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KS3 · Physics

K11 Forces and Motion

AQA AQA KS3 Science · Calculator allowed · about 75 minutes

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For each force below, state whether it is a contact force or a non-contact force. 1) Friction between a car's tyres and the road. 2) The gravitational force between the Earth and the Moon. 3) The normal contact force from a table pushing up on a book. 4) The magnetic force between two magnets. 5) Air resistance acting on a falling leaf. 6) The electrostatic force between a charged balloon and a wall (without touching).

(6)

A book rests on a table and is not moving.

(a)Name the force pulling the book downward.(1)

(b)Name the force pushing the book upward from the surface of the table.(1)

(c)The book stays still. Use the idea of resultant force to explain why.(2)

Use the Physics Equations Sheet: weight = mass x gravitational field strength (W = m g). On Earth, gravitational field strength g = 10 N/kg.

(a)A rucksack has a mass of 4.5 kg. Calculate its weight on Earth.(2)

(b)Explain why the rucksack's mass would stay the same on the Moon, but its weight would change.(2)

Use the Physics Equations Sheet: speed = distance / time. The table shows the distance travelled by a jogger. Time (s): 0, 10, 20, 30, 40. Distance (m): 0, 40, 80, 80, 120.

(a)Calculate the jogger's speed between 0 s and 20 s.(2)

(b)Describe what is happening to the jogger between 20 s and 30 s.(1)

(c)Calculate the jogger's speed between 30 s and 40 s.(2)

Two students push on opposite sides of a stuck door. Amir pushes with a force of 60 N. Beth pushes in the opposite direction with a force of 25 N.

(a)Calculate the resultant force on the door, including its direction.(3)

(b)State what would happen to the resultant force if Beth's force increased to exactly 60 N.(1)

Required practical: a student investigates how the extension of a spring depends on the force applied. She hangs a spring from a clamp stand, measures its natural length, then adds slotted masses one at a time, measuring the new length each time and calculating the extension. Her results are shown in the table. Force (N): 0, 1, 2, 3, 4, 5. Extension (mm): 0, 20, 40, 60, 80, 110.

(a)Identify the independent variable and the dependent variable in this investigation.(2)

(b)Describe the pattern shown by the results between 0 N and 4 N.(1)

(c)State the force at which the spring stops extending proportionally with the force applied (the limit of proportionality).(1)

(d)Use the Physics Equations Sheet: spring constant k = force / extension (k = F / e). Use the result at 2 N to calculate the spring constant of this spring. Give your answer in N/mm.(2)

(e)The student forgets to check the spring's natural length, with no masses attached, before starting. Suggest one improvement to her method and explain why it would make her results more accurate.(2)

Required practical: a group of students wants to find out how the roughness of a surface affects the force of friction needed to drag a shoe across it at a steady speed. They pull the shoe using a newtonmeter across four different surfaces: carpet, wood, ice and sandpaper.

(a)Identify one control variable the students should keep the same to make this a fair test.(1)

(b)Explain why the students should pull the shoe at a steady (constant) speed while taking each reading.(2)

(c)The students' results were: carpet 3.2 N, wood 1.8 N, ice 0.4 N, sandpaper 4.5 N. Put the surfaces in order of increasing friction force, and state how this relates to the roughness of each surface.(2)

(d)Suggest one way the students could improve the repeatability of their results.(1)

(e)Explain why testing the ice surface might give unreliable readings in a normal classroom.(1)

Stopping distance = thinking distance + braking distance.

(a)A car's thinking distance is 15 m and its braking distance is 38 m. Calculate the car's total stopping distance.(1)

(b)State two factors that could increase the driver's thinking distance.(2)

(c)State two factors that could increase the car's braking distance.(2)

Use the Physics Equations Sheet: pressure = force / area.

(a)A hiker wearing a snowshoe presses down on the snow with a force of 350 N. The snowshoe has an area of 0.25 m^2. Calculate the pressure exerted on the snow.(2)

(b)Without the snowshoe, the same hiker's boot has a contact area of only 0.02 m^2. Calculate the pressure exerted by the boot alone, using the same force of 350 N.(2)

(c)Use your answers to parts (a) and (b) to explain why wearing snowshoes stops the hiker sinking into soft snow.(1)

Use the Physics Equations Sheet: moment = force x distance (from the pivot). Priya weighs 400 N and sits 1.5 m from the pivot of a balanced seesaw.

(a)Calculate the moment Priya produces about the pivot.(2)

(b)Tomasz weighs 500 N. Calculate the distance from the pivot at which he must sit on the other side for the seesaw to balance.(3)

A bus is travelling forward at a steady speed. The driver suddenly brakes hard. A standing passenger is not holding on.

(a)Which row correctly describes what happens to the passenger, and why?(1)

A) The passenger is thrown backward, because their body wants to keep moving forward while the bus slows down
B) The passenger is thrown forward, because their body wants to keep moving forward while the bus slows down
C) The passenger stays perfectly still, because the forces on a passenger are always balanced
D) The passenger is thrown forward, because the brakes push the passenger forward

(b)Name a safety feature that provides a force on a passenger to slow them down safely in a crash, and explain how it helps to reduce injury.(2)

A cyclist's distance-time graph is a curve that becomes steeper as time goes on.

(a)State what this shape shows about the cyclist's speed.(1)

(b)The cyclist then travels at a constant (steady) speed. Describe how this part of the distance-time graph would look.(1)

Explain the factors that affect a car's overall stopping distance, and evaluate why speed limits are often lower near schools than on motorways. Refer to thinking distance, braking distance and the forces involved in your answer.

(6)

A swimmer floats motionless on the surface of a swimming pool.

(a)Name the two forces acting on the floating swimmer.(2)

(b)The swimmer is floating perfectly still. State the relationship between the sizes of these two forces.(1)

(c)The swimmer puts on a heavy weight belt and begins to sink. Explain, in terms of forces, why this happens.(2)

Cyclists in a race often wear tight-fitting clothing and helmets shaped to a point at the back.

(a)Name the force that acts on a cyclist, due to the air, as they move forward.(1)

(b)Explain how the streamlined shape of a cyclist's helmet and clothing helps them cycle faster for the same pedalling force.(2)

A rubber band and a piece of plasticine are each stretched gently, then the force is removed.

(a)State what happens to the rubber band when the force is removed.(1)

(b)State what happens to the plasticine when the force is removed.(1)

(c)Define the term 'elastic deformation'.(1)

Use the Physics Equations Sheet: resultant force = mass x acceleration (F = m a). A shopping trolley of mass 25 kg is pushed with a resultant force of 20 N.

(a)Calculate the trolley's acceleration.(3)

(b)The trolley's mass is doubled to 50 kg but the resultant force stays at 20 N. State and explain what happens to its acceleration.(2)

A go-kart of mass 60 kg is pushed forward by an engine force of 400 N. Air resistance and friction act backward on the go-kart with a combined force of 150 N.

(a)Calculate the resultant force on the go-kart.(2)

(b)Use the Physics Equations Sheet: resultant force = mass x acceleration (F = m a). Calculate the go-kart's acceleration.(3)

Use the Physics Equations Sheet: extension = force / spring constant (e = F / k). A spring has a spring constant of 25 N/m while it obeys Hooke's law.

(a)Calculate the extension produced by a force of 6 N.(2)

(b)When a force of 12 N is applied to the same spring, its extension is measured as 0.55 m, rather than the 0.48 m predicted by Hooke's law. Explain what this tells us about the spring at this force.(2)

Use the Physics Equations Sheet: moment = force x distance (from the pivot). A uniform plank is balanced on a central pivot. A 300 N weight is placed 2 m to the left of the pivot, and a 200 N weight is placed 1.5 m to the left of the pivot (the same side).

(a)Calculate the total anticlockwise moment produced by these two weights.(3)

(b)Calculate the distance from the pivot at which a single 450 N weight must be placed on the right-hand side of the plank to balance it.(2)

A train travels at an average speed of 108 km/h.

(a)Convert this speed into metres per second (m/s).(2)

(b)Use the Physics Equations Sheet: speed = distance / time. Calculate how long the train takes to travel 15 km at this average speed. Give your answer in seconds.(3)

Mark scheme · K11 Forces and Motion

Question 1

B1 1) contact
B1 2) non-contact
B1 3) contact
B1 4) non-contact
B1 5) contact
B1 6) non-contact
Answer: 1) contact, 2) non-contact, 3) contact, 4) non-contact, 5) contact, 6) non-contact

Question 2

(a) B1 weight
(a) Answer: Weight.
(b) B1 normal contact force (accept normal reaction force)
(b) Answer: Normal contact force (normal reaction force).
(c) M1 the two forces are balanced (equal in size, opposite in direction)
(c) A1 so the resultant force on the book is 0 N, meaning there is no change in its motion (it stays at rest) (oe)
(c) Answer: The weight and normal contact force are balanced, giving a resultant force of 0 N, so the book stays at rest.

Question 3

(a) M1 correct substitution W = 4.5 x 10
(a) A1 45 N cao
(a) Answer: 45 N
(b) B1 mass is the amount of matter in the rucksack, and this does not change wherever it is
(b) B1 weight depends on the gravitational field strength, which is smaller on the Moon, so the weight decreases (oe)
(b) Answer: Mass stays the same because the amount of matter is unchanged; weight decreases because the Moon's gravitational field strength is smaller than Earth's.

Question 4

(a) M1 correct substitution speed = 80 / 20
(a) A1 4 m/s cao
(a) Answer: 4 m/s
(b) B1 the jogger is stationary/resting, since the distance does not change (oe)
(b) Answer: The jogger is stationary (resting), since the distance does not change.
(c) M1 correct substitution speed = (120-80) / (40-30)
(c) A1 4 m/s cao
(c) Answer: 4 m/s

Question 5

(a) M1 correct subtraction 60 - 25
(a) A1 35 N
(a) A1 direction: in the direction of Amir's push cao
(a) Answer: 35 N in the direction of Amir's push.
(b) B1 resultant force = 0 N, forces balanced (oe)
(b) Answer: Resultant force = 0 N (the forces would be balanced).

Question 6

(a) B1 independent variable = force applied to the spring (oe)
(a) B1 dependent variable = extension of the spring (oe)
(a) Answer: Independent variable: force applied. Dependent variable: extension of the spring.
(b) B1 extension is directly proportional to force between 0 N and 4 N (oe, e.g. a straight line through the origin, extension increases by 20 mm for each extra 1 N)
(b) Answer: Extension is directly proportional to the force applied (a straight-line relationship through the origin).
(c) B1 4 N (accept between 4 N and 5 N) cao
(c) Answer: 4 N.
(d) M1 correct substitution k = 2 / 40
(d) A1 0.05 N/mm cao
(d) Answer: 0.05 N/mm
(e) B1 measure and record the spring's natural (unstretched) length before adding any masses (oe)
(e) B1 so that every extension is measured from the correct zero point, rather than being offset by an existing stretch (oe)
(e) Answer: Measure the spring's natural length with no force applied first, so every extension calculated afterwards is accurate.

Question 7

(a) B1 any one sensible control variable, e.g. the same shoe (same mass and sole), the same newtonmeter, the same pulling speed (any one)
(a) Answer: e.g. Use the same shoe (same mass and sole) for every surface.
(b) M1 at a constant/steady speed, the forces on the shoe are balanced
(b) A1 so the reading on the newtonmeter is equal to the friction force only, not also accelerating the shoe (oe)
(b) Answer: At constant speed the pulling force exactly balances friction, so the newtonmeter reading equals the friction force.
(c) B1 correct order: ice, wood, carpet, sandpaper
(c) B1 rougher surfaces produce a greater friction force (sandpaper is roughest and gives the most friction; ice is smoothest and gives the least) (oe)
(c) Answer: Ice (0.4 N), wood (1.8 N), carpet (3.2 N), sandpaper (4.5 N); rougher surfaces give more friction.
(d) B1 repeat each measurement, e.g. three times, and calculate a mean, ignoring any anomalous results (oe)
(d) Answer: Repeat each surface test several times and calculate a mean friction force.
(e) B1 the ice may start to melt/change surface during testing because room temperature is above 0 degC, so conditions are not the same for every reading (oe)
(e) Answer: The ice may melt slightly during testing, changing the surface and making readings inconsistent.

Question 8

(a) B1 53 m (15 + 38) cao
(a) Answer: 53 m
(b) B1 any one valid factor, e.g. tiredness
(b) B1 any second valid factor, e.g. distraction (phone use), alcohol or drugs, illness (any two different factors)
(b) Answer: e.g. Tiredness and being distracted (e.g. using a phone).
(c) B1 any one valid factor, e.g. a wet or icy road surface
(c) B1 any second valid factor, e.g. worn tyres, worn brakes, greater speed, heavier load (any two different factors)
(c) Answer: e.g. A wet or icy road and worn tyres.

Question 9

(a) M1 correct substitution P = 350 / 0.25
(a) A1 1400 Pa (accept 1400 N/m²) cao
(a) Answer: 1400 Pa
(b) M1 correct substitution P = 350 / 0.02
(b) A1 17500 Pa cao
(b) Answer: 17500 Pa
(c) B1 the snowshoe spreads the same force over a much larger area, giving a much lower pressure on the snow (ft from a and b) (oe)
(c) Answer: The larger area of the snowshoe spreads the same force out, producing a far lower pressure (1400 Pa) than the boot alone (17500 Pa), so the hiker is less likely to sink.

Question 10

(a) M1 correct substitution moment = 400 x 1.5
(a) A1 600 Nm cao
(a) Answer: 600 Nm
(b) M1 uses the balancing condition: Tomasz's moment = Priya's moment, 500 x d = 600 (ft from a)
(b) dM1 correct rearrangement d = 600 / 500
(b) A1 1.2 m cao
(b) Answer: 1.2 m

Question 11

(a) B1 B cao
(a) Answer: B
(b) B1 names a suitable safety feature, e.g. seatbelt (or airbag)
(b) B1 explains that it applies a force to slow the passenger down gradually over a longer time, rather than the passenger stopping suddenly by hitting something rigid (oe)
(b) Answer: e.g. Seatbelt: it stretches slightly and applies a force to slow the passenger down over a longer time than hitting the windscreen would.

Question 12

(a) B1 the cyclist's speed is increasing, the cyclist is accelerating/speeding up (oe)
(a) Answer: The cyclist's speed is increasing.
(b) B1 a straight line with a constant/unchanging gradient (oe)
(b) Answer: A straight line.

Question 13

Level 1 (1-2): Basic, mostly unlinked statements about stopping distance, thinking distance or braking distance, with little reference to forces. May be presented as a simple list.
Level 2 (3-4): A description of factors affecting thinking distance and/or braking distance, with some correct reference to forces (e.g. friction, reaction time), but the evaluation of speed limits near schools is limited or missing.
Level 3 (5-6): A clear, logically linked explanation of factors affecting both thinking distance and braking distance, correctly referring to forces (e.g. friction between tyres and road, reaction time), together with a well-reasoned evaluation of why lower speed limits near schools reduce the risk of collisions.
Indicative content:
- Thinking distance is the distance travelled during the driver's reaction time, before the brakes are applied.
- Thinking distance increases if the driver is tired, distracted, or under the influence of drink or drugs (slower reaction time).
- Braking distance is the distance travelled while the brakes are being applied until the vehicle stops.
- Braking distance increases in wet or icy conditions, with worn tyres or worn brakes, or with a greater vehicle mass or load, because the friction force between tyres and road is reduced or a larger braking force is needed.
- At higher speed, both thinking distance and braking distance increase, so total stopping distance is much greater at higher speeds.
- Near schools there is more risk of pedestrians, especially children, stepping into the road unexpectedly, so a shorter stopping distance is needed to avoid a collision.
- Lower speed limits near schools reduce both thinking distance and braking distance, giving drivers more time and distance to stop safely.

Question 14

(a) B1 weight
(a) B1 upthrust (the upward force from the water)
(a) Answer: Weight and upthrust.
(b) B1 they are equal in size, balanced (oe)
(b) Answer: They are equal in size (balanced forces).
(c) M1 the weight belt increases the swimmer's total weight (downward force)
(c) A1 weight is now greater than upthrust, giving a resultant (unbalanced) force downward, so the swimmer sinks (oe)
(c) Answer: The weight belt increases the swimmer's weight so it becomes greater than the upthrust, giving an unbalanced downward force that makes the swimmer sink.

Question 15

(a) B1 air resistance (drag)
(a) Answer: Air resistance (drag).
(b) M1 the streamlined/smooth shape reduces the air resistance (drag) acting on the cyclist
(b) A1 so a smaller force opposes their motion, giving a larger resultant forward force for the same pedalling force, allowing greater speed (oe)
(b) Answer: The streamlined shape reduces air resistance, so the resultant forward force, and therefore speed, is greater for the same pedalling force.

Question 16

(a) B1 it returns to its original length/shape (oe)
(a) Answer: It returns to its original shape and length.
(b) B1 it stays stretched, it does not return to its original shape (oe)
(b) Answer: It stays stretched and does not return to its original shape.
(c) B1 a deformation in which the object returns to its original shape and length once the force causing it is removed (oe)
(c) Answer: A deformation where the object returns to its original shape and length once the force is removed.

Question 17

(a) M1 correct rearrangement a = F / m
(a) A1 a = 20 / 25 = 0.8
(a) B1 correct unit, m/s² cao
(a) Answer: 0.8 m/s²
(b) M1 acceleration is halved, decreases
(b) A1 because acceleration is inversely proportional to mass for a constant force (a = F/m, so doubling m halves a; new a = 20/50 = 0.4 m/s²) (ft from a)
(b) Answer: Acceleration halves, decreasing from 0.8 m/s² to 0.4 m/s^2.

Question 18

(a) M1 correct subtraction 400 - 150
(a) A1 250 N, forward, in the direction of the engine force cao
(a) Answer: 250 N forward
(b) M1 correct rearrangement a = F / m (ft from a)
(b) A1 a = 250 / 60
(b) A1 4.17 m/s² (awrt 4.2 m/s²) (ft from a)
(b) Answer: awrt 4.2 m/s² (4.17 m/s²)

Question 19

(a) M1 correct substitution e = 6 / 25
(a) A1 0.24 m (accept 24 cm) cao
(a) Answer: 0.24 m (24 cm)
(b) M1 the spring has been stretched beyond its limit of proportionality
(b) A1 so its extension is no longer directly proportional to the force applied, Hooke's law no longer applies, and the spring may not return to its original length (oe)
(b) Answer: The spring has passed its limit of proportionality, so extension is no longer directly proportional to force.

Question 20

(a) M1 moment of 300 N weight = 300 x 2 = 600 Nm
(a) M1 moment of 200 N weight = 200 x 1.5 = 300 Nm
(a) A1 total anticlockwise moment = 600 + 300 = 900 Nm cao
(a) Answer: 900 Nm
(b) M1 uses balancing condition 450 x d = 900 (ft from a)
(b) A1 2 m cao
(b) Answer: 2 m

Question 21

(a) M1 correct method, e.g. 108 x 1000 / 3600 (or 108 / 3.6)
(a) A1 30 m/s cao
(a) Answer: 30 m/s
(b) M1 converts 15 km to 15000 m (or works consistently in km and hours)
(b) M1 correct rearrangement time = distance / speed = 15000 / 30 (ft from a)
(b) A1 500 s cao
(b) Answer: 500 s

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