Dynamics — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Newton’s three laws of motion, mass vs weight, linear momentum, conservation of momentum, impulse, force-time graphs, and elastic and inelastic collisions, aligned with the latest A-Level Physics syllabus.

You should already know: IGCSE Physics, basic algebra and trigonometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Physics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Dynamics?

Dynamics is the branch of mechanics that studies the effect of forces on the motion of objects, in contrast to kinematics which describes motion without reference to its causes. It forms 10-15% of marks across A-Level Physics Paper 1 (multiple choice), Paper 2 (AS structured) and Paper 4 (A2 structured), and is a prerequisite for all later mechanics topics including circular motion, gravitation, and field physics. You will frequently see dynamics combined with kinematics (SUVAT equations) or energy problems in multi-part exam questions.

2. Newton's three laws

Newton’s laws of motion are the foundational rules that describe the relationship between force and motion for all macroscopic, non-relativistic objects.

1. Newton’s First Law of Inertia: A body remains at rest or moves at constant velocity unless acted on by a net external unbalanced force. A net force is the vector sum of all forces acting on an object; if forces are balanced, net force is zero and velocity is constant. For example, a book resting on a table stays stationary because its downward weight is perfectly balanced by an upward normal contact force from the table.
2. Newton’s Second Law: The rate of change of linear momentum of a body is proportional to the net external force acting on it, and occurs in the direction of that force. The general form of the law is: $$F_{net}=\frac{\Delta p}{\Delta t}$$ For objects with constant mass, this simplifies to the more familiar $F=ma$, covered in the next section.
3. Newton’s Third Law: If body A exerts a force on body B, body B exerts a force of equal magnitude, opposite direction, and identical type on body A, acting on the two separate bodies. For example, when you push a wall, the wall pushes back on you with an equal normal force; the Earth pulls you down with gravitational weight, and you pull the Earth up with an equal gravitational force (the Earth’s very large mass makes its acceleration negligible).

Worked Example: A 3kg block slides across a rough table at constant velocity 2m/s, pulled by an applied force of 7N. What is the magnitude of the friction force acting on the block? Solution: Constant velocity means net force is zero (Newton’s First Law), so friction must be equal and opposite to the applied force: $F_{friction}=7N$. Exam tip: Examiners often test understanding of third law pairs: they never act on the same object, so weight and normal force on a stationary book are not a third law pair, as both act on the book.

3. Mass vs weight, $F=ma$ and $W=mg$

The two quantities are often confused in everyday language, but have distinct physical definitions:

• Mass: A scalar quantity measuring an object’s inertia (resistance to acceleration), with SI unit kilograms (kg). Mass is constant regardless of location, so a 5kg mass is 5kg on Earth and 5kg on the Moon.
• Weight: A vector quantity measuring the gravitational force acting on an object’s mass, with SI unit newtons (N). Weight varies with the strength of the local gravitational field $g$.

When an object has constant mass, Newton’s Second Law simplifies to: $$F_{net}=ma$$ where $a$ is the acceleration of the object, in the same direction as the net force. Weight is a special case of this formula, when the only force acting on an object is gravity, so its acceleration equals gravitational field strength $g$: $$W=mg$$ For A-Level exams, always use $g=9.81m/s^2$ unless explicitly told to use 10 $m/s^2$.

Worked Example: A 62kg astronaut travels to Mars, where the gravitational field strength is $3.72m/s^2$. Calculate (a) their mass on Mars, (b) their weight on Mars, (c) their weight on Earth. Solution: (a) Mass is constant: $m=62kg$ (b) Weight on Mars: $W=62\ast 3.72=230.64\approx 231N$ (3 significant figures, per A-Level convention) (c) Weight on Earth: $W=62\ast 9.81=608.22\approx 608N$ Exam tip: Always write units next to every numerical value to avoid mixing up mass (kg) and weight (N), a common mark-losing mistake.

4. Linear momentum and conservation in collisions

Linear momentum $p$ is defined as the product of an object’s mass and velocity, and is a vector quantity (direction matches velocity): $$p=mv$$ Its SI unit is $kg\cdot m/s$.

The principle of conservation of linear momentum states that for a closed system (no external unbalanced forces acting), total linear momentum remains constant in all directions. This rule is derived directly from Newton’s Third Law: when two objects interact, the forces they exert on each other are equal and opposite, acting for the same length of time, so their changes in momentum cancel out, leaving total momentum unchanged.

For two colliding objects with masses $m_1,m_2$, initial velocities $u_1,u_2$ and final velocities $v_1,v_2$, the conservation of momentum formula is: $$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$ Always define a positive direction (e.g., right = positive, left = negative) before using this formula, to correctly account for vector direction.

Worked Example: A 3kg trolley moving right at 5m/s collides with a 2kg trolley moving left at 2m/s. After collision, the 3kg trolley moves left at 1m/s. Find the velocity of the 2kg trolley after collision. Solution: Take right as positive. Initial total momentum: $(3\ast 5)+(2\ast (-2))=15-4=11kg\cdot m/s$ Final total momentum: $(3\ast (-1))+2v_2=-3+2v_2$ Equate initial and final momentum: $11=-3+2v_2\to 2v_2=14\to v_2=7m/s$ (positive, so 7m/s to the right)

5. Impulse and force-time graphs

Impulse $J$ is defined as the change in momentum of an object, and is equal to the product of the average force applied and the time interval the force acts for: $$J=F_{avg}\Delta t=\Delta p=m(v-u)$$ Impulse is a vector quantity, with SI unit $N\cdot s$, which is equivalent to $kg\cdot m/s$.

A force-time (F-t) graph plots force acting on an object on the y-axis against time on the x-axis. The area under a F-t graph is equal to the total impulse applied to the object, a relationship examiners test frequently across all papers. If the force is constant, the graph is a horizontal rectangle, and area is simply $F\ast \Delta t$. If force varies (e.g., during a ball collision where force peaks at impact), calculate area by summing geometric shapes (triangles, rectangles) or counting grid squares for rough estimates, which is accepted in mark schemes.

Worked Example: A 0.15kg tennis ball is thrown at a wall at 12m/s left, bounces off at 9m/s right, and is in contact with the wall for 0.015s. Calculate (a) impulse of the wall on the ball, (b) average force exerted by the wall, (c) peak force if the F-t graph is a triangle with base 0.015s. Solution: Take right as positive. (a) Impulse = $\Delta p=0.15\ast (9-(-12))=0.15\ast 21=3.15N\cdot s$ to the right (b) Average force = $J/\Delta t=3.15/0.015=210N$ (c) Area of triangle = $\frac{1}{2}\ast base\ast height=3.15=\frac{1}{2}\ast 0.015\ast F_{peak}\to F_{peak}=420N$

6. Elastic vs inelastic collisions

Collisions are classified based on whether total kinetic energy (KE, $KE=\frac{1}{2}m{v}^2$, a scalar quantity) is conserved:

• Elastic collision: Total kinetic energy is conserved, as well as total momentum. No energy is lost as heat, sound, or permanent deformation. For one-dimensional elastic collisions, an extra rule applies: relative speed of approach equals relative speed of separation: $$u_1-u_2=v_2-v_1$$ Examples include collisions between ideal gas molecules, or near-elastic collisions between snooker balls.
• Inelastic collision: Total kinetic energy is not conserved; some KE is converted to other forms (heat, sound, deformation). Total momentum remains conserved for closed systems. A perfectly inelastic collision is a special case where the two colliding objects stick together after collision, moving with the same final velocity, and maximum KE is lost.

Worked Example: A 2kg ball moving at 6m/s collides head-on with a 1kg ball at rest. (a) If the collision is elastic, find final velocities of both balls. (b) If the collision is perfectly inelastic, find final velocity and total KE lost. Solution: (a) Conservation of momentum: $2\ast 6+1\ast 0=2v_1+1v_2\to 12=2v_1+v_2$ Elastic collision rule: $6-0=v_2-v_1\to v_2=v_1+6$ Substitute: $12=2v_1+v_1+6\to 3v_1=6\to v_1=2m/s$ right, $v_2=8m/s$ right. (b) Perfectly inelastic: both balls move at same final velocity $v$. $12=(2+1)v\to v=4m/s$ right. Initial KE: $\frac{1}{2}\ast 2\ast 36+0=36J$ Final KE: $\frac{1}{2}\ast 3\ast 16=24J$ KE lost: $36-24=12J$ Exam tip: Examiners often ask you to prove if a collision is elastic or inelastic: always calculate total KE before and after, never assume based on context.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Mixing up mass and weight, using kg for weight or N for mass. Why it happens: everyday language uses "weight" to refer to mass. Correct move: write units next to every value, and double-check that mass is in kg, force/weight in N.
• Pitfall 2: Applying conservation of momentum to systems with external forces (e.g., a braking car, where friction is an external force). Why it happens: memorizing the rule without the "closed system, no external forces" condition. Correct move: first confirm no unbalanced external forces act during the interaction before using conservation of momentum.
• Pitfall 3: Treating momentum as a scalar, using only magnitudes in calculations. Why it happens: confusing momentum with kinetic energy, which is scalar. Correct move: define a positive direction at the start of every momentum question, and assign negative signs to velocities in the opposite direction.
• Pitfall 4: Assuming all collisions conserve kinetic energy. Why it happens: mixing up elastic and inelastic collision rules. Correct move: only use KE conservation if the question explicitly states the collision is elastic. For all other collisions, only momentum is conserved.
• Pitfall 5: Ignoring negative area on F-t graphs when force changes direction. Why it happens: forgetting impulse is a vector. Correct move: subtract areas below the x-axis from areas above the x-axis to calculate total impulse.

8. Practice Questions (A-Level Physics Style)

Question 1 (Paper 1 Multiple Choice)

A 0.4kg ball is dropped from rest, falls for 1.0s before hitting the ground, then bounces back up with a speed of 7 m/s. Air resistance is negligible, $g=9.81m/s^2$. What is the magnitude of the impulse exerted by the ground on the ball? A) 1.1 N·s B) 2.8 N·s C) 6.7 N·s D) 9.5 N·s

Solution: Velocity just before impact: $v=u+gt=0+9.81\ast 1=9.81m/s$ downwards. Take upwards as positive: initial velocity $u=-9.81m/s$, final velocity $v=+7m/s$. Impulse = $\Delta p=0.4\ast (7-(-9.81))=0.4\ast 16.81\approx 6.7N\cdot s$. Correct answer: C.

Question 2 (Paper 2 Structured, 7 marks)

Two trolleys, P of mass 5kg and Q of mass 3kg, are held at rest on a frictionless horizontal track with a compressed spring between them. When released, the spring pushes the trolleys apart, and P moves with a velocity of 2.4 m/s to the left. (a) State the principle of conservation of linear momentum. [2 marks] (b) Calculate the velocity of Q after release. [3 marks] (c) Calculate the total kinetic energy supplied by the spring. [2 marks]

Solution: (a) For a closed system with no external unbalanced forces acting, the total linear momentum remains constant in all directions. (2 marks: 1 for closed system condition, 1 for total momentum constant) (b) Take right as positive. Initial total momentum = 0 (both at rest). Final total momentum: $5\ast (-2.4)+3v_Q=0\to -12+3v_Q=0\to v_Q=4m/s$ to the right. (3 marks: 1 for sign convention, 1 for equating initial/final momentum, 1 for correct answer and direction) (c) Total KE = $\frac{1}{2}\ast 5\ast (2.4{)}^2+\frac{1}{2}\ast 3\ast (4{)}^2=14.4+24=38.4J$. (2 marks: 1 for correct KE formula applied to both trolleys, 1 for final answer)

Question 3 (Paper 4 A2 Structured, 6 marks)

The force-time graph for a 0.25kg hockey ball hit by a stick is a triangle with peak force 150N, base duration 0.008s, starting at t=0, peaking at t=0.004s, returning to zero at t=0.008s. The ball was initially at rest. (a) Calculate the impulse exerted on the ball. [2 marks] (b) Find the speed of the ball as it leaves the stick. [2 marks] (c) Calculate the maximum acceleration of the ball during contact. [2 marks]

Solution: (a) Impulse = area under F-t graph = $\frac{1}{2}\ast 0.008\ast 150=0.6N\cdot s$. (2 marks: 1 for area of triangle formula, 1 for correct answer) (b) Impulse = change in momentum: $0.6=0.25\ast (v-0)\to v=2.4m/s$. (2 marks: 1 for impulse = Δp, 1 for correct answer) (c) Maximum force = 150N, $F=ma\to a=150/0.25=600m/s^2$. (2 marks: 1 for F=ma, 1 for correct answer)

9. Quick Reference Cheatsheet

10. What's Next

Dynamics is the foundational building block for all subsequent mechanics topics in the A-Level Physics syllabus. You will apply these rules to circular motion, where centripetal force produces centripetal acceleration, to gravitational fields where gravitational force causes orbital motion, to electric and magnetic fields where charged particles experience force and accelerate, and even to materials physics when calculating forces on structures. Multi-part exam questions often combine dynamics with kinematics, work, energy, and power, so mastering these concepts now will save you significant time when studying later topics.

If you struggle with any of the concepts, practice questions, or past paper problems in this guide, you can ask Ollie our AI tutor for step-by-step explanations, custom practice questions, or clarification of any mark scheme point at any time on the homepage. We recommend following this guide with our study notes on work, energy, and power, then kinematics, to build a full mastery of AS Level mechanics before moving to A2 topics.

Aligned with the Cambridge International AS & A Level Physics 9702 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.