Conservation of linear momentum & center of mass

[yoni162]

Homework Statement

Two cars are driving along the road at constant speeds V1, V2. At t=0 they begin to accelerate with constant acceleration a1, a2. Under what circumstances is the center of mass frame of reference inertial?

Homework Equations

Fext=0 ==> change in momentum=0

The Attempt at a Solution

It's known that if there are no external forces acting on a system, the momentum is conserved. The same applies if there are external forces present, but they cancel each other so that the net external force is 0 (correct?). Now, from the center of mass (c.o.m) frame of reference, when the momentum is conserved, the total momentum is 0. Would that apply if despite the fact that there are external forces acting on the cars, if with their acceleration, the velocity of the c.o.m is kept the same (thus making it inertial)? Or is Fext=0 necessary in order for that to happen?

 

[kuruman]

You are on the right track, so I will help you see through this. Suppose you are on a skateboard moving along at the velocity of the center of mass before the cars start accelerating. Now at t = 0 they start accelerating. How must the accelerations of the cars be related for you to say that Newton's Laws are obeyed and you need no fictitious external force to explain what you see?

 

[yoni162]

You are on the right track, so I will help you see through this. Suppose you are on a skateboard moving along at the velocity of the center of mass before the cars start accelerating. Now at t = 0 they start accelerating. How must the accelerations of the cars be related for you to say that Newton's Laws are obeyed and you need no fictitious external force to explain what you see?

OK I don't know if that's what you were going for, but I said: in order for the c.o.m frame of reference to remain inertial, we need to have the the net external forces=0.
I know that the net external forces is the velocity of the c.o.m divided by the sum of the masses, and I want it to be 0, meaning (in a one-dimensional case):

m1a1+m2a2=0

Is that true?

Also, how do I explain the fact that the original frame of reference is inertial to begin with, considering there's friction acting on the cars? Or do I say that because V is constant (for each car), the net force is 0 so there's really no problem here, even though the forces are external?

 

[kuruman]

I know that the net external forces is the velocity of the c.o.m divided by the sum of the masses, and I want it to be 0, meaning (in a one-dimensional case):

m1a1+m2a2=0

Is that true?

It is. If the above condition is satisfied, the center of mass does not accelerate, i.e. it is an inertial frame. Furthermore, an observer in that frame will interpret the above condition to mean that the two cars exert equal and opposite forces on each other in accordance with Newton's 3rd Law, i.e. all forces appear to be internal to the two-car system.

Also, how do I explain the fact that the original frame of reference is inertial to begin with, considering there's friction acting on the cars? Or do I say that because V is constant (for each car), the net force is 0 so there's really no problem here, even though the forces are external?

Correct. A frame is inertial as long as it is not accelerating. When both cars are moving at constant velocity, their center of mass is not accelerating.

 

[yoni162]

One more thing I wanted to ask..is there conservation of linear momentum in this case? Since the forces acting on the cars aren't internal forces after all..

 

[kuruman]

This is a philosophical question. If you define, strictly, that momentum is conserved as long as the center of mass does not accelerate, yes, linear momentum is conserved. The non-conservation of linear momentum when external forces act on the system comes about when internal forces act in pairs. Here, if m₁a₁+m₂a₂=0, internal forces cancel in pairs. Therefore, by the strict definition, linear momentum is conserved. An observer in the CM frame will not be able to tell otherwise although he/she may have to account in some way or another for the action-reaction apparent force between the cars. In accelerating frames, one has to invent fictitious external forces; this is a special case where one has to invent a fictitious internal force.