Conservation of Linear Momentum And Impulse

PhysicsSystem Of Particles And Rotational MotionFor JEE aspirants

Linear momentum is the product of mass and velocity, $p = m v$ , and it is the natural quantity for describing how motion is transferred between bodies. Newton's second law in its most general form states $F = d p / d t$ . When the net external force on a system is zero, the total linear momentum stays constant - this is the law of conservation of linear momentum, one of the deepest laws of physics. Impulse, defined as $J = \int F d t$ , equals the change in momentum a force produces. These principles underlie collisions, recoil, rocket propulsion, and virtually every JEE/NEET problem where forces act briefly or systems change composition.

Key Formulas - Quick Reference

Linear momentum: $p = m v$ ; magnitude $p = m v = 2 m K$
Newton's 2nd law (general): $F = \frac{d p}{d t}$
Impulse: $J = \int_{t_{1}}^{t_{2}} F d t = Δ p = p_{f} - p_{i}$
Impulse from constant force: $J = F Δ t$
Impulse from F-t graph: $J =$ area under the force-time curve
Conservation: if $F_{e x t} = 0$ , then $P_{t o t a l} =$ constant
Rocket thrust: $F_{t} = v_{r e l} (- \frac{d m}{d t})$
Rocket velocity: $v = u - g t + v_{r} ln (\frac{m _{0}}{m})$

1. Linear Momentum

The linear momentum of a particle is defined as the product of its mass and its velocity:

p = m v

Momentum is a vector in the direction of velocity, with SI unit kg m/s (equivalently N s). Its magnitude relates to kinetic energy as

p^{2} = m^{2} v^{2} = 2 m (\frac{1}{2} m v^{2}) = 2 m K ⟹ p = 2 m K, K = \frac{p ^{2}}{2 m}

Newton's second law, in its most fundamental form, is written in terms of momentum:

F = \frac{d p}{d t}

For a particle of constant mass this reduces to $F = m a$ , but the momentum form is more general - it also handles systems where mass changes with time (like rockets).

2. Conservation of Linear Momentum

If the net external force on a particle (or system) is zero, then $d p / d t = 0$ , so $p$ is constant in time.

Law of conservation of linear momentum: When the net external force acting on a system of particles is zero, the total linear momentum of the system remains constant in both magnitude and direction.

F_{e x t} = 0 ⟹ P_{t o t a l} = p_{1} + p_{2} + \dots + p_{n} = constant

Since momentum is a vector, conservation applies to each component independently. This is a common exam trick: even when a net force acts in one direction (say vertically, due to gravity), momentum may still be conserved in the perpendicular direction (horizontally), provided no external force acts there.

Internal forces between particles of the system cancel in pairs by Newton's third law. So a system's total momentum is unaffected by internal explosions, collisions, or interactions - only external forces can change it.

Solved Example 1

A man of mass

m_{1}

stands on a plank of mass

m_{2}

resting on a smooth (frictionless) horizontal surface. The man starts walking on the plank with velocity

v_{r}

relative to the plank. Find the recoil velocity of the plank.

Solution:

Let $v$ be the plank's velocity (say to the left, i.e. negative) and $v_{r} - v$ be the man's absolute velocity (to the right). Taking the man + plank as the system, there is no external horizontal force, so horizontal momentum is conserved. Initially both are at rest, so:

0 = m_{1} (v_{r} - v) + m_{2} (- v) ⟹ m_{1} v_{r} = (m_{1} + m_{2}) v

v = \frac{m _{1} v _{r}}{m _{1} + m _{2}}

The plank recoils in the direction opposite to the man's walk. Note that only the relative velocity $v_{r}$ is prescribed; the absolute velocities of both bodies emerge from momentum conservation.

3. Impulse

When a force $F$ acts on a body over a time interval $[t_{1}, t_{2}]$ , the impulse of the force is defined as

J = \int_{t_{1}}^{t_{2}} F d t

Combining this with Newton's second law $F = d p / d t$ gives the impulse-momentum theorem:

J = \int_{t_{1}}^{t_{2}} \frac{d p}{d t} d t = p_{f} - p_{i} = Δ p

Impulse-momentum theorem: The impulse of the net force acting on a particle over a time interval equals the change in the particle's linear momentum over that interval.

For a constant force $J = F Δ t$ . For a variable force, the impulse equals the area under the force-time graph, which is often the only way to measure a brief, large force (like a bat hitting a ball).

Figure 1: Impulse

J

equals the area under the force-time graph, which in turn equals the change in momentum

Δ p

Instantaneous impulse

In many collisions (a bat striking a ball, a hammer hitting a nail) the force acts for a very short interval, but the impulse is finite. During such an interval, ordinary forces like gravity contribute negligible impulse because $Δ t \to 0$ , so they can be ignored while the collision is in progress. Only the large collision force matters.

Solved Example 2

A ball of mass

m

travelling with velocity

v_{i} = 2 \hat{i} + 3 \hat{j}

m/s receives an impulse

J = - 3 m \hat{i}

N s. Find the velocity of the ball just after the impulse.

Solution:

From the impulse-momentum theorem $J = m (v_{f} - v_{i})$ :

- 3 m \hat{i} = m v_{f} - m (2 \hat{i} + 3 \hat{j})

v_{f} = - 3 \hat{i} + 2 \hat{i} + 3 \hat{j} = - \hat{i} + 3 \hat{j} m/s

The impulse changes only the x-component of momentum. The y-component (3 m/s) is unchanged because the impulse has no y-component.

4. Variable-Mass Systems: Rocket Propulsion

So far we assumed the mass of the system is constant. Many practical problems involve systems whose mass changes with time - a rocket ejecting fuel, a hopper leaking sand, a conveyor being loaded. The general approach: apply momentum conservation (or Newton's second law) to a suitably chosen constant-mass system that includes both the main body and the mass gained/lost during the interval.

Thrust force

When mass is ejected from (or added to) a body with velocity $v_{r e l}$ relative to the body, an additional force called the thrust force acts on the body:

F_{t h r u s t} = v_{r e l} (\frac{d m}{d t})

For a rocket, mass is being ejected, so $d m / d t$ is negative and the exhaust moves backward relative to the rocket. The thrust force therefore points forward (in the direction opposite to $v_{r e l}$ ):

F_{t h r u s t} = v_{r} (- \frac{d m}{d t}) (upward for a vertical rocket)

where $v_{r}$ is the exhaust speed relative to the rocket.

Rocket equations

Let $m_{0}$ be the initial mass of the rocket (structure + fuel), $m$ its mass at time $t$ , $v_{r}$ the constant exhaust speed, and $- d m / d t$ the constant rate of fuel consumption. Applying Newton's second law to the rocket in the presence of gravity:

F_{n e t} = F_{t h r u s t} - m g ⟹ m \frac{d v}{d t} = v_{r} (- \frac{d m}{d t}) - m g

Rearranging and integrating from $u$ at $t = 0$ to $v$ at time $t$ :

v - u = v_{r} ln (\frac{m _{0}}{m}) - g t

This is the Tsiolkovsky rocket equation (with gravity). If $u = 0$ and gravity is negligible (deep space): $v = v_{r} ln (m_{0} / m)$ . To achieve high velocity you need either high exhaust speed or a large mass ratio.

Solved Example 3

A rocket weighs 50 kg empty and holds 450 kg of fuel. Exhaust speed is 2 km/s. Find (i) the minimum rate of fuel consumption to just lift off, and (ii) the rate to give an acceleration of 20 m/s² at launch. Take

g = 9.8

m/s².

Solution:

(i) Just lift off: Thrust must equal weight.

v_{r} (- \frac{d m}{d t}) = m g ⟹ - \frac{d m}{d t} = \frac{m g}{v _{r}} = \frac{500 \times 9.8}{2000} = 2.45 kg/s

(ii) Acceleration 20 m/s²: $F_{t h r u s t} - m g = ma$ , so

- \frac{d m}{d t} = \frac{m ( g + a )}{v _{r}} = \frac{500 \times 29.8}{2000} = 7.45 kg/s

5. Momentum Conservation in Component Form

Because momentum is a vector, conservation holds in each direction independently whenever the corresponding component of external force is zero. This is especially useful when one direction has a force (like gravity) but the perpendicular direction does not.

Standard example: A shell moving horizontally explodes into two fragments. Gravity acts vertically, so vertical momentum is not conserved during the explosion (though the explosion is brief, so it is approximately conserved). Horizontal momentum is conserved because no external horizontal force acts.

Common Mistakes to Avoid

Watch out

Applying momentum conservation across a period when an external force is present (gravity, normal reaction from a wall, hinge reaction). Momentum is only conserved when the net external force in the direction of interest is zero.
Confusing impulse (a change in momentum, $J = Δ p$ ) with force (rate of change of momentum, $F = d p / d t$ ). Impulse has SI units of N s = kg m/s; force has units of N.
Ignoring the vector nature of momentum. In two dimensions you must write two component equations, not a single scalar equation.
Applying $F = m a$ (with $m$ constant) to a variable-mass system like a rocket. The correct form is $F = d p / d t$ , which gives the thrust term.
Forgetting that during a very short collision, ordinary forces like gravity contribute negligible impulse and can usually be dropped for the duration of the collision.
In the rocket equation, using absolute exhaust velocity instead of the velocity relative to the rocket. The thrust depends on relative velocity only.

Frequently Asked Questions

Q1. What is the difference between momentum and kinetic energy?

Momentum $p = m v$ is a vector; kinetic energy $K = \frac{1}{2} m v^{2}$ is a scalar. Momentum is conserved when external forces are zero; kinetic energy is conserved only in elastic collisions. They relate by $K = p^{2} / (2 m)$ , so two bodies of equal momentum but different masses have different KE.

Q2. If a system's momentum is conserved, is its kinetic energy also conserved?

No. Momentum conservation and energy conservation are independent laws. In a perfectly inelastic collision, momentum is conserved but kinetic energy decreases (converted to heat, sound, deformation). Momentum conservation follows from Newton's third law; kinetic energy conservation requires the interaction to be elastic.

Q3. When can I ignore gravity during a collision?

When the collision duration $Δ t$ is very short. Gravity's impulse is $m g Δ t$ , which becomes negligible compared to the collision force's impulse when $Δ t$ is milliseconds or less. This is why we treat collisions as instantaneous and apply pre- and post-collision momentum conservation directly.

Q4. Why is momentum conservation more fundamental than $F = ma$ ?

The form $F = d p / d t$ automatically handles variable-mass systems (rockets, raindrops accumulating water) and is consistent with special relativity, while $F = m a$ is only valid when $m$ is constant. Conservation of momentum ultimately follows from the symmetry of physics under spatial translation - a very deep principle.

Q5. How does a rocket accelerate in empty space if there is nothing to push against?

It pushes against its own exhaust. When the rocket ejects gas backward, momentum conservation for the (rocket + gas) system means the rocket must gain forward momentum equal in magnitude to the gas's backward momentum. No external medium is needed - the reaction is internal to the (rocket + fuel) system.

Q6. If two bodies collide and stick together, what happens to the lost kinetic energy?

It is converted to internal energy: heat, sound, permanent deformation, and vibration of the combined body. Total energy is still conserved (that is a broader law), but the useful mechanical kinetic energy decreases. The maximum possible KE loss occurs in a perfectly inelastic collision.

Q7. Can impulse be negative?

Impulse is a vector, so its components can be negative depending on direction. The magnitude is always non-negative. A negative impulse component means the force acts opposite to the chosen positive direction and reduces the momentum component.

Previous year questions on Conservation of Linear Momentum And Impulse

1 question from past papers, each with a step-by-step solution.

JEE Main 2026 Jan 23 Shift 2, Physics Q2

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1. Linear Momentum

2. Conservation of Linear Momentum

3. Impulse

Instantaneous impulse

4. Variable-Mass Systems: Rocket Propulsion

Thrust force

Rocket equations

5. Momentum Conservation in Component Form

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between momentum and kinetic energy?

Q2. If a system's momentum is conserved, is its kinetic energy also conserved?

Q3. When can I ignore gravity during a collision?

Q4. Why is momentum conservation more fundamental than F=ma?

Q5. How does a rocket accelerate in empty space if there is nothing to push against?

Q6. If two bodies collide and stick together, what happens to the lost kinetic energy?

Q7. Can impulse be negative?

Previous year questions on Conservation of Linear Momentum And Impulse

Ready to master System Of Particles And Rotational Motion?

Q4. Why is momentum conservation more fundamental than $F = ma$ ?