Unit 4: Further Mechanics

This guide covers sub-topic 4.3 Further Mechanics, the first of three sub-topics in Unit 4: Further Mechanics, Fields and Particles, from the Pearson Edexcel International Advanced Level in Physics (YPH11), Issue 3 specification (first assessment January 2020). Unit 4 is a compulsory, externally assessed IA2 unit examined by a 1 hour 45 minute paper worth 90 marks, and this topic is commonly studied using applications such as modern rail transportation.

Before studying this

This topic builds directly on Unit 1’s mechanics content (SUVAT equations, Newton’s laws, momentum) and assumes comfort converting between degrees and radians.

Syllabus coverage

PEARSON EDEXCEL INTERNATIONAL A LEVEL PHYSICS (YPH11) — Sub-topic 4.3

Candidates will be assessed on their ability to: understand how to use the equation impulse = FΔt = Δp (Newton’s second law of motion); CORE PRACTICAL 9: investigate the relationship between the force exerted on an object and its change of momentum; understand how to apply conservation of linear momentum to problems in two dimensions; CORE PRACTICAL 10: use ICT to analyse collisions between small spheres, e.g. ball bearings on a table top; understand how to determine whether a collision is elastic or inelastic; derive and use the equation Ek = p²/2m for the kinetic energy of a non-relativistic particle; express angular displacement in radians and in degrees and convert between these units; understand what is meant by angular velocity and use the equations v = ωr and ω = 2π/T; use vector diagrams to derive the equations for centripetal acceleration a = v²/r = ω²r and understand how to use these equations; understand that a resultant force (centripetal force) is required to produce and maintain circular motion; and use the equations for centripetal force F = ma = mv²/r = mω²r.

Impulse and two-dimensional momentum

Impulse is the change in momentum produced by a force acting over time:

impulse = FΔt = Δp

— a restatement of Newton’s second law in terms of momentum change. CORE PRACTICAL 9 investigates the relationship between the force exerted on an object and its change of momentum.

Unlike Unit 1’s one-dimensional treatment, momentum conservation here extends to two dimensions: the total momentum vector before a collision or interaction equals the total momentum vector after, with horizontal and vertical (or x and y) components conserved independently. CORE PRACTICAL 10 uses ICT (e.g. video analysis software) to analyse collisions between small spheres such as ball bearings on a table top.

A collision is elastic if total kinetic energy is conserved, and inelastic if it is not (some kinetic energy is transferred to other forms, such as heat or sound) — determined by comparing kinetic energy before and after using Ek = p²/2m, a useful form when momentum, rather than velocity, is known directly.

Angular motion and circular motion

Angular displacement can be expressed in radians or degrees (2π radians = 360°). Angular velocity ω relates to linear speed v by v = ωr, and to the period T of a full revolution by:

ω = 2π / T

An object moving in a circle at constant speed continuously changes direction, so it accelerates towards the centre of the circle — this centripetal acceleration can be derived from a vector diagram of changing velocity, giving:

a = v²/r = ω²r

A resultant force — the centripetal force — is required to produce and maintain circular motion, directed towards the centre. Combining with Newton’s second law:

F = ma = mv²/r = mω²r

Worked example. A 0.15 kg ball moving at 8.0 m/s collides head-on with a stationary 0.25 kg ball. After the collision the first ball rebounds at 2.0 m/s. Find the velocity of the second ball, and state whether the collision is elastic.

Taking the initial direction as positive, conservation of momentum gives:

(0.15 × 8.0) + (0.25 × 0) = (0.15 × −2.0) + (0.25 × v)

1.2 = −0.30 + 0.25v

v = 1.5 / 0.25 = 6.0 m/s

Kinetic energy before: ½(0.15)(8.0)² = 4.8 J. Kinetic energy after: ½(0.15)(2.0)² + ½(0.25)(6.0)² = 0.3 + 4.5 = 4.8 J.

Kinetic energy is conserved, so the collision is elastic.

Common mistakes

Forgetting that momentum is a vector in two-dimensional collisions — components must be resolved and conserved separately, not just total speed. Confusing centripetal force with a separate, additional force — centripetal force is simply the name given to the resultant (net) force producing circular motion, not a new force acting alongside the others. Mixing radians and degrees within the same calculation without converting. Assuming all collisions conserve kinetic energy — only elastic collisions do; momentum is conserved in both elastic and inelastic collisions.

Quick revision checklist

• Use impulse = FΔt = Δp and describe CORE PRACTICAL 9.
• Apply conservation of momentum in two dimensions and describe CORE PRACTICAL 10.
• Distinguish elastic from inelastic collisions using kinetic energy conservation.
• Use Ek = p²/2m for a non-relativistic particle.
• Convert between radians and degrees, and use v = ωr and ω = 2π/T.
• Derive and use a = v²/r = ω²r for centripetal acceleration.
• Explain why circular motion requires a resultant centripetal force, and use F = ma = mv²/r = mω²r.

Related resources

• Unit 4: Electric and Magnetic Fields — the next sub-topic
• Unit 1: Mechanics — foundational mechanics content this topic builds on
• Pearson Edexcel International A Level Physics hub

This guide is intended to support, not replace, engagement with the official Pearson Edexcel specification and your own teacher’s guidance. Always check the current version of the specification for authoritative detail.