the equation: MX = (sum of terms) Σ $${m_i}{x_i}$$ where (M) is the total mass, (X) is the x component, and the i's denote individual masses, (similarly for the y and z components). Taking the derivative on both sides gives the total x momentum in terms of the velocity of the center of mass: M$$V_x$$ = Σ $${m_i}{v_{xi}}$$. In the center of mass system, V is zero. Hence the total momentum is zero in that system.

the collision equations in your case are

$$m_1\,(v_1-u_1)=dp\\ m_2\,(v_2-u_2)=-dp\\ v_2-v_1=-e\,(u_2-u_1)$$

you can solve those equation and obtain $$~v_1~,v_2~,dp~$$

with those solutions you can check the conservation of the linear momentum

$$m_1\,u_1+m_2\,u_2\overset{!}{=}m_1\,v_1+m_2\,v_2$$

hence the collision equations give you the conservation of the linear momentum but not conservation of the energy