Conservation of Linear Momentum

1. Combining the second and third laws

The conservation of linear momentum is not a new physical postulate — NCERT §4.7 derives it as a direct consequence of the second and third laws. The second law tells us that the change in a particle's momentum equals the impulse applied to it, \(\Delta\vec p = \vec F\,\Delta t\). The third law tells us that whenever a force \(\vec F_{AB}\) acts on body A due to B, an equal-and-opposite force \(\vec F_{BA}=-\vec F_{AB}\) acts on B. Apply the two together to a closed pair of bodies and the algebra forces the sum of their momenta to stay constant. We will do this twice — first for the textbook bullet-and-gun, then for a general collision.

2. Bullet-and-gun derivation

Take the classic NCERT setup: a gun of mass \(M\) holding a bullet of mass \(m\). Before firing, both are at rest, so the total linear momentum of the system is zero. When the trigger is pulled, the gun's internal mechanism exerts a forward force \(\vec F\) on the bullet for a brief contact time \(\Delta t\). By the third law, the bullet exerts an equal-and-opposite force \(-\vec F\) on the gun during the same interval.

Apply the impulse form of the second law to each body separately:

Both started at rest, so the changes are the final momenta themselves. Adding the two equations,

The total momentum of the bullet-plus-gun system is the same after firing as before — zero. The gun must therefore recoil backwards with momentum equal in magnitude to the bullet's forward momentum, even though no external force acted on the combined system. This is the entire idea of the conservation law, dressed in a familiar scene.

3. Two-body collision derivation

Now generalise to two bodies A and B with arbitrary initial momenta \(\vec p_A\) and \(\vec p_B\). They collide, interact for a short time \(\Delta t\), and separate with final momenta \(\vec p_A{}'\) and \(\vec p_B{}'\). During the collision the only forces acting are the mutual contact forces — call them \(\vec F_{AB}\) (on A by B) and \(\vec F_{BA}\) (on B by A). Newton's second law over the interval \(\Delta t\) gives

By the third law, \(\vec F_{AB} = -\vec F_{BA}\). Adding the two impulse equations and using this fact,

The vector sum of the two momenta is preserved through the collision. Notice the derivation never used the nature of the contact — soft, hard, sticky, elastic, magnetic, electrostatic. Whatever the interaction, as long as it is internal and Newton's third law applies, momentum survives.

4. Statement of the principle

The total momentum of an isolated system of interacting particles is conserved.

Two phrases carry all the weight. "Isolated system" means the net external force on the system is zero. Forces between members of the system — however violent — are internal and cancel pairwise by the third law. "Total momentum" is the vector sum \(\vec P = \sum_i \vec p_i\); it is preserved as a vector, not just in magnitude. If you write this out for an \(n\)-particle system, Newton's second law for the system reads \(\vec F_{\text{ext}} = d\vec P/dt\); setting \(\vec F_{\text{ext}}=0\) gives \(d\vec P/dt = 0\) and so \(\vec P = \) constant.

Crucially, the proof made no assumption about energy. Conservation of momentum holds for both elastic and inelastic collisions. In an elastic collision (treated in detail in Chapter 5) kinetic energy is conserved in addition; in an inelastic one, only momentum survives.

5. Component-wise conservation

Conservation of momentum is a vector statement. It splits into three independent scalar conservation laws — one per coordinate axis. If the net external force on the system has no \(x\)-component, then the total \(x\)-momentum is conserved even when the \(y\) and \(z\) momenta are not. This is a single line in NCERT, but it powers an entire family of NEET problems.

Consider a projectile of mass \(M\) launched at some angle. Mid-flight, an internal explosion splits it into two fragments. The only external force on the system is gravity — a vertical \(-Mg\,\hat j\). There is no horizontal external force, so the horizontal component of the system's momentum is conserved across the explosion. The vertical component, on the other hand, keeps changing under gravity as usual. The combined centre of mass therefore traces the same parabola the unexploded projectile would have followed.

6. NEET-favourite applications

6.1 Gun recoil

The bullet-and-gun problem of §2 is a single-unknown PYQ template. Total momentum before firing is zero. After firing, the bullet of mass \(m\) leaves at speed \(v\); the gun of mass \(M\) recoils at speed \(V\). Setting \(mv - MV = 0\) gives \(V = (m/M)\,v\), opposite to the bullet. Because \(M\gg m\), the recoil is small but never zero — and NCERT Exercise 4.19 makes you compute it explicitly.

6.2 Nuclear disintegration

A radioactive nucleus at rest splits into two fragments. The initial momentum is zero, so the final total must also be zero: the two fragments fly apart in exactly opposite directions, with \(m_1 v_1 = m_2 v_2\). This is NCERT Exercise 4.17 and a direct PYQ ingredient — the alpha decay model, the spontaneous fission model, and any "two-fragment from rest" stem all collapse to the same line of algebra.

6.3 Astronaut in space

NCERT Example 4.1 dramatises the principle. An astronaut floating at rest in interstellar space, with no rope to the spaceship, can return by throwing any object — a spanner, a pen — directly away from the spaceship. Total momentum of (astronaut + tool) was zero, so after the throw the astronaut acquires a small velocity directed back toward the spaceship. Without an internal "engine" he could not have manufactured this momentum — the law lets him borrow it from a tossed object.

6.4 Rocket propulsion

A rocket is a system of varying mass. Gas is ejected backward at speed \(u\) relative to the rocket; the rocket gains the corresponding forward momentum. The instantaneous thrust on the rocket is

where \(dm/dt\) is the rate at which the rocket loses mass to exhaust. NIOS §3.5 treats this only briefly; for NEET, knowing the formula and that the rocket needs no surrounding medium to push against is enough. The momentum it gains came from the mass it threw out.

6.5 Exploding projectile in mid-flight

A shell launched at angle \(\theta\) bursts at the top of its trajectory into two equal fragments. One fragment falls vertically down. The horizontal momentum just before the burst was \(M u\cos\theta\). After the burst, half the mass has zero horizontal velocity; the other half must therefore carry the whole horizontal momentum: \((M/2)\,v_x = M u\cos\theta\), giving \(v_x = 2u\cos\theta\) for the second fragment. The vertical equation is solved by ordinary kinematics, not by conservation.

7. Worked examples