Edexcel IGCSE Physics 1(c) Forces, Movement, Shape and Momentum — Notes
Written by an examiner for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular. The units for Topic 1 (N, N/kg, N m, kg m/s) are on the 1(b) page.
Which parts do I need?
Single Award (4SS0) — in your specification this sub-topic is called “Forces and movement”. You need: the effects of forces, types of force, friction, \(F=ma\), \(W=mg\) and stopping distance (1.11, 1.12, 1.16–1.20). Skip scalars and vectors and resultant force (1.13–1.15), falling objects and terminal velocity (1.21), the spring practical, Hooke's law and elastic behaviour (1.22–1.24), and everything on momentum, Newton's third law and moments (1.25P–1.33P).
Double Award (4SD0) — you need everything except the Physics-only sections on momentum, safety features, Newton's third law, moments, centre of gravity and beams (1.25P–1.33P).
What this page covers
| Spec | You need to be able to… | Route |
|---|---|---|
| 1.11 | describe how forces between objects can change their speed, direction of motion or shape | Single · Double · Physics |
| 1.12 | name different kinds of force, for example gravitational and electrostatic | Single · Double · Physics |
| 1.13 | explain the difference between a vector and a scalar | Double · Physics |
| 1.14 | know that a force has direction as well as size, so it is a vector | Double · Physics |
| 1.15 | combine forces acting along one line into a single resultant force | Double · Physics |
| 1.16 | know that friction acts against motion | Single · Double · Physics |
| 1.17 | recall and use \(F=ma\) for an unbalanced force | Single · Double · Physics |
| 1.18 | recall and use \(W=mg\) | Single · Double · Physics |
| 1.19 | know that stopping distance is thinking distance plus braking distance | Single · Double · Physics |
| 1.20 | describe how speed, mass, road surface and reaction time affect stopping distance | Single · Double · Physics |
| 1.21 | describe the forces on a falling object and explain why it reaches terminal velocity | Double · Physics |
| 1.22 | investigate how the extension of springs, metal wires and rubber bands depends on the force (practical) | Double · Physics |
| 1.23 | link the straight first part of a force–extension graph to Hooke's law | Double · Physics |
| 1.24 | describe elastic behaviour: returning to the original shape once the deforming forces are removed | Double · Physics |
| 1.25P | recall and use \(p=mv\) | Physics only · 2P |
| 1.26P | explain how safety features work using momentum | Physics only · 2P |
| 1.27P | use conservation of momentum to find an unknown mass, velocity or momentum | Physics only · 2P |
| 1.28P | use force = change in momentum ÷ time taken | Physics only · 2P |
| 1.29P | apply Newton's third law to pairs of interacting objects | Physics only · 2P |
| 1.30P | recall and use moment = force × perpendicular distance from the pivot | Physics only · 2P |
| 1.31P | know that weight acts through the centre of gravity | Physics only · 2P |
| 1.32P | apply the principle of moments to parallel forces in one plane | Physics only · 2P |
| 1.33P | explain how the support forces on a light beam change as a heavy load moves along it | Physics only · 2P |
Modular candidates: 1.11–1.33P are the same numbers in 4XPH1 Unit 1 (4WPH1); 1.11–1.24 are in 4XSD1 Physics Unit 5 (4WSD5).
In one minute
- Forces can change an object's speed, direction or shape. Friction always opposes motion.
- An unbalanced force causes acceleration: \(F=ma\). Weight is a force: \(W=mg\), with \(g=10\) N/kg on Earth.
- Stopping distance = thinking distance + braking distance.
- Force is a vector; forces along a line combine into a resultant. A falling object reaches terminal velocity when air resistance equals weight. Double · Physics
- In the straight first part of a force–extension graph, extension is proportional to force (Hooke's law). Double · Physics
- Momentum \(p=mv\) is conserved in collisions; \(F=\dfrac{mv-mu}{t}\); moment = force × perpendicular distance; in balance, clockwise moments = anticlockwise moments. Physics only · 2P
What can forces do, and what types of force are there? Single · Double · Physics
A force is a push or a pull that one object exerts on another. Forces always act between two objects, and they are measured in newtons (N).
Effects of a force
- Change of speed — a force can make an object speed up, slow down, start moving or stop.
- Change of direction — a force can make a moving object turn, as when a footballer heads a ball.
- Change of shape — a force can stretch, squash, bend or twist an object.
One force can do more than one of these at once: a racket hitting a tennis ball squashes the ball and reverses its direction.
Types of force
| Force | Acts between… | Example |
|---|---|---|
| gravitational force (weight) | any two masses; no contact needed | the Earth pulling you downwards |
| electrostatic force | charged objects; no contact needed | a charged balloon attracting small pieces of paper |
| magnetic force | magnets and magnetic materials; no contact needed | a magnet picking up an iron nail |
| friction | surfaces in contact that slide, or try to slide, over each other | a box being pushed across a floor |
| air resistance (drag) | an object and the air (or water) it moves through | a cyclist riding into the wind |
| normal reaction (contact force) | a surface and an object pressing on it | a table pushing up on a book |
| tension | a stretched string, rope or spring and the object it is attached to | a rope towing a car |
| upthrust | a fluid and an object in it | water pushing up on a floating boat |
Friction
- Friction between surfaces in contact heats them — rub your hands together.
- Air resistance and water resistance (drag) are forms of friction. They increase as the speed increases.
- Friction is not always unwanted: tyres need friction with the road to grip, and brakes use friction to slow the wheels.
How is a vector different from a scalar, and how do you find a resultant force? Double · Physics
Not required for Single Award (4SS0).
| Scalars (size only) | Vectors (size and direction) |
|---|---|
| distance, speed, mass, time, energy, temperature | force (including weight and friction), velocity, acceleration, displacement |
- Force is a vector: “20 N” is incomplete; “20 N to the left” or “20 N downwards” describes the force fully. On a diagram, the arrow's direction shows the direction and its length shows the size.
- Speed and velocity show the difference clearly: 15 m/s is a speed; 15 m/s due north is a velocity.
Resultant force along a line
- Choose one direction as positive. Add forces that act in the same direction; subtract forces that act in the opposite direction.
- Always give the direction of the resultant as well as its size.
- If the forces are balanced, the resultant is zero: the object stays at rest, or keeps moving at a constant velocity.
How are force, mass and acceleration related? Single · Double · Physics
An unbalanced (resultant) force makes an object accelerate in the direction of that force. A bigger force gives a bigger acceleration; a bigger mass gives a smaller acceleration for the same force.
\[ \text{force}=\text{mass}\times\text{acceleration}\qquad F=m\times a \]- \(F\) = unbalanced force (N), \(m\) = mass (kg), \(a\) = acceleration (m/s²).
- 1 N is the force that gives a mass of 1 kg an acceleration of 1 m/s².
- If the forces are balanced, \(a=0\): the object is either at rest or moving at constant velocity. It does not need a resultant force to keep moving.
- Rearranged: \(a=\dfrac{F}{m}\) and \(m=\dfrac{F}{a}\).
Single Award: you need \(F=ma\) with the unbalanced force; combining several forces into one resultant first (as in worked example 2) is Double and Physics content.
(a) An unbalanced force of 6.0 N acts on a trolley of mass 2.5 kg. Calculate the acceleration of the trolley.
(b) A car of mass 1200 kg accelerates at 1.5 m/s². Calculate the unbalanced force on the car.
(a) \(a=\dfrac{F}{m}=\dfrac{6.0}{2.5}=\) 2.4 m/s²
(b) \(F=m\times a=1200\times1.5=\) 1800 N
A motor boat of mass 1400 kg has the horizontal forces shown in the diagram acting on it.
(a) Calculate the resultant force on the boat. (b) Calculate the acceleration of the boat. (c) The engine thrust is reduced to 1450 N while the other forces stay the same. Explain what happens to the motion of the boat.
(a) Taking forwards as positive: 1800 − 1200 − 250 = 350 N forwards
(b) \(a=\dfrac{F}{m}=\dfrac{350}{1400}=\) 0.25 m/s² (forwards)
(c) 1450 − 1200 − 250 = 0, so the forces are balanced and the acceleration is zero. The boat keeps moving forwards at a constant speed — it does not stop.
What is the difference between mass and weight? Single · Double · Physics
- \(W\) = weight (N), \(m\) = mass (kg), \(g\) = gravitational field strength (N/kg).
- On Earth, \(g\) is about 10 N/kg. Pearson papers tell you to assume \(g=10\) N/kg; answers using 9.8 N/kg are also accepted.
- Mass does not change from place to place. Weight changes if \(g\) changes, for example on the Moon or another planet.
- Weight is measured with a newtonmeter (force meter); mass is measured with a balance.
An astronaut and her spacesuit have a total mass of 120 kg. The gravitational field strength on the Moon is 1.6 N/kg.
(a) Calculate the weight of the astronaut and suit on Earth. (b) Calculate their weight on the Moon. (c) State their mass on the Moon.
(a) \(W=m\times g=120\times10=\) 1200 N
(b) \(W=120\times1.6=192\ \text{N}\approx\) 190 N
(c) 120 kg — mass does not depend on where the object is.
What affects the stopping distance of a vehicle? Single · Double · Physics
- Thinking distance — the distance travelled during the driver's reaction time, between seeing a hazard and pressing the brake. The car is still moving at its original speed, so thinking distance = speed × reaction time.
- Braking distance — the distance travelled after the brakes are applied, while the car decelerates to rest.
| Factor | Affects thinking distance? | Affects braking distance? |
|---|---|---|
| greater speed | yes — more distance covered during the reaction time | yes — more kinetic energy to remove, so braking takes a longer distance |
| greater mass (passengers, luggage) | no | yes — for the same braking force, the deceleration is smaller (\(a=F/m\)) |
| poor road condition (wet, icy, loose gravel, oil) | no | yes — less friction between tyres and road |
| longer reaction time (tiredness, alcohol, drugs, using a phone) | yes | no |
| worn tyres or worn brakes | no | yes — less grip or a smaller braking force |
A car travels at 20 m/s. The driver's reaction time is 0.70 s and the braking distance is 30 m.
(a) Calculate the thinking distance. (b) Calculate the stopping distance. (c) The driver is tired and his reaction time becomes 1.2 s. Calculate the new stopping distance. (d) Explain why rain on the road would change the braking distance but not the thinking distance.
(a) thinking distance = speed × reaction time = 20 × 0.70 = 14 m
(b) stopping distance = 14 + 30 = 44 m
(c) thinking distance = 20 × 1.2 = 24 m; braking distance is unchanged, so stopping distance = 24 + 30 = 54 m
(d) A wet road reduces the friction between the tyres and the road, so the braking force is smaller and the car takes a greater distance to stop. Thinking distance depends only on the speed and the driver's reaction time, and the road surface affects neither.
Why does a falling object reach a terminal velocity? Double · Physics
Not required for Single Award (4SS0).
Two forces act on an object falling through air: its weight (downwards, constant) and air resistance (upwards, increasing with speed).
- A — just released: the speed is zero, so there is no air resistance. The resultant force is the weight, and the object accelerates at about 10 m/s².
- B — speeding up: as the speed increases, air resistance increases. The resultant force (weight − air resistance) gets smaller, so the acceleration decreases.
- C — terminal velocity: air resistance has grown until it equals the weight. The resultant force is zero, the acceleration is zero, and the object falls at a constant velocity — its terminal velocity.
- An object with a larger surface area (for example a skydiver with an open parachute) has more air resistance at a given speed, so it reaches a lower terminal velocity.
- At terminal velocity there are still forces acting — they are balanced, not absent.
Terminal velocity simulation: drop an object, change its mass and area, and watch air resistance grow until it balances the weight.
Practical: how does extension depend on force for springs, wires and rubber bands? Double · Physics
Not required for Single Award (4SS0).
The specification asks you to investigate three materials: a helical spring, a metal wire and a rubber band. The method is the same for each — add known forces and measure how much the sample stretches.
Investigating how the extension of a spring varies with the applied force
- Aim: find how the extension of a helical spring depends on the force stretching it.
- Apparatus: clamp stand, boss and clamp, G-clamp, helical spring, mass hanger and 100 g slotted masses, metre rule clamped vertically, pointer (or a set square), safety goggles.
- Independent variable: the force applied (the weight of the masses: each 100 g mass adds 0.1 kg × 10 N/kg = 1 N).
- Dependent variable: the extension of the spring.
- Control variables: the same spring, the same temperature, adding the masses gently so the spring is not jerked.
Method
- Hang the spring from the clamp and attach the pointer to its lower end. Record the pointer reading on the metre rule with no load — this gives the original length.
- Add the hanger plus one 100 g mass, wait for the spring to stop moving, and record the new reading.
- Continue adding masses one at a time, recording the reading each time, for at least six loads.
- Calculate each extension by subtracting the original reading.
- Remove the masses one at a time and check whether the spring returns to its original length.
- Plot force (vertical axis) against extension (horizontal axis) and draw a line or curve of best fit.
Adapting the method
- Metal wire: extensions are tiny, so use a long wire (two metres or more) clamped at one end, passing over a pulley with the masses hung from the other end. A small marker stuck to the wire is read against a fixed scale.
- Rubber band: hang it like the spring and measure its length directly with the metre rule. Rubber bands may not return exactly to their original length, so record readings while unloading as well.
Errors and improvements
- Parallax error: read the scale with your eye level with the pointer; use a set square to line the pointer up with the rule.
- Measure the extension, not the length — a very common slip when processing results.
- Wait for oscillations to stop before reading; repeat the whole set of readings and calculate means.
- Use smaller steps of force to locate where the graph stops being straight.
- Safety: wear safety goggles in case a spring, wire or band snaps; put a box of sand or a cushion under the masses; use a G-clamp so the stand cannot topple; do not stand with feet under the masses.
Some questions plot extension on the vertical axis and force on the horizontal axis. Either way, a straight line through the origin shows that extension is proportional to force.
What does a force–extension graph tell you? Hooke's law Double · Physics
- The initial straight-line region through the origin of a force–extension graph is where Hooke's law applies: double the force, double the extension.
- Beyond the end of the linear region the graph curves and extension is no longer proportional to force.
- Springs and metal wires have a clear linear region. A rubber band's graph is curved from the start, so it does not obey Hooke's law.
A student hangs loads from a spring and records its length.
| Force / N | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
|---|---|---|---|---|---|---|
| Length / mm | 40 | 48 | 56 | 64 | 72 | 83 |
(a) Determine whether the spring obeys Hooke's law for all the loads. (b) Calculate the length of the spring when a force of 2.5 N is applied.
(a) Extensions (length − 40 mm): 0, 8, 16, 24, 32, 43 mm. Up to 4.0 N the extension rises by 8 mm for every 1.0 N, so it is proportional to force. At 5.0 N, proportionality predicts 40 mm but the extension is 43 mm. The spring obeys Hooke's law up to 4.0 N but not at 5.0 N.
(b) 2.5 N is inside the linear region: extension = 2.5 × 8 = 20 mm, so length = 40 + 20 = 60 mm.
Common error: working with the lengths instead of the extensions. Lengths (40, 48, 56…) are not proportional to force, because the spring has a length of 40 mm with no load.
What is elastic behaviour? Double · Physics
- A spring stretched only a little behaves elastically: unload it and it goes back to its original length.
- If a spring or wire is stretched too far, it is permanently deformed — it stays longer than before when the load is removed. Its behaviour is then no longer elastic.
- The practical tests this directly: remove the loads one by one and check whether the extension returns to zero.
- “Obeys Hooke's law” and “behaves elastically” are different ideas: a rubber band behaves elastically over a wide range even though its graph is never straight.
What is momentum and when is it conserved? Physics only · 2P
International GCSE Physics only (Paper 2P). Not in Double or Single Award.
\[ \text{momentum}=\text{mass}\times\text{velocity}\qquad p=m\times v \]- \(p\) = momentum (kg m/s), \(m\) = mass (kg), \(v\) = velocity (m/s).
- Momentum is a vector: it has the same direction as the velocity. Choose a positive direction; momentum in the opposite direction is negative.
- Collisions: add up the momentum of every object before, then after, and set them equal.
- Objects that stick together move off with one common velocity; use their combined mass.
- Explosions and recoil: if everything starts at rest, the total momentum is zero before, so afterwards the two parts have equal momentum in opposite directions (a gun recoils as the bullet leaves).
Trolley A, mass 0.60 kg, moves at 0.50 m/s towards trolley B, mass 0.40 kg, which is at rest. The trolleys stick together.
(a) Calculate the momentum of trolley A before the collision. (b) Calculate the velocity of the trolleys after the collision.
(a) \(p=m\times v=0.60\times0.50=\) 0.30 kg m/s
(b) momentum before = momentum after: \(0.30=(0.60+0.40)\times v\)
\(v=\dfrac{0.30}{1.00}=\) 0.30 m/s in A's original direction
How is force linked to change in momentum, and how do safety features use this? Physics only · 2P
\[ \text{force}=\frac{\text{change in momentum}}{\text{time taken}}\qquad F=\frac{(mv-mu)}{t} \]- \(F\) = force (N), \(mv\) = final momentum, \(mu\) = initial momentum (kg m/s), \(t\) = time over which the change happens (s).
- Force is the rate of change of momentum. Because \(\dfrac{mv-mu}{t}=m\times\dfrac{v-u}{t}\), this is the same physics as \(F=ma\).
- For a fixed change in momentum, a longer time means a smaller force.
Safety features
In a crash, a passenger's momentum must fall to zero whatever happens. Safety features cannot change that change in momentum, but they make it happen over a longer time, so the force on the person is smaller and injuries are less serious.
| Feature | How it increases the stopping time |
|---|---|
| crumple zones | the front and back of the car crush gradually, so the car takes longer to stop |
| seat belts | stretch slightly, so the passenger slows down over a longer time than if they hit the windscreen |
| airbags | inflate and compress as the head hits them, so the head stops over a longer time |
| cycle and motorcycle helmets | the foam lining crushes on impact, lengthening the time for the head to stop |
| soft surfaces (gym mats, playground surfaces, collapsible crash barriers) | deform on impact, so the falling or moving body stops more gradually |
For example, a 60 kg passenger travelling at 12 m/s has 720 kg m/s of momentum to lose. Stopped by a seat belt in 0.080 s, the average force is 720 ÷ 0.080 = 9000 N; stopped by a dashboard in 0.008 s, it would be 90 000 N — ten times larger.
Full marks for a safety-feature explanation usually need the chain: the feature increases the time → for the same change in momentum → the rate of change of momentum is smaller → so the force is smaller.
Ball A, mass 0.20 kg, moving at 3.0 m/s, hits ball B, mass 0.50 kg, which is at rest. After the collision A moves back at 1.0 m/s. The balls are in contact for 0.040 s.
(a) Calculate the velocity of B after the collision. (b) Calculate the average force on B during the collision. (c) State the size and direction of the average force on A, and explain your answer.
(a) Right is positive, so A's velocity after is −1.0 m/s.
momentum before = 0.20 × 3.0 + 0 = 0.60 kg m/s
momentum after = 0.20 × (−1.0) + 0.50 × \(v\) = −0.20 + 0.50\(v\)
0.60 = −0.20 + 0.50\(v\) ⇒ \(v=\dfrac{0.80}{0.50}=\) 1.6 m/s to the right
(b) \(F=\dfrac{mv-mu}{t}=\dfrac{0.50\times1.6-0}{0.040}=\dfrac{0.80}{0.040}=\) 20 N to the right
(c) 20 N to the left. By Newton's third law, the force A exerts on B and the force B exerts on A are equal in size and opposite in direction. Check: change in momentum of A = 0.20 × (−1.0 − 3.0) = −0.80 kg m/s, and −0.80 ÷ 0.040 = −20 N.
Common error in (a): using +1.0 m/s for A after the collision, which gives 0.80 m/s for B. The minus sign for the reversed direction is essential.
What does Newton's third law say? Physics only · 2P
- The two forces of a third-law pair act on different objects, so they never cancel each other out.
- They are always the same type of force (both gravitational, both contact, both electrostatic…) and act for the same time.
- Examples: a swimmer pushes water backwards and the water pushes the swimmer forwards; a rocket pushes exhaust gas downwards and the gas pushes the rocket upwards; the Earth pulls you down with your weight and you pull the Earth up with an equal gravitational force.
- Third-law pairs explain conservation of momentum: in a collision the equal and opposite forces act for the same time, so the momentum gained by one object equals the momentum lost by the other.
A book resting on a table has its weight (Earth pulls book down) and the normal reaction (table pushes book up) acting on it. These are equal and opposite, but they act on the same object and are different types of force, so they are not a third-law pair. The partner of the book's weight is the book pulling the Earth upwards.
What is a moment and how do you use the principle of moments? Physics only · 2P
- Moment in N m, force in N, distance in m. Convert cm to m before substituting.
- The distance is measured from the pivot to the line of action of the force, at right angles to it. A force whose line of action passes through the pivot has no moment.
- A larger force, or the same force applied further from the pivot, gives a larger moment — which is why a long spanner makes a stiff nut easier to turn.
- For a balanced object, the total upward force also equals the total downward force.
- Write the equation in words first: “sum of clockwise moments = sum of anticlockwise moments”, then substitute force × distance for each term.
A uniform seesaw of weight 250 N is pivoted at its centre. Child A (weight 300 N) sits 1.6 m to the left of the pivot. Child B (weight 400 N) sits on the right so that the seesaw balances.
(a) Calculate the distance of child B from the pivot. (b) Calculate the upward force of the pivot on the seesaw.
(a) The seesaw's own weight acts at the pivot, so it has no moment about the pivot.
anticlockwise moment = 300 × 1.6 = 480 N m
clockwise moment = 400 × \(d\)
400\(d\) = 480 ⇒ \(d=\) 1.2 m
(b) Balanced, so upward force = total downward force = 300 + 400 + 250 = 950 N
The heavier child must sit closer to the pivot to give the same moment.
What is the centre of gravity, and how do the forces on a supported beam change? Physics only · 2P
- For a uniform object (a metre rule, a plank), the centre of gravity is at its centre.
- In moments problems, draw the whole weight of an object as one arrow from its centre of gravity. If the centre of gravity is at the pivot, the weight has no moment about it (as in worked example 8).
- A light beam is one whose weight is small enough to ignore.
A light beam supported at both ends
When a heavy object sits on a light beam resting on two supports, A and B, each support pushes up on the beam. Because the beam is balanced:
- the two upward forces add up to the weight of the object, \(R_A+R_B=W\);
- the support closer to the object provides the larger upward force;
- with the object exactly in the middle, each support provides half the weight;
- as the object moves from A towards B, \(R_A\) decreases steadily and \(R_B\) increases steadily.
To find the forces, take moments about one support: the force at that support then has no moment, leaving only one unknown.
A light beam 3.0 m long rests on supports A and B at its ends. A box of weight 900 N is placed 1.0 m from A.
(a) Calculate the upward force from support B. (b) Calculate the upward force from support A. (c) The box is moved to 2.4 m from A. Calculate the new support forces. (d) Describe how the two support forces change as the box is moved slowly from A to B.
(a) Take moments about A, so \(R_A\) has no moment.
clockwise moment of the box = 900 × 1.0 = 900 N m; anticlockwise moment of \(R_B\) = \(R_B\) × 3.0
\(R_B=\dfrac{900}{3.0}=\) 300 N
(b) upward forces = downward forces: \(R_A=900-300=\) 600 N
(c) \(R_B\times3.0=900\times2.4\) ⇒ \(R_B=\) 720 N; \(R_A=900-720=\) 180 N
(d) \(R_A\) decreases steadily from 900 N to zero while \(R_B\) increases steadily from zero to 900 N; the two forces always add up to 900 N and are equal (450 N each) when the box is at the centre.
Examiner tips for 1(c)
When several forces act, the \(F\) in \(F=ma\) is the resultant force. Find it first, with its direction, then divide by the mass. Using the driving force alone is the most common way to lose the final mark.
For a falling object, build the answer from forces: what acts, which one changes with speed, what that does to the resultant force, and so to the acceleration. Answer only the situation in the question — if it says the object does not reach terminal velocity, statements about air resistance equalling weight earn nothing.
Say which part of the stopping distance a factor changes. “Alcohol increases the stopping distance” is weaker than “alcohol increases the reaction time, so the thinking distance increases”.
Write down which direction is positive before you start. Any velocity in the other direction goes into the equation with a minus sign. State the direction of every velocity or force you calculate.
State the principle of moments, list each moment as force × distance with the distance in metres from the chosen pivot, and set clockwise equal to anticlockwise. Take moments about the point where an unknown force acts to remove it from the equation.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A moving object needs a resultant force to keep moving. | A resultant force changes the motion; it is not needed to maintain it. | Balanced forces: at rest or constant velocity. |
| Mass and weight are the same thing. | Mass is matter (kg); weight is a force (N). | \(W=mg\); weight changes with \(g\), mass does not. |
| At terminal velocity there are no forces on the object. | Weight and air resistance still act. | They are equal and opposite, so the resultant force is zero. |
| Extension is the length of the spring. | The spring already has a length with no load. | extension = stretched length − original length. |
| Weight and the normal reaction on a resting book are a third-law pair. | They act on the same object and are different types of force. | A third-law pair acts on two different objects and is the same type of force. |
| Safety features reduce the change in momentum. | The person still goes from moving to rest. | They increase the time, so the force is smaller. |
| Moment = force × any distance from the pivot. | Only the distance at right angles to the line of action counts. | Use the perpendicular distance from the pivot. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| \(F=m\times a\) | unbalanced force, mass or acceleration | All | yes — all papers |
| \(W=m\times g\) | weight from mass | All | yes — all papers |
| stopping distance = thinking distance + braking distance | total distance to stop | All | no — learn it |
| \(p=m\times v\) | momentum | Physics only · 2P | yes — Physics 2P only |
| \(F=\dfrac{(mv-mu)}{t}\) | force from a change in momentum | Physics only · 2P | yes — Physics 2P only |
| moment = force × perpendicular distance from the pivot | turning effect of a force | Physics only · 2P | yes — Physics 2P only |
| sum of clockwise moments = sum of anticlockwise moments | an unknown force or distance on a balanced object | Physics only · 2P | no — learn it |
Pearson includes the Equation List with the question papers, but the first mark in many questions is for writing the relationship yourself — learn them anyway.
Check your understanding
1. A cyclist stops pedalling on a level road and gradually slows down. Name two forces that slow her down and state their direction. All
Friction (between the moving parts and between tyres and road) and air resistance. Both act backwards, opposite to her direction of motion.
2. An unbalanced force of 0.90 N acts on a ball of mass 0.15 kg. Calculate the acceleration of the ball. All
\(a=\dfrac{F}{m}=\dfrac{0.90}{0.15}=\) 6.0 m/s²
3. A driver uses a mobile phone while driving on an icy road. State which part of the stopping distance each of these factors increases, and why. All
Using the phone increases the reaction time, so the thinking distance increases. Ice reduces the friction between tyres and road, so the braking distance increases.
4. A crate has a force of 65 N to the right and a force of 40 N to the left acting on it. Calculate the resultant force, and explain why force is a vector quantity. Double · Physics
65 − 40 = 25 N to the right. Force is a vector because it has both size and direction — the direction affects how forces combine.
5. A small ball is dropped from a tall building and reaches terminal velocity. Explain why. Double · Physics
At first only weight acts, so the ball accelerates. As its speed increases, air resistance increases (1), so the resultant force and acceleration decrease (1). Eventually air resistance equals weight, so the resultant force is zero (1) and the ball falls at constant (terminal) velocity (1).
6. A cyclist's head (mass 4.5 kg) moving at 6.0 m/s is stopped by a helmet in 0.015 s. Without a helmet it would stop in 0.0030 s. Calculate the average force on the head in each case and explain how the helmet protects the cyclist. Physics only · 2P
Change in momentum = 4.5 × 6.0 = 27 kg m/s. With helmet: 27 ÷ 0.015 = 1800 N. Without: 27 ÷ 0.0030 = 9000 N. The helmet's lining crushes, increasing the time for the same change in momentum, so the rate of change of momentum — the force — is smaller.
7. A uniform metre rule is balanced on a pivot at its 50 cm mark. A 2.0 N weight hangs at the 20 cm mark. Where must a 1.5 N weight hang to balance the rule? Physics only · 2P
Anticlockwise moment = 2.0 × 0.30 = 0.60 N m. Clockwise: 1.5 × \(d\) = 0.60 ⇒ \(d\) = 0.40 m to the right of the pivot, so at the 90 cm mark. (The rule's weight acts at the pivot, so it has no moment.)
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