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Momentum Conservation and Centre of Mass Motion

We now consider the case when no net external force acts on the system. The particles of the system, however, move only under the influence of their mutual internal forces. 

In this case, setting Ftot = 0,
MaCM = 0
Thus aCM = 0
VCM = 0
VCM = constant
Where VCM is the velocity of the centre of mass of the system. 

Thus in the absence of external forces (or if the external forces balance) the velocity of the centre of mass remains constant, i. e.,
RCM(t) = RCM(0) + VCMt
and it moves uniformly in a straight line. This is Newton's first law of motion.
The case of an isolated system (i. e. a system on which no external forces act) is significant because, as will be evident from the following, the total linear momentum of such a system is conserved. we find that if Ftot = 0, it implies that
F1 + F2 + F3 + ..... + FN = 0
Or m1a1 + m2a2 + m3a3 + ............. + mNaN = 0
Or (m1V1 + m2V2 + m3V3 + ......... + mNVN) = 0
Where V1, V2, V3, ....... VN are the velocities of the particles of the system.
Thus
m1V1 + m2V2 + m3V3 + .......... + mNVN = constant
P = p1 + p2 + p3 + .......... + pN = constant
where P is the total linear momentum of the system.
Now, differentiating with respect to time, we have
=
VCM = (m1V1 + m2V2 + m3V3 + ……….. +mNVN)
where VCM is the velocity of the centre of mass and V1, V2, etc. are the velocities of the individual particles of the system.

which is an equivalent definition of the momentum of a system of particles. It states that the total linear momentum of a system of particles is equal to the product of the mass of the system and the velocity of its centre of mass.

If no net external force acts on the system, the total linear momentum of the system remains constant. This simple and general result is called the principle of conservation of linear momentum which we have analyze the problem of collisions.

We will also consider systems of particles, which are not rigid bodies.

For example, the Earth-Moon system has two bodies, which exert forces on each other, and the distance between them does change.

In such situations, some properties of a system of particles can be calculated, which are very useful in understanding its motion. Other than the familiar concepts of energy and momentum another important property, which is a measure of the overall rotational motion of the system is angular momentum.

Suppose that at a fireworks display, a rocket is launched on a parabolic path. At a certain point, it explodes into fragments. If the explosion had not occurred, the rocket would have continued along the trajectory shown in the figure. The forces of the explosion are internal to the system (the rocket or its fragments).

They are forces exerted by one part of the system on another part. If air resistance is ignored, the total external force forces Fext = F1 + F2 + F3 + F4…. Fn acting on the system is the weight Mg of the system, regardless of whether the rocket explodes.

The acceleration of the center of mass of the fragments (while they are in flight) remains equal to g, and the center of mass of the fragments follows the same parabolic trajectory that the unexploded rocket would have followed.
A fireworks rocket explodes in flight. In the absence of air drag, the center of mass of the fragments would continue to follow the original parabolic path, until fragments began to hit the ground.




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