# Linear momentum

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u3-linear-momentum/

This sub-topic covers linear momentum, impulse, the principle of conservation of momentum, and classification of collisions. You will learn to solve common one-dimensional collision and explosion problems for CIE A-Level exams.

**Prerequisites:** [Newton's laws of motion](https://www.owlsprep.com/study/cie-9702-u3-newtons-laws-of-motion/); [Vector kinematics](https://www.owlsprep.com/study/cie-9702-u1-kinematics/)

## Learning objectives

- Define linear momentum and impulse, and relate them to Newton's second law
- State and apply the principle of conservation of linear momentum to one-dimensional interactions
- Classify collisions as elastic or inelastic by checking kinetic energy conservation
- Solve standard exam problems involving collisions and rebounds

## Linear Momentum and Impulse

**Linear momentum** — Momentum is the product of an object's mass and velocity. It is a vector quantity, with direction matching the velocity of the object.

*Notation:* \vec{p}

*Example:* A 3 kg block moving at 2 m s⁻¹ right has momentum $p = 3 \times 2 = 6$ kg m s⁻¹ right.

Impulse describes the effect of a force acting over a period of time to change an object's momentum.

**Impulse** — Impulse equals the product of the average force acting on an object and the contact time, and is equal to the change in momentum of the object: $J = F_{\text{avg}} \Delta t = \Delta p$.

*Notation:* $J$

**Worked example:** A 0.15 kg ball travels at 10 m s⁻¹ towards a wall, and rebounds at 8 m s⁻¹ along the same line. Calculate the impulse exerted on the ball by the wall.

1. Define the positive direction as away from the wall. Initial velocity $u = -10$ m s⁻¹, final velocity $v = +8$ m s⁻¹.
2. Impulse equals change in momentum $\Delta p = m(v - u)$:
3. $$J = 0.15 \left(8 - (-10)\right) = 0.15 \times 18 = 2.7 N s$$
4. The impulse exerted on the ball is 2.7 N s away from the wall.

> **Exam tip:** Always define your positive direction first to avoid sign errors in rebound impulse calculations.

## Momentum and Newton's Second Law

Newton's original formulation of the second law of motion is written in terms of momentum, rather than acceleration. This form works for both constant and changing mass.

**Newton's Second Law (momentum form)** — The net force acting on an object equals the rate of change of its linear momentum.

*Example:* If mass is constant: $F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$, which matches the standard $F=ma$ form.

**Worked example:** A constant net force of 12 N acts on a 4 kg object initially at rest. What is the momentum of the object after 3 seconds?

1. Rearrange Newton's second law to solve for change in momentum:
2. $$\Delta p = F_{\text{net}} \Delta t = 12 \times 3 = 36 kg m s^{-1}$$
3. Initial momentum is 0, so final momentum equals the change in momentum.
4. Final momentum = $36$ kg m s⁻¹

> **tip**
>
> This momentum form is used for problems with changing mass, such as rocket propulsion, which can appear in extended CIE questions.

## Conservation of Linear Momentum

The principle of conservation of momentum is one of the most widely tested concepts in CIE A-Level dynamics.

**Principle of Conservation of Linear Momentum** — For a closed system with no net external force acting, the total linear momentum of the system is constant. This means total momentum before an interaction equals total momentum after.

**Worked example:** A 5000 kg truck moving at 12 m s⁻¹ collides with a stationary 1000 kg car. After collision, they move together along the same line. Find their common speed after collision.

1. Calculate total momentum before collision:
2. $$p_{before} = m_{truck}u_{truck} + m_{car}u_{car} = (5000 \times 12) + (1000 \times 0) = 60000 kg m s^{-1}$$
3. By conservation of momentum, $p_{after} = p_{before} = 60000$ kg m s⁻¹. Combined mass $= 5000 + 1000 = 6000$ kg. Let $v$ = common speed:
4. $$6000 v = 60000 \implies v = 10 m s^{-1}$$
5. Common speed after collision is 10 m s⁻¹.

> **Exam tip:** Always check you add masses correctly for collisions where objects stick together after impact.

## Elastic and Inelastic Collisions

Collisions are classified based on whether kinetic energy is conserved during the interaction:

- **Elastic collision**: Total momentum *and* total kinetic energy are both conserved.
- **Inelastic collision**: Only total momentum is conserved; kinetic energy is converted to heat, sound or other forms, so total kinetic energy decreases.
- **Perfectly inelastic collision**: Objects stick together after collision, maximum kinetic energy is lost.

**Worked example:** Show that the truck-car collision in the previous example is inelastic by calculating the change in kinetic energy.

1. Calculate total initial kinetic energy before collision:
2. $$KE_{initial} = \frac{1}{2}m_{truck}u_{truck}^2 = 0.5 \times 5000 \times 12^2 = 360000 J$$
3. Calculate total final kinetic energy after collision:
4. $$KE_{final} = \frac{1}{2}(m_{truck} + m_{car})v^2 = 0.5 \times 6000 \times 10^2 = 300000 J$$
5. 60000 J of kinetic energy is lost during the collision, so kinetic energy is not conserved. Therefore the collision is inelastic.

## Common pitfalls

- **Wrong:** Forgetting momentum is a vector, so ignoring sign when adding momenta for opposite directions
  - Why it fails: Failing to account for direction gives an incorrect total momentum for the system
  - Correct: Always define a positive direction before starting calculations, assign negative signs to momentum moving in the opposite direction
- **Wrong:** Assuming kinetic energy is conserved in all collisions, just like momentum
  - Why it fails: Many students mix up the conservation rules for momentum and kinetic energy
  - Correct: Momentum is always conserved (no external forces). Kinetic energy is only conserved for elastic collisions
- **Wrong:** Calculating rebound impulse as $m(v + u)$ with both speeds positive without direction convention
  - Why it fails: This can lead to magnitude errors if the direction is not accounted for correctly
  - Correct: Always use $\Delta p = m(v - u)$ with correct signed values for initial and final velocity
- **Wrong:** Applying conservation of momentum when external forces like friction are acting on the system
  - Why it fails: The principle only holds for closed systems with no net external force
  - Correct: Only use conservation of momentum if no external forces act, or the interaction time is too short for external forces to change total momentum

## Cheatsheet

| Quantity/Concept | Formula | Key Fact |
| --- | --- | --- |
| Linear momentum | $p = mv$ | Vector, units kg m s⁻¹ |
| Impulse | $J = F\Delta t = \Delta p$ | Equal to change in momentum |
| Newton's 2nd Law | $F_{net} = \frac{\Delta p}{\Delta t}$ | Original, general form of the law |
| Conservation of momentum | $\sum p_{before} = \sum p_{after}$ | Holds for closed systems, no external forces |
| Elastic collision | - | Momentum *and* kinetic energy conserved |
| Inelastic collision | - | Only momentum conserved; KE lost |

## What's next

Linear momentum is a core foundational concept for dynamics that underpins almost all further topics in CIE A-Level Physics, from circular motion and energy to particle physics and collision experiments. Mastering the sign conventions and application of conservation of momentum will help you solve a wide range of structured and multiple choice questions across all units. After completing this sub-topic, you can extend your knowledge to more advanced interactions and related topics in dynamics and energy.

- [Conservation of momentum](https://www.owlsprep.com/study/cie-9702-u3-conservation-of-momentum/)
- [Collisions](https://www.owlsprep.com/study/cie-9702-u3-collisions/)
- [Mass and weight](https://www.owlsprep.com/study/cie-9702-u3-mass-and-weight/)

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