Kinematics describes motion without asking what causes it. With constant acceleration, three equations connect displacement, velocity, acceleration and time.
📐 The equations of motion
For constant acceleration a, starting at velocity u and reaching v after time t with displacement s:
v = u + at
s = ut + ½at²
v² = u² + 2as
A car starts from rest and accelerates at 2 m/s² for 5 s
v = 0 + 2·5 = 10 m/s
s = ½·2·5² = 25 m
Braking: u = 20 m/s, v = 0, a = −5 m/s²
s = (0 − 400) / (2·(−5)) = 40 m
Graphs: on a velocity–time graph the slope is acceleration and the area under the line is displacement.
🖼️ See it
Range = v² sin 2θ ÷ g = 40 m, and the flight lasts about 2.8 s (g = 10 m/s²).
🎯 Projectile motion
Split the motion into two independent parts. Horizontally there is no force, so the velocity is constant. Vertically the acceleration is g = 10 m/s² downward.
Ball thrown horizontally at 20 m/s from a 45 m cliff
vertical: 45 = ½·10·t² → t² = 9 → t = 3 s
horizontal: range = 20 · 3 = 60 m
Thrown straight up at 20 m/s:
time to top = u/g = 2 s, total flight = 4 s
max height = u²/2g = 400/20 = 20 m
Key idea: the time to fall depends only on the height, not on the horizontal speed.
🧪 Worked examples: projectiles at an angle
Resolve the velocity. A ball is kicked at 50 m/s at an angle whose sin is 0.6 and cos is 0.8 (about 37°).
horizontal vₓ = 50 × 0.8 = 40 m/s (constant)
vertical vᵧ = 50 × 0.6 = 30 m/s (decelerated by g)
time to the top t = vᵧ ÷ g = 30 ÷ 10 = 3 s
maximum height h = vᵧ² ÷ 2g = 900 ÷ 20 = 45 m
total flight time = 2 × 3 = 6 s (returns to the same height)
range = vₓ × 6 = 40 × 6 = 240 m
Using v² = u² + 2as. Choose a direction as positive and keep signs consistent.
a train slows from 30 m/s to 10 m/s over 100 m
a = (v² − u²) ÷ 2s = (100 − 900) ÷ 200 = −4 m/s²
(deceleration of 4 m/s²)
Speed and direction at any moment. After 4 s of the flight above: vᵧ = 30 − 40 = −10 m/s (falling), vₓ = 40 m/s, so the speed is √(40² + 10²) = √1700 ≈ 41 m/s.
Independence. The horizontal and vertical motions are independent. They share only the time.
🎯 Exam focus: setting out an analysis
Write a table with two columns: horizontal (a = 0) and vertical (a = −g). Fill in u, v, s, a and t for each, then link them with the common time.
at the highest point: vᵧ = 0 but vₓ ≠ 0
the speed at the top is the horizontal component only
maximum range on level ground (no air resistance) occurs at 45°
Common mistakes. Using the full launch speed in the vertical equation. Saying that the acceleration is zero at the top (it is still g downward). Forgetting the sign of g when the object is thrown up. Using the range formula when the landing height differs from the launch height. Mixing degrees and radians on the calculator.
Air resistance. It reduces both the range and the maximum height, and makes the path asymmetric: the descent is steeper than the rise.
✏️ Quick check · 1 of 11
Constant in flight
Ignoring air resistance, which quantity stays constant during projectile motion?
Which direction has zero force?
✏️ Quick check · 2 of 11
Top of the path
A ball is thrown straight up. At the very top of its path:
Is gravity switched off at the top?
✍️ Type your answer · 3 of 11
A car starts from rest and accelerates at 2 m/s².
What is its speed after 5 s, in m/s?
Enter an exact answer, without rounding.
m/s
v = u + at.
✍️ Type your answer · 4 of 11
A car starts from rest and accelerates at 2 m/s² for 5 s.
How far does it travel, in m?
Enter an exact answer, without rounding.
m
s = ut + ½at² with u = 0.
✍️ Type your answer · 5 of 11
A car moving at 20 m/s brakes at a constant 5 m/s² until it stops.
What is the stopping distance, in m?
Enter an exact answer, without rounding.
m
Use v² = u² + 2as with a = −5.
✍️ Type your answer · 6 of 11
A stone is dropped from rest from a height of 80 m (g = 10 m/s²).
How long does it take to land, in s?
Enter an exact answer, without rounding.
s
s = ½gt².
✍️ Type your answer · 7 of 11
A ball is thrown horizontally at 20 m/s from a 45 m high cliff.
How far from the base of the cliff does it land, in m?
Enter an exact answer, without rounding.
m
Find the fall time first.
✍️ Type your answer · 8 of 11
A ball is kicked at 50 m/s with sin θ = 0.6 and cos θ = 0.8. Use g = 10 m/s².
What is the maximum height, in metres?
Enter an exact answer, without rounding.
m
Use v² = u² − 2gh with v = 0 at the top.
✍️ Type your answer · 9 of 11
The same ball is kicked from level ground and lands on level ground.
What is the horizontal range, in metres?
Enter an exact answer, without rounding.
m
Horizontal speed × total time.
✍️ Type your answer · 10 of 11
A train slows from 30 m/s to 10 m/s while travelling 100 m with constant deceleration.
What is the size of the deceleration, in m/s²?
Enter an exact answer, without rounding.
m/s²
Use v² = u² + 2as.
✏️ Quick check · 11 of 11
At the top of the path
At the highest point of a projectile’s path (no air resistance), which statement is correct?
Does gravity switch off?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The slope of a velocity–time graph gives:
Question 2 of 5
The area under a velocity–time graph gives:
Question 3 of 5
A ball is thrown straight up at 20 m/s (g = 10 m/s²). Its total time in the air is:
Question 4 of 5
A ball is thrown straight up at 20 m/s. The maximum height reached is:
Question 5 of 5
For a given launch speed on level ground, the range is greatest at an angle of:
Lesson 2 · Mechanics
💪 Newton's Laws, Friction & Inclines
Forces cause changes in motion. Newton's three laws let you predict acceleration from the forces acting, including friction and the pull of gravity along a slope.
⚖️ Newton's laws in action
1st law: with zero net force, velocity stays constant. 2nd law: ΣF = ma. 3rd law: forces come in equal and opposite pairs acting on different objects.
weight W = mg (m = 8 kg → W = 80 N)
friction f = μN (opposes motion)
10 kg block on a floor, pushed with 50 N, μ = 0.2
N = mg = 100 N
f = 0.2 · 100 = 20 N
net force = 50 − 20 = 30 N
a = 30 / 10 = 3 m/s²
Mass vs weight: mass (kg) is the amount of matter and never changes. Weight (N) is the gravitational force and depends on g.
🖼️ See it
Resolve the weight into a part along the slope (mg sin θ, which accelerates the object) and a part into the slope (mg cos θ, which sets the normal force and friction).
⛰️ Forces on a slope
On an incline of angle θ, split the weight into a component along the slope and one into the slope.
along the slope: mg sin θ
into the slope: mg cos θ (= normal force N)
Frictionless 30° slope: a = g sin 30° = 10 · 0.5 = 5 m/s²
With friction: a = g sin θ − μ g cos θ
Terminal velocity: a falling object speeds up until air drag equals its weight. Then the net force is zero and the speed stops increasing.
🧪 Worked examples: connected bodies, slopes and lifts
Two bodies pulled together. Treat them as one system to find a, then isolate one body for the tension.
5 kg and 3 kg masses joined by a light string on a smooth surface, pulled by 40 N on the 5 kg mass
system: a = 40 ÷ 8 = 5 m/s²
the 3 kg mass is pulled only by the tension: T = 3 × 5 = 15 N
Slope with friction. Use sin θ = 0.6, cos θ = 0.8, μ = 0.25, m = 10 kg.
along the slope: mg sin θ = 100 × 0.6 = 60 N down
normal force N = mg cos θ = 80 N, friction = 0.25 × 80 = 20 N up
net = 60 − 20 = 40 N → a = 40 ÷ 10 = 4 m/s² down the slope
pulling UP at constant speed: F = 60 + 20 = 80 N
Apparent weight in a lift. The reading on scales is the normal force.
60 kg person, lift accelerating upward at 2 m/s²
N − mg = ma → N = m(g + a) = 60 × 12 = 720 N
accelerating downward at 2 m/s²: N = 60 × 8 = 480 N
Free-fall feeling. If the lift cable breaks, a = g downward, N = 0, and you feel weightless, though gravity still acts.
🎯 Exam focus: free-body diagrams and sign conventions
Draw a free-body diagram for each body, with every force labelled (weight, normal, tension, friction, drag). Choose a positive direction and write ΣF = ma for each body.
static friction f ≤ μₛ N (adjusts up to a maximum)
kinetic friction f = μₖ N (constant once sliding; usually smaller)
terminal velocity: drag = driving force, so a = 0
Common mistakes. Using N = mg on a slope (it is mg cos θ). Assuming that friction is always μN (static friction can be smaller). Forgetting that the tension is the same throughout a light string over a smooth pulley. Using the wrong direction for friction (it opposes the motion or the tendency to move). Treating the third-law pair as a balanced pair on a single body.
Angle check. The angle between the weight and the normal to the slope equals the angle of the slope.
✏️ Quick check · 1 of 11
Third law
A book rests on a table. Which force is the Newton's-third-law partner of the book's weight (Earth pulling the book down)?
The partner force has the same type and swaps the two objects.
✏️ Quick check · 2 of 11
Zero net force
An object moves at constant velocity. The net force on it is:
Constant velocity means a = 0.
✍️ Type your answer · 3 of 11
A 5 kg object accelerates at 4 m/s².
What net force acts on it, in N?
Enter an exact answer, without rounding.
N
F = ma.
✍️ Type your answer · 4 of 11
Use g = 10 m/s².
What is the weight of an 8 kg mass, in N?
Enter an exact answer, without rounding.
N
W = mg.
✍️ Type your answer · 5 of 11
A 10 kg block is pushed with 50 N across a floor with μ = 0.2.
What is its acceleration, in m/s²?
Enter an exact answer, without rounding.
m/s²
Find friction first: f = μmg.
✍️ Type your answer · 6 of 11
The normal force on a crate is 200 N and μ = 0.25.
What is the friction force, in N?
Enter an exact answer, without rounding.
N
f = μN.
✍️ Type your answer · 7 of 11
A block slides down a frictionless 30° slope (sin 30° = 0.5).
What is its acceleration, in m/s²?
Enter an exact answer, without rounding.
m/s²
The component of weight along the slope is mg sin θ.
✍️ Type your answer · 8 of 11
Masses of 5 kg and 3 kg are joined by a light string on a smooth surface, and a 40 N force pulls the 5 kg mass directly away from the 3 kg mass.
What is the tension in the string, in newtons?
Enter an exact answer, without rounding.
N
Find a for the whole system first.
✍️ Type your answer · 9 of 11
A 10 kg block slides down a slope with sin θ = 0.6 and cos θ = 0.8. The coefficient of kinetic friction is 0.25.
What is its acceleration, in m/s²?
Enter an exact answer, without rounding.
m/s²
Resolve along the slope.
✍️ Type your answer · 10 of 11
The same 10 kg block is pulled up the same slope at constant speed.
What force is needed along the slope, in newtons?
Enter an exact answer, without rounding.
N
Friction now acts down the slope.
✍️ Type your answer · 11 of 11
A 60 kg person stands on scales in a lift that accelerates upward at 2 m/s².
What force do the scales show, in newtons?
Enter an exact answer, without rounding.
N
N − mg = ma.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Which has more inertia?
Question 2 of 5
A 2 kg object rests on a level table. The normal force on it is:
Question 3 of 5
Terminal velocity is reached when:
Question 4 of 5
Friction on a sliding object acts:
Question 5 of 5
Doubling the net force on an object (same mass) will:
Lesson 3 · Mechanics
🌍 Circular Motion & Gravitation
An object moving in a circle is always accelerating towards the centre, even at constant speed. Gravity supplies this inward force for planets and satellites.
🔄 Uniform circular motion
The speed is constant but the direction keeps changing, so the velocity changes. The acceleration points to the centre of the circle.
v = 2πr / T
a_c = v² / r
F_c = m v² / r
r = 5 m, T = 3.14 s (take π = 3.14)
v = 2·3.14·5 / 3.14 = 10 m/s
a_c = 100 / 5 = 20 m/s²
m = 2 kg: F_c = 2·100 / 5 = 40 N
Centripetal force is not a new force: it is the name for whatever force (tension, friction, gravity, a normal force) points towards the centre.
🖼️ See it
Twice as far from the centre gives a quarter of the pull. Circular motion also needs a centripetal force F = mv²/r, which gravity provides for satellites.
🪐 Newton's law of gravitation
Every pair of masses attracts with a force that falls off with the square of the distance.
F = G M m / r² g = G M / r²
Double r: F becomes 1/4
Triple r: F becomes 1/9
Same planet mass but twice the radius:
g becomes 10 / 4 = 2.5 m/s²
A satellite in orbit is in free fall: gravity provides exactly the centripetal force needed. A geostationary satellite orbits once every 24 h above the equator, so it stays above the same point on Earth.
🧪 Worked examples: bends, vertical circles and orbits
A car on a flat bend. Friction provides the centripetal force: μ m g = m v² ÷ r.
r = 20 m, μ = 0.5
v² = μ g r = 0.5 × 10 × 20 = 100 → v = 10 m/s maximum
(the mass cancels)
Vertical circle. At the top, the weight and tension together provide the centripetal force. The minimum speed is when T = 0: m g = m v² ÷ r.
bucket on a string, r = 0.9 m
v²min = g r = 10 × 0.9 = 9 → v = 3 m/s at the top
Orbits and Kepler’s third law. For circular orbits around the same body, T² ∝ r³, so T ∝ r^1.5.
orbit A has radius r and period 2 h; orbit B has radius 4r
T_B = 2 × 4^1.5 = 2 × 8 = 16 h
Gravitational field strength. g = GM ÷ r². A planet with twice the mass and twice the radius of a planet where g = 10 N/kg has g = 10 × 2 ÷ 2² = 5 N/kg.
Weightlessness. Astronauts in orbit are in free fall around the Earth, with the same acceleration as their spacecraft, so there is no contact force between them and the floor.
🎯 Exam focus: naming the force that supplies the centripetal force
Never write “centripetal force” as a separate force on a free-body diagram. Name the real force: friction (car on a bend), tension (string), gravity (orbit), normal reaction (loop-the-loop).
v = ωr ω = 2π ÷ T = 2πf a = v² ÷ r = ω² r
F = m v² ÷ r = m ω² r
geostationary orbit: T = 24 h, above the equator, moving in the same direction as Earth’s spin
Common mistakes. Saying that a body in circular motion has no acceleration because its speed is constant. Pointing the acceleration outwards. Confusing the radius of the orbit (from the centre) with the height above the surface. Using r from the surface in GM ÷ r². Forgetting to convert rev/min to rad/s.
Escape speed. v = √(2GM ÷ r), independent of the direction and the mass of the projectile.
✏️ Quick check · 1 of 11
Direction
For an object moving in a circle at constant speed, the acceleration points:
The object is pulled towards the middle.
✏️ Quick check · 2 of 11
Inverse square
If the distance between two masses doubles, the gravitational force between them:
Square the factor, then invert.
✍️ Type your answer · 3 of 11
An object travels around a circle of radius 5 m at 10 m/s.
What is its centripetal acceleration, in m/s²?
Enter an exact answer, without rounding.
m/s²
a = v²/r.
✍️ Type your answer · 4 of 11
A 2 kg mass moves at 10 m/s in a circle of radius 5 m.
What centripetal force is needed, in N?
Enter an exact answer, without rounding.
N
F = ma with a = v²/r.
✍️ Type your answer · 5 of 11
The distance between two masses is tripled.
The gravitational force becomes 1/n of its original value. What is n?
Enter an exact answer, without rounding.
Square the factor 3.
✍️ Type your answer · 6 of 11
A 10 kg object is on a planet where g = 5 m/s².
What is its weight, in N?
Enter an exact answer, without rounding.
N
W = mg with the planet's g.
✍️ Type your answer · 7 of 11
A planet has the same mass as Earth but twice the radius (Earth g = 10 m/s²).
What is g on its surface, in m/s²?
Enter an exact answer, without rounding.
m/s²
Double the radius → divide g by 4.
✍️ Type your answer · 8 of 11
A car rounds a flat bend of radius 20 m. The coefficient of friction is 0.5.
What is the maximum speed, in m/s?
Round your answer as the question asks.
m/s
Friction provides the centripetal force.
✍️ Type your answer · 9 of 11
A bucket of water is swung in a vertical circle of radius 0.9 m.
What is the minimum speed at the top for the water to stay in the bucket, in m/s?
Enter an exact answer, without rounding.
m/s
Set the tension to zero at the top.
✍️ Type your answer · 10 of 11
A satellite in a circular orbit has a period of 2 h. A second satellite orbits the same planet at 4 times the radius.
What is the period of the second satellite, in hours?
Enter an exact answer, without rounding.
h
Use T² ∝ r³.
✏️ Quick check · 11 of 11
Gravitational field
A planet has twice the mass and twice the radius of Earth. Compared with the value on Earth, its surface g is:
Mass doubles, but the radius is squared.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A satellite in a stable orbit is best described as:
Question 2 of 5
The orbital period of a geostationary satellite is:
Question 3 of 5
A car goes round a bend at twice the speed. The centripetal force needed becomes:
Question 4 of 5
During uniform circular motion, which quantity is constant?
Question 5 of 5
Mass and weight differ because:
Lesson 4 · Energy
⚡ Work, Energy & Power
Energy is the ability to do work. It changes form but is never created or destroyed, which makes many problems far simpler than using forces.
🔧 Work and energy
Work is force times the distance moved in the direction of the force. Energy is measured in joules (J).
W = F d cos θ
KE = ½ m v²
GPE = m g h
A 40 N force pushes a box 5 m along the force: W = 200 J
4 kg at 5 m/s: KE = ½·4·25 = 50 J
3 kg lifted 10 m: GPE = 3·10·10 = 300 J
Perpendicular force = no work. A centripetal force, or the normal force on a level surface, does zero work because there is no motion along it.
🖼️ See it
Work is force × distance moved in the direction of the force. On a force-displacement graph it is the area under the line, and dividing by the time gives power.
♻️ Conservation of energy and power
With no friction, mechanical energy is conserved: GPE lost = KE gained.
Dropped from 20 m: mgh = ½mv²
v = √(2gh) = √(2·10·20) = 20 m/s (the mass cancels)
P = W / t = F v
600 J in 3 s → P = 200 W
efficiency = useful output / input × 100%
400 J out from 500 J in → 80%
🧪 Worked examples: friction, power and springs
Energy lost to friction. The work done against friction equals the energy that did not become kinetic.
4 kg block slides from rest down a rough slope, a vertical drop of 3 m, and reaches 6 m/s
GPE lost = 4 × 10 × 3 = 120 J
KE gained = ½ × 4 × 36 = 72 J
work done against friction = 120 − 72 = 48 J (transferred as heat)
Power and speed. P = F v. A car at a constant 25 m/s against a resistive force of 800 N needs P = 800 × 25 = 20 000 W (20 kW).
Elastic energy. The area under a force–extension graph is the work done: E = ½ F x = ½ k x².
spring constant 800 N/m compressed by 0.10 m: E = ½ × 800 × 0.01 = 4 J
launches a 0.50 kg ball: ½ m v² = 4 → v² = 16 → v = 4 m/s
Work from a graph. For a variable force, the work done is the area under the force–distance graph (count squares or use a trapezium).
🎯 Exam focus: energy conservation arguments
Start with the energy statement: “loss in GPE = gain in KE + work done against resistance”. Then substitute. Marks are given for the statement, even if the arithmetic goes wrong.
W = F d cos θ P = W ÷ t = F v KE = ½ m v²
the net work done on a body = its change in kinetic energy (work–energy theorem)
conservative forces (gravity, springs): the work done is independent of the path
Common mistakes. Using the length of the slope for the vertical height. Forgetting cos θ when the force is not along the motion. Using P = F v with the average speed. Counting the work done by gravity as zero on a slope. Saying that the work done by a centripetal force is non-zero: the force is perpendicular to the motion, so the work is zero.
Efficiency. Always compare useful output energy (or power) with the total input.
✏️ Quick check · 1 of 12
Zero work
A force does no work on an object when it is:
cos 90° = 0.
✏️ Quick check · 2 of 12
Doubling speed
If an object's speed doubles, its kinetic energy:
v is squared.
✍️ Type your answer · 3 of 12
A 40 N force pushes a box 5 m in the direction of the force.
How much work is done, in J?
Enter an exact answer, without rounding.
J
W = Fd cos 0°.
✍️ Type your answer · 4 of 12
A 4 kg ball moves at 5 m/s.
What is its kinetic energy, in J?
Enter an exact answer, without rounding.
J
KE = ½mv².
✍️ Type your answer · 5 of 12
A 3 kg mass is raised 10 m (g = 10 m/s²).
How much gravitational potential energy does it gain, in J?
Enter an exact answer, without rounding.
J
GPE = mgh.
✍️ Type your answer · 6 of 12
A ball is dropped from a height of 20 m (no air resistance).
What is its speed just before landing, in m/s?
Enter an exact answer, without rounding.
m/s
GPE lost equals KE gained.
✍️ Type your answer · 7 of 12
A motor does 600 J of work in 3 s.
What is its power, in W?
Enter an exact answer, without rounding.
W
P = W/t.
✍️ Type your answer · 8 of 12
A machine takes in 500 J and delivers 400 J of useful work.
What is its efficiency, in %?
Enter an exact answer, without rounding.
%
useful ÷ input × 100.
✍️ Type your answer · 9 of 12
A 4 kg block slides from rest down a rough slope with a vertical drop of 3 m, and reaches 6 m/s at the bottom.
How much energy is transferred to heat by friction, in joules?
Enter an exact answer, without rounding.
J
Energy lost = GPE lost − KE gained.
✍️ Type your answer · 10 of 12
A car travels at a constant 25 m/s against a total resistive force of 800 N.
What power does the engine deliver, in watts?
Enter an exact answer, without rounding.
W
Constant speed means the driving force equals the resistance.
✍️ Type your answer · 11 of 12
A spring of constant 800 N/m is compressed by 0.10 m and launches a 0.50 kg ball horizontally.
What is the launch speed, in m/s?
Enter an exact answer, without rounding.
m/s
Elastic energy becomes kinetic energy.
✏️ Quick check · 12 of 12
Zero work
In which case is the work done by the force zero?
Look for a force perpendicular to the displacement.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The unit of power is the:
Question 2 of 5
A roller coaster at the top of a hill has mostly:
Question 3 of 5
Energy lost to friction is converted mainly into:
Question 4 of 5
Doubling the height an object is raised, the GPE gained:
Question 5 of 5
Which statement about efficiency is true?
Lesson 5 · Mechanics
🎱 Momentum & Collisions
Momentum is mass in motion. In any isolated system the total momentum is conserved, which lets you analyse collisions and explosions.
📦 Momentum and impulse
Momentum p = mv is a vector measured in kg·m/s. An impulse (force × time) changes it.
p = m v
impulse J = F Δt = Δp
5 kg moving at 4 m/s: p = 20 kg·m/s
F = 50 N acting for 0.2 s: J = 10 N·s
1500 kg car at 20 m/s stops in 4 s
Δp = 30 000 F = 30 000 / 4 = 7500 N
Crumple zones and airbags make Δt longer, so for the same Δp the force is smaller.
🖼️ See it
Momentum is conserved (6 before, 6 after) but kinetic energy is not: it falls from 9 J to 6 J as the objects stick together.
💥 Conservation in collisions
Momentum is conserved in every collision when no external force acts. Kinetic energy is conserved only in elastic collisions.
total momentum before = total momentum after
Sticky (perfectly inelastic) collision:
2 kg at 6 m/s hits a 4 kg mass at rest and they stick
12 = (2 + 4) v → v = 2 m/s
KE before = ½·2·36 = 36 J
KE after = ½·6·4 = 12 J (24 J lost as heat and sound)
🧪 Worked examples: elastic collisions, explosions and impacts
Elastic collision in one dimension. Both momentum and kinetic energy are conserved. Use the relative speed of approach = relative speed of separation.
2 kg at 5 m/s hits a 3 kg mass at rest, elastically
momentum: 2 × 5 = 2 v₁ + 3 v₂ → 10 = 2 v₁ + 3 v₂
relative speeds: 5 = v₂ − v₁
solve: v₂ = 4 m/s, v₁ = −1 m/s (rebounds)
Explosion (recoil). The total momentum before is zero, so the fragments move in opposite directions with equal momenta.
10 kg shell at rest splits into 6 kg and 4 kg; the 4 kg piece moves at 15 m/s
6 × v = 4 × 15 → v = 10 m/s the other way
kinetic energy increases (it comes from the chemical energy of the explosive)
Impact force.
0.20 kg ball hits a wall at 10 m/s and rebounds at 6 m/s
Δp = 0.20 × (6 − (−10)) = 3.2 N s
contact time 0.010 s → F = 3.2 ÷ 0.010 = 320 N
Rebound doubles the change. Because velocity reverses, Δp is the sum of the two momenta, not the difference.
🎯 Exam focus: conservation arguments and sign conventions
Pick a positive direction, write “momentum before = momentum after” as an equation, and keep the signs. Then test the kinetic energy: if it is conserved, the collision is elastic.
perfectly inelastic the bodies stick together; maximum KE lost (but not all)
elastic KE conserved; relative speed of separation = of approach
force–time graph the area under it is the impulse (change in momentum)
Common mistakes. Forgetting that momentum is a vector (a rebound has a negative velocity). Saying that kinetic energy is lost in the sense of “destroyed” (it becomes heat, sound or deformation). Assuming that every collision is elastic. Mixing up the impulse (N s) with the energy (J). Writing that momentum is not conserved when the Earth is involved (it is conserved if you include the Earth, which has a tiny change in velocity).
Crumple zones. A longer stopping time reduces the average force for the same change in momentum.
✏️ Quick check · 1 of 11
Conserved quantity
In a collision between two carts on a frictionless track, which quantity is always conserved?
It is conserved for the system, not for each object.
✏️ Quick check · 2 of 11
Airbag
An airbag reduces injury in a crash because it:
Impulse = FΔt.
✍️ Type your answer · 3 of 11
A 5 kg trolley moves at 4 m/s.
What is its momentum, in kg·m/s?
Enter an exact answer, without rounding.
kg·m/s
p = mv.
✍️ Type your answer · 4 of 11
A constant force of 50 N acts for 0.2 s.
What impulse is delivered, in N·s?
Enter an exact answer, without rounding.
N·s
J = FΔt.
✍️ Type your answer · 5 of 11
A 1500 kg car travelling at 20 m/s is brought to rest in 4 s.
What average braking force is needed, in N?
Enter an exact answer, without rounding.
N
Δp = mΔv, then divide by Δt.
✍️ Type your answer · 6 of 11
A 2 kg cart at 6 m/s hits a 4 kg cart at rest and they stick together.
What is their common speed, in m/s?
Enter an exact answer, without rounding.
m/s
Total momentum before = total momentum after.
✍️ Type your answer · 7 of 11
Two carts of mass 2 kg and 3 kg sit at rest, joined by a compressed spring. They fly apart and the 2 kg cart moves at 3 m/s.
What is the speed of the 3 kg cart, in m/s?
Enter an exact answer, without rounding.
m/s
Total momentum stays zero.
✍️ Type your answer · 8 of 11
A 2 kg mass at 5 m/s collides elastically with a 3 kg mass at rest.
What is the speed of the 3 kg mass after the collision, in m/s?
Enter an exact answer, without rounding.
m/s
Use conservation of momentum and relative speeds.
✍️ Type your answer · 9 of 11
A 10 kg shell at rest explodes into a 6 kg fragment and a 4 kg fragment. The 4 kg fragment moves at 15 m/s.
What is the speed of the 6 kg fragment, in m/s?
Enter an exact answer, without rounding.
m/s
Total momentum stays zero.
✍️ Type your answer · 10 of 11
A 0.20 kg ball hits a wall at 10 m/s and rebounds at 6 m/s. The contact time is 0.010 s.
What is the average force on the ball, in newtons?
Enter an exact answer, without rounding.
N
The velocity changes from −10 to +6 (a change of 16).
✏️ Quick check · 11 of 11
Types of collision
In an inelastic collision between two trolleys, which quantity is conserved?
Which quantity is always conserved?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
In an elastic collision:
Question 2 of 5
Impulse is equal to:
Question 3 of 5
The unit of momentum is:
Question 4 of 5
A heavy truck and a small car have the same momentum. Which has more kinetic energy?
Question 5 of 5
When a rocket expels gas backwards, it moves forwards because:
Lesson 6 · Thermodynamics
🌡️ Thermal Physics & Gas Laws
Temperature measures the average kinetic energy of particles; heat is energy transferred because of a temperature difference. Gases follow simple laws linking pressure, volume and temperature.
🔥 Heat, specific heat and latent heat
Use Q = mcΔT while the temperature changes, and Q = mL while the substance changes state at constant temperature.
c(water) = 4200 J/(kg·K) L(melting ice) = 334 000 J/kg
Heat 2 kg of water by 10 K:
Q = 2 · 4200 · 10 = 84 000 J
Melt 0.5 kg of ice at 0 °C:
Q = 0.5 · 334 000 = 167 000 J
Kelvin: T(K) = T(°C) + 273. Gas laws always use Kelvin, so 27 °C = 300 K.
🖼️ See it
Pressure is proportional to temperature in kelvin, so the line passes through the origin. Always convert °C to K by adding 273.
🎈 The gas laws
For a fixed amount of gas:
Boyle (T constant): P₁V₁ = P₂V₂
Charles (P constant): V₁/T₁ = V₂/T₂
Combined: P₁V₁/T₁ = P₂V₂/T₂ (PV = nRT)
100 kPa and 2 L, expanded to 4 L (T constant)
P₂ = 100 · 2 / 4 = 50 kPa
3 L at 300 K heated to 600 K (constant P):
V₂ = 3 · 600 / 300 = 6 L
Gas pressure comes from particles colliding with the container walls: hotter particles hit harder and more often.
🧪 Worked examples: mixing, gas laws and kinetic theory
Mixing liquids. Heat lost by the hot liquid = heat gained by the cold one (no losses).
0.2 kg of water at 80 °C + 0.3 kg of water at 20 °C
0.2 × (80 − T) = 0.3 × (T − 20)
16 − 0.2 T = 0.3 T − 6 → 22 = 0.5 T → T = 44 °C
Combined gas law. Convert to kelvin.
2 L at 100 kPa and 300 K is heated to 600 K and expands to 8 L
p₂ = p₁ V₁ T₂ ÷ (V₂ T₁) = 100 × 2 × 600 ÷ (8 × 300) = 50 kPa
A bubble rising. At 10 m depth in water, the pressure is 1.0 × 10⁵ + 1000 × 10 × 10 = 2.0 × 10⁵ Pa, which is twice atmospheric.
bubble of 2 cm³ at 10 m depth rises to the surface at constant temperature
V₂ = p₁V₁ ÷ p₂ = 2.0 × 2 ÷ 1.0 = 4 cm³
Kinetic theory. The average kinetic energy of the molecules is proportional to the absolute temperature: ½ m⟨c²⟩ = (3/2) k T. If T is quadrupled, the rms speed doubles.
🎯 Exam focus: explaining pressure and ideal gas assumptions
Explain gas pressure in four steps: molecules move randomly; they collide with the walls; each collision changes the momentum (so exerts a force); many collisions per second give a steady average pressure.
assumptions of the ideal gas model
• molecules are point-like, with negligible volume
• no intermolecular forces except during collisions
• collisions are elastic and the motion is random
• p V = n R T (R = 8.31 J mol⁻¹ K⁻¹)
Common mistakes. Using °C in gas law equations. Forgetting to add atmospheric pressure to the pressure from a liquid column. Saying that a hotter gas has molecules that “all” move faster (it is the average that increases). Mixing up n (moles) with N (molecules). Explaining pressure as molecules “hitting each other”.
Absolute zero. At 0 K, −273 °C, the kinetic energy of the molecules would be minimal; it is the zero of the Kelvin scale.
✏️ Quick check · 1 of 11
Conduction
Heat transfer through a solid, with no movement of the material itself, is called:
It happens in a metal spoon in hot soup.
✏️ Quick check · 2 of 11
Temperature
The temperature of a gas is a measure of:
Not the total energy, the average.
✍️ Type your answer · 3 of 11
2 kg of water is heated by 10 K (c = 4200 J/(kg·K)).
How much heat is needed, in J?
Enter an exact answer, without rounding.
J
Q = mcΔT.
✍️ Type your answer · 4 of 11
0.5 kg of ice at 0 °C melts (L = 334 000 J/kg).
How much heat is absorbed, in J?
Enter an exact answer, without rounding.
J
Q = mL.
✍️ Type your answer · 5 of 11
A gas is at 27 °C.
What is its temperature in kelvin?
Enter an exact answer, without rounding.
K
Add 273.
✍️ Type your answer · 6 of 11
A gas at 100 kPa and 2 L is expanded to 4 L at constant temperature.
What is its new pressure, in kPa?
Enter an exact answer, without rounding.
kPa
Boyle's law.
✍️ Type your answer · 7 of 11
3 L of gas at 300 K is heated to 600 K at constant pressure.
What is the new volume, in L?
Enter an exact answer, without rounding.
L
Charles's law with kelvin temperatures.
✍️ Type your answer · 8 of 11
A gas at 100 kPa and 300 K occupies 2 L. It is heated to 600 K and expands to 8 L.
What is the final pressure, in kPa?
Enter an exact answer, without rounding.
kPa
Use p₁V₁/T₁ = p₂V₂/T₂ with kelvin.
✍️ Type your answer · 9 of 11
0.2 kg of water at 80 °C is mixed with 0.3 kg of water at 20 °C. Ignore heat losses.
What is the final temperature, in °C?
Enter an exact answer, without rounding.
°C
Heat lost = heat gained.
✍️ Type your answer · 10 of 11
A bubble of volume 2 cm³ rises from a depth of 10 m in water (density 1000 kg/m³) to the surface. Atmospheric pressure is 1.0 × 10⁵ Pa, and the temperature is constant.
What is the volume of the bubble at the surface, in cm³?
Enter an exact answer, without rounding.
cm³
Add atmospheric pressure to ρ g h.
✏️ Quick check · 11 of 11
Kinetic theory
The absolute temperature of a gas is multiplied by 4. What happens to the root-mean-square speed of its molecules?
KE depends on v².
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
During melting, the temperature of a pure substance:
Question 2 of 5
What happens to the pressure of a fixed mass of gas if its volume is halved at constant temperature?
Question 3 of 5
Absolute zero is:
Question 4 of 5
A substance with a high specific heat capacity:
Question 5 of 5
Which method of heat transfer needs no medium?
Lesson 7 · Waves
🌊 Waves & Sound
Waves carry energy without carrying matter. Their speed, frequency and wavelength are linked, and waves can overlap to reinforce or cancel each other.
📡 Wave basics
Transverse waves oscillate perpendicular to the direction of travel (light, water). Longitudinal waves oscillate along it (sound).
v = f λ T = 1/f
sound in air: v = 340 m/s
f = 170 Hz → λ = 340 / 170 = 2 m
Echo: sound returns after 2 s
distance to wall = 340 · 2 / 2 = 340 m
Beats: two notes of 500 Hz and 504 Hz
beat frequency = |504 − 500| = 4 Hz
🖼️ See it
Wave speed v = f × λ. Doubling the frequency halves the wavelength at constant speed.
🎻 Standing waves and the Doppler effect
Waves that reflect and overlap form standing waves with fixed nodes and antinodes. A string of length L fixed at both ends resonates at f_n = n·v/(2L).
L = 1.2 m, wave speed 240 m/s
f₁ = 240 / (2 · 1.2) = 100 Hz
f₂ = 200 Hz, f₃ = 300 Hz
The Doppler effect: if a source moves towards you, the wavefronts bunch up and you hear a higher frequency.
f′ = f · v / (v − v_source) (source approaching)
f = 600 Hz, v_source = 40 m/s, v = 340 m/s
f′ = 600 · 340 / 300 = 680 Hz
🧪 Worked examples: pipes, Doppler shift and interference
Closed pipe. There is a node at the closed end and an antinode at the open end. Only odd harmonics occur: f = v ÷ 4L, then 3f, 5f.
closed pipe L = 0.85 m, v = 340 m/s
f₁ = 340 ÷ (4 × 0.85) = 100 Hz
next resonance = 3 × 100 = 300 Hz
an open pipe of the same length: f₁ = v ÷ 2L = 200 Hz, then 400 Hz, 600 Hz
Doppler effect. For a source moving towards a stationary observer, f′ = f v ÷ (v − vₛ).
f = 600 Hz, vₛ = 40 m/s, v = 340 m/s
f′ = 600 × 340 ÷ (340 − 40) = 680 Hz
Path difference. Constructive interference when the path difference is nλ, destructive when it is (n + ½)λ.
λ = 0.50 m, path difference = 0.75 m = 1.5 λ → destructive (quiet)
String tension. The wave speed is v = √(T ÷ μ). A fundamental of 100 Hz becomes 200 Hz if the tension is multiplied by 4, since v doubles.
Nodes and antinodes. On a string fixed at both ends, the nth harmonic has n antinodes and n + 1 nodes (including the ends).
🎯 Exam focus: explaining standing waves and superposition
A standing wave forms when two waves of the same frequency and amplitude travelling in opposite directions superpose. Energy is not transferred along it.
node zero amplitude, always at rest
antinode maximum amplitude
distance between neighbouring nodes = λ ÷ 2
points between two adjacent nodes oscillate in phase;
points in neighbouring loops oscillate in antiphase
Common mistakes. Using the wrong pipe formula (closed pipes have only odd harmonics). Saying that a standing wave “transfers” energy. Counting the antinodes as the harmonic number on a closed pipe. Using the Doppler formula with the source moving away (the denominator becomes v + vₛ). Forgetting to convert path difference into wavelengths.
Intensity. The intensity is proportional to the amplitude squared, and for a point source it falls with the inverse square of the distance.
✏️ Quick check · 1 of 11
Sound is...
Sound waves in air are:
They need a medium and compress the air.
✏️ Quick check · 2 of 11
Doppler
As an ambulance drives towards you, the siren sounds:
It drops in pitch only after it passes.
✍️ Type your answer · 3 of 11
A sound of frequency 170 Hz travels at 340 m/s.
What is its wavelength, in m?
Enter an exact answer, without rounding.
m
v = fλ.
✍️ Type your answer · 4 of 11
You clap and hear an echo from a wall after 2 s (v = 340 m/s).
How far away is the wall, in m?
Enter an exact answer, without rounding.
m
Remember the round trip.
✍️ Type your answer · 5 of 11
A string 1.2 m long, fixed at both ends, carries waves at 240 m/s.
What is the fundamental frequency, in Hz?
Enter an exact answer, without rounding.
Hz
f₁ = v/(2L).
✍️ Type your answer · 6 of 11
Two tuning forks sound at 500 Hz and 504 Hz together.
What is the beat frequency, in Hz?
Enter an exact answer, without rounding.
Hz
Subtract the two frequencies.
✍️ Type your answer · 7 of 11
A 600 Hz siren moves towards you at 40 m/s (v = 340 m/s).
What frequency do you hear, in Hz?
Enter an exact answer, without rounding.
Hz
f′ = f·v/(v − v_s).
✍️ Type your answer · 8 of 11
A closed pipe of length 0.85 m resonates in air where the speed of sound is 340 m/s.
What is the frequency of the second resonance (the first overtone), in hertz?
Enter an exact answer, without rounding.
Hz
A closed pipe has only odd harmonics.
✍️ Type your answer · 9 of 11
A siren of frequency 600 Hz moves towards a stationary listener at 40 m/s. The speed of sound is 340 m/s.
What frequency does the listener hear, in hertz?
Enter an exact answer, without rounding.
Hz
f′ = f v ÷ (v − vₛ).
✍️ Type your answer · 10 of 11
A stretched string has a fundamental frequency of 100 Hz. The tension is multiplied by 4.
What is the new fundamental frequency, in hertz?
Enter an exact answer, without rounding.
Hz
Use v = √(T/μ).
✏️ Quick check · 11 of 11
Interference
Two loudspeakers emit sound of wavelength 0.50 m in phase. At a point, the path difference is 0.75 m. What do you hear?
Convert the path difference to wavelengths.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The period of a wave with frequency 50 Hz is:
Question 2 of 5
When two waves meet and add up to a larger wave, this is:
Question 3 of 5
At a node of a standing wave the displacement is:
Question 4 of 5
Light is different from sound because light:
Question 5 of 5
If the frequency of a wave doubles while its speed stays the same, the wavelength:
Lesson 8 · Electricity
🔌 Electric Circuits
Electric current is the flow of charge. Ohm's law links voltage, current and resistance, and simple rules let you analyse series and parallel circuits.
⚡ Ohm's law and power
Current I (amperes) flows when a voltage V (volts) is applied across a resistance R (ohms).
V = I R P = V I = I² R = V² / R
V = 12 V, R = 4 Ω → I = 3 A
P = 12 · 3 = 36 W
Energy = P t: 100 W bulb on for 2 h
E = 0.1 kW · 2 h = 0.2 kWh
🖼️ See it
In parallel, 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω. The 12 V cell drives 12 ÷ 2 = 6 A, with 2 A through the 6 Ω resistor and 4 A through the 3 Ω one.
🔗 Series and parallel
In series the same current flows through every component and the voltages add. In parallel every branch has the same voltage and the currents add.
Series: R_total = R₁ + R₂ + … 3 Ω + 6 Ω = 9 Ω
Parallel: 1/R_total = 1/R₁ + 1/R₂ + …
6 Ω ∥ 3 Ω : 1/6 + 1/3 = 1/2 → R = 2 Ω
Kirchhoff: currents entering a junction = currents leaving
5 A in, 2 A out one way → 3 A out the other
Parallel resistance is always smaller than the smallest branch, because adding a branch gives the current more paths.
🧪 Worked examples: internal resistance, dividers and resistivity
Internal resistance. A real cell has an emf ε and an internal resistance r: ε = I (R + r), and the terminal pd is V = ε − I r.
ε = 12 V, r = 1.0 Ω, load R = 5.0 Ω
I = 12 ÷ 6 = 2.0 A
terminal pd V = 12 − 2.0 × 1.0 = 10 V (lost volts = 2 V)
check: V = I R = 2.0 × 5.0 = 10 V
Potential divider. The output across R₂ is V_out = V × R₂ ÷ (R₁ + R₂).
Resistivity. R = ρ L ÷ A: it increases with length and decreases with cross-sectional area.
wire L = 2.0 m, A = 0.50 × 10⁻⁶ m², ρ = 1.0 × 10⁻⁶ Ω m
R = 1.0 × 10⁻⁶ × 2.0 ÷ 0.50 × 10⁻⁶ = 4.0 Ω
halving the diameter makes the area ¼, so R × 4
Series lamps. The lamp with the larger resistance dissipates more power (P = I² R, same I), so it is brighter in series. In parallel the lamp with the smaller resistance is brighter (P = V² ÷ R).
🎯 Exam focus: Kirchhoff’s laws and circuit analysis
Kirchhoff’s first law: the sum of the currents into a junction equals the sum out (conservation of charge). Second law: around a closed loop, the sum of emfs equals the sum of potential differences (conservation of energy).
current–voltage graphs
ohmic conductor straight line through the origin
filament lamp curve flattening: R rises with temperature
diode conducts one way only; ~0.6 V threshold
thermistor (NTC) R falls as the temperature rises
Common mistakes. Saying that the terminal pd equals the emf under load. Forgetting the internal resistance in the series total. Using the diameter instead of the area in R = ρL ÷ A. Writing that current “is used up”. Forgetting that the resistance of a metal increases with temperature while that of a thermistor falls.
Maximum power transfer. The power in the load is greatest when the load resistance equals the internal resistance.
✏️ Quick check · 1 of 11
Series current
In a series circuit, the current is:
There is no junction to split it.
✏️ Quick check · 2 of 11
Parallel voltage
Components connected in parallel to a battery all have:
Think of household sockets.
✍️ Type your answer · 3 of 11
A 12 V battery is connected across a 4 Ω resistor.
What current flows, in A?
Enter an exact answer, without rounding.
A
V = IR.
✍️ Type your answer · 4 of 11
The same circuit carries 3 A from a 12 V supply.
What power is used, in W?
Enter an exact answer, without rounding.
W
P = VI.
✍️ Type your answer · 5 of 11
A 3 Ω and a 6 Ω resistor are connected in series.
What is the total resistance, in Ω?
Enter an exact answer, without rounding.
Ω
Add them.
✍️ Type your answer · 6 of 11
A 6 Ω and a 3 Ω resistor are connected in parallel.
What is the total resistance, in Ω?
Enter an exact answer, without rounding.
Ω
Add the reciprocals.
✍️ Type your answer · 7 of 11
A 100 W bulb is on for 2 hours.
How much energy does it use, in kWh?
Enter an exact answer, without rounding.
kWh
Convert watts to kilowatts first.
✍️ Type your answer · 8 of 11
A cell has an emf of 12 V and an internal resistance of 1.0 Ω. It is connected to a 5.0 Ω resistor.
What is the terminal potential difference, in volts?
Enter an exact answer, without rounding.
V
V = ε − I r.
✍️ Type your answer · 9 of 11
A 12 V supply is connected across a 4 kΩ resistor in series with an 8 kΩ resistor.
What is the potential difference across the 8 kΩ resistor, in volts?
Enter an exact answer, without rounding.
V
Share the voltage in proportion to the resistances.
✍️ Type your answer · 10 of 11
A wire is 2.0 m long, has a cross-sectional area of 0.50 × 10⁻⁶ m² and a resistivity of 1.0 × 10⁻⁶ Ω m.
What is its resistance, in ohms?
Enter an exact answer, without rounding.
Ω
Use R = ρL/A.
✏️ Quick check · 11 of 11
Lamps in series
A 24 Ω lamp and a 6 Ω lamp are connected in series to a battery. Which statement is correct?
Which quantity is the same for both?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Resistance is measured in:
Question 2 of 5
5 A flows into a junction. 2 A leaves along one wire. The current in the other wire is:
Question 3 of 5
Adding another resistor in parallel makes the total resistance:
Question 4 of 5
For a fixed voltage, doubling the resistance makes the current:
Question 5 of 5
An ammeter should be connected:
Final Challenge
🏆 16-Question Challenge Quiz
Questions from every topic in this course. Pick an answer for each one, then hit Check My Answers.
0 / 16
💪 Keep practicing — go back through the lessons and retake the quiz!
Q1 · Kinematics & Projectile Motion
A ball is dropped and another is thrown horizontally from the same height at the same moment. Which hits the ground first?
Q2 · Kinematics & Projectile Motion
For a given launch speed on level ground, the range is greatest at an angle of:
Q3 · Newton's Laws, Friction & Inclines
A 2 kg mass accelerates at 3 m/s². The net force on it is:
Q4 · Newton's Laws, Friction & Inclines
Doubling the net force on an object (same mass) will:
Q5 · Circular Motion & Gravitation
A car takes a curve at double the speed. The centripetal force it needs is multiplied by:
Q6 · Circular Motion & Gravitation
Mass and weight differ because:
Q7 · Work, Energy & Power
A 2 kg object moves at 6 m/s. Its kinetic energy is:
Q8 · Work, Energy & Power
Which statement about efficiency is true?
Q9 · Momentum & Collisions
A 3 kg ball moving at 4 m/s hits a wall and bounces back at 4 m/s. The size of its change in momentum is:
Q10 · Momentum & Collisions
When a rocket expels gas backwards, it moves forwards because:
Q11 · Thermal Physics & Gas Laws
Which Celsius temperature equals 300 K?
Q12 · Thermal Physics & Gas Laws
Which method of heat transfer needs no medium?
Q13 · Waves & Sound
A wave has a frequency of 5 Hz and a wavelength of 4 m. Its speed is:
Q14 · Waves & Sound
If the frequency of a wave doubles while its speed stays the same, the wavelength:
Q15 · Electric Circuits
A 2 Ω resistor carries a current of 3 A. The voltage across it is: