AQA GCSE Physics coverage

Forces

Section 4.5
25 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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4.5.1.1

Scalar and vector quantities

  • A scalar quantity has magnitude only, whereas a vector quantity has both magnitude and an associated direction.
  • Classify a stated quantity by checking whether its direction is needed to specify it completely; force, weight and velocity are vectors, while mass, energy and speed are scalars.
  • A vector can be shown by an arrow: its length represents the magnitude and its arrowhead shows the direction.
  • Do not call a quantity a vector merely because it can be large or negative; a vector must include direction.

Tier 1 · Easy

1 mark
ORIGINAL

A trolley has a mass of 6.0kg6.0\,\text{kg} and moves with a velocity of 2.5m/s2.5\,\text{m/s} east. State which of these two quantities is a vector.

Tier 2 · Standard

2 marks
ORIGINAL

A scale drawing uses 1.0cm1.0\,\text{cm} to represent 4.0N4.0\,\text{N}. Describe the arrow that represents a force of 20N20\,\text{N} acting vertically downwards.

Tier 3 · Hard

3 marks
ORIGINAL

A cyclist's speedometer reads 11m/s11\,\text{m/s} while the cyclist travels south-west. Give the cyclist's speed and velocity, and explain why these are different quantities even though their magnitudes match.

4.5.1.2

Contact and non-contact forces

  • A force is a push or pull caused by an interaction between objects; force is a vector quantity.
  • Decide whether the objects must touch: friction, air resistance, tension and normal contact force are contact forces, while gravitational, electrostatic and magnetic forces are non-contact forces.
  • An interaction produces a force on each object; represent each force with a vector arrow on the object that experiences it.
  • Do not describe air resistance as non-contact just because air is hard to see: collisions with air particles make it a contact force.

Tier 1 · Easy

1 mark
ORIGINAL

State whether friction and gravitational force are contact or non-contact forces.

Tier 2 · Standard

2 marks
ORIGINAL

A magnet attracts an iron pin across a small gap. Describe the force interaction between the magnet and the pin.

Tier 3 · Hard

3 marks
ORIGINAL

A parachutist is falling through the air while attached to an open parachute by cords. Name one non-contact force and two different contact forces acting in this situation, giving the object on which each named force acts.

4.5.1.3

Gravity

  • Weight is the force on an object due to gravity, while mass measures the amount of matter and does not change when gravitational field strength changes.
  • Calculate weight using W=mgW=mg, with WW in newtons, mm in kilograms and gg in newtons per kilogram; use the value of gg supplied in the question.
  • For example, a 3.0kg3.0\,\text{kg} object where g=9.8N/kgg=9.8\,\text{N/kg} has weight 3.0×9.8=29.4N3.0\times9.8=29.4\,\text{N}, acting through its centre of mass.
  • Do not give weight in kilograms or assume it is constant everywhere; weight changes with gg and is measured with a calibrated newtonmeter.

Tier 1 · Easy

1 mark
ORIGINAL

A 7.5kg7.5\,\text{kg} bag is in a region where g=9.8N/kgg=9.8\,\text{N/kg}. Calculate the bag's weight.

Tier 2 · Standard

3 marks
ORIGINAL

A sample has mass 2.4kg2.4\,\text{kg}. Its weight is 23.5N23.5\,\text{N} at location A and 3.8N3.8\,\text{N} at location B. Calculate the gravitational field strength at each location.

Tier 3 · Hard

4 marks
ORIGINAL

An explorer has mass 68kg68\,\text{kg}. A newtonmeter would show 666N666\,\text{N} for the explorer on planet P and 177N177\,\text{N} on moon Q. Determine gg at P and Q, then explain what happens to the explorer's mass during the journey.

4.5.1.4

Resultant forces

  • The resultant force is the single force that has the same effect as all the forces acting together.
  • For forces along one straight line, choose a positive direction, give opposite forces opposite signs and add them.
  • For example, 18N18\,\text{N} right and 11N11\,\text{N} left give a resultant of 1811=7N18-11=7\,\text{N} to the right; equal opposing forces give zero resultant.
  • Do not add magnitudes when forces oppose, and do not omit the direction of a non-zero resultant because force is a vector.

Tier 1 · Easy

1 mark
ORIGINAL

Two horizontal forces act on a crate: 16N16\,\text{N} east and 9N9\,\text{N} west. Calculate the resultant force.

Tier 2 · Standard

2 marks
ORIGINAL

A model boat is pulled forwards by 42N42\,\text{N}. Water resistance is 27N27\,\text{N} and air resistance is 6N6\,\text{N}, both backwards. Determine the resultant force on the boat.

Tier 3 · Hard

3 marks
ORIGINAL

Forces on a rail cart are 85N85\,\text{N} east, 34N34\,\text{N} west and 19N19\,\text{N} west. Calculate the resultant, then state the additional single force needed to make the forces balanced.

4.5.2

Work done and energy transfer

  • Work is done when a force causes a displacement, transferring energy between stores; work done against friction raises temperature.
  • Use W=FsW=Fs, where ss is the distance moved along the force's line of action; WW is in joules, FF in newtons and ss in metres.
  • For example, a 25N25\,\text{N} force moving an object 3.0m3.0\,\text{m} along its line of action does 25×3.0=75J25\times3.0=75\,\text{J} of work.
  • Do not multiply by a distance perpendicular to the force, and remember that 1J=1N m1\,\text{J}=1\,\text{N m}.

Tier 1 · Easy

1 mark
ORIGINAL

A horizontal force of 35N35\,\text{N} moves a box 4.0m4.0\,\text{m} horizontally. Calculate the work done by the force.

Tier 2 · Standard

3 marks
ORIGINAL

A winch lifts a load vertically through 6.5m6.5\,\text{m} using a constant upward force of 480N480\,\text{N}. Calculate the work done and identify the energy store that increases.

Tier 3 · Hard

5 marks
ORIGINAL

A powered trolley moves 12m12\,\text{m} along a level floor. Its motor provides a forward force of 180N180\,\text{N} while friction is 140N140\,\text{N}. Calculate the work done by the motor, the energy transferred thermally by friction and the remaining energy transferred to the trolley's kinetic energy store.

4.5.3

Forces and elasticity

  • Elastic deformation is reversed when the force is removed; inelastic deformation leaves the object permanently changed.
  • Up to the limit of proportionality use F=keF=ke, measuring extension ee from the original length; a straight force-extension graph through the origin represents direct proportion.
  • For example, a spring with k=160N/mk=160\,\text{N/m} extended by 0.050m0.050\,\text{m} needs F=160×0.050=8.0NF=160\times0.050=8.0\,\text{N} and stores Ee=12ke2=0.20JE_e=\frac12ke^2=0.20\,\text{J}.
  • Do not substitute the spring's total length for extension, and do not apply the linear relationship beyond the limit of proportionality.

Tier 1 · Easy

1 mark
ORIGINAL

A spring's length changes from 0.18m0.18\,\text{m} to 0.23m0.23\,\text{m}. Calculate its extension.

Tier 2 · Standard

3 marks
ORIGINAL

Within its linear region, a spring extends by 0.075m0.075\,\text{m} when a force of 18N18\,\text{N} is applied. Calculate the spring constant and predict the extension produced by 12N12\,\text{N}.

Tier 3 · Hard

5 marks
ORIGINAL

A spring of constant 240N/m240\,\text{N/m} is stretched by 0.18m0.18\,\text{m} without exceeding its limit of proportionality. Calculate the applied force and the elastic potential energy stored. State what observation after unloading would show that the spring had instead been inelastically deformed.

4.5.4

Moments, levers and gears (physics only)

  • A moment is the turning effect of a force about a pivot, and its size is M=FdM=Fd where dd is the perpendicular distance to the force's line of action.
  • For a balanced object, choose one pivot and set the total clockwise moment equal to the total anticlockwise moment.
  • For example, 15N15\,\text{N} acting 0.40m0.40\,\text{m} from a pivot produces a moment of 6.0N m6.0\,\text{N m}; a lever or gear system transmits such rotational effects.
  • Do not use a sloping distance measured to the point where the force is applied; the equation requires the perpendicular distance to the line of action.

Tier 1 · Easy

1 mark
ORIGINAL

A force of 28N28\,\text{N} acts perpendicular to a handle 0.35m0.35\,\text{m} from its pivot. Calculate the moment.

Tier 2 · Standard

3 marks
ORIGINAL

A seesaw is balanced. A child of weight 360N360\,\text{N} sits 1.5m1.5\,\text{m} to the left of the pivot. Calculate how far to the right of the pivot a child of weight 450N450\,\text{N} must sit.

Tier 3 · Hard

4 marks
ORIGINAL

A uniform 4.0m4.0\,\text{m} beam of weight 220N220\,\text{N} is supported at its left end and at a point 3.2m3.2\,\text{m} from the left end. A 300N300\,\text{N} load is placed at the right end. Calculate the upward force from the support at 3.2m3.2\,\text{m}.

4.5.5.1.1

Pressure in a fluid 1 (physics only)

  • A fluid is a liquid or a gas, and fluid pressure produces a force normal, or at right angles, to a surface.
  • Calculate pressure using p=F/Ap=F/A, with normal force FF in newtons, surface area AA in square metres and pressure pp in pascals.
  • For example, a normal force of 90N90\,\text{N} on 0.030m20.030\,\text{m}^2 produces p=90÷0.030=3000Pap=90\div0.030=3000\,\text{Pa}.
  • Do not use an area in cm2\text{cm}^2 without converting it to m2\text{m}^2, and use only the component of force normal to the surface.

Tier 1 · Easy

1 mark
ORIGINAL

A gas pushes normally on a hatch with force 240N240\,\text{N}. The hatch area is 0.080m20.080\,\text{m}^2. Calculate the pressure.

Tier 2 · Standard

3 marks
ORIGINAL

A liquid exerts a normal force of 54N54\,\text{N} on a sensor of area 30cm230\,\text{cm}^2. Calculate the pressure on the sensor.

Tier 3 · Hard

5 marks
ORIGINAL

A sealed chamber produces the same normal force of 1260N1260\,\text{N} on either of two removable panels. Panel X measures 0.30m0.30\,\text{m} by 0.20m0.20\,\text{m} and panel Y measures 0.45m0.45\,\text{m} by 0.35m0.35\,\text{m}. Calculate the pressure on each panel and determine which panel experiences the lower pressure.

4.5.5.1.2

Pressure in a fluid 2 (physics only) (HT only)

  • In a liquid, pressure increases with the height of liquid above the point and with the liquid's density.
  • Use p=hρgp=h\rho g for pressure due to a liquid column, with hh in metres, ρ\rho in kg/m3\text{kg/m}^3 and gg in N/kg\text{N/kg}; subtract depths before using it for a pressure difference.
  • For example, in water with ρ=1000kg/m3\rho=1000\,\text{kg/m}^3 and g=9.8N/kgg=9.8\,\text{N/kg}, a 0.50m0.50\,\text{m} depth change gives Δp=0.50×1000×9.8=4900Pa\Delta p=0.50\times1000\times9.8=4900\,\text{Pa}.
  • Do not include horizontal position in p=hρgp=h\rho g; upthrust arises because the bottom of a submerged object is at greater pressure than its top.

Tier 1 · Easy

2 marks
ORIGINAL

Oil has density 820kg/m3820\,\text{kg/m}^3. Calculate the pressure due to a 1.5m1.5\,\text{m} column of the oil when g=9.8N/kgg=9.8\,\text{N/kg}.

Tier 2 · Standard

3 marks
ORIGINAL

Two pressure sensors are 0.40m0.40\,\text{m} and 2.10m2.10\,\text{m} below the surface of a liquid of density 960kg/m3960\,\text{kg/m}^3. Calculate the pressure difference when g=9.8N/kgg=9.8\,\text{N/kg}.

Tier 3 · Hard

5 marks
ORIGINAL

A fully submerged cuboid is 0.30m0.30\,\text{m} high and has horizontal top and bottom areas of 0.012m20.012\,\text{m}^2. It is in water of density 1000kg/m31000\,\text{kg/m}^3, where g=9.8N/kgg=9.8\,\text{N/kg}. Calculate the pressure difference between its bottom and top, use this to calculate the upthrust, and predict its initial motion if its weight is 30N30\,\text{N}.

4.5.5.2

Atmospheric pressure (physics only)

  • Atmospheric pressure is produced by air molecules colliding with surfaces; the atmosphere is a thin layer of air around Earth.
  • As altitude increases, there are fewer air molecules above a surface and the air is less dense, so atmospheric pressure decreases.
  • A pressure difference across a surface produces a resultant normal force; calculate it by rearranging p=F/Ap=F/A to F=pAF=pA.
  • Do not say that atmospheric pressure becomes zero on a mountain; it decreases with height because there is less air above, but an atmosphere remains.

Tier 1 · Easy

1 mark
ORIGINAL

State what microscopic event produces atmospheric pressure on a window.

Tier 2 · Standard

3 marks
ORIGINAL

Explain why a barometer records a lower atmospheric pressure at the top of a tall mountain than at sea level.

Tier 3 · Hard

4 marks
ORIGINAL

At high altitude, the pressure inside a sealed case is 101kPa101\,\text{kPa} and the atmospheric pressure outside is 75kPa75\,\text{kPa}. A flat lid has area 0.012m20.012\,\text{m}^2. Calculate the resultant force on the lid and state its direction.

4.5.6.1.1

Distance and displacement

  • Distance is the total length of the path travelled and is scalar; displacement is the straight-line change from start to finish and includes direction, so it is vector.
  • Add every part of a route to find distance, but use only the start and finish positions to find displacement.
  • For example, travelling 9m9\,\text{m} east and then 4m4\,\text{m} west gives distance 13m13\,\text{m} and displacement 5m5\,\text{m} east.
  • Do not report displacement without a direction, and do not assume distance and displacement are equal unless the path is straight without reversing.

Tier 1 · Easy

2 marks
ORIGINAL

A walker travels 14m14\,\text{m} north and then 5m5\,\text{m} south. Determine the distance and displacement.

Tier 2 · Standard

4 marks
ORIGINAL

A robot moves 6.0m6.0\,\text{m} east and then 8.0m8.0\,\text{m} north. Calculate its distance and the magnitude and direction of its displacement.

Tier 3 · Hard

5 marks
ORIGINAL

A survey drone flies 600m600\,\text{m} east, 250m250\,\text{m} west and then 120m120\,\text{m} north. Calculate the total distance and the magnitude and direction of its displacement from launch.

4.5.6.1.2

Speed

  • Speed is a scalar rate of change of distance; typical values are about 1.5m/s1.5\,\text{m/s} for walking, 3m/s3\,\text{m/s} for running, 6m/s6\,\text{m/s} for cycling and 330m/s330\,\text{m/s} for sound in air.
  • Use s=vts=vt for constant speed and rearrange to v=s/tv=s/t; for non-uniform motion, average speed is total distance divided by total time.
  • For example, 450m450\,\text{m} travelled in 30s30\,\text{s} gives an average speed of 450÷30=15m/s450\div30=15\,\text{m/s}.
  • Do not average two speeds unless the time spent at each speed is equal; instead use the complete distance and complete time, including stops when appropriate.

Tier 1 · Easy

1 mark
ORIGINAL

A toy car travels 24m24\,\text{m} in 6.0s6.0\,\text{s} at constant speed. Calculate its speed.

Tier 2 · Standard

3 marks
ORIGINAL

A runner covers 300m300\,\text{m} in 48s48\,\text{s}, rests for 12s12\,\text{s} and then covers another 200m200\,\text{m} in 32s32\,\text{s}. Calculate the average speed for the whole interval.

Tier 3 · Hard

5 marks
ORIGINAL

A delivery drone travels 1.8km1.8\,\text{km} to a site in 2min 20s2\,\text{min}\ 20\,\text{s}. It waits for 30s30\,\text{s}, then returns along the same route at a constant 15m/s15\,\text{m/s}. Calculate its average speed for the complete trip from departure to return.

4.5.6.1.3

Velocity

  • Velocity is speed in a stated direction, so it is a vector quantity; speed is scalar.
  • For motion over an interval, use displacement rather than distance when finding average velocity, then give the direction of the displacement.
  • For example, a displacement of 60m60\,\text{m} west in 15s15\,\text{s} gives an average velocity of 4.0m/s4.0\,\text{m/s} west.
  • Do not use total path length to calculate average velocity, and do not omit direction from a velocity value.

Tier 1 · Easy

1 mark
ORIGINAL

A train moves north at a speed of 18m/s18\,\text{m/s}. State its velocity.

Tier 2 · Standard

3 marks
ORIGINAL

A cart moves 50m50\,\text{m} east and then 20m20\,\text{m} west in a total time of 10s10\,\text{s}. Calculate its average velocity.

Tier 3 · Hard

5 marks
ORIGINAL

A rescue vehicle travels 90m90\,\text{m} east and then 120m120\,\text{m} north in 30s30\,\text{s}. Calculate the magnitude and direction of its average velocity, and compare this with its average speed.

4.5.6.1.4

The distance–time relationship

  • On a distance–time graph, the vertical coordinate is the distance travelled and the horizontal coordinate is time; a horizontal section means the object is stationary.
  • Calculate speed from the gradient: v=ΔsΔtv=\dfrac{\Delta s}{\Delta t}. A steeper straight section represents a greater speed.
  • For example, a rise of 75m75\,\text{m} in 15s15\,\text{s} gives v=75/15=5.0m s1v=75/15=5.0\,\text{m s}^{-1}. Higher-tier students can find instantaneous speed by drawing a tangent to a curve.
  • A common error is to use the total coordinates instead of the changes between two points; use the rise and run of the chosen section.

Tier 1 · Easy

2 marks
ORIGINAL

A runner travels 156m156\,\text{m} in 48s48\,\text{s} at constant speed. Calculate the gradient of the distance–time graph.

Tier 2 · Standard

3 marks
ORIGINAL

A distance–time graph is a straight line from (0s,0m)(0\,\text{s},0\,\text{m}) to (24s,120m)(24\,\text{s},120\,\text{m}). It is then horizontal for 8s8\,\text{s}. State what happens during the horizontal section and calculate the speed during the first section.

Tier 3 · Hard

5 marks
ORIGINAL

A robot moves along a straight track. Its distance from the start is 0m0\,\text{m} at 0s0\,\text{s}, 54m54\,\text{m} at 12s12\,\text{s}, 150m150\,\text{m} at 28s28\,\text{s} and 150m150\,\text{m} at 40s40\,\text{s}. The graph joins these points with straight lines. Calculate the speed in each time interval and the average speed over all 40s40\,\text{s}.

4.5.6.1.5

Acceleration

  • Average acceleration is the change in velocity per unit time: a=Δvt=vuta=\dfrac{\Delta v}{t}=\dfrac{v-u}{t}, measured in m s2\text{m s}^{-2}.
  • The gradient of a velocity–time graph gives acceleration; a negative gradient represents deceleration when the object is moving in the positive direction. Higher-tier students also find distance or displacement from the area under the graph, using shapes or counting squares.
  • For example, changing velocity from 4m s14\,\text{m s}^{-1} to 16m s116\,\text{m s}^{-1} in 6s6\,\text{s} gives a=(164)/6=2.0m s2a=(16-4)/6=2.0\,\text{m s}^{-2}.
  • A common error is to divide the final velocity by time. First find the change vuv-u; Higher-tier area-under-graph work is a separate distance calculation.

Tier 1 · Easy

2 marks
ORIGINAL

A scooter increases its velocity from 3.0m s13.0\,\text{m s}^{-1} to 15.0m s115.0\,\text{m s}^{-1} in 8.0s8.0\,\text{s}. Calculate its average acceleration.

Tier 2 · Standard

4 marks
ORIGINAL

A velocity–time graph rises uniformly from 00 to 12m s112\,\text{m s}^{-1} in 4.0s4.0\,\text{s}, stays horizontal for 6.0s6.0\,\text{s}, then falls uniformly to 00 in 3.0s3.0\,\text{s}. Determine the acceleration in each section.

Tier 3 · Hard

5 marks
ORIGINAL

A train accelerates uniformly from 8.0m s18.0\,\text{m s}^{-1} to 20.0m s120.0\,\text{m s}^{-1} while travelling 168m168\,\text{m}. Calculate its acceleration and the time taken.

4.5.6.2.1

Newton's First Law

  • Newton's First Law states that an object remains at rest, or continues at constant velocity, when the resultant force on it is zero.
  • For a vehicle moving steadily, the driving force balances the resistive forces; this is dynamic equilibrium, not an absence of forces.
  • For example, a 900N900\,\text{N} driving force and 900N900\,\text{N} resistance give zero resultant force, so the velocity does not change.
  • A common error is to claim that a forward force is needed to maintain motion. A resultant force is needed only to change velocity; Higher-tier students name resistance to change as inertia.

Tier 1 · Easy

2 marks
ORIGINAL

A boat moves at constant velocity. Its propeller provides a forward force of 680N680\,\text{N}. Determine the total resistive force.

Tier 2 · Standard

3 marks
ORIGINAL

A van travels in a straight line at a steady 18m s118\,\text{m s}^{-1}. The engine force is 920N920\,\text{N}. Explain the motion in terms of the forces and state the resultant force.

Tier 3 · Hard

5 marks
ORIGINAL

A lift moves upward. The motor force is 6200N6200\,\text{N} upward, its weight is 5800N5800\,\text{N} and friction is 400N400\,\text{N} downward. Describe its motion. The motor force then falls to 5000N5000\,\text{N} while the lift is still moving upward. Calculate the new resultant force and describe the change in motion.

4.5.6.2.2

Newton's Second Law

  • Newton's Second Law gives F=maF=ma: acceleration is proportional to resultant force and inversely proportional to mass.
  • First combine all forces with directions to find the resultant force, then substitute SI units into a=F/ma=F/m.
  • For example, a 10N10\,\text{N} pull opposed by 4N4\,\text{N} friction gives F=6NF=6\,\text{N}; for m=2kgm=2\,\text{kg}, a=3m s2a=3\,\text{m s}^{-2}. Higher-tier students define inertial mass as m=F/am=F/a, a measure of how difficult it is to change velocity.
  • A common error is to substitute the applied force instead of the resultant force. In the required practical, vary force at constant total mass or vary mass at constant force, measure acceleration and repeat readings.

Tier 1 · Easy

2 marks
ORIGINAL

A resultant force accelerates a 12kg12\,\text{kg} object at 1.5m s21.5\,\text{m s}^{-2}. Calculate the resultant force.

Tier 2 · Standard

5 marks
ORIGINAL

A student uses a wheeled cart, a pulley and slotted masses to investigate how force affects acceleration while total mass stays constant. Describe how the student should change the force, measure the acceleration and improve the reliability of the results. State the expected relationship.

Tier 3 · Hard

5 marks
ORIGINAL

A 1250kg1250\,\text{kg} car increases its velocity from 8.0m s18.0\,\text{m s}^{-1} to 26.0m s126.0\,\text{m s}^{-1} in 9.0s9.0\,\text{s}. The resistive forces total 700N700\,\text{N}. Calculate the acceleration, the resultant force and the engine force.

4.5.6.2.3

Newton's Third Law

  • Newton's Third Law states that whenever two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.
  • Name both objects when identifying a pair: the force of A on B pairs with the force of B on A.
  • For example, a swimmer pushes water backwards and the water pushes the swimmer forwards with an equal force.
  • A common error is to treat balanced forces on one object as a third-law pair. Third-law forces act on different objects, so they do not cancel on either object.

Tier 1 · Easy

2 marks
ORIGINAL

A hammer exerts a downward force on a nail. State the corresponding Newton's Third Law force.

Tier 2 · Standard

3 marks
ORIGINAL

A book rests on a table. The Earth pulls the book downward. Identify the Newton's Third Law partner to this force and explain why it is not the table's upward force on the book.

Tier 3 · Hard

5 marks
ORIGINAL

A propeller pushes water backward with a force of 4.2kN4.2\,\text{kN}. The water resistance on the boat is 1.8kN1.8\,\text{kN} backward and the boat's mass is 1200kg1200\,\text{kg}. State the third-law force exerted by the water because of the propeller interaction, then calculate the boat's acceleration.

4.5.6.3.1

Stopping distance

  • Stopping distance is the sum of thinking distance and braking distance: sstop=sthinking+sbrakings_{\text{stop}}=s_{\text{thinking}}+s_{\text{braking}}.
  • Thinking distance can be calculated from s=vts=vt when speed is constant during the reaction time; braking distance begins once the brakes act.
  • For example, a 9m9\,\text{m} thinking distance and 27m27\,\text{m} braking distance give a 36m36\,\text{m} stopping distance.
  • A common error is to add the reaction time directly to a distance. Convert it to thinking distance first and use consistent units.

Tier 1 · Easy

1 mark
ORIGINAL

A driver's thinking distance is 12m12\,\text{m} and the braking distance is 28m28\,\text{m}. Calculate the stopping distance.

Tier 2 · Standard

3 marks
ORIGINAL

A car travels at 15m s115\,\text{m s}^{-1}. The driver's reaction time is 0.64s0.64\,\text{s} and the braking distance is 19m19\,\text{m}. Calculate the thinking distance and the stopping distance.

Tier 3 · Hard

5 marks
ORIGINAL

A delivery van has a reaction time of 0.75s0.75\,\text{s}. At 20m s120\,\text{m s}^{-1} its braking distance is 32m32\,\text{m}; at 25m s125\,\text{m s}^{-1} its braking distance is 50m50\,\text{m}. Calculate the stopping distance at each speed and the increase in stopping distance.

4.5.6.3.2

Reaction time

  • Human reaction times vary; typical values are about 0.2s0.2\,\text{s} to 0.9s0.9\,\text{s}.
  • Tiredness, alcohol, some drugs and distractions can increase reaction time, so a vehicle travels farther before braking begins.
  • A ruler-drop test can compare reactions: keep release conditions constant, repeat readings, identify anomalies and compare mean results.
  • A common error is to change several variables at once or use one reading. Control the method and use repeats before judging an effect.

Tier 1 · Easy

2 marks
ORIGINAL

State the typical range of human reaction times and give one factor that can increase a driver's reaction time.

Tier 2 · Standard

2 marks
ORIGINAL

A driver travels at 22m s122\,\text{m s}^{-1} and reacts in 0.48s0.48\,\text{s}. Calculate the thinking distance.

Tier 3 · Hard

5 marks
ORIGINAL

A student measures reaction time four times without a distraction and obtains 0.310.31, 0.290.29, 0.300.30 and 0.89s0.89\,\text{s}. With a distraction the results are 0.440.44, 0.470.47, 0.450.45 and 0.46s0.46\,\text{s}. Identify the anomalous result, calculate suitable mean times and evaluate the effect of the distraction.

4.5.6.3.3

Factors affecting braking distance 1

  • Braking distance increases when road grip is reduced by wet or icy conditions, or when tyres or brakes are in poor condition.
  • Reduced friction produces a smaller braking force and deceleration, so the vehicle travels farther before stopping.
  • If conditions and braking force are comparable, greater initial speed produces a much larger braking distance; stopping-distance data may be used for estimates.
  • A common error is to say tiredness increases braking distance. Tiredness changes reaction time and thinking distance, whereas grip and vehicle condition change braking distance.

Tier 1 · Easy

2 marks
ORIGINAL

Give one example of poor vehicle condition that increases braking distance and explain why it does so.

Tier 2 · Standard

3 marks
ORIGINAL

The average braking force on a car is 6200N6200\,\text{N} on a dry road and 3100N3100\,\text{N} on a wet road. The car has the same initial speed in both tests and stops in 18m18\,\text{m} on the dry road. Estimate the wet-road braking distance.

Tier 3 · Hard

4 marks
ORIGINAL

For one car in fixed conditions, measured braking distances are 6.2m6.2\,\text{m} at 10m s110\,\text{m s}^{-1}, 13.9m13.9\,\text{m} at 15m s115\,\text{m s}^{-1} and 24.8m24.8\,\text{m} at 20m s120\,\text{m s}^{-1}. Show that the data are consistent with braking distance being proportional to speed squared, then estimate the distance at 25m s125\,\text{m s}^{-1}.

4.5.6.3.4

Factors affecting braking distance 2

  • During braking, friction does work and reduces the vehicle's kinetic energy; energy is transferred to the thermal energy stores of the brakes and surroundings.
  • A greater initial speed means more kinetic energy, so a greater braking force is needed to stop in the same distance.
  • For a given mass, a greater braking force produces a greater deceleration. Very large decelerations can overheat brakes or cause loss of control.
  • A common error is to say kinetic energy is destroyed. Track the energy transfer and distinguish a large deceleration from a long stopping time.

Tier 1 · Easy

2 marks
ORIGINAL

Describe the main energy transfer when friction in a vehicle's brakes brings the vehicle to rest.

Tier 2 · Standard

4 marks
ORIGINAL

A 1100kg1100\,\text{kg} car travels at 18m s118\,\text{m s}^{-1}. Its average braking force is 6600N6600\,\text{N}. Calculate its initial kinetic energy and the braking distance, assuming all of this energy is removed by the braking force.

Tier 3 · Hard

6 marks
ORIGINAL

A 1500kg1500\,\text{kg} car travelling at 22m s122\,\text{m s}^{-1} is stopped by an average braking force of 8250N8250\,\text{N}. Calculate the braking distance and the magnitude of the deceleration. Explain one danger of increasing the braking force substantially.

4.5.7.1

Momentum is a property of moving objects (HT only)

  • Momentum is defined by p=mvp=mv, where pp is in kg m s1\text{kg m s}^{-1}, mm is in kilograms and vv is in m s1\text{m s}^{-1}.
  • Momentum is a vector, so choose a positive direction and give momenta in the opposite direction negative signs.
  • For example, a 0.40kg0.40\,\text{kg} ball moving at 6.0m s16.0\,\text{m s}^{-1} has momentum 2.4kg m s12.4\,\text{kg m s}^{-1} in its direction of travel.
  • A common error is to use mass in grams or omit direction. Convert mass to kilograms before applying p=mvp=mv.

Tier 1 · Easy

2 marks
ORIGINAL

A 0.18kg0.18\,\text{kg} ball moves at 12m s112\,\text{m s}^{-1}. Calculate its momentum.

Tier 2 · Standard

2 marks
ORIGINAL

A 1350kg1350\,\text{kg} car travels west at 16m s116\,\text{m s}^{-1}. Calculate its momentum, including direction.

Tier 3 · Hard

5 marks
ORIGINAL

Vehicle A has mass 720kg720\,\text{kg} and travels east at 18m s118\,\text{m s}^{-1}. Vehicle B has mass 1080kg1080\,\text{kg} and travels west at 11m s111\,\text{m s}^{-1}. Taking east as positive, calculate each momentum and determine which has the greater momentum magnitude and by how much.

4.5.7.2

Conservation of momentum (HT only)

  • In a closed system, total momentum before an event equals total momentum after it.
  • Choose a positive direction, write pbefore=pafter\sum p_{\text{before}}=\sum p_{\text{after}}, and include a negative sign for motion in the opposite direction.
  • If objects stick together, their final momentum is (m1+m2)v(m_1+m_2)v because they share one final velocity.
  • A common error is to conserve kinetic energy in every collision. Momentum is conserved in a closed system, but kinetic energy need not be.

Tier 1 · Easy

3 marks
ORIGINAL

A 2.0kg2.0\,\text{kg} trolley moving at 3.0m s13.0\,\text{m s}^{-1} collides with a stationary 1.0kg1.0\,\text{kg} trolley. They stick together. Calculate their common velocity.

Tier 2 · Standard

4 marks
ORIGINAL

A 0.75kg0.75\,\text{kg} trolley moving right at 6.4m s16.4\,\text{m s}^{-1} catches a 1.25kg1.25\,\text{kg} trolley moving right at 1.6m s11.6\,\text{m s}^{-1}. The trolleys lock together. Determine their final velocity.

Tier 3 · Hard

4 marks
ORIGINAL

A launcher of mass 3.8kg3.8\,\text{kg} and a 0.20kg0.20\,\text{kg} projectile are initially at rest. The projectile is fired horizontally at 32m s132\,\text{m s}^{-1}. Calculate the launcher's recoil velocity.

4.5.7.3

Changes in momentum (physics only) (HT only)

  • Force equals the rate of change of momentum: F=ΔpΔt=mΔvΔtF=\dfrac{\Delta p}{\Delta t}=\dfrac{m\Delta v}{\Delta t} for constant mass.
  • The product FΔtF\Delta t (force ×\times time) equals the change in momentum; use signed velocities when the object reverses direction.
  • For the same momentum change, increasing collision time reduces the average force, which is the principle behind airbags, helmets and crash mats.
  • A common error is to use speed change when direction reverses. Choose a positive direction and calculate Δv=vu\Delta v=v-u.

Tier 1 · Easy

3 marks
ORIGINAL

A 0.35kg0.35\,\text{kg} ball moving at 8.0m s18.0\,\text{m s}^{-1} is brought to rest in 0.10s0.10\,\text{s}. Calculate the magnitude of the average force.

Tier 2 · Standard

4 marks
ORIGINAL

A 0.16kg0.16\,\text{kg} ball travels toward a wall at 12m s112\,\text{m s}^{-1} and rebounds at 8.0m s18.0\,\text{m s}^{-1}. Contact lasts 0.050s0.050\,\text{s}. Calculate the magnitude and direction of the average force on the ball.

Tier 3 · Hard

6 marks
ORIGINAL

A 950kg950\,\text{kg} vehicle travelling at 14m s114\,\text{m s}^{-1} stops in a collision. A rigid structure would stop it in 0.080s0.080\,\text{s}, while a crumple zone increases the stopping time to 0.32s0.32\,\text{s}. Calculate the average force magnitude in each case and explain the safety benefit of the crumple zone.