# Momentum and Impulse (Extended Only)

> Physics · CIE IGCSE
> Source: https://www.owlsprep.com/study/cie-0625-u1-momentum-and-impulse/

This Extended-only guide covers momentum, impulse, and conservation of momentum for CIE IGCSE Physics 0625 Unit 1, with exam-aligned worked examples and common pitfalls to avoid.

**Prerequisites:** [Newton's laws of motion](https://www.owlsprep.com/study/cie-0625-u1-newtons-laws/); [Physical quantities, unit conversion and vector direction rules](https://www.owlsprep.com/study/cie-0625-u1-physical-quantities/)

## Learning objectives

- Calculate momentum using the formula $p = mv$ and account for its vector nature
- Define impulse as change in momentum, and calculate impulse using $Ft = mv - mu$
- Apply the principle of conservation of momentum to collision and explosion scenarios for isolated systems
- Relate impulse to real-world applications including car safety features

## 1. Momentum: Definition and Calculation

**Momentum** — A vector quantity equal to the product of an object's mass and its velocity, with direction matching the velocity of the object.

*Notation:* $p$

*Example:* A 2kg ball moving right at 3m/s has a momentum of 6 kg m/s to the right.

The standard unit of momentum is kilogram-metres per second (kg m/s). As a vector quantity, momentum can be positive or negative depending on direction: for example, if right is assigned as positive, motion left will have a negative velocity and negative momentum.

**Worked example:** Take north as the positive direction. Calculate the momentum of (a) a 1200kg car travelling north at 15 m/s, and (b) a 0.05kg bullet travelling south at 400 m/s.

1. Use the momentum formula:

   $$p = m \times v$$
2. Calculate momentum of the car (north = positive velocity):

   $$p_{car} = 1200 \times 15 = 18000 \text{ kg m/s (north direction)}$$
3. Calculate momentum of the bullet (south = negative velocity = -400 m/s):

   $$p_{bullet} = 0.05 \times (-400) = -20 \text{ kg m/s (south direction)}$$

> **Exam tip:** Always assign a positive direction for velocity at the start of every momentum question to avoid sign errors for opposite-direction motion.

## 2. Impulse and the Force-Momentum Relationship

**Impulse** — The change in momentum of an object, equal to the product of the average force applied to the object and the time the force acts for.

*Notation:* $\Delta p$

*Example:* A force of 10N applied for 2s gives an impulse of 20 Ns, equal to a 20 kg m/s change in momentum.

Impulse can be measured in either newton-seconds (Ns) or kg m/s, as the two units are equivalent. There are two equivalent formulas for impulse:

$$\Delta p = F \times t = m(v - u)$$

**Worked example:** A 0.4kg football is kicked from rest, reaching a velocity of 15 m/s after 0.02s of contact with the boot. Calculate (a) the impulse of the kick, and (b) the average force exerted on the ball.

1. Calculate impulse as the change in momentum, where initial velocity $u=0$:

   $$\Delta p = m(v - u) = 0.4(15 - 0) = 6 \text{ Ns}$$
2. Rearrange the impulse formula to solve for average force:

   $$F = \frac{\Delta p}{t} = \frac{6}{0.02} = 300 \text{ N}$$

> **Exam tip:** Impulse questions often reference car safety features: crumple zones and airbags increase collision time, reducing the force exerted on passengers for the same change in momentum.

## 3. Conservation of Momentum for Collisions

**Conservation of Momentum** — For an isolated system (no external forces acting on the system), the total momentum before a collision or explosion is equal to the total momentum after the event.

*Example:* Two trolleys colliding on a frictionless track form an isolated system, so their total momentum before collision equals their total momentum after collision.

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

**Worked example:** A 2kg trolley moving right at 3 m/s collides with a stationary 1kg trolley on a frictionless track. After collision, the 1kg trolley moves right at 3.5 m/s. Calculate the velocity of the 2kg trolley after collision, taking right as positive.

1. Apply conservation of momentum: total momentum before = total momentum after

   $$(2 \times 3) + (1 \times 0) = (2 \times v_1) + (1 \times 3.5)$$
2. Simplify and rearrange to solve for $v_1$:

   $$6 = 2v_1 + 3.5$$
3. $$2v_1 = 2.5 \implies v_1 = 1.25 \text{ m/s (right direction)}$$

> **Exam tip:** If two objects stick together after collision, they share the same final velocity, so combine their masses when calculating total post-collision momentum.

## 4. Conservation of Momentum for Explosions

Explosions are the reverse of collisions: all components are stationary before the explosion, so total initial momentum is zero. The total final momentum must also equal zero, meaning objects move in opposite directions with equal magnitude of momentum.

**Worked example:** A stationary 4kg cannon fires a 0.1kg cannonball horizontally at 120 m/s. Calculate the recoil velocity of the cannon.

1. Total initial momentum = 0, so total final momentum = 0

   $$m_cv_c + m_bv_b = 0$$
2. Substitute values (take cannonball direction as positive):

   $$4v_c + (0.1 \times 120) = 0$$
3. Rearrange to solve for recoil velocity:

   $$4v_c = -12 \implies v_c = -3 \text{ m/s (opposite direction to cannonball)}$$

**Summary**

- Momentum is a vector quantity, so signs for direction are required for all calculations
- Impulse equals both change in momentum and the product of average force and time of application
- Conservation of momentum only applies to isolated systems with no external forces acting

## Common pitfalls

- **Wrong:** Forgetting momentum is a vector and omitting direction/signs for velocity
  - Why it fails: Leads to incorrect total momentum calculations for objects moving in opposite directions
  - Correct: Assign a positive direction at the start of every question, use negative values for motion in the opposite direction
- **Wrong:** Using instantaneous force instead of average force in the impulse formula $Ft$
  - Why it fails: Force varies during collisions, so only average force will give the correct impulse value
  - Correct: Always use the stated average force, or calculate impulse via change in momentum if average force is not given
- **Wrong:** Applying conservation of momentum when external forces (friction, gravity) act on the system
  - Why it fails: External forces change the total momentum of the system, so the principle no longer holds
  - Correct: Only use conservation of momentum if the question explicitly states no external forces act, or specifies an isolated system
- **Wrong:** Using mass in grams instead of kilograms for momentum calculations
  - Why it fails: The standard unit of momentum is kg m/s, so mass must be in kilograms to get a correct result
  - Correct: Convert all masses to kg before substituting into any momentum or impulse formula

## Cheatsheet

| Concept | Formula | Units | Key Note |
| --- | --- | --- | --- |
| Momentum | $p = m \times v$ | kg m/s | Vector, direction matches velocity |
| Impulse | $\Delta p = Ft = m(v-u)$ | Ns or kg m/s | Equal to change in momentum |
| Conservation of Momentum | Total $p_{before}$ = Total $p_{after}$ | kg m/s | Applies only to isolated systems with no external forces |

## What's next

Now that you have mastered momentum and impulse for CIE IGCSE Physics 0625 Extended, you can apply these concepts to solve complex force and motion problems, including car safety feature questions that frequently appear in Paper 4 extended theory exams. You will also use these rules when studying momentum changes in radioactive decay later in the course. Next, move on to work, power and energy to complete your Unit 1 (Motion, Forces and Energy) preparation, and practice extended past paper questions to reinforce your understanding of sign conventions and system boundaries for conservation of momentum.

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