Vectors and Newton laws as a system: resultant forces, equilibrium, and force as the rate of change of momentum
By the end of this lesson you will be able to
- Distinguish scalar and vector quantities and calculate the resultant of two vectors in a line or at right angles, with its size and direction.
- Construct a scale drawing to add two vectors at any angle, and resolve a vector into two perpendicular components.
- State Newton’s three laws in their exam form and apply the first law to objects in equilibrium.
- Explain why the second law is written as force = rate of change of momentum, and derive F = ma from it for constant mass.
- Calculate the force exerted by or on a jet of water, a stream of air or a rocket exhaust from the mass leaving per second and its speed.
- Identify Newton third-law pairs in a system and use them to explain how rockets, walkers and balloon rockets accelerate.
You met this before
- A scalar has size only (distance, speed, mass, time, energy); a vector has size and direction (displacement, velocity, acceleration, force, momentum). Unit 2, Lesson 1.
- The resultant force is the single force that has the same effect as all the forces acting together. Newton’s second law uses it: F = ma, with F in N, m in kg and a in m/s². Unit 2, Lesson 3; Year 3.
- Momentum p = mv, in kg m/s, in the direction of the velocity.
- Impulse: force × time, FΔt, equals the change in momentum Δp. Unit 2, Lesson 5.
- Newton’s third law: when A pushes on B, B pushes on A with a force equal in size, opposite in direction and of the same type. The two forces act on different objects. Unit 2, Lesson 3.
- Weight W = mg, with g = 9.8 N/kg near the Earth’s surface.
A 1 200 kg car speeds up from 10 m/s to 25 m/s in 5.0 s. Calculate its change in momentum and the average resultant force on it.
Reveal
Δp = m(v − u)
Δp = 1 200 kg × (25 − 10) m/s
Δp = 18 000 kg m/s
F = Δp ÷ Δt
F = 18 000 kg m/s ÷ 5.0 s
F = 3 600 N
Check with F = ma: a = 15 m/s ÷ 5.0 s, which is 3.0 m/s², and 1 200 kg × 3.0 m/s² gives the same 3 600 N. This lesson explains why those two routes must always agree.
Revise it in full: Year 3 · Physics · Unit 9 · Lesson 2 — Forces that change motion: F = ma, friction, air resistance and redesign from data · Year 2 · Unit 1 · Lesson 4 — Adding the arrows: balanced forces, resultant force and terminal velocity · Year 3 · Physics · Unit 9 · Lesson 1 — Describing motion
The puzzle
Watch a crew fight a large fire and you will see two or three people leaning into one hose. The water is not pushing on them from outside — it is leaving them. So why does a hose kick backwards hard enough to knock a single firefighter off balance?
A big attack line can deliver about 15 kg of water every second at 30 m/s. Every second, the nozzle gives 15 kg of water a momentum of 450 kg m/s that it did not have before.
Newton’s second law, in the form he actually wrote it, says a force is a rate of change of momentum: 450 kg m/s every second is 450 N. By the third law the water pushes back on the nozzle with 450 N — about the weight of a 46 kg person, pushing the crew backwards.
This lesson rebuilds Newton’s laws around momentum and vectors, so that hoses, rockets, helicopters and walking all become one calculation.
You aim a garden hose at a wall. In one case the water splashes and dribbles down the wall. In another, a curved metal plate on the wall sends the jet straight back towards you at the same speed. Predict which case pushes harder on the wall, and by roughly how much.
Compare your answer
The bounced jet pushes about twice as hard. When the water stops, each kilogram loses its momentum mv. When it bounces straight back, each kilogram goes from +mv to −mv: a change of 2mv. Twice the change in momentum per second means twice the force.
Engineers use exactly this in the Pelton water wheel of hydroelectric schemes: its split, curved buckets turn the jet almost straight back, which roughly doubles the force the water exerts on each bucket.
The science — structured notes
1.1 Scalars, vectors and adding them
A scalar quantity has magnitude (size) only. A vector quantity is described completely only when both its magnitude and its direction are given. Those are the IB’s own definitions and the ones to write in an exam.
| Scalar | Its vector partner | Why the difference matters in dynamics |
|---|---|---|
| distance (m) | displacement (m) | A lap of a track: distance 400 m, displacement 0. |
| speed (m/s) | velocity (m/s) | A car rounding a bend at a steady 20 m/s has a changing velocity, so it is accelerating (Lesson 3). |
| mass (kg) | weight (N) | Weight is a force, pointing towards the centre of the Earth. |
| energy, work, power (J, J, W) | momentum (kg m/s), force (N), acceleration (m/s²) | Momentum can cancel (two equal trolleys moving towards each other have zero total momentum); kinetic energy never cancels, because it is never negative. |
Adding vectors in a line. Choose a positive direction and give every vector a sign. A 30 N push to the right and a 12 N friction force to the left give a resultant of +30 − 12 = +18 N, that is 18 N to the right.
Adding vectors at right angles. Draw them tip to tail; the resultant runs from the start of the first to the tip of the second. Its size comes from Pythagoras and its direction from tan θ.
Two workers drag a crate across a smooth warehouse floor. One pulls with 120 N due east, the other with 50 N due north. Calculate the resultant force and its direction, and the acceleration of the 85 kg crate.
Resultant² = 120² + 50²
Resultant² = 14 400 + 2 500
Resultant² = 16 900 N²
Resultant = 130 N
tan θ = 50 ÷ 120
θ = 23° north of east (22.6°)
a = F ÷ m
a = 130 N ÷ 85 kg
a = 1.5 m/s² in the direction of the resultant
The acceleration is a vector too; it always points along the resultant force, not along either rope.
Any angle: the scale drawing. When the vectors are not at right angles, draw them tip to tail with a ruler and protractor at a stated scale (for example 1 cm to 10 N), draw the resultant, measure its length and angle, and convert back. Use a scale that makes the diagram at least 8 cm across, because a 1 mm error then costs only about 1 %.
Resolving: running the addition backwards. Any vector F at an angle θ to a chosen axis can be replaced by two perpendicular components: F cos θ along the axis and F sin θ across it.
A 200 N pull on a suitcase handle at 30° above the horizontal has a horizontal component of 200 cos 30°, which is 173 N, pulling the case along, and a vertical component of 200 sin 30°, which is 100 N, lifting some of its weight off the wheels.
An aircraft points due north with an airspeed of 250 m/s while a steady wind blows due east at 30 m/s. Calculate its velocity relative to the ground.
Reveal answer
v² = 250² + 30²
v² = 62 500 + 900
v² = 63 400 m²/s²
v = 252 m/s
tan θ = 30 ÷ 250
θ = 6.8° east of north
A velocity without its direction scores only half the marks. Pilots correct for this “drift” by pointing slightly into the wind.
“Forces of 6 N and 8 N act on a ball, so the resultant is 14 N.” Explain when this is right, when it is wrong, and the range of possible answers.
Reveal answer
It is right only if the two forces point the same way. If they point in opposite directions the resultant is 8 − 6, which is 2 N. At right angles it is √(6² + 8²), which is 10 N.
At other angles it lies between 2 N and 14 N. Vectors add by direction as well as size, so the direction must be known before you add.
1.2 The first law: equilibrium is a vector statement
Newton’s first law, in the IB wording: in the absence of a net external force, a body is either at rest or moves in a straight line with constant velocity. “Net” means the vector sum. “External” means forces from outside the object or system — the forces the parts of a system exert on each other always cancel in pairs, by the third law.
An object with zero resultant force is in equilibrium. For three forces this has a neat test: drawn tip to tail, they close into a triangle, because they must add up to nothing.
A 300 N shop sign hangs from two cables, each at 30° above the horizontal, as in the diagram. Find the tension in each cable. Then find it when the cables are fixed flatter, at 10° above the horizontal.
Horizontally: T cos 30° to the left cancels T cos 30° to the right, so the two tensions are equal.
Vertically, upward components balance the weight: 2T sin θ = W
T = W ÷ (2 sin θ)
T = 300 N ÷ (2 × 0.500)
T = 300 N in each cable at 30°
At 10°: T = 300 N ÷ (2 × 0.174)
T = 860 N (864 N)
Nearly three times the weight of the sign. A cable can never be pulled perfectly straight under a load, because its upward component would then be zero — the tension would have to be infinite. That is why washing lines always sag.
“A spacecraft coasting between the planets must keep its engines running, or it will slow down and stop.”
Reveal answer
False. With no net external force, the first law says it carries on at constant velocity. Engines are needed only to change the velocity. (Gravity from the Sun is an external force, which is why probes follow curved paths.)
The May 2025 physics report noted “a widespread misconception that satellites need continuous thrust” — this is that misconception.
1.3 The second law as Newton wrote it: force = rate of change of momentum
The IB states the second law as: the rate of change of momentum is proportional to the applied force. In SI units the constant of proportionality is 1, which is how the newton is defined:
F = Δp ÷ Δt
F = (mv − mu) ÷ t
The force here is the resultant (net) force, and it acts in the direction of the change in momentum.
Getting F = ma back. If the mass does not change, the change in momentum is just the mass times the change in velocity:
F = m(v − u) ÷ t
(v − u) ÷ t = a
F = ma
So F = ma is a special case. It is the one you use for a trolley or a car. F = Δp ÷ Δt is the one you need when the mass itself is moving in or out of the system — a hose, a rocket, a conveyor belt being loaded with sand, a helicopter throwing air downwards.
A 58 g tennis ball arrives at a racket at 20 m/s and leaves in the opposite direction at 40 m/s. The contact lasts 5.0 ms. Calculate the average force of the racket on the ball.
Take the direction the ball leaves as positive: u = −20 m/s, v = +40 m/s
Δp = m(v − u)
Δp = 0.058 kg × (40 − (−20)) m/s
Δp = 0.058 kg × 60 m/s
Δp = 3.48 kg m/s
F = Δp ÷ Δt
F = 3.48 kg m/s ÷ 0.0050 s
F = 700 N (696 N), in the direction the ball leaves
Forgetting that velocity is a vector gives Δp = 0.058 × 20, which is only 1.16 kg m/s and a third of the true force. The reversal is what makes a return hard on the arm.
Δp = m(v − u)
Δp = 0.058 kg × (+40 − (−20)) m/s
Δp = 3.48 kg m/s
Show that the units kg m/s², N and kg m/s ÷ s are all the same, and explain why this means F = Δp ÷ Δt and F = ma cannot disagree for a trolley of fixed mass.
Reveal answer
kg m/s ÷ s is kg m/s², and the newton is defined as the force that gives 1 kg an acceleration of 1 m/s², so 1 N is 1 kg m/s².
For a fixed mass, Δp is mΔv, so Δp ÷ Δt is m × (Δv ÷ Δt), which is ma — the same quantity written two ways, not two different laws.
1.4 Jets, hoses and helicopters: force from mass flow
When fluid flows steadily, the same mass leaves every second. If a mass m leaves each second with a speed v, having started at rest, the momentum given to it each second is mv. So the force on the fluid — and, by the third law, the push back on whatever launched it — is:
F = (mass per second) × (change in velocity)
F = (Δm ÷ Δt) × Δv
A 1 200 kg helicopter hovers. Its rotor drives air straight down at 12 m/s from rest. Calculate the mass of air it must push down every second. The air density is 1.2 kg/m³; calculate the volume of air per second too.
Hovering: net force zero, so upward push of air on rotor = weight
W = mg
W = 1 200 kg × 9.8 N/kg
W = 11 760 N
Force on the air = (Δm ÷ Δt) × Δv, so Δm ÷ Δt = F ÷ Δv
Δm ÷ Δt = 11 760 N ÷ 12 m/s
Δm ÷ Δt = 980 kg/s
Volume per second = 980 kg/s ÷ 1.2 kg/m³
Volume per second = 820 m³/s (817 m³/s)
Nearly a tonne of air every second, a volume larger than several classrooms — which is why hovering burns so much fuel and why the downwash flattens grass. Rotor pushes air down; air pushes rotor up: a third-law pair.
Sand falls vertically onto a horizontal conveyor belt at 40 kg every second. The belt moves at a steady 2.0 m/s. Calculate the extra horizontal force the motor must supply to keep the belt at constant speed.
Reveal answer
Each second, 40 kg of sand must be given a horizontal velocity of 2.0 m/s from zero.
F = (Δm ÷ Δt) × Δv
F = 40 kg/s × 2.0 m/s
F = 80 N
The belt’s speed never changes, so F = ma (with a = 0) would wrongly predict zero. The force is needed because the mass on the belt keeps growing — exactly what the momentum form handles.
1.5 The third law in a system: rockets, walkers and the horse that pulls a cart
Newton’s third law, in the IB wording: action and reaction are opposite and equal. The two forces of a pair act on different objects, are the same type of force, and last for exactly the same time.
That last point is why they give equal and opposite impulses — the key to Lesson 2.
A rocket pushes its exhaust gas backwards; the gas pushes the rocket forwards. Nothing outside is needed to “push against”, which is why rockets work best in the vacuum of space. A walker pushes backwards on the ground with friction; the ground pushes forwards on the walker. On ice, where friction is tiny, that forward force is tiny too.
A rocket of total mass 5.0 × 10⁵ kg burns 2 500 kg of propellant every second and ejects it at 3 000 m/s relative to the rocket. Calculate the thrust, the resultant force and the acceleration at lift-off. Then find the acceleration 100 s later, ignoring air resistance and the small change in g.
Thrust = (Δm ÷ Δt) × Δv
Thrust = 2 500 kg/s × 3 000 m/s
Thrust = 7.5 × 10⁶ N
Weight = 5.0 × 10⁵ kg × 9.8 N/kg
Weight = 4.9 × 10⁶ N
Resultant = 7.5 × 10⁶ N − 4.9 × 10⁶ N
Resultant = 2.6 × 10⁶ N upwards
a = F ÷ m
a = 2.6 × 10⁶ N ÷ 5.0 × 10⁵ kg
a = 5.2 m/s²
After 100 s: mass = 5.0 × 10⁵ − (2 500 × 100)
mass = 2.5 × 10⁵ kg
weight = 2.5 × 10⁵ kg × 9.8 N/kg
weight = 2.45 × 10⁶ N
Resultant = 7.5 × 10⁶ N − 2.45 × 10⁶ N
Resultant = 5.05 × 10⁶ N
a = 5.05 × 10⁶ N ÷ 2.5 × 10⁵ kg
a = 20 m/s²
Same thrust, half the mass, and the weight has fallen too: the acceleration has almost quadrupled (20 ÷ 5.2 ≈ 3.9). This is why crewed rockets throttle back or shut down engines late in a burn, to keep the acceleration the astronauts feel within safe limits.
“The horse pulls the cart forwards, and by the third law the cart pulls the horse backwards with an equal force. So the forces cancel and they can never move.” Explain what is wrong.
Reveal answer
The two forces of a third-law pair act on different objects, so they never cancel each other. To find whether the horse accelerates, add only the forces on the horse.
Those are the cart pulling it back and the ground pushing it forwards (friction on its hooves). The horse pushes back hard on the ground, so the ground’s forward push exceeds the cart’s pull.
For the cart, the horse’s pull exceeds the friction on its wheels. Both have a forward resultant and accelerate together.
A student stands on a bathroom scale in a lift. The scale reads 690 N while the lift is still, and 760 N as it starts moving. Find her mass, the lift’s acceleration and its direction, and identify the third-law partner of the scale’s push on her.
Reveal answer
At rest the scale reads her weight: m = 690 N ÷ 9.8 N/kg
m = 70.4 kg
Resultant = 760 N − 690 N
Resultant = 70 N upwards
a = 70 N ÷ 70.4 kg
a = 0.99 m/s² upwards (the lift is starting to rise, or slowing while going down)
The partner of “scale pushes up on student” is “student pushes down on scale” (760 N). Her weight is not its partner: the partner of her weight is her gravitational pull on the Earth.
1.6 Testing force = rate of change of momentum with light gates
A trolley system of total mass 1.00 kg is pulled by a constant resultant force of 1.0 N along a friction-compensated track. A card on it passes a light gate at 0.447 m/s and a second gate 0.40 m further on at 1.000 m/s. The time between the gates is 0.553 s.
Δp = m(v₂ − v₁)
Δp = 1.00 kg × (1.000 − 0.447) m/s
Δp = 0.553 kg m/s
Δp ÷ Δt = 0.553 kg m/s ÷ 0.553 s
Δp ÷ Δt = 1.00 N
The rate of change of momentum equals the applied resultant force. Double the force and the momentum rises twice as fast — which is what the lab below lets you test.
The same experiment is done on a level track that is not friction-compensated; friction is 0.20 N. With the same 1.0 N pull, what rate of change of momentum would you expect, and how would a graph of Δp ÷ Δt against the applied pull look?
Reveal answer
Resultant = 1.0 N − 0.20 N
Resultant = 0.80 N, so Δp ÷ Δt = 0.80 kg m/s per second
Plotted against the applied pull the line is still straight with gradient 1, but shifted: it cuts the pull axis at 0.20 N instead of passing through the origin. A systematic error, which is exactly why the track is tilted until the trolley runs at a steady speed with no pull.
Try it — force and the rate of change of momentum: a trolley and two light gates
The lab is a friction-compensated track with a trolley pulled by a string over a pulley. The slotted masses that make the pull are moved between the trolley and the hanger, so the total mass accelerated never changes. Two light gates 0.40 m apart give the speeds; the lab reports the acceleration a.
You set the resultant force (0.2–2.0 N). Use rate of change of momentum = M × a, where M is the total mass.
Common mistakes to avoid
- Adding vectors as numbers. 6 N and 8 N make 14 N only if they point the same way; at right angles they make 10 N.
- A vector answer with no direction. “252 m/s” loses the direction mark; write “252 m/s at 6.8° east of north”.
- Ignoring the sign of a velocity. A ball that reverses changes its momentum by m(v + u), not m(v − u) of the speeds.
- Using F = ma when the mass is changing. Hoses, rockets, conveyor belts: use force = mass per second × change in velocity.
- “The forces cancel” for a third-law pair. They act on different objects; only forces on the same object can cancel.
- Calling weight the partner of the normal force. The partner of the floor pushing you up is you pushing the floor down.
- Forgetting “net” and “external” in the first law. A moving object with balanced forces keeps moving; it does not stop.
- A scale drawing that is too small. Use a stated scale and make the diagram at least 8 cm across.
Study smart: how this is examined
Study routine. Today: write the three laws in the IB wording from memory, then redo the crate and tennis EXAMPLES with the book closed. In two days: the helicopter and rocket EXAMPLES, and explain the horse-and-cart paradox aloud in under a minute. In a week: run the trolley lab as a full investigation, plot Δp ÷ Δt against F, and take the quiz.
How we found out — from “quantity of motion” to the moon rocket
René Descartes, in his Principles of Philosophy (1644), argued that God had put a fixed “quantity of motion” into the universe, measured by size times speed. His rules for collisions were mostly wrong, because he treated speed as a scalar: two bodies moving in opposite directions could not, in his scheme, cancel.
In 1668 the Royal Society in London asked for a proper account of collisions. John Wallis, Christopher Wren and Christiaan Huygens answered in 1668–1669, and between them showed that the quantity conserved is mass times velocity with direction — what we call momentum.
Huygens added that in hard, elastic collisions a second quantity, mass times speed squared, is also conserved. That quantity grew into kinetic energy (Lesson 4).
Isaac Newton published his three laws in the Principia in 1687. His second law says that the change of motion is proportional to the motive force impressed, and his “motion” was defined as quantity of matter times velocity — our momentum. The familiar F = ma was written in that form later, in the eighteenth century, notably by Leonhard Euler.
Newton gave the third law, and the collision results of Wallis, Wren and Huygens as its evidence, in the same book.
Rockets tested the laws at a new scale. In 1903 Konstantin Tsiolkovsky, a Russian schoolteacher, worked out how a rocket’s final speed depends on its exhaust speed and how much of its mass is fuel. In the United States Robert Goddard launched the first liquid-fuelled rocket on 16 March 1926 in Auburn, Massachusetts.
In 1920 a New York Times editorial had mocked Goddard, claiming a rocket could not work in a vacuum because there would be nothing to push against. The newspaper printed a correction on 17 July 1969, while Apollo 11 was on its way to the Moon — the third law had been right all along.
Use Newton’s third law to explain why the 1920 editorial was wrong, and why a rocket actually works slightly better in a vacuum than in air.
Reveal answer
The rocket does not push on the air; it pushes on its own exhaust gas, forcing it backwards. By the third law the gas pushes the rocket forwards with an equal force. Nothing outside the rocket–exhaust system is involved, so a vacuum is no obstacle.
In air, the surrounding pressure pushes back on the exhaust as it leaves the nozzle and slightly reduces the thrust. In space that back-pressure is gone, so rocket engines deliver a little more thrust there.
Where it is used now
- Firefighting and dust control. High-pressure water cannons on fire engines and at mine sites are mounted on heavy bases or vehicles because the jet’s reaction force, mass per second × speed, can reach hundreds or thousands of newtons.
- Hydroelectric turbines. A Pelton wheel’s cupped buckets reverse the water jet, roughly doubling the momentum change and the force; they suit the high heads of water found in mountain schemes, for example in the Alps and Norway.
- Mining conveyors. Conveyor belts carry ore at the large open-pit diamond mines of Botswana. Every tonne dropped onto a moving belt must be accelerated to belt speed, and the drive motors are sized with F = (mass per second) × (belt speed) plus friction.
- Rockets and satellites. Botswana’s first satellite, BOTSAT-1, built with the Botswana International University of Science and Technology, was launched in 2025 on a rideshare rocket; every launch vehicle is designed by working out the thrust from its propellant flow rate and exhaust speed.
- Sport science. Force plates record the force on a sprinter’s foot many hundreds of times a second; the area under the force–time graph is the impulse, the change in momentum, and it predicts the start speed.
- Drones. A delivery drone hovers only if its rotors give enough air enough downward momentum each second to match its weight — the helicopter EXAMPLE, scaled down, and the reason battery life falls when it carries a heavy package.
Science now — and the scientist you need to be
What is happening in 2026
Newton’s laws are the working equations of the reusable-rocket era. Orbital-class boosters now return and land upright on their own engines, and some individual boosters have flown more than twenty times. Every landing is a live solution of the second law with a changing mass: the guidance computer throttles the thrust so the booster’s momentum reaches zero exactly at the landing pad.
Much smaller, the same physics runs medical-delivery drones, which in several African countries fly blood and vaccines to rural clinics, and the autonomous tractors and crop-spraying drones entering commercial farming.
Engineers increasingly design these vehicles with simulation software and machine-learning controllers, but the controllers are trained and checked against Newton’s laws.
Careers here include aerospace and mechanical engineering, robotics, drone operation, sports biomechanics and physics teaching. All of them start with a clean free-body diagram and a sign convention.
The mindshift
What you need to be able to do
Give every vector a size and a direction. Draw one object at a time with every force on it named. Write a sign convention before any momentum change. Choose F = ma for a fixed mass and “mass per second × change in velocity” for a flow. And name the law you are using, because the examiners give marks for the name as well as the physics.
See an example
First law: before release the balloon is at rest, with no net force. Second law: once the neck opens, the resultant forward force makes it accelerate; as the air runs out the thrust falls and friction on the straw slows it. Third law: the balloon squeezes air backwards out of the neck; the air pushes the balloon forwards with an equal and opposite force.
A typical run: 3.0 m in about 1.5 s. The AI answer should mention both forces on the horse (ground forwards, cart backwards); if it says the pair “cancels”, it is wrong.
Key terms
| Scalar quantity | A quantity that has magnitude (size) only, such as mass, distance, speed, energy or time. |
| Vector quantity | A quantity described completely only if both its magnitude and its direction are given, such as displacement, velocity, acceleration, force or momentum. |
| Resultant (net) force | The vector sum of all the forces on an object; the single force with the same effect. |
| Components | Two perpendicular vectors, F cos θ and F sin θ, that together have the same effect as a vector F. |
| Equilibrium | The state of an object with zero resultant force (at rest or at constant velocity); for three forces, a closed vector triangle. |
| Newton first law | In the absence of a net external force, a body is either at rest or moves in a straight line with constant velocity. |
| Newton second law | The rate of change of momentum is proportional to the applied force: F = Δp ÷ Δt, which gives F = ma for constant mass. |
| Newton third law | Action and reaction are opposite and equal: the two forces act on different objects, are of the same type and act for the same time. |
| Momentum | The product of an object’s mass and its velocity, p = mv, in kg m/s; a vector. |
| Thrust | The forward force on a rocket, jet or propeller, equal to the mass ejected per second × the change in its velocity. |
Think like a scientist
Criterion C A student investigates the force of a water jet. Water leaves a nozzle of cross-sectional area 2.0 × 10⁻⁵ m² and strikes a flat plate standing on a top-pan balance. It spreads sideways and runs off, so each kilogram loses all its downward velocity. She sets the mass flow rate Q with a tap, measures it by collecting water for 30 s, and records how much the balance reading rises.
| Mass flow rate Q / kg/s | Rise in balance reading trial 1 / g | trial 2 / g | trial 3 / g |
|---|---|---|---|
| 0.020 | 2.0 | 2.1 | 2.0 |
| 0.040 | 8.1 | 8.3 | 8.2 |
| 0.060 | 18.6 | 18.2 | 18.5 |
| 0.080 | 32.5 | 33.1 | 38.9 |
| 0.100 | 51.2 | 50.6 | 51.0 |
Her hypothesis: “If the flow rate is doubled, the force of the jet will double, because force is the rate of change of momentum.”
(a) Identify the anomalous result, suggest a specific cause, and calculate the mean rise for each flow rate. [3]
(b) Convert each mean into the force of the jet in N (g = 9.8 N/kg). [2]
(c) The jet speed is v = Q ÷ (ρA), with a water density ρ of 1 000 kg/m³. Show that the force should be F = Q² ÷ (ρA), and explain which graph you would plot to test this. [3]
(d) Calculate F ÷ Q² for each row, and the gradient of the graph from (c) with its unit, and compare it with 1 ÷ (ρA). [3]
(e) Evaluate the student’s hypothesis using the data. [2]
(f) Explain one systematic error in this method and its effect on the gradient, and suggest an improvement. [3]
Model answer
(a) Trial 3 at 0.080 kg/s (38.9 g) is anomalous — about 6 g above the other two [1]. Cause: water pooled on the plate for that run (for example a blocked drain edge), adding its weight to the reading; or the tap was nudged to a higher flow [1].
Means [1]: 2.03 g; 8.20 g; 18.43 g; 32.80 g (anomaly excluded); 50.93 g.
(b) F = (mean rise in kg) × 9.8 N/kg [1]:
2.03 × 10⁻³ kg × 9.8 N/kg = 0.0199 N
8.20 × 10⁻³ kg × 9.8 N/kg = 0.0804 N
18.43 × 10⁻³ kg × 9.8 N/kg = 0.181 N
32.80 × 10⁻³ kg × 9.8 N/kg = 0.321 N
50.93 × 10⁻³ kg × 9.8 N/kg = 0.499 N [1]
(c) Force = mass per second × change in velocity, and the water is brought to rest, so F = Qv [1].
v = Q ÷ (ρA)
F = Q × Q ÷ (ρA)
F = Q² ÷ (ρA) [1]
F against Q would curve upwards, so plot F (vertical, N) against Q² (horizontal, kg²/s²): a straight line through the origin if the theory holds [1].
(d) Q² values: 4.0 × 10⁻⁴, 1.6 × 10⁻³, 3.6 × 10⁻³, 6.4 × 10⁻³, 1.0 × 10⁻² kg²/s². F ÷ Q²: 49.8, 50.2, 50.2, 50.2, 49.9 [1].
Gradient from (0, 0) and (1.0 × 10⁻², 0.499) = 0.499 ÷ 0.010
Gradient = 50 m/kg (N s²/kg² reduces to m/kg) [1]
1 ÷ (ρA) = 1 ÷ (1 000 × 2.0 × 10⁻⁵)
1 ÷ (ρA) = 50 m/kg — agreement within 1 % [1]
(e) Only partly supported [1]: doubling Q from 0.040 to 0.080 kg/s raised the force from 0.0804 N to 0.321 N — four times, not twice. The force is proportional to Q², because doubling the flow doubles both the mass per second and the speed. The reasoning (rate of change of momentum) was right; the prediction was not [1].
(f) Water that stays on the plate adds its weight to the reading [1]. A larger flow leaves more water standing, so the high-Q readings are too large and the gradient is overestimated [1].
Improvement: use a plate with raised edges and a drain spout off the balance, or a cone that turns the jet sideways and away; zero the balance with the plate wet [1]. (Also accepted: water splashing back upwards increases Δv beyond v and the force above Qv.)
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Select the two vectors from this list and state why the others are scalars: speed · momentum · kinetic energy · mass · force · time.
Reveal answer
Momentum and force [1]. Speed, kinetic energy, mass and time have a size but no direction, so they are scalars [1].
Forces of 5.0 N due east and 12 N due north act on a puck on ice. Calculate the resultant force and its direction.
Reveal answer
F² = 5.0² + 12²
F² = 169 N²
F = 13 N [1]
tan θ = 12 ÷ 5.0
θ = 67° north of east [1] (67.4°)
Direction in words with a reference [1]; “67°” alone scores 0 for that mark.
A 1 500 kg car speeds up from 12 m/s to 27 m/s in 6.0 s. Calculate the change in momentum and the average resultant force.
Reveal answer
Δp = 1 500 kg × (27 − 12) m/s
Δp = 22 500 kg m/s [1]
F = Δp ÷ Δt
F = 22 500 ÷ 6.0 [1]
F = 3 750 N ≈ 3 800 N [1]
When the neck of a balloon threaded on a string is released, the balloon accelerates along the string. Explain its motion using Newton’s laws.
Reveal answer
The balloon squeezes air backwards out of its neck; the air pushes the balloon forwards with an equal and opposite force [1]. This is Newton’s third law — an action–reaction pair [1]. The forward thrust is larger than the friction on the string, so there is a resultant force and, by the second law (F = ma), the balloon accelerates [1].
A water jet delivers 3.0 kg of water per second at 8.0 m/s onto a wall, where the water stops. Calculate the force on the wall, and state how it would change if the water bounced straight back at 8.0 m/s.
Reveal answer
F = (Δm ÷ Δt) × Δv
F = 3.0 kg/s × 8.0 m/s
F = 24 N [1]
Bouncing back, Δv is 16 m/s, so the force doubles to 48 N [1].
True or false, with a reason: “A skydiver falling at terminal velocity has no forces acting on her.”
Reveal answer
False [1]. Her weight and the air resistance both act; they are equal and opposite, so the resultant force is zero and, by the first law, her velocity is constant [1].
7. A 2.4 kg drone hovers by driving air straight down from rest. Its rotors sweep a total area of 0.30 m² and the air density is 1.2 kg/m³. Taking the air leaving at speed v, the mass per second is ρAv. Find v, and the mass of air pushed down each second.
Reveal answer
Force on air = (mass per second) × v, and mass per second is ρAv [1].
ρAv × v = weight
ρAv² = W
W = 2.4 kg × 9.8 N/kg
W = 23.5 N
v² = W ÷ (ρA)
v² = 23.5 ÷ (1.2 × 0.30)
v² = 65.3 m²/s² [1]
v = 8.1 m/s [1]
mass per second = 1.2 × 0.30 × 8.1
mass per second = 2.9 kg/s [1]
Check: 2.9 kg/s × 8.1 m/s gives 23.5 N, the weight.
8. A small rocket of initial mass 1 200 kg ejects 8.0 kg of gas per second at 2 000 m/s. Calculate its acceleration at launch and when half its mass has been used, ignoring air resistance. Explain why the acceleration is not constant even though the thrust is.
Reveal answer
Thrust = 8.0 kg/s × 2 000 m/s
Thrust = 16 000 N
Launch: resultant = 16 000 N − (1 200 kg × 9.8 N/kg)
resultant = 16 000 N − 11 760 N
resultant = 4 240 N
a = 4 240 N ÷ 1 200 kg
a = 3.5 m/s² [1]
At 600 kg: resultant = 16 000 N − 5 880 N
resultant = 10 120 N
a = 10 120 N ÷ 600 kg
a = 17 m/s² [1]
The thrust is fixed by the flow rate and exhaust speed, but the rocket’s mass keeps falling as fuel is burned [1]. A smaller mass needs less force for the same acceleration, and the weight to overcome is smaller too, so the resultant rises and the acceleration grows throughout the burn [1].
Summary
- A scalar has size only; a vector needs size and direction. Add vectors tip to tail: in a line with signs, at right angles with Pythagoras and tan θ, at any angle by a scale drawing. Resolve F into F cos θ and F sin θ.
- First law: with no net external force a body stays at rest or at constant velocity. Equilibrium means zero resultant; three forces in equilibrium close into a triangle.
- Second law: force is the rate of change of momentum, F = Δp ÷ Δt.
- For a fixed mass this becomes F = ma; for a flow, force is the mass per second × the change in velocity.
- Third law: action and reaction are opposite and equal, act on different objects, are of the same type and last for the same time — so they never cancel.
- Rockets, hoses, helicopters and conveyor belts are all “momentum per second” problems; a reversal (bounce) doubles the change in momentum and the force.
Next time → Conservation of momentum: why the total momentum of a system is the same before and after any collision or explosion, sticky and bouncy collisions, recoil, and the new collision lab.