Further Mechanics (Edexcel IAL Physics Unit 4)

Physics· 2018 spec Issue 3, Unit 4 (WPH14) statements 81-91· 45 min read

1. Impulse and Momentum Relationships★★☆☆☆⏱ 10 min

📘 Definition

Impulse

$J = F Δ t = Δ p = m (v - u)$

Impulse is the change in momentum of an object, equal to the product of the average force applied to the object and the time interval the force acts for. The area under a force-time graph equals total impulse.

Impulse is derived directly from Newton's second law of motion, rearranged without calculus. You are also required to derive the kinetic energy formula $E_{k} = \frac{p ^{2}}{2 m}$ algebraically for exam questions.

🔬 Derivation

Goal:

Derive $E_{k} = \frac{p ^{2}}{2 m}$

1
Start with the standard non-relativistic kinetic energy formula:
2
$E_k = \frac{1}{2}mv^2$
3
Recall linear momentum $p = m v$ , so rearrange to get $v = \frac{p}{m}$
4
Substitute $v = \frac{p}{m}$ into the kinetic energy equation:
5
$E_k = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}$

Result:

This formula lets you relate kinetic energy directly to momentum without calculating velocity first.

📐 Worked Example

A 0.15 kg ball travelling horizontally at 12 m s⁻¹ is hit by a bat and rebounds horizontally at 18 m s⁻¹. The bat is in contact with the ball for 0.002 s. Calculate the average force exerted by the bat on the ball.

1
Step 1: Define the initial direction of the ball as positive, so final velocity is negative
2
Step 2: Calculate change in momentum:
3
$\Delta p = m(v - u) = 0.15(-18 - 12) = -4.5 \text{ kg m s}^{-1}$
4
Step 3: Rearrange the impulse formula to solve for force:
5
$F = \frac{\Delta p}{\Delta t} = \frac{-4.5}{0.002} = -2250 \text{ N}$
6
The negative sign indicates force acts opposite to the initial direction, so magnitude is 2250 N.

Exam tip:

Always explicitly define your positive direction when calculating change in momentum to avoid sign errors, especially for rebound scenarios. Marks are awarded for correct sign convention as well as numerical answer.

2. 2D Conservation of Momentum and Collisions★★★☆☆⏱ 15 min

For any collision with no external resultant force, total linear momentum is conserved in all directions. For 2D collisions, resolve velocity vectors into horizontal (x) and vertical (y) components, apply conservation of momentum separately to each axis, then combine components to find final velocity magnitude and direction.

📘 Definition

Elastic vs Inelastic Collisions

Elastic collisions conserve both total momentum and total kinetic energy. Inelastic collisions conserve total momentum but not total kinetic energy (energy is lost as heat, sound or deformation).

Example:

A billiard ball collision is nearly elastic; a car collision is highly inelastic.

📐 Worked Example

A 2 kg snooker ball travelling at 3 m s⁻¹ along the x-axis collides with a stationary 2 kg snooker ball. After collision, the first ball moves at 1.5 m s⁻¹ at 60° above the x-axis. Calculate the velocity (magnitude and direction) of the second ball, and state if the collision is elastic.

1
Step 1: Apply conservation of momentum in the x-direction:
2
$(2 \times 3) + 0 = (2 \times 1.5\cos60°) + (2 \times v_{2x}) \implies v_{2x} = 2.25 \text{ m s}^{-1}$
3
Step 2: Apply conservation of momentum in the y-direction (initial total y momentum = 0):
4
$0 = (2 \times 1.5\sin60°) + (2 \times v_{2y}) \implies v_{2y} = -1.30 \text{ m s}^{-1} (below x-axis)$
5
Step 3: Calculate magnitude of second ball's velocity:
6
$v_2 = \sqrt{2.25^2 + 1.30^2} = 2.60 \text{ m s}^{-1}$
7
Step 4: Calculate direction: $θ = arctan (∣1.30/2.25∣) = 30°$ below the x-axis
8
Step 5: Check KE conservation: Initial $E_{k} = 9$ J, Final $E_{k} = 2.25 + 6.76 = 9.01$ J (rounding error), so collision is elastic.

Exam tip:

Never apply conservation of momentum to resultant velocity directly without resolving into components. Always state the direction of final velocity with reference to a given axis (e.g. 30° below the horizontal).

3. Circular Motion Kinematics★★☆☆☆⏱ 8 min

Circular motion describes an object moving at constant speed around a fixed circular path. Angular displacement is the angle rotated by the object around the centre, measured in degrees or radians. Convert between units using $1 rad = \frac{180°}{π}$ .

📘 Definition

Angular Velocity

$ω = \frac{Δ θ}{Δ t} = \frac{2 π}{T} = 2 π f$ , $v = ω r$

Angular velocity is the rate of change of angular displacement, measured in radians per second (rad s⁻¹). $T$ is the time period for one full rotation, $f$ is frequency of rotation, $r$ is path radius, $v$ is linear speed.

📐 Worked Example

A car travels around a circular roundabout of radius 25 m at a constant linear speed of 12 m s⁻¹. Calculate its angular velocity and time taken to complete one full lap.

1
Step 1: Use $v = ω r$ to calculate angular velocity:
2
$\omega = \frac{v}{r} = \frac{12}{25} = 0.48 \text{ rad s}^{-1}$
3
Step 2: Use $T = \frac{2 π}{ω}$ to calculate time period:
4
$T = \frac{2\pi}{0.48} = 13.1 \text{ s (3 s.f.)}$

Exam tip:

Always convert angular displacement to radians before calculating angular velocity or using other circular motion formulae. Answers in degrees per second will not be awarded marks unless specifically requested.

4. Centripetal Acceleration and Force★★★★☆⏱ 12 min

Even if an object moves at constant speed in a circular path, it accelerates because its direction (and hence velocity vector) constantly changes. This acceleration is directed towards the centre of the circle, called centripetal acceleration. You must derive this using vector diagrams, not calculus.

🔬 Derivation

Goal:

Derive centripetal acceleration $a = \frac{v ^{2}}{r}$ using vector change in velocity

1
Consider an object moving at constant speed $v$ around a circle of radius $r$ , moving small angular displacement $Δ θ$ in time $Δ t$ .
2
Initial velocity $v_{1}$ and final velocity $v_{2}$ have equal magnitude $v$ , separated by angle $Δ θ$ .
3
Draw a vector triangle for $Δ v = v_{2} - v_{1}$ . For small $Δ θ$ , magnitude of $Δ v \approx v Δ θ$ .
4
Acceleration $a = \frac{Δ v}{Δ t} = \frac{v Δ θ}{Δ t}$ .
5
Substitute $ω = \frac{Δ θ}{Δ t}$ and $ω = \frac{v}{r}$ :
6
$a = v\omega = \frac{v^2}{r} = r\omega^2$

Result:

Centripetal acceleration is always directed towards the centre of the circular path, perpendicular to instantaneous velocity.

A resultant force is required to produce centripetal acceleration, called centripetal force. This force also acts towards the centre of the circle, and can be provided by friction, tension, normal contact force, or other forces depending on the scenario. Centripetal force formulae: $F = \frac{m v ^{2}}{r} = m r ω^{2}$ .

📐 Worked Example

A 1200 kg car travels around a flat circular bend of radius 50 m at 15 m s⁻¹. Calculate the minimum coefficient of friction between tyres and road required to prevent skidding.

1
Step 1: Centripetal force is provided by friction between tyres and road:
2
$F_f = F_c = \frac{mv^2}{r}$
3
Step 2: Maximum friction force $F_{f} = μ R$ , where normal contact force $R = m g$ on a flat road
4
Step 3: Equate and cancel mass $m$ from both sides:
5
$\mu = \frac{v^2}{rg} = \frac{15^2}{50 \times 9.81} = 0.459 (3 s.f.)$

Exam tip:

Centripetal force is not a new type of force: it is the name given to the resultant force that causes circular motion. Always state what force provides the centripetal force in a given scenario (e.g. friction, tension) to gain full marks.

5. Common Pitfalls

Wrong move:

Using the same sign for initial and final velocity in rebound impulse calculations

Why:

Velocity is a vector, so direction changes require a sign change if you defined a positive axis

Correct move:

Explicitly define a positive direction at the start of any momentum calculation, and apply signs consistently to velocity values

Wrong move:

Applying conservation of momentum to resultant velocity in 2D collisions instead of resolving into components

Why:

Momentum is conserved separately in each perpendicular direction, not in the resultant direction directly

Correct move:

Resolve all velocity vectors into x and y components, apply conservation of momentum to each axis separately, then combine components if needed

Wrong move:

Using degrees instead of radians in circular motion formulae involving $ω$

Why:

All standard circular motion formulae require $ω$ to be in radians per second, not degrees per second

Correct move:

Convert all angular displacement values to radians before calculating angular velocity or using circular motion formulae

Wrong move:

Treating centripetal force as an additional separate force on free-body diagrams

Why:

Centripetal force is the resultant of real forces (friction, tension, normal force) acting on the object, not a new force itself

Correct move:

Draw all real forces acting on the object, then state that the resultant of these forces provides the required centripetal force

Wrong move:

Assuming all collisions are elastic unless stated otherwise

Why:

Most real collisions are inelastic, with some kinetic energy lost as heat, sound or deformation

Correct move:

Always calculate total initial and final kinetic energy to confirm if a collision is elastic, rather than assuming it is

6. Quick Reference Cheatsheet

Concept	Formulae	Key Exam Tip
Impulse	$F Δ t = Δ p = m (v - u)$	Define positive direction to avoid sign errors
KE-Momentum Relation	$E_{k} = \frac{p ^{2}}{2 m}$	Use to link KE and momentum without velocity
2D Momentum	Conserved in x and y axes separately	Resolve vectors into components first
Angular Kinematics	$ω = \frac{2 π}{T}$ , $v = ω r$	Use radians only for $ω$
Centripetal Acceleration	$a = \frac{v ^{2}}{r} = r ω^{2}$	Derive via vector triangles, not calculus
Centripetal Force	$F = \frac{m v ^{2}}{r} = m r ω^{2}$	State what force provides centripetal force

7. Frequently Asked

Do I need to use calculus to derive centripetal acceleration for Edexcel IAL Physics Unit 4?

No. You must derive centripetal acceleration using vector change-in-velocity triangles only. Calculus-based derivations are not required and will not be awarded marks in your exam.

What is the difference between elastic and inelastic collisions in 2D?

For both collision types, total linear momentum is conserved in both x and y directions. For elastic collisions only, total kinetic energy is also conserved. Most real collisions are inelastic, with some kinetic energy lost as heat, sound or deformation energy.

Going deeper

What's Next

Now that you have mastered Further Mechanics for Edexcel IAL Physics Unit 4, you are ready to move on to Electric and Magnetic Fields, where you will apply your force and motion knowledge to charged particles in fields. Practice past paper questions focused on this topic to reinforce your understanding, and review Core Practicals 9 and 10 in detail, as they often appear in 3-6 mark extended response questions. Your circular motion knowledge will also be built on later in Unit 5 when studying orbital motion under gravitational fields.