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| Motion of Centre of Mass |
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| Consider two particles A and B of masses m1 and m2, respectively. Take the line joining A and B as the X-axis. Let the coordinates of the particles at time 't' be x1 and x2. Suppose no external force acts on the system. The particles A and B, however, exert forces on each other and these particles accelerate along the line joining them. Suppose the particles are initially at rest and the force between them is attractive. The particles will then move along the line AB as shown in the figure. |
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| As time passes, x1 and x2 change and hence, X changes and the centre of mass moves along the X-axis. Velocity of the centre of mass at time t is, |
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| The acceleration of the centre of mass is |
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Suppose the magnitude of the forces between the particles is F. As
the only force acting on A towards B, is F, its acceleration is 
The force on B is (-F) and hence,
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| Substituting this in equation (1) |
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This means, the velocity of the centre of mass does not change with time. But as we assumed initially, the particles are at rest. Thus, v1 =  then Vcm has to be zero. Hence, the centre of mass remains
fixed and does not change with time. |
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| Thus, if no external force acts on a two-particle system and its centre of mass is at rest, initially it remains fixed even when the particles individually move and accelerate. |
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| If the external forces do not add up to zero, the centre of mass is accelerated and is given by |
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If we have a single particle of mass m on which a force
 acts,
its acceleration would be the same as  . Thus,
the motion of the centre of mass of a system is identical to the motion of a single particle of mass equal to the mass of given system, acted upon by the same external forces that act on the system. |
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| Let us consider the case of a projectile, which explodes in flight. The force of gravity continues to act on the total mass of the projectile, now scattered into fragments. |
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| The centre of mass of the fragments continues along the parabolic trajectory. It is the same, as was followed by the intact projectile. |
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| The forces of explosion are all internal forces. These forces are exerted by one part of the system on other parts of the system. These forces may change the momenta of all the individual fragments from the values they had when they made up the projectile. But the internal forces cannot change the total vector momentum of the system. It is only the external force that can change the total momentum of the system. Here, the only external force is gravity. The change in total momentum of the system due to gravity is the same, whether the shell explodes or not. |
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| We know that the moon revolves around the Earth in a circular orbit and the Earth revolves around the sun in an elliptical orbit. |
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| The Earth and the moon exert gravitational forces of attraction on each other, which are internal forces only. To rule out the possibility of the Earth and the moon colliding against each other under the internal gravitational forces, we imagine that both the Earth and moon are revolving about their common centre of mass, such that they are always on opposite sides of the common centre of mass. The much larger mass of the Earth makes their centre of mass very close to the Earth. We can consider the common centre of mass at the Earth itself. By this approximation, we often talk of the moon revolving around the Earth. |
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| In fact, the Earth-moon system moves in such a way that its centre of mass follows the elliptical path around the sun and the forces of attraction due to the sun is the external force. |
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| When a radioactive nucleus (initially at rest) decays, the fragments fly in different directions, obeying the principles of momentum and energy conservation. As the decay occurs spontaneously, no external forces are involved. The momentum of the system is conserved. As the parent nucleus is initially at rest, its momentum before decay is zero. The fragments fly in different directions with different velocities so that the vector sum of the linear momentum of all fragments is zero. The momentum of the centre of mass of all fragments must also be zero after decay. Hence, the centre of mass of the entire system of fragments continues to be at rest after decay. The decay of a parent nucleus into one light and one heavy fragment. Heavy fragments move with a smaller velocity and lighter ones with a greater velocity in opposite directions. |
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