Linear Momentum Of A System Of Particles

Let us recall that the linear momentum of a particle is defined as

$\mathbf{p}=m\mathbf{v}$

Let us also recall that Newton’s second law written in symbolic form for a single particle is

$\mathbf{F}=\frac{d\mathbf{p}}{dt}$

where $\mathbf{F}$ is the force on the particle. Let us consider a system of n particles with masses $m_1$, $m_2$,...$m_n$ respectively and velocities $\mathbf{v_1}$, $\mathbf{v_2}$, ... $\mathbf{v_n}$ respectively. The particles may be interacting and have external forces acting on them. The linear momentum of the first particle is $m_1\mathbf{v_1}$, of the second particle is $m_2\mathbf{v_2}$ and so on.

For the system of n particles, the linear momentum of the system is defined to be the vector sum of all individual particles of the system,

$\mathbf{P}=\mathbf{p_1}+\mathbf{p_2}+...+\mathbf{p_n}$

$\mathbf{P}=m_1\mathbf{v_1}+m_2\mathbf{v_2}+...+m_n\mathbf{v_n}$

Comparing this with Eq. (7.8)

$\mathbf{P}=M\mathbf{V}$

Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (7.15) with respect to time,

$\frac{d\mathbf{P}}{dt}=M\frac{d\mathbf{V}}{dt}$

Comparing Eq.(7.16) and Eq. (7.11),

$\frac{d\mathbf{P}}{dt}=\mathbf{F_{ext}}$

This is the statement of Newton’s second law of motion extended to a system of particles. Suppose now, that the sum of external forces acting on a system of particles is zero. Then from Eq.(7.17)

$\frac{d\mathbf{P}}{dt}=0$ or $\mathbf{P}$ = Constant

Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum of a system of particles. Because of Eq. (7.15), this also means that when the total external force on the system is zero the velocity of the centre of mass remains constant. (We assume throughout the discussion on systems of particles in this chapter that the total mass of the system remains constant.)

Note that on account of the internal forces, i.e. the forces exerted by the particles on one another, the individual particles may have complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle.

The vector Eq. (7.18a) is equivalent to three scalar equations,

$p_x=c_1$; $p_y=c_2$ and $p_z=c_3$

Here $P_x$, $P_y$ and $P_z$ are the components of the total linear momentum vector P along the x–, y– and z–axes respectively; $c_1$, $c_2$ and $c_3$ are constants.

Fig.1

Fig.1: (a) A heavy nucleus radium (Ra) splits into a lighter nucleus radon (Rn) and an alpha particle (nucleus of helium atom). The CM of the system is in uniform motion. (b) The same spliting of the heavy nucleus radium (Ra) with the centre of mass at rest. The two product particles fly back to back.

As an example, let us consider the radioactive decay of a moving unstable particle, like the nucleus of radium. A radium nucleus disintegrates into a nucleus of radon and an alpha particle. The forces leading to the decay are internal to the system and the external forces on the system are negligible. So the total linear momentum of the system is the same before and after decay. The two particles produced in the decay, the radon nucleus and the alpha particle, move in different directions in such a way that their centre of mass moves along the same path along which the original decaying radium nucleus was moving [Fig.1(a)].

If we observe the decay from the frame of reference in which the centre of mass is at rest, the motion of the particles involved in the decay looks particularly simple; the product particles move back to back with their centre of mass remaining at rest as shown in Fig.1(b).

In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference.

Fig.2

Fig.2: (a) Trajectories of two stars, $S_1$ (dotted line) and $S_2$ (solid line) forming a binary system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest.

In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star moves like a free particle, as shown in Fig.2(a). The trajectories of the two stars of equal mass are also shown in the figure; they look complicated. If we go to the centre of mass frame, then we find that there the two stars are moving in a circle, about the centre of mass, which is at rest. Note that the position of the stars have to be diametrically opposite to each other [Fig.2(b)]. Thus in our frame of reference, the trajectories of the stars are a combination of (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.

As can be seen from the two examples, separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.

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