Linear Momentum of a System of Particles

Total linear momentum of a system

For a single particle of mass \(m\) and velocity \(\vec v\), momentum is \(\vec p = m\vec v\), and Newton's second law reads \(\vec F = \dfrac{d\vec p}{dt}\). NCERT begins §6.4 by recalling exactly these two definitions, then asks the natural question: what is the momentum of a whole system of \(n\) interacting particles?

The answer is a definition. The total linear momentum of the system is the vector sum of the individual momenta of all its particles:

$$\vec P = \vec p_1 + \vec p_2 + \cdots + \vec p_n = m_1\vec v_1 + m_2\vec v_2 + \cdots + m_n\vec v_n = \sum_{i} m_i\vec v_i$$

The particles may be interacting with one another and may have external forces acting on them — none of that changes the definition. \(\vec P\) is simply the running vector total of every \(m_i\vec v_i\) in the system.

Why P equals M·v_cm

The definition becomes powerful when compared with the centre-of-mass velocity. From the motion-of-centre-of-mass result, \(M\vec V = \sum_i m_i\vec v_i\), where \(M\) is the total mass and \(\vec V\) the velocity of the centre of mass. The right-hand side is exactly the sum we just called \(\vec P\). Therefore

$$\boxed{\;\vec P = M\vec V\;}$$

In NCERT's words: "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." This is not an approximation — it is an algebraic identity. A spinning, tumbling, exploding cloud of particles carries linear momentum exactly as if its entire mass \(M\) were a single point at the centre of mass moving with velocity \(\vec V\).

Newton's second law for a system

Differentiate \(\vec P = M\vec V\) with respect to time. With the total mass constant, \(\dfrac{d\vec P}{dt} = M\dfrac{d\vec V}{dt} = M\vec A\), where \(\vec A\) is the acceleration of the centre of mass. But the motion-of-centre-of-mass theorem already told us \(M\vec A = \vec F_{\text{ext}}\). Combining,

$$\frac{d\vec P}{dt} = \vec F_{\text{ext}}$$

NCERT calls this "the statement of Newton's second law of motion extended to a system of particles." The crucial word is external. Internal forces — the forces the particles exert on one another — occur in equal-and-opposite third-law pairs and cancel in the sum, so they contribute nothing to \(\dfrac{d\vec P}{dt}\). Only forces from outside the system can change its total momentum.

This single equation is the engine of the whole topic. It says the total momentum responds only to the net external push, exactly as a single particle's momentum responds only to the net force on it. The internal complexity — collisions between parts, chemical explosions, mutual gravitation — is invisible to \(\vec P\).

Conservation when F_external is zero

Set the net external force to zero. Then \(\dfrac{d\vec P}{dt} = 0\), so

$$\vec P = \text{constant} \qquad (\text{when } \vec F_{\text{ext}} = 0)$$

This is the law of conservation of total linear momentum of a system of particles. Because \(\vec P = M\vec V\) and \(M\) is fixed, an equivalent statement is that the velocity of the centre of mass remains constant: the centre of mass moves uniformly in a straight line, like a free particle, however complicated the individual trajectories may be.

The vector statement is equivalent to three independent scalar conservation laws, one per axis:

$$P_x = c_1,\qquad P_y = c_2,\qquad P_z = c_3$$

This component form is what you actually use in problems: each direction is conserved separately, so a perpendicular split of fragments lets you handle \(x\) and \(y\) independently.

System momentum vs individual momenta

The most-tested subtlety is the gap between the total and the parts. NCERT states it plainly: "on account of the internal forces, 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."

So conservation of \(\vec P\) does not mean each particle's momentum is conserved. Internal forces continuously transfer momentum from one part to another; what they cannot do is change the running total. A gun's bullet gains forward momentum while the gun gains exactly the opposite — individuals change dramatically, the sum stays put.

Recoil, explosion and collision

Three classic problem families are all the same conservation law in different costumes. In each, the external impulse is negligible over the short interaction, so \(\vec P_{\text{before}} = \vec P_{\text{after}}\).

• Recoil. A gun-and-bullet (or rocket-and-gas) system starts at rest, so \(\vec P = 0\). After firing, bullet and gun carry equal and opposite momenta; the gun's backward recoil exactly balances the bullet's forward momentum.
• Explosion. A shell or nucleus breaks into fragments by internal forces. The vector sum of fragment momenta equals the momentum just before bursting. NCERT's radium decay is the textbook case; a bomb bursting in flight is the projectile version — the centre of mass stays on the original parabola.
• Collision. Two bodies interact through internal contact forces; with no external impulse, total momentum before equals total momentum after, whether the collision is elastic or inelastic.

The same machinery — choose a system, confirm the external force is zero, equate \(\vec P\) before and after component by component — solves all three.

Worked example 1 — recoil of a gun

Worked example 2 — exploding shell into perpendicular fragments

Worked example 3 — two-body interaction on a smooth surface

Notice the common thread across all three examples: the system starts at rest, the interaction is internal, the external force is zero, so \(\vec P = 0\) is preserved and the centre of mass never moves. Reading off the lighter or unknown part is then a one-line vector balance.