150 free, high-quality past-paper-style questions on Work, energy and power for IB Physics SL, each with a full mark scheme. Covers: Work done by a force; Conservation of energy; Sankey diagrams; and more. For when you have finished the past papers, or want to practise one topic at a time.A.3 at Higher Level →Read the A.3 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.765 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-09Work done by a forcePaper 1AEasy1 mark
A box is dragged 4.0 m along horizontal ground by a force of 25 N applied at 60° to the horizontal.
What is the work done on the box by this force?
Diagram not to scale
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Step 1Only the component of the force along the direction of motion does work, so W = Fs cos θ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2W = 25 × 4.0 × cos 60° = 25 × 4.0 × 0.500 = 50 J.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is Fs, ignoring the angle. Only the horizontal component of the force does work here.
BThis uses sin 60°. The angle in W = Fs cos θ is measured between the force and the displacement.
CCorrect: W = Fs cos θ = 25 × 4.0 × cos 60° = 50 J.
DThe work would be zero only if the force were perpendicular to the displacement, that is at 90°.
A ball is released from rest at a height of 5.0 m above the ground. Air resistance is negligible. Take g = 9.81 m s−2.
What is its speed just before it lands?
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Step 1With no resistive forces the mechanical energy is conserved, so the loss of gravitational potential energy equals the gain in kinetic energy: mgh = ½mv². The mass cancels, giving v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis is √(gh). The factor of 2 has been lost — v² = 2gh, not gh.
BCorrect: mgh = ½mv², so v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.
CThis is gh. Check the algebra — you need the square root of 2gh.
DThis is v² = 2gh = 98, left unrooted. The units give it away: m² s⁻², not m s⁻¹.
A skier of mass 70 kg slides 50 m down a straight slope inclined at 20° to the horizontal.
What is the work done on the skier by the gravitational force?
Diagram not to scale
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Step 1The angle between the weight (vertically down) and the displacement (along the slope) is 90° − 20° = 70°; equivalently the skier descends 50 sin 20° = 17.1 m.
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The petrol engine of a leaf blower has an efficiency of 25 %. Each Sankey diagram is drawn to scale for 100 J of chemical energy supplied to the engine.
Which Sankey diagram represents the energy transfers in the engine?
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Step 1Useful output = 0.25 × 100 J = 25 J, so the wasted energy is 100 − 25 = 75 J: the two output bands must together be as wide as the input band. Only diagram D shows 25 J useful and 75 J thermal.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThe useful and wasted bands are swapped: this diagram shows an efficiency of 75/100 = 75 %.
BThe useful band is right, but the outputs add up to only 75 J. Energy is conserved, so the band widths must add up to the 100 J input.
CThis takes efficiency as useful ÷ wasted (20/80 = 0.25). Efficiency is useful output ÷ total input, which here is 20/100 = 20 %.
DCorrect: 25 J of the 100 J input is useful (25 %) and the other 75 J is dissipated, so the widths add up to the input.
A solar-powered boat moves across a lake at a constant speed of 3.0 m s−1. The useful output power of its motor is 1.2 kW.
What is the total resistive force on the boat?
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Step 1At constant speed the driving force equals the total resistive force, and P = Fv.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2F = P/v = 1.2 × 103/3.0 = 4.0 × 102 N.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis multiplies the power by the speed. P = Fv, so the force is P ÷ v.
BCorrect: the driving force balances the resistive force, and F = P/v = 1200/3.0 = 400 N.
CThe power has been left in kilowatts: 1.2 ÷ 3.0 = 0.40. The power must be in watts, 1.2 × 103 W.
DThis confuses the resultant force, which is zero at constant speed, with the resistive force. The motor must keep pushing to balance a non-zero resistive force.
A child bounces vertically on a pogo stick, which contains a spring. Air resistance and friction are negligible. At the lowest point of a bounce the spring has its maximum compression.
Which row describes the child-and-pogo-stick system at the lowest point?
Kinetic energyTotal mechanical energy
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Step 1At maximum compression the spring stops the child, so the speed, and the kinetic energy, is zero.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Mechanical energy is the sum of kinetic, gravitational potential and elastic potential energy; with no friction or air resistance it is conserved, so it is the same as at the highest point: the gravitational potential energy lost has been transferred to elastic potential energy.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe kinetic energy is right, but the energy stored in the compressed spring is elastic potential energy, which is part of the mechanical energy.
BThe child moves fastest during the compression, not at the lowest point. At maximum compression the child is momentarily at rest.
CThis adds the spring's energy to the energy the child had at the top, as if the spring were a source of energy. The elastic potential energy comes from the kinetic and gravitational potential energy.
DCorrect: at maximum compression the child is momentarily at rest, and with no resistive forces the mechanical energy (kinetic + gravitational potential + elastic potential) equals its value at the top.
A cyclist and her bicycle have a total mass of 80 kg. She rides 12 km along a mountain-pass road that rises through a vertical height of 600 m. The climb takes 50 minutes.
What is the average rate at which the cyclist and bicycle gain gravitational potential energy?
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Step 1Gain in gravitational potential energy = mgΔh = 80 × 9.81 × 600 = 4.7 × 105 J.
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Step 2Rate = 4.7 × 105 J ÷ (50 × 60 s) = 157 W ≈ 160 W.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe time has been left in minutes: 4.7 × 105 ÷ 50. It must be in seconds, 3000 s.
BThis uses the 12 km length of the road as the height. Only the vertical rise, 600 m, determines the change in gravitational potential energy.
CCorrect: mgΔh = 80 × 9.81 × 600 = 4.7 × 105 J gained in 3000 s, about 160 W.
DThis leaves out g: 80 × 600 ÷ 3000 = 16, which is not an energy per second. ΔEp = mgΔh.
A Sankey diagram for an LED desk lamp is drawn to scale. The input electrical energy arrow is 40 mm wide. It splits into two arrows: one for light, 10 mm wide, and one for thermal energy transferred to the surroundings. What is the efficiency of the lamp?
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In a scaled Sankey diagram, arrow width is proportional to energy transferred. Efficiency = useful output ÷ total input = 10 mm ÷ 40 mm = 0.25 = 25 %. 33 % comes from dividing useful output by wasted output (10 ÷ 30). 75 % is the fraction that is wasted. 10 % treats the width in millimetres as a percentage.
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Answer B
Answer: B
Common error: Efficiency always divides useful output by total input, never by the wasted output.
10IBP2S-005-01-02Conservation of energyPaper 1AEasy1 mark
An archer holds a drawn bow and then releases an arrow vertically upwards. Air resistance is negligible. Which sequence describes the main energy transfers from the moment of release until the arrow reaches its highest point?
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At release the energy is stored as elastic potential energy in the bent bow. This becomes kinetic energy of the arrow. As the arrow rises, its kinetic energy becomes gravitational potential energy, which is greatest at the highest point. Chemical energy is used by the archer to draw the bow, but that happens before release. The other two sequences put the stages in the wrong order.
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Answer A
Answer: A
Exam tip: Check the start and end points given in the question so that you include only the transfers in that interval.
11IBP2S-005-01-09Conservation of energyPaper 1AEasy1 mark
A skydiver falls vertically at constant terminal speed. Which statement is correct for this part of the fall?
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At constant speed the kinetic energy does not change. So the gravitational potential energy lost cannot become kinetic energy. Instead, the work done against air resistance transfers it to internal (thermal) energy of the air and the skydiver. Total energy is always conserved, so D is wrong.
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Answer C
Answer: C
Exam tip: Mechanical energy can decrease, but total energy is always conserved. Say where the 'lost' energy goes.
12IBP2S-005-03-01Elastic potential energyPaper 1AEasy1 mark
A spring in a hydraulic door-closer obeys Hooke's law. When the door is opened further, the extension of the spring increases from x to 2x. By what factor does the elastic potential energy stored in the spring increase?
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Elastic potential energy is \(E_p = \frac{1}{2}kx^2\), so it is proportional to the square of the extension. Doubling the extension multiplies the energy by 22 = 4. Choosing 2 treats energy as proportional to extension, which is how the force behaves. Choosing 8 or √2 uses the wrong power of the extension.
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Answer C
Answer: C
Common error: Force is proportional to extension, but elastic energy is proportional to extension squared. Do not carry the force relationship over to the energy.
13IBP2S-005-03-02Elastic potential energyPaper 1AEasy1 mark
A sports scientist records a force–extension graph for a human Achilles tendon as it is stretched during a sprint start. Which quantity represents the work done in stretching the tendon?
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Work done is force × displacement in the direction of the force. For a varying force this sum is the area under the force–extension graph, which is the area between the curve and the extension axis. The gradient gives the stiffness, not the energy. The product Fmaxxmax is the area of a rectangle, which overestimates the work. The area against the force axis has no energy meaning here.
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Answer B
Answer: B
Exam tip: 'Area under the graph' means the area down to the extension (horizontal) axis. This rule works even when the graph is not a straight line.
14IBP2S-005-03-03Elastic potential energyPaper 1AEasy1 mark
A spring in a robotic gripper has spring constant 200 N m−1. It is compressed by 0.10 m. What is the elastic potential energy stored?
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\(E_p = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times 0.10^2 = 1.0\) J. Each distractor comes from a specific slip: • 2.0 J omits the ½. • 10 J does not square the compression. • 20 J is kx, which is the force in newtons, not an energy.
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Answer A
Answer: A
Common error: Check the units of your result. kx has units of N, while ½kx² has units of N m = J.
15IBP2S-005-02-01Work done by a forcePaper 1AEasy1 mark
A communications satellite moves at constant speed in a circular orbit of radius r. The gravitational force on the satellite has constant magnitude F. What is the work done by the gravitational force on the satellite during one complete orbit?
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The gravitational force always points towards the centre of the orbit, so it is perpendicular to the satellite's velocity and to every small displacement. Work = F s cos 90° = 0 for each small step, so the total work is zero. This agrees with the constant speed: the kinetic energy does not change. Option B multiplies the force by the path length and ignores direction. Option D treats the radius as a displacement along the force. Option C halves the path length for no reason.
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Answer A
Answer: A
Common error: Work depends on the component of displacement along the force, not on the total distance travelled. A force perpendicular to the motion does no work, however far the object moves.
16IBP2S-005-02-02Work done by a forcePaper 1AEasy1 mark
A hospital porter pulls a bed trolley along a level corridor with a force of 50 N directed at 60° above the horizontal. The trolley moves 10 m along the corridor. How much work is done on the trolley by the porter's force?
Diagram not to scale
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Only the component of the force along the displacement does work: W = Fs cos θ = 50 × 10 × cos 60° = 50 × 10 × 0.5 = 250 J. Option A uses sin 60°, which gives the vertical component. Option C ignores the angle completely. Option D wrongly assumes that a force at an angle to the motion does no work at all.
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Answer B
Answer: B
Exam tip: Use cos θ when θ is the angle between the force and the displacement. Check that the angle in the question is measured from the direction of motion.
17IBP2S-005-02-09Kinetic energy & momentumPaper 1AEasy1 mark
A brake pad does work against friction as it slows a bicycle wheel. Which of the following gives the unit of this work in SI base units?
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Work = force × distance, and 1 N = 1 kg m s−2. Multiplying by metres gives 1 J = 1 kg m2 s−2. Option A is the unit of force. Option B is the unit of power, because it has an extra division by seconds. Option C is the unit of momentum.
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Answer D
Answer: D
Exam tip: Build derived units from defining equations: W = Fs, F = ma, P = W/t.
18IBP2S-005-04-01Energy density of fuelsPaper 1AEasy1 mark
Which is a correct SI unit for the energy density of a fuel?
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Energy density is the energy released per unit volume of fuel, so its unit is J m−3. J kg−1 is the unit of specific energy (energy per unit mass). W m−3 is power per unit volume, not energy. J m−2 would be energy per unit area.
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Answer B
Answer: B
Common error: Do not mix up energy density (per m³) with specific energy (per kg). Check the unit given in the data before you use a value.
19IBP2S-005-04-02Energy density of fuelsPaper 1AEasy1 mark
Hydrogen gas has a specific energy roughly three times that of petrol. However, at atmospheric pressure a fuel tank filled with hydrogen gas stores far less energy than the same tank filled with petrol. The best explanation is that hydrogen gas
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Energy stored in a tank = energy density × volume, and energy density = specific energy × density. Although hydrogen's specific energy is high, its density at atmospheric pressure is tiny, so its energy per unit volume is very small. Option A contradicts the data. Flame temperature and efficiency do not affect how much chemical energy the tank stores.
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Answer B
Answer: B
Exam tip: Remember the link: energy density = specific energy × density. A fuel can be excellent per kilogram and poor per cubic metre.
20IBP2S-005-04-06Energy density of fuelsPaper 1AEasy1 mark
Which fuel source has the greatest specific energy?
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Fission of uranium releases energy from nuclear binding, which is about a million times larger per kilogram than the chemical energy released when wood, coal or petrol burns. Of the chemical fuels, petrol is highest, then coal, then wood, but all are far below nuclear fuel.
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Answer D
Answer: D
Exam tip: Know rough orders of magnitude: chemical fuels around 10⁷ J kg⁻¹, nuclear fission fuel around 10¹³–10¹⁴ J kg⁻¹.
21IBP2S-005-00-01Kinetic energy & momentumPaper 1AEasy1 mark
Object X has mass m and object Y has mass 4m. Both have the same momentum. What is \(\frac{\text{kinetic energy of X}}{\text{kinetic energy of Y}}\)?
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Kinetic energy can be written as Ek = p2/2m. With p the same, Ek ∝ 1/m, so the lighter object X has 4 times the kinetic energy of Y. Option A reverses the dependence on mass. Option B assumes equal momentum means equal energy. Option C takes a square root that should not be there.
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Answer D
Answer: D
Common error: Equal momentum does not mean equal kinetic energy. For a fixed momentum, the lighter object always carries more kinetic energy.
22IBP2S-005-00-02Sankey diagramsPaper 1AEasy1 mark
A medical X-ray tube converts 1.0% of its electrical input power into X-rays. The rest becomes internal energy in the anode. The electrical input power is 2.0 kW. What is the rate at which internal energy is transferred to the anode?
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Useful output = 0.010 × 2000 W = 20 W. Wasted power = input − useful = 2000 − 20 = 1980 W, and this goes to the anode. Option A is the useful X-ray power. Option C ignores the energy carried away as X-rays. Option D adds the useful power instead of subtracting it.
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Answer B
Answer: B
Exam tip: Input power = useful output power + wasted power. Write this out each time to avoid adding when you should subtract.
23IBP2S-005-00-03Power & efficiencyPaper 1AEasy1 mark
Which of the following expresses the unit of power in SI base units?
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1 W = 1 J s−1 and 1 J = 1 N m = 1 kg m2 s−2. So 1 W = 1 kg m2 s−3. Option A is the newton. Option B is the joule. Option D is the unit of J s, which comes from multiplying by time instead of dividing.
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Answer C
Answer: C
Exam tip: Build derived units step by step: force, then energy, then power. Each step adds or removes one factor.
25A-1A-11Conservation of energyPaper 1AMedium1 mark
A crate is pulled up a rough ramp by a rope parallel to the ramp, and its speed increases. Over a certain distance the rope does work W on the crate, the crate gains kinetic energy ΔEk and gravitational potential energy ΔEp, and the work done against friction is Wf.
Which equation is correct?
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Step 1Energy is conserved: the energy transferred to the crate by the rope becomes kinetic energy, gravitational potential energy and, through the work done against friction, internal energy of the crate and ramp, so W = ΔEk + ΔEp + Wf.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis treats friction as a source of energy. Work done against friction is energy the rope must supply in addition, dissipated as internal energy.
BThis treats the gain in height as releasing energy. Moving up the ramp increases the gravitational potential energy, which the rope must supply.
CThis treats the rope's work as the work done by the resultant force. The resultant force also includes the weight component and friction, so the rope does more work than the gain in kinetic energy.
DCorrect: all the work done by the rope is accounted for as kinetic energy, gravitational potential energy and the energy dissipated against friction.
27A-1A-13Elastic potential energyPaper 1AMedium1 mark
A block is held against a spring on a frictionless horizontal surface, compressing the spring by x. When the block is released, the spring pushes it away. The spring obeys Hooke's law.
The spring is now compressed by 2x instead. What happens to the elastic potential energy stored before release and to the speed of the block after it leaves the spring?
Elastic potential energy storedSpeed of the block
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Step 1EH = ½k(Δx)², so doubling the compression stores 2² = 4 times as much energy.
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Step 2With no friction all of it becomes kinetic energy, ½mv², so v ∝ √EH and the speed doubles.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis takes the stored energy to be proportional to the compression, as the force is. The energy is the area under the force–compression graph, ½k(Δx)².
BThis takes the stored energy to be proportional to the compression and then correctly takes v ∝ √EH.
CThe energy is right, but the speed has been scaled like the energy. Kinetic energy ∝ v², so four times the energy gives twice the speed.
DCorrect: ½k(Δx)² gives × 4 for the energy, and ½mv² = EH gives × √4 = × 2 for the speed.
29A-1A-34Energy density of fuelsPaper 1AMedium1 mark
A diesel generator is 40 % efficient. The energy density of its fuel is 45 MJ kg−1.
What mass of fuel is used when the generator supplies 9.0 MJ of electrical energy?
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Step 1Energy that must be released by the fuel: Einput = Eoutput/η = 9.0/0.40 = 22.5 MJ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Mass of fuel = 22.5 MJ ÷ 45 MJ kg−1 = 0.50 kg.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis multiplies by the efficiency instead of dividing: 9.0 × 0.40/45. Only 40 % of the fuel energy is useful, so more fuel is needed than 9.0 MJ alone would suggest, not less.
BThis ignores the efficiency, assuming all the fuel's energy becomes electrical energy: 9.0/45.
CThis divides by the wasted fraction, 0.60, instead of the useful fraction, 0.40: 9.0/0.60/45.
DCorrect: the fuel must release 9.0/0.40 = 22.5 MJ, which needs 22.5/45 = 0.50 kg.
A puck on a smooth horizontal table is attached by a string to a fixed pin. The puck moves in a horizontal circle around the pin at constant speed.
Which statement about the work done on the puck by the tension in the string is correct?
Plan view of the smooth table
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Step 1In W = Fs cos θ, the tension points to the centre and every small displacement is along the tangent, so θ = 90° and no work is done at any stage of the motion — consistent with the constant speed (no change in kinetic energy).
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the tension is perpendicular to every small displacement, so cos 90° = 0 and no energy is transferred; this is why the speed stays constant.
BA force does work only if it has a component along the displacement. A non-zero force acting on a moving body can still do no work.
CNo work is done over any part of a revolution, not only over a whole one. The work depends on the angle between the force and each small displacement, not on the overall displacement.
DIf the tension did work, the kinetic energy of the puck would change. The speed is constant, so no energy is transferred.
A box is pushed at constant velocity across a rough horizontal floor.
Which statement is correct?
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Step 1The kinetic energy is constant, and the work done by the resultant force equals the change in kinetic energy, so the total work done by all the forces is zero: the positive work of the push equals the negative work of friction.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThe push acts along the displacement, so it does positive work. It is the total work, not the work of the push alone, that equals the change in kinetic energy.
BCorrect: the resultant force is zero, so the total work done is zero and the kinetic energy does not change.
CFriction acts opposite to the displacement, so cos θ = −1 and the work done by friction is negative.
DMotion at constant velocity needs no resultant force: the push equals the friction, so the two amounts of work are equal in magnitude.
A lift cabin of mass M hangs from a cable that passes over a drive wheel at the top of the lift shaft. A counterweight of mass m, where m < M, hangs from the other end of the cable. A motor turns the wheel so that the cabin rises at a constant speed v. Friction is negligible.
What is the useful output power of the motor?
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Step 1Each second the cabin gains gravitational potential energy Mgv and the counterweight, moving down at the same speed, loses mgv; the kinetic energy is constant.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Power = rate of increase of energy of the system = (M − m)gv.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the counterweight's loss of gravitational potential energy supplies part of the cabin's gain, so the motor supplies only (M − m)gv.
BThis ignores the counterweight. As it descends it loses potential energy, which reduces the work the motor must do.
CThis treats the counterweight as an extra load to be lifted. It moves downwards, so its potential energy decreases.
DThis is the kinetic energy of the moving cabin and counterweight. It is constant, and it is an energy, not a rate of energy transfer.
34A-1A-66Energy density of fuelsPaper 1AMedium1 mark
A hiker of mass 75 kg eats a 50 g cereal bar. The energy density of the bar is 18 MJ kg−1. Her body converts chemical energy into gravitational potential energy with an efficiency of 20 %.
What is the greatest height she could climb using the energy from the bar?
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Step 1Energy from the bar = 0.050 kg × 18 × 106 J kg−1 = 9.0 × 105 J; useful energy = 0.20 × 9.0 × 105 = 1.8 × 105 J.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2h = 1.8 × 105 ÷ (75 × 9.81) = 2.4 × 102 m.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis divides the useful energy by the mass only, leaving out g: 1.8 × 105/75 is not a height.
BThis ignores the efficiency, as if all of the chemical energy became gravitational potential energy.
CCorrect: 20 % of 9.0 × 105 J is 1.8 × 105 J, and mgΔh = 1.8 × 105 J gives Δh ≈ 240 m.
35A-1A-105Conservation of energyPaper 1AMedium1 mark
A hummingbird hovers at rest in front of a flower. Its wings continuously push air downwards.
Which statement about the energy transfers while the bird hovers is correct?
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Step 1The bird's displacement is zero, so no work is done on the bird and its kinetic and gravitational potential energies stay constant. But the wings move and push the air, so they do work on the air: chemical energy is transferred to kinetic energy of the air (and, in the muscles and the air, to internal energy). Energy is conserved throughout.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
ANo work is done on the bird, but the wings are moving and push on the air, so work is done on the air. Chemical energy must be used to supply it.
BWork needs a displacement along the force. The bird does not rise, so the force of the air on it does no work and its gravitational potential energy does not change.
CCorrect: the moving wings exert a force on the air along the air's displacement, so they do work on it; the chemical energy becomes kinetic energy of the air and internal energy.
DEnergy is transferred, not destroyed. The chemical energy that is used ends up as kinetic and internal energy of the air and the bird, so the total is unchanged.
36A-1A-106Work done by a forcePaper 1AMedium1 mark
A shopper of mass m stands still on a step of an escalator that carries her upwards at constant speed. She rises through a vertical height h.
Which row gives the work done on the shopper by the contact force from the step and the work done on her by the gravitational force?
Work done by contact force from stepWork done by gravitational force
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Step 1The velocity is constant, so the contact force from the step is vertical and equal to mg. Her displacement has a vertical component h in the direction of this force, so W = Fs cos θ = +mgh.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The weight acts opposite to the vertical component of the displacement, so the gravitational force does −mgh; the total work is zero, matching the constant kinetic energy.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the upward contact force has a component of displacement h along it (+mgh); the weight is opposed to that component (−mgh), so the total work, and the change in kinetic energy, is zero.
BThis assumes a normal force can never do work. The step's force is vertical but the shopper moves along the slope, so the angle between them is less than 90° and the force does work.
CThis assumes that no force does work when the speed is constant. It is the total work that is zero; the two forces each do work, of opposite signs.
DThe signs are reversed. The weight acts downwards while she moves up, so the gravitational force does negative work; her gain in gravitational potential energy is supplied by the work done by the step.
37A-1A-108Kinetic energy & momentumPaper 1AMedium1 mark
In a particle accelerator, a deuteron and a proton are given the same kinetic energy. The mass of the deuteron is twice the mass of the proton, and both speeds are much less than the speed of light.
Which row gives (momentum of deuteron ÷ momentum of proton) and (speed of deuteron ÷ speed of proton)?
Momentum ratioSpeed ratio
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Step 1Ek = p²/2m, so p = √(2mEk): at equal kinetic energy p ∝ √m and the momentum ratio is √2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = p/m, so the speed ratio is √2/2 = 1/√2 (equivalently Ek = ½mv² gives v ∝ 1/√m).
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: p = √(2mEk) gives √2 for the momenta, and v = √(2Ek/m) gives 1/√2 for the speeds.
BThis assumes the two particles have the same speed, so that p ∝ m. With the same kinetic energy, the heavier particle must move more slowly.
CThis assumes that equal kinetic energies mean equal momenta. Ek = p²/2m, so for the same Ek the larger mass has the larger momentum.
DThe mass ratio has been inverted in p = √(2mEk), giving the lighter proton the larger momentum and the deuteron the larger speed.
An electric go-kart starts from rest and accelerates along a straight, level track. The driving force on it is constant and resistive forces are negligible.
Which graph shows how the power P developed by the motor varies with time t?
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Step 1The force is constant, so the acceleration is uniform and v = at.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = Fv = Fat, so P ∝ t: a straight line through the origin.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes that a constant force means a constant power. P = Fv, and v increases, so the rate of energy transfer increases.
BThis is the shape of the kinetic energy, ½m(at)² ∝ t², against time. Power is the rate of energy transfer, the gradient of that graph, which is proportional to t.
CThis is the shape of P against the distance travelled: from rest v = √(2as), so P = Fv ∝ √s. Against time, v ∝ t.
DCorrect: with a constant force the speed increases uniformly, so P = Fv increases in proportion to t.
39A-1A-112Elastic potential energyPaper 1AMedium1 mark
A gym resistance band obeys Hooke's law. Work W is done in stretching it from its natural length to an extension x.
What is (work done in stretching it further, from extension x to extension 2x) ÷ W?
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Step 1EH = ½k(Δx)², so the energy stored is W at extension x and 4W at extension 2x.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Work done from x to 2x = 4W − W = 3W, so the ratio is 3.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis treats the force as constant, so that equal increases in extension need equal work. The force grows with extension, so the second stretch needs more work.
BThis takes the energy stored as proportional to the extension. It is proportional to the square of the extension, ½k(Δx)².
CCorrect: the stored energy rises from W to 4W, so the extra work done is 3W.
DThis is the total energy stored at extension 2x, 4W. The work done from x to 2x is the increase, 4W − W.
40A-1A-114Conservation of energyPaper 1AMedium1 mark
A diver leaves a springboard moving vertically upwards and rises to her highest point. Air resistance is negligible.
Which statement about the work done on the diver by the gravitational force during the rise is correct?
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Step 1The gravitational force mg acts downwards while the diver rises through h, so θ = 180° and W = Fs cos θ = −mgh: the work done is negative.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2With no resistive force, mechanical energy is conserved: the kinetic energy lost (mgh) equals the gravitational potential energy gained (mgh), so the magnitude of the work is the energy transformed from one form to the other.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis confuses conservation of mechanical energy with zero work. The force acts along the line of the displacement, so it does work (−mgh); that work transforms kinetic energy into gravitational potential energy without changing the total.
BThe sign is wrong. The force acts downwards while the diver moves upwards (θ = 180°), so the gravitational force does negative work and removes kinetic energy from her.
CCorrect: W = mgh cos 180° = −mgh, and mgh is exactly the kinetic energy that becomes gravitational potential energy.
DThis counts the same energy twice. The kinetic energy lost and the gravitational potential energy gained are each mgh; the work done equals the energy transformed, not 2mgh.
42IBP2S-005-01-03Sankey diagramsPaper 1AMedium1 mark
Three statements are made about Sankey diagrams. I. The width of each arrow is proportional to the energy (or power) it represents. II. The total width of the output arrows equals the width of the input arrow. III. Arrows representing wasted energy continue in the same direction as the useful output arrow. Which statements are correct?
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Arrow width, not length, represents the size of the energy transfer, so I is correct. Energy is conserved, so the output widths must add up to the input width, so II is correct. By convention, wasted energy arrows bend away (usually downwards) from the main useful flow, so III is wrong. The answer is I and II only.
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Answer B
Answer: B
Exam tip: When you describe or draw a Sankey diagram, check that the output widths add up to the input width. This is conservation of energy.
43IBP2S-005-01-04Conservation of energyPaper 1AMedium1 mark
An electric car of mass 1200 kg brakes from 20 m s−1 to rest on a level road. Its regenerative braking system stores 30 % of the car's initial kinetic energy in the battery. How much energy is stored?
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Ek = ½mv² = ½ × 1200 × 20² = 240 kJ. Energy stored = 0.30 × 240 kJ = 72 kJ. 240 kJ ignores the efficiency. 168 kJ is the energy that is not recovered. 3.6 kJ comes from using v instead of v².
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Answer B
Answer: B
Common error: Square the speed before multiplying when you calculate kinetic energy.
44IBP2S-005-01-05Elastic potential energyPaper 1AMedium1 mark
A toy dart launcher uses a spring that obeys Hooke's law. When the spring is compressed by a distance x and released, a dart leaves horizontally with speed v. Friction is negligible. The same dart is launched again with the spring compressed by 2x. What is the new launch speed?
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Elastic energy ½kx² becomes kinetic energy ½mv², so v ∝ x. Doubling the compression doubles the speed, giving 2v. √2 v wrongly assumes the energy doubles. 4v treats the speed as proportional to the energy. v wrongly assumes the launch speed does not depend on compression.
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Answer B
Answer: B
Exam tip: Equate the two energy expressions first. The proportionality then follows without any numbers.
45IBP2S-005-01-06Power & efficiencyPaper 1AMedium1 mark
In a small hydroelectric scheme, water falls through a vertical height of 50 m and flows through a turbine at 200 kg s−1. The overall efficiency of turning gravitational potential energy into electrical energy is 80 %. What is the electrical power output?
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The rate of loss of gravitational potential energy is (m/t)gh = 200 × 9.8 × 50 = 98 kW. Electrical output = 0.80 × 98 kW = 78 kW. 98 kW ignores the efficiency. 123 kW divides by the efficiency instead of multiplying. 8.0 kW leaves out g.
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Answer B
Answer: B
Common error: Useful output is smaller than input, so multiply the input by the efficiency. Do not divide.
46IBP2S-005-01-07Power & efficiencyPaper 1AMedium1 mark
A water pump in an irrigation system is driven by an electric motor. The motor converts electrical energy to kinetic energy of its shaft with an efficiency of 80 %. The pump converts this shaft energy to gravitational potential energy of the water with an efficiency of 60 %. The electrical input power is 2.0 kW. At what rate does the water gain gravitational potential energy?
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For stages in series, the efficiencies multiply: 0.80 × 0.60 = 0.48. The useful power is 0.48 × 2.0 kW = 0.96 kW. 1.6 kW includes only the motor. 1.4 kW uses the average of the two efficiencies. 0.40 kW uses their difference.
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Answer B
Answer: B
Exam tip: For a chain of devices, the overall efficiency is the product of the individual efficiencies. It is always lower than the lowest one.
47IBP2S-005-03-04Kinetic energy & momentumPaper 1AMedium1 mark
A car seat is supported by two identical light springs, each of spring constant k. In design X the springs are joined end to end (in series). In design Y they are side by side (in parallel) and share the load equally. The same total force F is applied to each design. What is the ratio \(\frac{\text{elastic energy stored in X}}{\text{elastic energy stored in Y}}\)?
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Design X (series): each spring carries the full force F, so each extends by F/k. Each spring stores F²/2k, giving a total of F²/k. Design Y (parallel): each spring carries F/2, so each extends by F/2k. Each spring stores F²/8k, giving a total of F²/4k. The ratio X : Y is therefore 4. Choosing ¼ inverts the ratio. Choosing 1 assumes the energy depends only on the applied force. Choosing 2 comes from correctly halving the stiffness for series but leaving the parallel stiffness as k instead of 2k.
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Answer D
Answer: D
Exam tip: For a fixed force, \(E_H = \frac{F^2}{2k}\). The less stiff arrangement stores more energy because it stretches further.
48IBP2S-005-03-05Elastic potential energyPaper 1AMedium1 mark
Two trampoline springs P and Q obey Hooke's law. The spring constant of P is four times that of Q. When a gymnast lands, both springs store the same elastic potential energy. What is \(\frac{\text{extension of P}}{\text{extension of Q}}\)?
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For equal energy, \(\frac{1}{2}k_Px_P^2 = \frac{1}{2}k_Qx_Q^2\), so \(\frac{x_P}{x_Q} = \sqrt{\frac{k_Q}{k_P}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\). Choosing ¼ forgets the square root. Choosing 2 or 4 assumes the stiffer spring stretches more.
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Answer B
Answer: B
Common error: Equal energies do not mean equal forces. Write out the energy equation for both springs before cancelling anything.
49IBP2S-005-03-06Elastic potential energyPaper 1AMedium1 mark
A pinball is launched horizontally by a plunger spring that obeys Hooke's law. Friction and the mass of the spring are negligible. The player triples the compression of the spring. By what factor does the launch speed of the ball change?
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Setting \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\) gives \(v = x\sqrt{\frac{k}{m}}\), so v is directly proportional to x. Tripling the compression therefore triples the speed. Choosing 9 gives the factor for the energy, not the speed. Choosing √3 assumes the energy is proportional to x.
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Answer B
Answer: B
Exam tip: Write the relationship between the two quantities you are asked about before applying any factor. Here the x² and v² cancel to give v ∝ x.
50IBP2S-005-03-07Elastic potential energyPaper 1AMedium1 mark
A dynamic climbing rope is stretched within its linear (Hooke's law) region. At extension x, the tension in the rope is F. Which expression gives the elastic potential energy stored in the rope?
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The force rises linearly from 0 to F as the rope extends from 0 to x. The area under the force–extension graph is therefore a triangle of area ½Fx. Fx would be correct only for a constant force F throughout. ½Fx² confuses the force F with the spring constant k, and its units are not joules.
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Answer A
Answer: A
Common error: \(\frac{1}{2}kx^2\) and \(\frac{1}{2}Fx\) are the same result, since F = kx. Do not mix F into the k version.
51IBP2S-005-03-08Elastic potential energyPaper 1AMedium1 mark
The foam midsole of a running shoe is compressed and then released in a test machine. On the force–compression graph, the unloading curve lies below the loading curve. What does the area enclosed between the two curves represent?
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The area under the loading curve is the work done on the foam. The area under the unloading curve is the energy the foam gives back. Their difference, the enclosed area, is energy that is not returned. It is dissipated as internal energy, warming the foam. Option A is the area under the unloading curve only. Options C and D both describe the area under the loading curve.
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Answer B
Answer: B
Exam tip: For a loading–unloading loop, the fraction of energy returned equals (area under the unloading curve) ÷ (area under the loading curve).
52IBP2S-005-02-03Friction & inclined planesPaper 1AMedium1 mark
A parcel slides down a rough sorting ramp at constant speed. Which statement about the parcel's motion down the ramp is correct?
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The speed is constant, so the kinetic energy does not change and the net work done on the parcel is zero. Gravity does positive work equal to the loss of GPE, and friction does an equal amount of negative work, transferring that energy to internal energy. Option B confuses the work done by gravity alone with the net work. Option C wrongly links constant speed to zero friction work. Option D is wrong because the normal force is perpendicular to the ramp and does no work.
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Answer A
Answer: A
Common error: Constant speed means the net work is zero. It does not mean that each individual force does zero work.
53IBP2S-005-02-04Work done by a forcePaper 1AMedium1 mark
A removal worker pushes a 20 kg crate at constant speed up a rough ramp of length 4.0 m into a van. The top of the ramp is 1.0 m higher than the bottom. The push acts parallel to the ramp, and the frictional force on the crate is 30 N. How much work does the worker do?
Diagram not to scale
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At constant speed, the work done by the worker equals the gain in GPE plus the work done against friction. GPE gain = mgh = 20 × 9.8 × 1.0 = 196 J. Work against friction = 30 × 4.0 = 120 J. Total = 196 + 120 = 316 J. Option A ignores friction. Option B subtracts the friction work instead of adding it. Option D uses the ramp length instead of the vertical height in mgh.
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Answer C
Answer: C
Exam tip: In mgh, use the vertical height gained. Friction acts along the ramp, so its work uses the distance along the slope.
54IBP2S-005-02-05Work done by a forcePaper 1AMedium1 mark
A rescue sled of mass m, released from rest, slides down a snowy slope inclined at angle θ at constant speed. The sled is then hauled back up the same slope at constant speed through a distance L, using a rope parallel to the slope. The frictional force has the same magnitude in both directions. What is the work done by the rope?
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Sliding down at constant speed means the friction force balances the component of weight along the slope, so friction = mg sin θ. When the sled is hauled up, the rope must balance both the weight component (mg sin θ) and friction (mg sin θ), which now acts down the slope. The rope tension is 2mg sin θ, so the work done is 2mgL sin θ. Option A ignores friction. Option B assumes the friction equals mg cos θ, which is the normal force, not the friction. Option C swaps sine and cosine.
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Answer D
Answer: D
Exam tip: Read 'slides down at constant speed' as the clue that friction equals mg sin θ, which also means μ = tan θ.
55IBP2S-005-02-06Conservation of energyPaper 1AMedium1 mark
A crate rests on the flat bed of a delivery truck. The truck accelerates forwards along a level road and the crate does not slip. In the frame of the road, what is the work done on the crate by the static frictional force?
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The only horizontal force on the crate is static friction, and it acts forwards, which is the direction in which the crate accelerates. In the road frame the crate moves forwards, so the friction force and the displacement are in the same direction. The work done is therefore positive, and it equals the crate's gain in kinetic energy. Option C confuses opposing relative motion with opposing motion. Options A and D wrongly assume that no slipping means no work in the ground frame.
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Answer B
Answer: B
Common error: Friction opposes relative sliding between surfaces, not motion in general. Here friction is what speeds the crate up.
56IBP2S-005-02-07Work done by a forcePaper 1AMedium1 mark
An archer draws a bowstring back through 0.50 m. The force needed increases uniformly from zero to 160 N over this distance. What is the work done by the archer?
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The force varies linearly, so the work done is the area under the force–extension line, which is a triangle: W = ½ × 160 × 0.50 = 40 J. This is the same as using the average force (80 N) multiplied by the distance. Option A uses the maximum force for the whole distance. Option C divides the force by the distance. Option D halves the answer twice.
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Answer B
Answer: B
Exam tip: For a force that varies with displacement, work = area under the force–displacement graph. For a linear increase from zero, use the average force, which is half the maximum.
57IBP2S-005-02-10Work done by a forcePaper 1AMedium1 mark
At an airport, a suitcase of mass m is placed from rest onto a horizontal conveyor belt moving at constant speed v. The suitcase slides on the belt until it moves at speed v with the belt. In the frame of the airport floor, what is the work done by friction on the suitcase?
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Friction is the only horizontal force on the suitcase. By the work–energy principle, the work it does equals the suitcase's gain in kinetic energy, which is ½mv2 − 0 = ½mv2. The work is positive because friction acts forwards while the suitcase moves forwards. Option B is the total energy the belt motor supplies during the slipping, half of which is dissipated as internal energy. Option D assumes friction must do negative work. Option C ignores the gain in kinetic energy.
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Answer A
Answer: A
Common error: Do not mix up the work done on the suitcase with the energy supplied by the motor. During slipping, the motor also supplies the energy that is dissipated as internal energy.
58IBP2S-005-04-03Energy density of fuelsPaper 1AMedium1 mark
Ethanol has a specific energy of 30 MJ kg−1 and a density of 790 kg m−3. What is the energy density of ethanol?
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Energy density = specific energy × density = (3.0×107 J kg−1)(790 kg m−3) = 2.4×1010 J m−3. Option A divides instead of multiplying. Option B is 24 MJ, the energy in one litre, wrongly quoted per cubic metre. Option C is simply the specific energy given a different unit.
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Answer D
Answer: D
Exam tip: Check units: (J kg⁻¹) × (kg m⁻³) = J m⁻³. If the units do not cancel to J m⁻³, you have divided the wrong way.
59IBP2S-005-04-04Energy density of fuelsPaper 1AMedium1 mark
A camera drone is powered by a 0.40 kg battery pack with a specific energy of 0.90 MJ kg−1. In flight the battery supplies an average power of 120 W. Assuming the battery can be fully discharged, what is the maximum flight time?
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Stored energy = 0.40 × 0.90×106 = 3.6×105 J. Time = energy ÷ power = 3.6×105 / 120 = 3000 s = 50 min. Option A confuses minutes with seconds. Option C ignores the battery mass (it uses the specific energy as if it were the total energy). Option D divides by the mass instead of multiplying.
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Answer B
Answer: B
Common error: Specific energy is per kilogram. Multiply it by the mass to find the total energy stored before dividing by the power.
60IBP2S-005-04-05Energy density of fuelsPaper 1AMedium1 mark
A car travelling at constant speed needs a useful output power of 15 kW from its engine. The engine has an efficiency of 25% and burns petrol with an energy density of 34 MJ L−1. What volume of petrol is used in one hour?
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Input power = 15 kW / 0.25 = 60 kW. Energy in one hour = 60×103 × 3600 = 2.16×108 J. Volume = 2.16×108 / 34×106 ≈ 6.4 L. Option A uses 60 s instead of 3600 s. Option B ignores the efficiency. Option D multiplies by the efficiency instead of dividing.
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Answer C
Answer: C
Exam tip: The fuel must supply the input power, which is always larger than the useful output. Dividing by the efficiency must make the number bigger.
61IBP2S-005-04-07Kinetic energy & momentumPaper 1AMedium1 mark
Two liquid fuels P and Q have the same energy density. The specific energy of P is twice the specific energy of Q. What is \(\frac{\text{density of P}}{\text{density of Q}}\)?
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Energy density = specific energy × density. Equal energy densities give ePρP = eQρQ, so ρP/ρQ = eQ/eP = 1/2. Option B is therefore correct. Option D assumes density is proportional to specific energy. Option C assumes equal energy density means equal density. Option A wrongly squares the ratio.
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Answer B
Answer: B
Common error: For a fixed product, the two factors are inversely proportional. Doubling one factor halves the other; it does not double it.
62IBP2S-005-04-09Power & efficiencyPaper 1AMedium1 mark
A 70 kg climber gains 800 m in height. Her muscles convert food energy into gravitational potential energy with an efficiency of 20%. Her energy bars have a specific energy of 2.0×107 J kg−1. What is the minimum mass of energy bar she must eat to supply this?
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ΔEp = mgh = 70 × 9.8 × 800 = 5.49×105 J. Food energy needed = 5.49×105 / 0.20 = 2.74×106 J. Mass = 2.74×106 / 2.0×107 = 0.14 kg ≈ 140 g. Option B ignores the efficiency. Option A multiplies by the efficiency. Option C divides by 0.80, the fraction wasted, instead of the fraction made useful.
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Answer D
Answer: D
Common error: Efficiency is the useful fraction. Divide the useful energy by 0.20, not by the wasted fraction 0.80.
63IBP2S-005-00-04Power & efficiencyPaper 1AMedium1 mark
A cyclist moves at constant speed on level ground. The total resistive force on the cyclist is proportional to the square of the speed. The cyclist doubles the speed and keeps it constant. By what factor does the power output of the cyclist increase?
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At constant speed the driving force equals the resistive force, F = kv2. Power is P = Fv = kv3, so doubling v multiplies P by 23 = 8. Option A treats the force as constant. Option B gives only the factor by which the force increases. Option D squares the power relationship a second time.
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Answer C
Answer: C
Common error: Remember that P = Fv contains a factor of v as well as the force. If the force depends on speed, the power rises faster than the force does.
64IBP2S-005-00-05Kinetic energy & momentumPaper 1AMedium1 mark
The kinetic energy of a drone of constant mass is doubled. What happens to the magnitude of its momentum?
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From p = √(2mEk), momentum is proportional to √Ek when the mass is constant. Doubling Ek therefore multiplies p by √2. Option C assumes p ∝ Ek. Option D squares the factor when it should take the square root. Option A wrongly assumes momentum is fixed.
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Answer B
Answer: B
Exam tip: Learn both forms of the link: Ek = p²/2m and p = √(2mEk).
65IBP2S-005-00-06Power & efficiencyPaper 1AMedium1 mark
A hospital lift and its passengers have a total mass of 700 kg. The lift rises at a constant speed of 2.0 m s−1. The efficiency of the motor system is 0.70. What is the electrical input power to the motor?
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At constant speed the motor supplies a force equal to the weight, so useful power = mgv = 700 × 9.8 × 2.0 = 13 720 W. Input power = useful power ÷ efficiency = 13 720 / 0.70 = 19 600 W = 19.6 kW, which is option D. Option C is the useful output only. Option A multiplies by the efficiency instead of dividing. Option B leaves out the speed and uses only mg/0.70.
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Answer D
Answer: D
Common error: Input power is always larger than useful power. If your input value comes out smaller, you have multiplied by the efficiency instead of dividing.
66IBP2S-005-00-08Kinetic energy & momentumPaper 1AMedium1 mark
A stationary firework shell bursts into two fragments of masses 1.0 kg and 3.0 kg, which fly apart in opposite directions. What is the ratio \(\frac{\text{kinetic energy of 1.0 kg fragment}}{\text{kinetic energy of 3.0 kg fragment}}\)?
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The total momentum is zero, so the two fragments have equal and opposite momenta. Using Ek = p2/2m with equal p, kinetic energy is inversely proportional to mass. The ratio is therefore 3.0/1.0 = 3 : 1. Option A assumes Ek ∝ m. Option B assumes the energy is shared equally. Option D uses the ratio of speeds squared without dividing by the ratio of masses.
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Answer C
Answer: C
Common error: In an explosion from rest the momenta are equal in size, but the kinetic energies are not. The lighter fragment always gets the larger share of the energy.
67IBP2S-005-00-09Kinetic energy & momentumPaper 1AMedium1 mark
A squash ball hits a wall at right angles and rebounds with half the speed it had before the impact. What fraction of its initial kinetic energy is transferred to other forms during the impact?
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Kinetic energy is proportional to v2. Halving the speed leaves (½)2 = 0.25 of the kinetic energy. The fraction transferred to other forms is 1 − 0.25 = 0.75. Option A is the fraction that remains, not the fraction lost. Option B assumes Ek ∝ v. Option D uses (½)3.
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Answer C
Answer: C
Exam tip: Read whether the question asks for the energy remaining or the energy lost. These two fractions add up to 1.
68IBP2S-005-00-10Energy density of fuelsPaper 1AMedium1 mark
A car engine delivers a useful output power of 23 kW with an efficiency of 25%. The fuel has an energy density of 46 MJ kg−1. What mass of fuel is used in one hour?
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Input power = 23 kW / 0.25 = 92 kW. Energy used in one hour = 9.2 × 104 × 3600 = 3.3 × 108 J. Mass = 3.3 × 108 / 4.6 × 107 = 7.2 kg. Option C ignores the efficiency. Option B multiplies by the efficiency instead of dividing. Option A gives the mass used in one minute.
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Answer D
Answer: D
Exam tip: Energy density links energy to mass: mass of fuel = energy supplied ÷ energy density. Always use the input energy here, not the useful output.
A car travels at a constant speed along a level road. The total resistive force on the car is proportional to the square of its speed. The useful output power of the engine is P1 at speed v and P2 at speed 2v.
What is P2/P1?
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Step 1At constant speed the resultant force is zero, so the driving force equals the resistive force, which increases by 2² = 4.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = Fv: the force is 4 times larger and the speed 2 times larger, so P2/P1 = 4 × 2 = 8.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis doubles the speed in P = Fv but keeps the driving force the same. The resistive force, and therefore the driving force, also increases.
BThis multiplies the force by 4 but leaves the speed unchanged in P = Fv.
CCorrect: F ∝ v² and P = Fv, so P ∝ v³ and doubling the speed gives 2³ = 8.
DThis squares both factors, 2² × 2². The force goes as v², but the speed enters P = Fv only once.
70A-1A-65Elastic potential energyPaper 1AHard1 mark
A hand-grip exerciser can be fitted with spring X or spring Y. Both springs obey Hooke's law, and the spring constant of Y is twice that of X. Each spring in turn is stretched by the same force.
What is (elastic potential energy stored in X) ÷ (elastic potential energy stored in Y)?
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Step 1The same force gives x = F/k, so X stretches twice as far as Y.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2EH = ½kx² = F²/2k, so at the same force EH ∝ 1/k: the ratio is 2.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis squares the ratio of the extensions (2² = 4) but forgets that k for X is half that of Y.
BCorrect: at the same force EH = F²/2k, so the softer spring X stores twice as much energy.
CThis assumes the same force means the same energy. The energy stored is the work done, which depends on how far each spring stretches.
DThis assumes the two springs have the same extension, so that EH ∝ k. With the same force, X stretches further.
A pump lifts water from a well through a vertical height h. The water starts at rest and leaves the outlet pipe with speed v. The pump delivers a mass R of water each second, and its efficiency is η.
What is the input power to the pump?
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Step 1Each second the water gains gravitational potential energy Rgh and kinetic energy ½Rv², so the useful output power is R(gh + ½v²).
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Input power = output power ÷ η = R(gh + ½v²)/η.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the useful power is the rate of gain of gravitational potential energy and kinetic energy of the water, and the input power is this divided by η.
BThis multiplies the useful power by the efficiency. The input must be larger than the useful output, so the useful power is divided by η.
CThis leaves out the kinetic energy given to the water. The water leaves at speed v, so the pump must also supply ½Rv² each second.
DThe factor ½ in the kinetic energy has been dropped: the kinetic energy gained each second is ½Rv², not Rv².
An electric motor on a building-site crane lifts a load of weight W vertically at a constant speed v. The efficiency of the motor is 80 %. All other energy losses are negligible.
At what rate is energy dissipated in the motor?
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Step 1Useful output power = Fv = Wv, so the input power = Wv/0.80 = 1.25Wv.
—
All 2 steps must be completed — there is no mark for a part-answer.
AThis takes 20 % of the useful output power. The 20 % that is dissipated is 20 % of the input power, 0.20 × 1.25Wv.
BCorrect: the input power is Wv/0.80 = 1.25Wv, and the difference between input and output is 0.25Wv.
CThis multiplies the useful power by the efficiency. That is smaller than the useful power, so it cannot be the input, and it is not the power dissipated either.
DThis is the input power to the motor. The rate of dissipation is the input power minus the useful output, Wv.
73A-1A-117Conservation of energyPaper 1AHard1 mark
A hailstone falls vertically from rest through still air. The air resistance on it increases with its speed, and it reaches its terminal speed before it lands.
Which graph shows how the total mechanical energy E of the hailstone varies with time t before it lands?
Sketch graphs, not to scaleShow mark scheme
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Notes
Step 1The change in mechanical energy equals the work done by the air resistance, so E decreases at the rate Fdv. At the start v = 0, so the gradient is zero.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2As the speed rises Fdv increases, so the graph curves downwards; at terminal speed Fdv = mgvT is constant, so the graph becomes a straight line of constant negative gradient.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AMechanical energy is conserved only when no resistive force acts. Here air resistance does negative work on the hailstone throughout its fall.
BCorrect: the rate of loss of mechanical energy, Fdv, rises from zero and then stays constant at mgvT, so the curve starts flat, steepens and becomes a straight line.
CA constant rate of loss from the start would need Fdv to be constant. At release v = 0 and the air resistance is zero, so no energy is lost at first.
DThis has the rate of loss falling to zero, as if no energy were dissipated at terminal speed. At terminal speed the hailstone loses gravitational potential energy at mgvT with no gain in kinetic energy.
74A-1A-118Elastic potential energyPaper 1AHard1 mark
A load of weight W is attached to the lower end of a vertical spring that obeys Hooke's law. The load is supported by hand and lowered slowly, from the position where the spring is unstretched, until it hangs at rest at its equilibrium position, a distance e lower.
Which statement is correct?
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Step 1At equilibrium ke = W, so EH = ½ke² = ½We, while the load loses gravitational potential energy We. The load ends at rest, so the remaining ½We is transferred to the hand, which pushes up while the load moves down (negative work on the load).
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the spring stores ½ke² = ½We of the We lost; the hand does −½We of work on the load, so mechanical energy of the load–spring system is not conserved.
BThis uses the final spring force W as if it acted over the whole extension. The spring force rises from zero to W, so the energy stored is the average force × e = ½We.
CThe load is lowered slowly and ends at rest, so it has no kinetic energy. This would describe a load released without the hand as it passes the equilibrium position.
DThe load falls through e with its full weight W acting, so it loses We of gravitational potential energy, not ½We.
75IBP2S-005-01-08Elastic potential energyPaper 1AHard1 mark
A bungee jumper of mass 70 kg steps off a platform. The cord has an unstretched length of 15 m. At the lowest point the jumper is momentarily at rest, 40 m below the platform. Air resistance and the mass of the cord are negligible. Take g = 9.8 m s−2. What is the elastic potential energy stored in the cord at the lowest point?
Diagram not to scale
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The jumper starts at rest and is at rest again at the lowest point, so there is no kinetic energy at either end. All the gravitational potential energy lost over the full 40 m fall is therefore stored elastically: mgh = 70 × 9.8 × 40 = 2.7 × 104 J. 1.7 × 104 J uses only the 25 m extension. It ignores the kinetic energy the jumper already had when the cord became taut. 1.0 × 104 J is only that kinetic energy, gained during the first 15 m of free fall. 4.5 × 104 J adds the 40 m and 25 m results, which counts energy twice.
✓ 1
Answer C
Answer: C
Common error: Use the total fall from rest to rest, not just the extension of the cord. The jumper is already moving when the cord becomes taut.
76IBP2S-005-01-10Work done by a forcePaper 1AHard1 mark
A geologist throws a rock from the top of a cliff 20 m high. The rock leaves the hand at 15 m s−1 at an angle above the horizontal. Air resistance is negligible. What is the speed of the rock just before it hits the ground at the base of the cliff?
Diagram not to scale
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Energy is a scalar, so the direction of projection does not affect the result: ½mv² = ½mu² + mgh. This gives v = √(15² + 2 × 9.8 × 20) = √617 = 24.8 m s−1. 19.8 m s−1 ignores the initial speed. 34.8 m s−1 adds the speeds instead of the energies. The angle affects only the path, not the final speed.
✓ 1
Answer B
Answer: B
Exam tip: When a question asks only for a speed, an energy method avoids having to split the motion into components.
77IBP2S-005-02-08Work done by a forcePaper 1AHard1 mark
A skier descends through the same vertical height h on two different straight slopes, one inclined at 30° to the horizontal and the other at 60°. The coefficient of dynamic friction is the same on both slopes. What is the ratio \(\frac{\text{work done against friction on the 30° slope}}{\text{work done against friction on the 60° slope}}\)?
Same vertical height h on both slopes; diagram not to scale
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Friction = μmg cos θ and the slope length = h/sin θ. The work against friction is therefore μmgh cos θ / sin θ = μmgh/tan θ. The ratio is tan 60° / tan 30° = √3 ÷ (1/√3) = 3. The shallower slope is longer and has a larger normal force, so more work is done against friction. Option A assumes friction work depends only on the height. Option B accounts for only one of the two effects. Option D inverts the ratio.
✓ 1
Answer C
Answer: C
Exam tip: Write the friction work in terms of the angle before substituting values. Both the normal force and the path length change with θ.
78IBP2S-005-04-08Energy density of fuelsPaper 1AHard1 mark
A delivery van has a fuel tank of fixed volume. Its owner switches from fuel A to fuel B. Fuel B has 1.2 times the specific energy of fuel A but only 0.75 times its density. A modified engine for fuel B also raises the efficiency from 30% to 36%. The van's load and driving conditions are unchanged. What is \(\frac{\text{range on a full tank of B}}{\text{range on a full tank of A}}\)?
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Energy in a full tank of fixed volume is proportional to energy density, which equals specific energy × density. The tank-energy ratio is 1.2 × 0.75 = 0.90. Useful work is proportional to efficiency, a ratio of 36/30 = 1.2. The resistive forces are unchanged, so range is proportional to useful work: 0.90 × 1.2 = 1.08. Option A (0.90) ignores the efficiency change. Option C (1.20) uses only the efficiency, or only the specific energy. Option D (1.44) ignores the lower density.
✓ 1
Answer B
Answer: B
Exam tip: For a tank of fixed volume, use energy density (specific energy × density), not specific energy alone. Then multiply by the efficiency to get the useful work, which sets the range.
79IBP2S-005-04-10Sankey diagramsPaper 1AHard1 mark
Petrol has a specific energy of 46 MJ kg−1 and is used in an engine with an efficiency of 25%. A lithium-ion battery has a specific energy of 0.72 MJ kg−1 and drives an electric motor; the overall efficiency from battery to motor output is 90%. What mass of battery provides the same useful work as 1.0 kg of petrol?
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Useful work from petrol = 46 × 0.25 = 11.5 MJ. Useful work per kg of battery = 0.72 × 0.90 = 0.648 MJ. Mass = 11.5 / 0.648 ≈ 18 kg. Option C ignores both efficiencies. Option A ignores the electric-system efficiency. Option D applies the two efficiencies the wrong way round.
✓ 1
Answer B
Answer: B
Exam tip: Compare fuels on useful energy per kilogram: specific energy × efficiency for each system. Then take the ratio.
80IBP2S-005-00-07Kinetic energy & momentumPaper 1AHard1 mark
An ion thruster on a space probe ejects ions backwards at a speed of 3.0 × 104 m s−1 relative to the probe. This produces a thrust of 0.20 N. Assume the probe's own speed change is negligible. What is the minimum power needed to give the ions their kinetic energy?
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Thrust is the rate of change of momentum, μv, where μ is the mass ejected per second. So μ = 0.20 / (3.0 × 104) = 6.7 × 10−6 kg s−1. The power given to the ions is the kinetic energy ejected per second: ½μv2 = ½Fv = ½ × 0.20 × 3.0 × 104 = 3.0 × 103 W = 3.0 kW, which is option B. Option C uses P = Fv. That formula applies to a force pushing an object at constant speed v, not to accelerating ejected mass from rest. Option A applies the factor of ½ twice, giving ¼Fv. Option D uses ½Fv2, confusing the force with the mass flow rate in the kinetic energy expression.
✓ 1
Answer B
Answer: B
Exam tip: When mass is accelerated from rest to speed v, the force is μv but the power is ½μv². The factor of ½ appears in the power and not in the force.
A small electric motor winds up a thread to lift a load of mass m vertically through 0.800 ± 0.005 m at a steady speed. The potential difference V across the motor and the current I in it are read from meters, and the time t for the lift is measured with a hand-held stopwatch that reads to 0.01 s. The uncertainty in t is estimated as ±0.2 s.
m / kg
V / V
I / A
t / s
Efficiency / %
0.100
6.02
0.21
3.10
20.0
0.200
5.98
0.30
3.42
25.6
0.300
5.95
0.39
3.81
0.400
5.91
0.49
4.35
24.9
0.500
5.86
0.60
5.12
21.8
Diagram not to scale. The load is lifted at a steady speed.
(a)
Explain why the uncertainty in t is much larger than the resolution of the stopwatch.
(1)
(b)
(i)
For the 0.300 kg load, calculate the useful output power of the motor.
(1)
(ii)
Hence complete the table.
(2)
(c)
Determine the percentage uncertainty in the output power for the 0.300 kg load. The uncertainty in m is negligible.
(2)
(d)
The student claims that the efficiency of the motor increases as the load increases. Comment on this claim.
(1)
(e)
Suggest one change to the method that would reduce the percentage uncertainty in t.
(1)
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Part (a)
the stopwatch is started and stopped by hand, so the human reaction time «≈ 0.2 s» dominates
✓ 1
OWTTE.
Part (b)(i)
P = mgh/t = 0.300 × 9.81 × 0.800 / 3.81 = 0.618 W
✓ 1
Accept 0.62 W.
Part (b)(ii)
input power = VI = 5.95 × 0.39 = 2.32 W
✓ 1
efficiency = 0.618/2.32 = 26.6 %
✓ 1
Accept 27 %. Allow ECF from (b)(i).
Part (c)
0.005/0.800 + 0.2/3.81 «= 0.0062 + 0.0525»
✓ 1
= 5.9 % «≈ 6 %»
✓ 1
Accept 5–6 %. Award [1] max if the percentage uncertainties are not added.
Part (d)
the efficiency rises to a maximum of about 27 % near 0.3 kg and then falls «to 22 % at 0.500 kg», by more than the uncertainty, so the claim is not supported
✓ 1
Allow ECF from (b)(ii).
Part (e)
use light gates / a motion sensor to time the lift OR increase the height of the lift so t is longer
✓ 1
Do not accept 'repeat the readings' alone.
Answers: (b)(i) 0.618 W · (b)(ii) 26.6 % · (c) 5.9 % (the remaining parts are explanations — see the table above)
82A-1B-16Energy density of fuelsPaper 1BEasy9 marks
A candle stands on a top-pan balance that reads to 0.01 g, under a metal can containing 0.150 kg of water. Every 2 minutes the mass of the candle and the temperature of the water are recorded. Δm is the mass of wax burnt and ΔT is the rise in temperature of the water since the start. The specific heat capacity of water is 4180 J kg−1 K−1.
Time / min
Mass of candle / g
Temperature / °C
Δm / g
ΔT / K
0
24.63
19.4
0.00
0.0
2
24.44
23.8
0.19
4.4
4
24.22
28.2
0.41
8.8
6
24.02
32.9
0.61
13.5
8
23.81
37.2
0.82
17.8
10
23.62
41.7
1.01
22.3
Graph drawn to scale
(a)
State the absolute uncertainty in a value of Δm.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line.
(2)
(d)
Determine the energy transferred to the water per kilogram of wax burnt.
(2)
(e)
The accepted energy density of candle wax is 43 MJ kg−1. Calculate the percentage of the energy released by the wax that reaches the water.
(1)
(f)
Suggest two improvements to the apparatus that would make the result closer to the accepted value.
(2)
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Notes
Part (a)
±0.02 g «two readings, each ±0.01 g, are subtracted»
✓ 1
Accept ±0.01 g only with the reasoning that the balance reads half a division.
Part (b)
single straight line through the points «and the origin»
✓ 1
Part (c)
gradient from a large triangle on the line
✓ 1
22 K g−1
✓ 1
Accept 21–23 K g−1.
Part (d)
energy per gram = mc × gradient = 0.150 × 4180 × 21.9 = 1.38 × 104 J g−1
✓ 1
Allow ECF from (c).
= 14 MJ kg−1 «1.38 × 107 J kg−1»
✓ 1
Accept 13–15 MJ kg−1.
Part (e)
13.8/43 × 100 = 32 %
✓ 1
Allow ECF from (d).
Part (f)
any one: a draught shield round the flame; a lid on the can / lagging round the can; place the can closer to the flame «so more of the hot gas reaches it»
✓ 1
any second, different improvement, eg a can with a blackened base or a copper can; stir the water before each reading
✓ 1
Do not accept 'repeat the experiment'.
Answers: (c) 22 K g−1 · (d) 14 MJ kg−1 · (e) 32 % (the remaining parts are explanations — see the table above)
A student of mass 62.0 kg runs up a staircase from a standing start. Each floor of the building is 18 steps high, and one step is measured as 17.0 ± 0.2 cm high. A partner starts a stopwatch on the command “go” and stops it when the student reaches the floor being timed. The student climbs one, two, three, four and five floors on different days, running as fast as she can each time.
Each time t is the mean of three runs and is uncertain by ±0.2 s. The graph shows the height h climbed against t.
Floors
h / m
t / s
1
3.06
3.6
2
6.12
6.4
3
9.18
9.4
4
12.24
12.2
5
15.30
15.2
Graph drawn to scale
(a)
Calculate the absolute uncertainty in the height of one floor.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line.
(2)
(d)
Determine the useful power developed by the student in raising her body.
(2)
(e)
The line does not pass through the origin. Suggest one reason for this.
(1)
(f)
The student claims that her power output becomes smaller when she climbs more floors, because she gets tired. Discuss whether the graph supports her claim.
(1)
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Part (a)
18 × 0.2 cm = ±3.6 cm «±0.036 m», so one floor is 3.06 ± 0.04 m
✓ 1
Accept ±0.04 m.
Part (b)
single straight line through all error bars, meeting the t axis at a small positive value
✓ 1
Do not accept a line forced through the origin.
Part (c)
two well-separated points on the line used
✓ 1
gradient = 1.06 «m s−1»
✓ 1
Accept 1.03–1.09 m s−1.
Part (d)
P = rate of gain of gravitational potential energy = mg × gradient
✓ 1
P = 62.0 × 9.81 × 1.06 = 645 «W»
✓ 1
Accept 626–663 W. Allow ECF from (c). Award [2] for CNA.
Part (e)
the student starts from rest and takes time to accelerate / the partner's reaction time in starting the stopwatch / the time to reach the first step
✓ 1
Any one. OWTTE.
Part (f)
the points lie on a straight line «within the error bars», so the rate of climbing and the power are constant: the graph does not support the claim
✓ 1
OWTTE. A curve with decreasing gradient would be needed to support it.
Answers: (a) ±0.04 m · (c) 1.06 m s−1 · (d) 645 W (the remaining parts are explanations — see the table above)
84A-1B-48Conservation of energyPaper 1BEasy9 marks
A small wooden block of mass 0.150 kg is released from rest on a curved chute coated with PTFE, on which friction is negligible. The block leaves the bottom of the chute horizontally and slides along a horizontal rubber mat until it stops. The release height h above the mat is measured with a metre rule, and the distance d that the block slides along the mat is measured three times for each height.
The graph shows the mean value of d against h. The error bars show half the range of the three readings (with a minimum of ±0.005 m).
h / m
d1 / m
d2 / m
d3 / m
mean d / m
0.050
0.069
0.059
0.064
0.064
0.100
0.140
0.152
0.151
0.148
0.150
0.238
0.224
0.234
0.200
0.319
0.304
0.315
0.313
0.250
0.392
0.406
0.395
0.398
0.300
0.488
0.473
0.484
0.482
Graph drawn to scale
(a)
For h = 0.150 m, calculate the mean value of d and its absolute uncertainty.
(2)
(b)
The frictional force F exerted by the mat on the block is constant. Show that, if no energy is dissipated on the chute, d = (mg/F) h, where m is the mass of the block.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine F using the graph.
(2)
(e)
Pulling the block across the mat at a slow, steady speed with a force sensor gives F = 0.90 ± 0.05 N. Comment on whether this value agrees with your answer to (d).
(1)
(f)
The line of best fit does not pass through the origin. Suggest a reason for this, and state whether it affects your value of F.
(2)
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Notes
Part (a)
mean d = (0.238 + 0.224 + 0.234)/3 = 0.232 «m»
✓ 1
Accept 0.2317 m.
uncertainty = ½ × (0.238 − 0.224) = ±0.007 «m»
✓ 1
Do not accept the resolution of the metre rule as the uncertainty.
Part (b)
loss of gravitational potential energy on the chute = work done against friction on the mat: mgh = Fd, so d = (mg/F)h
✓ 1
Must see the energy equation. Kinetic energy is zero at the start and at the end.
Part (c)
single straight line through all error bars, NOT forced through the origin
✓ 1
Do not accept a line joined point to point.
Part (d)
gradient = 1.67 «no unit»
✓ 1
Accept 1.59–1.75.
F = mg/gradient = 0.150 × 9.81/1.67 = 0.88 «N»
✓ 1
Accept 0.84–0.93 N. Allow ECF from their gradient. Award [2] for CNA.
Part (e)
0.88 N lies within 0.85–0.95 N, so the two values agree
✓ 1
Allow ECF from (d). Must refer to the uncertainty range.
Part (f)
the line meets the h axis at about 0.012 m: every height is too large by the same amount, eg h measured to the top of the block instead of to its base OR from the wrong reference point on the mat
✓ 1
Accept any constant (zero) error in h or d. Accept an intercept of 0.006–0.018 m. OWTTE.
no: a constant error shifts the line but does not change its gradient, from which F is found
✓ 1
OWTTE.
Answers: (a) 0.232 ± 0.007 m · (d) gradient ≈ 1.67; F ≈ 0.88 N (the remaining parts are explanations — see the table above)
A builders' hoist raises a platform loaded with bricks, of total mass 360 kg, from rest at ground level to rest 12.0 ± 0.1 m higher. A data logger records the electrical power P supplied to the hoist motor every 2.0 s during the lift. The data are shown in the table and on the graph.
t / s
P / kW
0.0
0.0
2.0
6.1
4.0
5.2
6.0
5.1
8.0
5.1
10.0
5.2
12.0
5.1
14.0
5.0
16.0
5.1
18.0
5.1
20.0
3.0
22.0
0.0
Graph drawn to scale
(a)
Calculate the gain in gravitational potential energy of the platform and bricks.
(1)
(b)
Estimate the total electrical energy supplied to the motor during the lift.
(2)
(c)
Calculate the efficiency of the hoist.
(1)
(d)
Outline why the power supplied is approximately constant between t = 4.0 s and t = 18.0 s.
(1)
(e)
The uncertainty in your estimate in (b) is about ±4 %, and the uncertainty in the mass is negligible. Determine the absolute uncertainty in the efficiency.
(2)
(f)
The manufacturer states that the efficiency of the hoist is at least 50 %. Comment on this statement.
(1)
(g)
Suggest one change to the data collection that would give a more reliable value for the energy supplied.
(1)
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Notes
Part (a)
ΔEp = 360 × 9.81 × 12.0 = 4.24 × 104 J
✓ 1
Accept 42 kJ.
Part (b)
energy = area under the P–t graph «trapezium rule or counting squares»
✓ 1
≈ 100 kJ «1.0 × 105 J»
✓ 1
Accept 96–104 kJ. Award [2] for CNA.
Part (c)
η = 42.4 kJ/100 kJ = 0.42 «42 %»
✓ 1
Allow ECF from (a) and (b).
Part (d)
the load rises at constant speed, so the force needed «weight + friction» is constant and the power P = Fv «and the rate of energy loss in the hoist» is constant
✓ 1
OWTTE.
Part (e)
fractional uncertainty = 0.04 + 0.1/12.0 = 0.048
✓ 1
Δη = 0.048 × 0.42 = ±0.02
✓ 1
Allow ECF from (c).
Part (f)
the largest value consistent with the data is 0.42 + 0.02 = 0.44 «44 %», which is less than 50 %, so the statement is not supported
✓ 1
Allow ECF from (c) and (e). Must refer to the uncertainty.
Part (g)
record the power at shorter intervals «eg every 0.1 s» so the shape of the graph is followed more closely OR measure the energy directly with a joulemeter / energy meter
✓ 1
Do not accept 'repeat the lift' alone.
Answers: (a) 4.24 × 104 J · (b) ≈ 100 kJ · (c) 0.42 · (e) ±0.02 (the remaining parts are explanations — see the table above)
86A-1B-05Elastic potential energyPaper 1BMedium11 marks
A trolley of mass 0.800 ± 0.004 kg on a horizontal track is pushed against a spring fixed to the end of the track, compressing it by a distance x. When the trolley is released the spring pushes it away, and a light gate placed just beyond the point where the trolley leaves the spring measures its speed v.
The uncertainty in x is ±0.1 cm and the uncertainty in v is ±0.02 m s−1.
x / cm
2.0
3.0
4.0
5.0
6.0
7.0
v / m s−1
0.34
0.48
0.66
0.82
1.00
1.15
Graph drawn to scale
(a)
The spring constant is k. Show that, if no energy is dissipated, v = x√(k/m).
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line in s−1.
(2)
(d)
Determine k.
(2)
(e)
The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in k.
(2)
(f)
Loading the spring with known weights gives k = 250 ± 5 N m−1. Compare this with your answer, and suggest a reason for any difference.
(2)
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Notes
Part (a)
elastic potential energy is transferred to kinetic energy: ½kx² = ½mv²
✓ 1
v² = kx²/m so v = x√(k/m)
✓ 1
Part (b)
single straight line through all error bars, passing through «or very close to» the origin
✓ 1
Part (c)
gradient from a large triangle with x converted to m
The gradient is squared, so its fractional uncertainty is doubled.
Δk = 0.065 × 218 ≈ 14 N m−1
✓ 1
Accept 13–16 N m−1.
Part (f)
range 204–232 N m−1 lies below 245–255 N m−1, so the launch value is significantly smaller
✓ 1
Allow ECF.
some elastic energy is not transferred to the trolley's kinetic energy «friction, kinetic energy of the spring itself, vibration, sound», so v is systematically low
✓ 1
OWTTE.
Answers: (c) 16.5 s−1 · (d) 218 N m−1 · (e) ±14 N m−1(the remaining parts are explanations — see the table above)
A cycling-science database lists the steady power P that one rider must deliver, measured with a power meter, to ride at constant speed v on a level road in still air.
A student tests the hypothesis that the total resistive force on the rider is proportional to v², so that P = kvn with n = 3.
v / km h−1
v / m s−1
P / W
lg(v / m s−1)
lg(P / W)
25
6.94
78.8
0.842
1.897
30
8.33
124
0.921
2.093
35
9.72
192
0.988
2.283
40
11.1
272
45
12.5
387
1.097
2.588
50
13.9
514
1.143
2.711
Graph drawn to scale
(a)
Explain why n = 3 if the resistive force is proportional to v².
(2)
(b)
Complete the table.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine n.
(2)
(e)
Discuss whether the data support the hypothesis.
(2)
(f)
Determine the total resistive force on the rider at 40 km h−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at constant speed the driving force equals the resistive force, and P = Fv
✓ 1
if F ∝ v² then P ∝ v³ «lg P = 3 lg v + lg k, gradient 3»
✓ 1
Part (b)
1.046 and 2.435
✓ 1
Both needed. Accept 2 or 3 d.p.
Part (c)
single straight line with points evenly scattered on either side
✓ 1
Part (d)
gradient from a large triangle on the line
✓ 1
n = 2.7
✓ 1
Accept 2.6–2.85.
Part (e)
n is clearly less than 3 «the difference is larger than the scatter allows», so the hypothesis is not fully supported
✓ 1
Allow ECF from (d).
part of the resistive force «rolling resistance, friction in the drive» does not increase as v², so P rises less steeply than v³
✓ 1
OWTTE.
Part (f)
v = 40/3.6 = 11.1 m s−1
✓ 1
F = P/v = 272/11.1 = 24.5 N
✓ 1
Accept 24–26 N.
Answers: (b) 1.046, 2.435 · (d) n = 2.7 · (f) 24 N (the remaining parts are explanations — see the table above)
A pulley system lifts a load of weight W. The effort F is applied to the free end of the string with a newton-meter, and F is read while the load rises at a slow, steady speed. Measurements with a ruler show that the effort moves 4 times as far as the load. Each reading of F is uncertain by ±0.05 N; the uncertainty in W is negligible.
The student's hypothesis is that the efficiency of the pulley system does not depend on the load. The graph shows F against W.
W / N
F / N
Efficiency
2.0
0.95
0.53
4.0
1.45
0.69
6.0
2.00
8.0
2.50
0.80
10.0
3.00
12.0
3.50
0.86
Graph drawn to scale
(a)
Show that the efficiency of the pulley system is W/(4F).
(1)
(b)
Complete the table.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the intercept of the line on the F axis, and outline what it represents.
(2)
(e)
Determine the efficiency for W = 2.0 N, together with its absolute uncertainty.
(2)
(f)
Discuss whether the data support the student's hypothesis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
useful work = Wh and work done by the effort = F × 4h, so efficiency = Wh/(4Fh) = W/(4F)
✓ 1
Must see the distances.
Part (b)
6.0/(4 × 2.00) = 0.75
✓ 1
10.0/(4 × 3.00) = 0.83
✓ 1
2 s.f. Accept 83 % for 0.83.
Part (c)
single straight line through all error bars, with a positive intercept on the F axis
✓ 1
Part (d)
intercept = 0.44 «N»
✓ 1
Accept 0.35–0.53 N.
the effort needed with no load, to lift the moving part of the pulley system «lower pulley block» and to overcome friction
89A-1B-50Energy density of fuelsPaper 1BMedium10 marks
A student collects published specifications for the battery packs of eight electric cars. For each pack the table gives its total mass m and the energy E it can store. Every pack contains cells of the same type, together with a casing, a cooling system and control electronics.
1 kWh = 3.6 × 106 J.
m / kg
E / kWh
180
30.6
230
38.4
290
53.3
340
62.0
400
77.2
450
85.2
510
99.3
560
107.9
Graph drawn to scale
(a)
Draw the line of best fit on the graph.
(1)
(b)
Determine the gradient of the line in MJ kg−1.
(2)
(c)
The line meets the m axis at a positive value m0. Determine m0, and suggest what it represents.
(2)
(d)
Calculate the energy stored per kilogram of the whole 450 kg pack, and explain why it is less than your answer to (b).
(2)
(e)
The electric car with the 450 kg pack transfers 90 % of the stored energy to useful work. A petrol car transfers 30 % of the energy released by its fuel to useful work, and the energy density of petrol is 46 MJ kg−1. Calculate the mass of petrol that would provide the same useful work as a full battery pack.
(2)
(f)
Suggest why the data points are scattered about the line, although every value of E is quoted to 0.1 kWh.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
single straight line with the points evenly scattered on both sides, NOT forced through the origin
✓ 1
Part (b)
gradient = 0.208 kWh kg−1 «from a large triangle»
✓ 1
Accept 0.202–0.214 kWh kg−1.
× 3.6 = 0.75 MJ kg−1
✓ 1
Accept 0.73–0.77 MJ kg−1. Allow ECF from their gradient. Award [1] max for an answer in kWh kg−1 only.
Part (c)
m0 = 37 kg
✓ 1
Accept 25–49 kg.
the mass of the parts of a pack that store no energy «casing, cooling system, electronics»
✓ 1
OWTTE.
Part (d)
85.2 × 3.6/450 = 0.68 MJ kg−1
✓ 1
Accept 0.67–0.69 MJ kg−1.
part of the mass of the pack «m0» stores no energy, so the energy per kilogram of the whole pack is less than that of the cells «the gradient»
mass of petrol = 2.76 × 108/(0.30 × 46 × 106) = 20 kg
✓ 1
Award [2] for CNA. Award [1] max if either efficiency is omitted or inverted.
Part (f)
the packs are different designs: the casing/cooling mass is not the same fraction for every pack OR the specifications are rounded or measured under different conditions by different manufacturers
✓ 1
OWTTE. Do not accept 'measurement error' alone.
Answers: (b) 0.75 MJ kg−1 · (c) m0 ≈ 37 kg · (d) 0.68 MJ kg−1 · (e) 20 kg (the remaining parts are explanations — see the table above)
90A-1B-52Work done by a forcePaper 1BMedium10 marks
A trolley of mass 0.750 ± 0.001 kg starts from rest on a level track. It is pulled by a thread attached to a force sensor on the trolley; the thread passes over a pulley to a falling mass. The sensor shows that the tension in the thread stays at 0.950 N while the trolley moves.
A card of width 50.0 ± 0.5 mm on the trolley passes through a light gate after the trolley has travelled a distance s, and the timer records the time t for which the beam is blocked, to ±0.1 ms. The speed of the trolley at the gate is taken as (card width)/t. The point for s = 0.400 m has not been plotted.
s / m
t / ms
v / m s−1
Ek / J
0.200
75.6
0.661
0.164
0.300
60.9
0.821
0.253
0.400
53.4
0.500
47.3
1.057
0.419
0.600
43.7
1.144
0.491
0.700
40.0
1.250
0.586
Graph drawn to scale
(a)
Complete the table for s = 0.400 m, and plot the point on the graph.
(2)
(b)
Determine the percentage uncertainty in Ek for s = 0.400 m.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line, and state what it represents.
(2)
(e)
Determine the average frictional force on the trolley.
(1)
(f)
Outline how the student could modify the experiment so that the gain in kinetic energy is equal to the work done by the tension alone, and state the gradient that would then be expected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 0.0500/0.0534 = 0.936 «m s−1»
✓ 1
Ek = ½ × 0.750 × 0.936² = 0.329 «J» AND point plotted at (0.400, 0.329)
✓ 1
Allow ECF from their v. Plotting to within half a small square.
The speed is squared, so its percentage uncertainty is doubled.
= 2.5 %
✓ 1
Accept 2.4–2.6 %. Award [1] max if the speed uncertainty is not doubled.
Part (c)
single straight line through all error bars, passing through or very close to the origin
✓ 1
Part (d)
gradient = 0.83 N «J m−1»
✓ 1
Accept 0.81–0.86 N. Unit required.
the resultant force on the trolley «work done by the resultant force = gain in kinetic energy»
✓ 1
Accept 'tension minus friction'.
Part (e)
f = 0.950 − 0.83 = 0.12 N
✓ 1
Accept 0.09–0.14 N. Allow ECF from (d).
Part (f)
tilt the track slightly «raise the end where the trolley starts» until the trolley, given a push with no thread attached, moves at constant speed: the component of the weight down the slope then balances friction
✓ 1
OWTTE.
gradient = 0.950 N «equal to the tension»
✓ 1
Answers: (a) 0.936 m s−1; 0.329 J · (b) 2.5 % · (d) 0.83 N · (e) 0.12 N (the remaining parts are explanations — see the table above)
In a school model of a hydroelectric scheme, water flows from a header tank, whose water level is kept constant, down a pipe to a small turbine that drives a generator. The water surface in the tank is 1.50 ± 0.01 m above the turbine. The generator is connected to a resistor, and the potential difference V across the resistor and the current I in it are measured.
A valve controls the flow. For each setting the water leaving the turbine is collected in a measuring jug for 20.0 ± 0.2 s, and its volume is read to ±25 cm3. Density of water = 1000 kg m−3.
Volume collected in 20.0 s / cm3
V / V
I / A
Efficiency / %
1200
1.19
0.119
16
1800
1.82
0.182
25
2400
2.30
0.230
3000
2.53
0.253
29
3600
2.52
0.252
24
Diagram not to scale; the generator is also connected to a resistor
(a)
Calculate the percentage uncertainty in the rate of flow of water for the smallest volume collected.
(2)
(b)
Show that, when 2400 cm3 is collected, the rate at which the water loses gravitational potential energy is about 1.8 W.
(1)
(c)
Determine the efficiency of the model when 2400 cm3 is collected, together with its absolute uncertainty. The uncertainties in V and I are ±0.01 V and ±0.001 A.
(3)
(d)
The student claims that the efficiency of the model increases as the rate of flow increases. Comment on this claim.
(1)
(e)
Sketch a labelled Sankey diagram, with band widths approximately to scale, for the energy transfers each second when 2400 cm3 is collected.
(2)
(f)
Suggest one change to the method that would reduce the percentage uncertainty in the rate of flow.
MP3 is for matching the precision of value and uncertainty. Allow ECF.
Part (d)
the efficiency rises from 16 % to a maximum of about 30 % at 2400 cm3 and then falls to 24 %, by much more than the uncertainty, so the claim is not supported over the whole range
✓ 1
OWTTE. Allow ECF from (c).
Part (e)
input band labelled gravitational potential energy of water «1.77 W» AND useful output band labelled electrical energy «0.53 W», about 0.3 of the input width
✓ 1
wasted band «1.24 W» about 0.7 of the input width, labelled with at least one valid destination: kinetic energy of the water leaving OR thermal energy in the pipe, turbine and generator OR sound
✓ 1
Widths must add up to the input width. Allow ECF from (c).
Part (f)
collect the water for a longer time «or use a larger container» so the volume and time are larger OR use a flow meter / weigh the water on a balance instead of reading the jug
✓ 1
Do not accept 'repeat the readings' alone.
Answers: (a) 3.1 % · (b) 1.77 W · (c) 0.30 ± 0.01 (the remaining parts are explanations — see the table above)
The manufacturer of a sports slingshot publishes the force F needed to hold its rubber band at each draw distance x. The published data are shown in the table and plotted on the graph.
A student draws the slingshot to x = 0.40 m and launches a projectile of mass 15.0 ± 0.1 g. A chronograph records the launch speed as 44.0 ± 0.5 m s−1.
x / m
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
F / N
0
13
25
39
52
67
83
101
119
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Explain why the energy stored in the band at full draw is equal to the area under the graph.
(1)
(c)
Estimate the energy stored in the band when x = 0.40 m.
(2)
(d)
Show that the kinetic energy of the projectile is about 15 J.
(1)
(e)
Determine the absolute uncertainty in the kinetic energy of the projectile.
(2)
(f)
The uncertainty in your estimate in (c) is about ±5 %. Determine the efficiency of the slingshot together with its absolute uncertainty.
(2)
(g)
The manufacturer claims an efficiency of 75 %. Comment on this claim.
(1)
(h)
Suggest where the energy that does not reach the projectile goes.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
smooth curve through all points, rising slightly more steeply at large x
✓ 1
Do not accept point-to-point joining or a single straight line.
Part (b)
the work done by the drawing force is ∫F dx «force × distance for each small step», and this work is stored as elastic energy
✓ 1
OWTTE. Do not accept ½Fx or ½kx² — the graph is not linear.
Part (c)
area by counting squares OR trapezium rule on the tabulated values
✓ 1
Trapezium rule gives 22.0 J.
22 J
✓ 1
Accept 21–23 J.
Part (d)
½ × 0.0150 × 44.0² = 14.5 J
✓ 1
Must see conversion of g to kg and full substitution OR answer to 3 s.f.
93A-1B-15Conservation of energyPaper 1BHard11 marks
A glider is released from rest on an inclined air track. A card of width 20.0 ± 0.2 mm on the glider passes through a light gate lower down the track, and the timer records the time t for which the beam is blocked. The vertical height h through which the glider falls before reaching the gate is varied. The student calculates the speed v = 0.0200/t and tests whether mechanical energy is conserved.
The error bars show an uncertainty of ±2 % in each value of v².
h / m
t / ms
v / m s−1
v² / m² s−2
0.050
22.54
0.887
0.787
0.100
15.31
1.31
1.71
0.150
12.27
1.63
2.66
0.200
10.60
0.250
9.39
2.13
4.54
0.300
8.59
2.33
5.42
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Show that, if mechanical energy is conserved, the graph of v² against h is a straight line of gradient 2g.
(1)
(c)
Outline why the uncertainty in v² is ±2 %.
(1)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the gradient of the line and its absolute uncertainty, using lines of maximum and minimum gradient.
(3)
(f)
Discuss whether the data show that mechanical energy is conserved.
(1)
(g)
The line of best fit meets the h axis at a positive value. Suggest a systematic error that would cause this, and state whether it affects your conclusion in (f).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 0.0200/0.01060 = 1.89 «m s−1»
✓ 1
v² = 3.56 «m² s−2»
✓ 1
Accept 3.54–3.58. Allow ECF from their v.
Part (b)
mgh = ½mv², so v² = 2gh: proportional to h with gradient 2g
✓ 1
Part (c)
the fractional uncertainty in the card width is 0.2/20.0 = 1 % «timing uncertainty negligible», and v is squared, so 2 × 1 % = 2 %
✓ 1
Part (d)
single straight line through all error bars, not forced through the origin
✓ 1
Part (e)
gradient of best line = 18.6 m s−2
✓ 1
Accept 18.3–19.0 m s−2. Unit required at least once.
maximum and minimum gradient lines drawn through all error bars
✓ 1
uncertainty = ½(max − min) ≈ 0.4 m s−2, so gradient = 18.6 ± 0.4 m s−2
✓ 1
Accept 0.3–0.7 m s−2. MP3 is for matching precision.
Part (f)
2g = 19.6 m s−2 lies above the range of the gradient, so some mechanical energy is dissipated «air resistance, residual friction»; energy is not conserved
✓ 1
Allow ECF from (e).
Part (g)
h systematically measured too large by about 8 mm «eg measured from the wrong point on the glider or gate»
✓ 1
Accept a zero error of the ruler or a mis-placed reference mark.
no: a constant error in h shifts the line sideways but does not change its gradient
✓ 1
Answers: (a) 1.89 m s−1, 3.56 m² s−2 · (e) 18.6 ± 0.4 m s−2(the remaining parts are explanations — see the table above)
94A-1B-28Conservation of energyPaper 1BHard10 marks
A rubber ball is released from rest 1.500 m above a concrete floor and bounces several times. A video recorded against a vertical scale gives the maximum height h reached after each bounce n (n = 0 is the release height). Each height is uncertain by ±0.010 m. Air resistance is negligible.
A student's hypothesis is that the ball keeps the same fraction f of its kinetic energy at every bounce. The graph shows lg h against n, with error bars.
n
h / m
lg(h / m)
0
1.500
0.176
1
1.084
0.035
2
0.773
3
0.563
−0.249
4
0.399
−0.399
5
0.293
6
0.207
−0.684
Graph drawn to scale
(a)
Explain why, if the hypothesis is correct, the graph of lg h against n is a straight line of gradient lg f.
(2)
(b)
Complete the table.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine f.
(2)
(e)
Discuss whether the data support the hypothesis.
(1)
(f)
(i)
Calculate the percentage uncertainty in h for n = 6.
(1)
(ii)
Suggest one improvement to the method that would reduce the uncertainty in the values of h for the later bounces.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at each highest point the energy is all gravitational potential energy, mgh, so if the same fraction f of energy is kept at each bounce, hn = h0fn
✓ 1
Accept: each height is f times the previous one.
lg h = lg h0 + n lg f, which is linear in n with gradient lg f
✓ 1
Part (b)
lg 0.773 = −0.112
✓ 1
3 d.p.; negative sign required.
lg 0.293 = −0.533
✓ 1
3 d.p.; negative sign required.
Part (c)
single straight line through all error bars
✓ 1
Part (d)
gradient = −0.143
✓ 1
Accept −0.145 to −0.140.
f = 10−0.143 = 0.72
✓ 1
Accept 0.71–0.73. Allow ECF from their gradient. Award [1] for f = |gradient|.
Part (e)
the points lie on a straight line within their error bars, so the data support the hypothesis
✓ 1
OWTTE. Do not accept 'the line is straight' without reference to the data/error bars.
Part (f)(i)
0.010/0.207 × 100 = 4.8 %
✓ 1
Accept 5 %.
Part (f)(ii)
any one specific improvement, eg: use a higher frame rate so the highest point is not missed; place the scale in the same plane as the ball «to avoid parallax»; zoom the camera on the lower region; drop the ball from a greater height so later bounces are higher
✓ 1
Do not accept 'repeat' without stating what is averaged.
Answers: (b) −0.112, −0.533 · (d) f ≈ 0.72 · (f)(i) 4.8 % (the remaining parts are explanations — see the table above)
95A-1B-49Elastic potential energyPaper 1BHard12 marks
A steel collar of mass 0.200 kg can slide on a vertical rod. A spring that obeys Hooke's law stands on the base of the rod. The collar is released from rest at a height h above the top of the spring, falls onto the spring and compresses it. The maximum compression x is found from a high-speed video recorded against a millimetre scale.
The uncertainty in h is ±0.002 m and the uncertainty in x is ±0.5 mm. The spring constant is k. The point for h = 0.200 m has not been plotted.
h / m
x / mm
h/x
uncertainty in h/x
0.050
26.5
1.89
0.11
0.100
34.5
2.90
0.10
0.150
41.5
3.61
0.09
0.200
47.0
0.250
52.5
4.76
0.08
0.300
56.5
5.31
0.08
Graph drawn to scale
(a)
Show that, if no energy is dissipated, h/x = (k/2mg) x − 1, where m is the mass of the collar.
(2)
(b)
For h = 0.200 m, calculate h/x and its absolute uncertainty, and plot this point with its error bar on the graph.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine k.
(2)
(e)
Lines of maximum and minimum gradient drawn through the error bars have gradients 113 m−1 and 108 m−1. Determine the absolute uncertainty in your value of k.
(2)
(f)
Hanging known loads from the spring gives k = 400 ± 8 N m−1. Compare this with your answer to (d) and (e).
(1)
(g)
Suggest one way in which energy is dissipated in this experiment, and explain its effect on the value of k found in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the collar is at rest at the start and at maximum compression, and it falls a total height h + x: mg(h + x) = ½kx²
✓ 1
Do not award if the fall is taken as h only.
dividing by mgx: h/x + 1 = (k/2mg)x, so h/x = (k/2mg)x − 1
✓ 1
Must see the rearrangement.
Part (b)
h/x = 0.200/0.0470 = 4.26
✓ 1
Accept 3 s.f.
uncertainty = 4.26 × (0.5/47.0 + 0.002/0.200) = ±0.09 AND point plotted correctly at x = 0.0470 m with its error bar
✓ 1
Accept ±0.08–0.10. Plotting to within half a small square.
Part (c)
single straight line through all error bars, extended to x = 0
the range 426–446 N m−1 lies above 392–408 N m−1: the ranges do not overlap, so the dynamic value is significantly larger
✓ 1
Allow ECF from (d) and (e). Must refer to both ranges.
Part (g)
friction between the collar and the rod OR energy transferred to kinetic energy / vibration of the spring and sound when the collar strikes it
✓ 1
Any one.
less elastic potential energy is stored for each h, so x is smaller and h/x larger: the gradient is too large and k is overestimated «consistent with (f)»
✓ 1
OWTTE. MP2 needs the direction of the effect.
Answers: (b) 4.26 ± 0.09 · (d) k ≈ 436 N m−1 · (e) ±10 N m−1(the remaining parts are explanations — see the table above)
A battery-powered toy truck of mass 0.850 kg climbs a straight ramp at constant speed. The angle θ of the ramp to the horizontal is varied. Two light gates on the ramp, 0.800 ± 0.002 m apart, record the time t taken by the truck to travel between them, and the speed is v = 0.800/t. Each value of t is the mean of several runs and is uncertain by ±0.02 s.
The student's hypothesis is that the motor delivers the same useful output power P on every slope. The total resistive force f on the truck is assumed to be the same on every slope.
θ / °
sin θ
t / s
(1/v) / s m−1
5
0.087
0.44
0.550
10
0.174
0.68
0.850
15
0.259
0.92
1.150
20
1.15
25
0.423
1.37
1.713
30
0.500
1.60
2.000
Graph drawn to scale
(a)
Show that, if the hypothesis is correct, 1/v = (mg/P) sin θ + f/P, where m is the mass of the truck.
(2)
(b)
Complete the table for θ = 20°.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine P.
(2)
(e)
Determine f.
(2)
(f)
Discuss whether the data support the hypothesis.
(1)
(g)
Suggest why the first light gate is placed some distance up the ramp rather than at its bottom.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at constant speed the driving force = mg sin θ + f «component of weight along the slope plus resistive force»
✓ 1
P = Fv = (mg sin θ + f)v, so 1/v = (mg sin θ + f)/P = (mg/P) sin θ + f/P
✓ 1
Must see the rearrangement.
Part (b)
sin 20° = 0.342
✓ 1
Accept 0.34.
1/v = 1.15/0.800 = 1.438 «s m−1»
✓ 1
Accept 1.44. Do not accept 0.800/t.
Part (c)
single straight line with the points close to it and evenly scattered, NOT forced through the origin
✓ 1
Do not accept a line joined point to point.
Part (d)
gradient = 3.50 «s m−1»
✓ 1
Accept 3.37–3.63 s m−1.
P = mg/gradient = 0.850 × 9.81/3.50 = 2.38 W
✓ 1
Accept 2.30–2.47 W. Allow ECF from their gradient.
Part (e)
intercept on the 1/v axis = 0.24 «s m−1»
✓ 1
Accept 0.21–0.28 s m−1.
f = intercept × P = 0.24 × 2.38 = 0.57 N
✓ 1
Accept 0.48–0.70 N «0.58 N from the unrounded intercept 0.243». Allow ECF from (d).
Part (f)
the points lie on a straight line «within the uncertainty of ±0.02 s in t», as the equation predicts, so the data support «a constant power and constant f»
✓ 1
OWTTE. Must refer to the linear relationship.
Part (g)
the truck accelerates when it starts, so the gates must be where it has already reached its constant «steady» speed
✓ 1
OWTTE.
Answers: (b) 0.342; 1.438 s m−1 · (d) P ≈ 2.38 W · (e) f ≈ 0.57 N (the remaining parts are explanations — see the table above)
A weightlifter raises a barbell of mass 120 kg from rest on the floor to rest above his head, a vertical rise of 1.9 m, in 2.4 s. He holds it stationary above his head and then lets go, so that it falls 1.9 m back to the floor. Air resistance is negligible.
(a)
The lift.
(i)
Calculate the work done by the weightlifter on the barbell.
(2)
(ii)
Calculate the average power developed by the weightlifter in raising the barbell.
(1)
(iii)
State the work done on the barbell by the gravitational force during the lift.
(1)
(iv)
Explain, by considering the work done by the resultant force, why your answers to (a)(i) and (a)(iii) are equal in magnitude.
(2)
(b)
State why the weightlifter does no work on the barbell while he holds it stationary.
(1)
(c)
(i)
Calculate the speed of the barbell just before it reaches the floor.
(2)
(ii)
A rubber mat on the floor is compressed by 4.0 cm as it brings the barbell to rest. Determine the average resultant force on the barbell while it is being stopped.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the barbell starts and ends at rest, so the work done equals the gain in gravitational potential energy mgΔh
✓ 1
120 × 9.81 × 1.9 = 2.24 × 103 J
✓ 1
Award [2] for CNA.
Part (a)(ii)
P = 2.24 × 103/2.4 = 930 W
✓ 1
Accept 0.93 kW. Allow ECF from (a)(i).
Part (a)(iii)
−2.24 × 103 J
✓ 1
The negative sign is required.
Part (a)(iv)
work done by the resultant force = change in kinetic energy of the barbell
✓ 1
the barbell is at rest at the start and the end, so the change in kinetic energy is zero and the two works add to zero
✓ 1
OWTTE.
Part (b)
the displacement of the barbell is zero, so W = Fs cos θ = 0
✓ 1
Accept “the barbell does not move”.
Part (c)(i)
½mv² = mgh so v = √(2 × 9.81 × 1.9)
✓ 1
= 6.1 m s−1
✓ 1
Accept 6.11 m s−1.
Part (c)(ii)
kinetic energy at impact = 2.24 × 103 J «= mgh»
✓ 1
Allow ECF from (c)(i).
work done by the resultant force over the stopping distance = loss of kinetic energy: F × 0.040 = 2.24 × 103
✓ 1
F = 5.6 × 104 N
✓ 1
Award [2] max for 5.7 × 104 N «force exerted by the mat, not the resultant force».
Answers: (a)(i) 2.24 × 103 J · (a)(ii) 930 W · (a)(iii) −2.24 × 103 J · (c)(i) 6.1 m s−1 · (c)(ii) 5.6 × 104 N (the remaining parts are explanations — see the table above)
A chairlift carries skiers from the bottom station to the top station of a ski slope, which is 320 m higher. The chairs are fixed to a continuous cable loop: chairs move up full on one side and come down empty on the other. When the lift is busy it carries 1200 skiers per hour. The average mass of a skier with equipment is 85 kg.
Diagram not to scale
(a)
(i)
Calculate the gain in gravitational potential energy of one skier carried from the bottom to the top.
(1)
(ii)
Show that the useful power needed to raise the skiers is about 90 kW.
(1)
(b)
Explain why the weight of the chairs does not need to be included in (a)(ii).
(2)
(c)
When the lift is busy, the input power to its electric motor is 130 kW. The cable moves at 2.5 m s−1.
(i)
Calculate the efficiency of the lift.
(1)
(ii)
Calculate the force that the drive wheel must exert on the cable to provide the useful power.
(2)
(d)
The operator wants to carry 1500 skiers per hour. The maximum input power of the motor is 150 kW. Determine whether this is possible at the same efficiency.
(3)
(e)
Suggest why the motor still needs an input of power when the lift runs with no skiers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
85 × 9.81 × 320 = 2.67 × 105 J
✓ 1
Part (a)(ii)
2.67 × 105 × 1200/3600 = 8.89 × 104 «W»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
for every chair moving up there is a chair «of the same mass» moving down
✓ 1
the gravitational potential energy gained by the rising chairs equals that lost by the falling chairs, so no net energy is needed for the chairs
✓ 1
OWTTE.
Part (c)(i)
η = 8.89 × 104/1.3 × 105 = 0.68 «68 %»
✓ 1
Allow ECF.
Part (c)(ii)
P = Fv so F = P/v
✓ 1
F = 8.89 × 104/2.5 = 3.6 × 104 N
✓ 1
Award [2] for CNA.
Part (d)
useful power needed = 8.89 × 104 × 1500/1200 = 1.11 × 105 W
✓ 1
input power needed = 1.11 × 105/0.684 = 163 kW
✓ 1
163 kW > 150 kW, so it is not possible
✓ 1
ALTERNATIVE: greatest useful power = 0.684 × 150 = 103 kW ✓ < 111 kW needed ✓ so not possible ✓. MP3 requires a comparison. Allow ECF.
Part (e)
energy is still dissipated by friction in the wheels, bearings and cable «and by air resistance and heating in the motor»
✓ 1
Do not accept “to lift the chairs”.
Answers: (a)(i) 2.67 × 105 J · (a)(ii) 8.89 × 104 W · (c)(i) 68 % · (c)(ii) 3.6 × 104 N · (d) 163 kW needed; not possible (the remaining parts are explanations — see the table above)
99A-2-32Friction & inclined planesPaper 2Easy10 marks
In a warehouse, parcels slide down a straight chute inclined at 25° to the horizontal. A parcel of mass 1.6 kg is released from rest at the top and slides 3.5 m to the bottom of the chute, where its speed is 3.6 m s−1.
Diagram NOT accurately drawn
(a)
(i)
Calculate the vertical height through which the parcel falls.
(1)
(ii)
Calculate the loss of gravitational potential energy of the parcel.
(1)
(b)
Determine the average frictional force on the parcel.
(3)
(c)
(i)
Show that the normal force on the parcel is about 14 N.
(1)
(ii)
Calculate the coefficient of dynamic friction between the parcel and the chute.
(1)
(d)
Calculate the time taken by the parcel to slide down the chute.
(2)
(e)
Parcels sometimes arrive at the bottom too quickly. Suggest one change to the chute that would reduce their speed at the bottom.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
h = 3.5 sin 25° = 1.48 m
✓ 1
Part (a)(ii)
ΔEp = 1.6 × 9.81 × 1.48 = 23.2 J
✓ 1
Allow ECF from (a)(i).
Part (b)
Ek at the bottom = ½ × 1.6 × 3.6² = 10.4 J
✓ 1
energy dissipated = 23.2 − 10.4 = 12.8 J
✓ 1
Ff = 12.8/3.5 = 3.7 N
✓ 1
Award [3] for CNA. Allow ECF.
Part (c)(i)
FN = mg cos 25° = 1.6 × 9.81 × cos 25° = 14.2 «N»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
μd = 3.67/14.2 = 0.26
✓ 1
Allow ECF from (b).
Part (d)
the forces on the parcel are constant, so its acceleration is uniform and s = (u + v)t/2: 3.5 = ½ × (0 + 3.6) × t
✓ 1
t = 1.9 s
✓ 1
Accept 1.94 s. Award [2] for CNA. ALTERNATIVE: a = v²/2s = 1.85 m s−2 ✓ and t = v/a = 1.9 s ✓.
Part (e)
reduce the angle of the chute OR give the chute a rougher surface «larger coefficient of friction» OR make the chute shorter from a lower starting point
✓ 1
Accept any one valid change.
Answers: (a)(i) 1.48 m · (a)(ii) 23.2 J · (b) 3.7 N · (c)(i) 14.2 N · (c)(ii) 0.26 · (d) 1.9 s (the remaining parts are explanations — see the table above)
A body of mass 1.5 kg is initially at rest on a horizontal, frictionless surface. A horizontal force is applied to it. The graph shows how the force varies with the displacement of the body.
Graph drawn to scale
(a)
State what is represented by the area under a force–displacement graph.
(1)
(b)
(i)
Calculate the work done by the force over the first 2.0 m.
(1)
(ii)
Show that the total work done by the force over 5.0 m is 105 J.
(1)
(c)
Calculate the speed of the body after it has moved 5.0 m.
(2)
(d)
Draw, on the graph, a line representing the constant force that would do the same work on the body over the 5.0 m.
(2)
(e)
Outline why the acceleration of the body decreases between 2.0 m and 5.0 m.
(1)
(f)
The same force–displacement graph applies to a body of mass 3.0 kg. Deduce the speed of this body after it has moved 5.0 m.
(2)
Show mark scheme
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Mark
Notes
Part (a)
the work done by the force «the energy transferred»
✓ 1
Part (b)(i)
30 × 2.0 = 60 J
✓ 1
Part (b)(ii)
60 + ½ × 3.0 × 30 = 60 + 45 = 105 «J»
✓ 1
Must see the area of the triangle added.
Part (c)
½ × 1.5 × v² = 105
✓ 1
v = 11.8 m s−1
✓ 1
Award [2] for CNA.
Part (d)
horizontal line at 21 N
✓ 1
Allow ± 1 N.
drawn from 0 to 5.0 m «so that the area under it is 105 J»
✓ 1
Part (e)
the force «the resultant force» decreases, and a = F/m with m constant
✓ 1
Part (f)
the same work is done, so the body gains the same kinetic energy «105 J»
✓ 1
v = √(2 × 105/3.0) = 8.4 m s−1
✓ 1
Accept 11.8/√2. Allow ECF from (c).
Answers: (b)(i) 60 J · (b)(ii) 105 J · (c) 11.8 m s−1 · (d) 21 N · (f) 8.4 m s−1(the remaining parts are explanations — see the table above)
101A-2-51Energy density of fuelsPaper 2Easy13 marks
A hiker uses a small stove that burns a gas fuel of energy density 46 MJ kg−1. In a test, the stove heats 0.75 kg of water from 12 °C to 100 °C in 4.0 minutes, and 11.0 g of fuel is burned.
Specific heat capacity of water = 4180 J kg−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
(a)
Outline why a hiker chooses a stove fuel with a large energy density.
(1)
(b)
Show that the energy transferred to the water is about 2.8 × 105 J.
(1)
(c)
Calculate the efficiency of the stove.
(2)
(d)
Calculate the average rate at which the burning fuel releases energy.
(1)
(e)
On a 5-day winter trek the hiker must obtain 2.0 kg of drinking water each day by melting snow at 0 °C and heating the water to 100 °C. A full gas canister contains 230 g of fuel. Determine whether one full canister is enough for the trek. Assume that the efficiency of the stove is the same as in the test.
(3)
(f)
Suggest two changes to the way the stove is used that would increase its efficiency.
(2)
(g)
An electric heater of efficiency 95 % powered by a battery of energy density 0.90 MJ kg−1 could be used instead. Calculate the mass of battery needed for one day of melting and heating snow, and comment on the result.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
less mass of fuel needs to be carried for the same energy released «energy density = energy released per unit mass»
✓ 1
OWTTE. Do not accept “it gives more energy” without reference to the mass carried.
Part (b)
Q = mcΔT = 0.75 × 4180 × 88 = 2.76 × 105 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
energy released by fuel = 0.0110 × 4.6 × 107 = 5.06 × 105 J
✓ 1
η = 2.76 × 105/5.06 × 105 = 0.55 «55 %»
✓ 1
Award [2] for CNA.
Part (d)
P = 5.06 × 105/240 = 2.1 × 103 W
✓ 1
Allow ECF from (c).
Part (e)
energy per day = 2.0 × (3.34 × 105 + 4180 × 100) = 1.50 × 106 J
✓ 1
Award [0] for MP1 if the latent heat is omitted.
fuel per day = 1.50 × 106/(0.55 × 4.6 × 107) = 60 g
✓ 1
Allow ECF from (c). Accept 59–60 g.
5 days need about 300 g > 230 g, so one canister is not enough
✓ 1
Accept 295–300 g. MP3 needs a comparison.
Part (f)
put a lid on the pan «to reduce energy transfer to the air by convection and evaporation»
✓ 1
shield the flame and pan from the wind / use a pan with a heat-exchanger base / keep the flame under the pan
✓ 1
Any two valid changes, one mark each. Do not accept “use more fuel”.
this is about 30 times the mass of gas needed per day «60 g», because the battery’s energy density is much smaller, even though the heater is more efficient: gas is far better for carrying
✓ 1
Comment must refer to mass or energy density.
Answers: (b) 2.76 × 105 J · (c) 55 % · (d) 2.1 × 103 W · (e) 300 g needed: not enough · (g) 1.8 kg (the remaining parts are explanations — see the table above)
A ramp is built so that a wheelchair user can reach a doorway 0.48 m above the pavement. The ramp rises 1.0 m for every 12 m travelled along it. A carer pushes the wheelchair and its user, of total mass 105 kg, up the ramp at a constant speed of 0.90 m s−1, with a force parallel to the ramp. The total resistive force on the wheelchair (rolling resistance and friction in the wheel bearings) is 22 N and does not depend on speed.
Figure 1 shows the wheelchair and user on the ramp. The dot marks their centre of mass.
Figure 1 — Diagram NOT accurately drawn (the slope is exaggerated)
(a)
Draw, on Figure 1, labelled arrows to represent the forces acting on the wheelchair and user as they move up the ramp.
(2)
(b)
Show that the force exerted by the carer is about 110 N.
(1)
(c)
Calculate the work done by the carer on the wheelchair over the whole length of the ramp.
(2)
(d)
For the whole ascent, calculate the work done on the wheelchair and user by
(i)
the gravitational force.
(2)
(ii)
the normal force from the ramp.
(1)
(e)
Calculate the power developed by the carer.
(1)
(f)
To save space, an architect proposes a shorter, steeper ramp to the same doorway, rising 1.0 m for every 8.0 m along it. The resistive force and the speed are unchanged. Deduce whether the carer does more or less work on the steeper ramp, and explain why the steeper ramp is nevertheless harder for the carer to use.
(3)
Show mark scheme
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Mark
Notes
Part (a)
weight «gravitational force» vertically downwards AND normal force perpendicular to the ramp surface, away from it
✓ 1
Allow the normal force drawn as one arrow at each wheel. Labels required.
push parallel to the ramp, up the slope, AND resistive force parallel to the ramp, down the slope, with the push arrow clearly longer than the resistive-force arrow
✓ 1
Award [1] max if any extra force «e.g. a “force of motion”» is drawn.
Part (b)
component of the weight along the ramp = 105 × 9.81 × (1.0/12) = 85.8 N, so F = 85.8 + 22 = 108 «N»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept sin θ with θ = 4.78°.
Part (c)
length of the ramp = 0.48 × 12 = 5.76 m
✓ 1
W = Fs = 108 × 5.76 = 621 J
✓ 1
Accept 634 J from 110 N. Allow ECF from (b). Award [2] for CNA.
Part (d)(i)
the angle between the weight and the displacement is 90° + θ, so W = mgs cos(90° + θ) OR only the vertical rise matters: W = −mgΔh
✓ 1
W = −105 × 9.81 × 0.48 = −494 J
✓ 1
The negative sign is required for MP2. Accept −490 J.
Part (d)(ii)
zero, because the normal force is perpendicular to the displacement «cos 90° = 0»
✓ 1
Part (e)
P = Fv = 108 × 0.90 = 97 W
✓ 1
Accept 99 W from 110 N. Allow ECF from (b).
Part (f)
steeper ramp: length 3.84 m, force = 105 × 9.81/8.0 + 22 = 151 N, work = 151 × 3.84 = 579 J
✓ 1
ALTERNATIVE: the gain in Ep is the same «494 J» but the work against the resistive force falls from 127 J to 84 J.
so the carer does less work on the steeper ramp «579 J < 621 J» because the resistive force acts over a shorter distance
✓ 1
Allow ECF from (c).
but the force needed is about 40 % larger «151 N against 108 N» and, at the same speed, the power is larger «136 W against 97 W», which the carer may not be able to provide
✓ 1
MP3 needs a reference to the larger force or power.
Answers: (b) 108 N · (c) 621 J · (d)(i) −494 J · (d)(ii) 0 · (e) 97 W · (f) 579 J, 151 N (the remaining parts are explanations — see the table above)
A hand-crank charger is used to charge a mobile phone where there is no mains supply. The user turns a handle at the end of a crank arm, so that the handle moves in a circle of radius 0.12 m. The user exerts a force of 18 N on the handle, always in the direction in which the handle is moving, and turns it at a steady rate of 1.5 revolutions per second. The generator in the charger supplies a current of 1.8 A to the phone at a potential difference of 5.0 V.
Figure 1 — Sankey diagram drawn to scale: the side of each small square represents 1 W
(a)
(i)
Show that the work done by the user in one revolution of the handle is about 14 J.
(1)
(ii)
The handle returns to its starting position after each revolution. Explain why the user nevertheless does work on the handle.
(2)
(b)
Show that the mean power developed by the user is about 20 W.
(1)
(c)
Calculate the efficiency of the generator.
(2)
(d)
Figure 1 is an incomplete Sankey diagram for the energy transfers each second. In the phone, 80 % of the electrical energy supplied is stored in the battery; the rest is transferred as thermal energy in the charging circuit and the battery. Complete Figure 1 by drawing and labelling the branches for the phone.
(2)
(e)
The battery of the phone stores 5.5 × 104 J when fully charged. Calculate the time for which the user must turn the handle to charge the battery fully from empty.
(2)
(f)
The user’s muscles convert chemical energy into work on the handle with an efficiency of 25 %. Determine the mass of food, of energy density 1.6 × 107 J kg−1, that supplies the chemical energy used during one full charge.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
W = F × 2πr = 18 × 2π × 0.12 = 13.6 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
at every instant the force is in the direction of motion «θ = 0 in W = Fs cos θ», so the work done in every small displacement is positive
✓ 1
the work depends on the distance moved along the force «the circumference 2πr», not on the overall displacement, so the work over one revolution is not zero
✓ 1
OWTTE. Do not accept “the handle moves” alone.
Part (b)
P = 13.6 × 1.5 = 20.4 «W»
✓ 1
Must see full substitution OR answer to 3 s.f. Allow ECF from (a)(i): 14 × 1.5 = 21 «W».
Part (c)
output power = IV = 1.8 × 5.0 = 9.0 W
✓ 1
η = 9.0/20.4 = 0.44 «44 %»
✓ 1
Allow ECF from (b). Award [2] for CNA.
Part (d)
the 9.0 W band split into a branch of width 7.2 W «7.2 squares» labelled energy stored in the battery «chemical energy» AND a branch of width 1.8 W «1.8 squares» labelled thermal energy
✓ 1
Allow ± half a square. Allow ECF from (c).
the thermal-energy branch turned away from the main direction, and the two new widths add up to the width of the 9.0 W band
✓ 1
Part (e)
power stored = 0.80 × 9.0 = 7.2 W
✓ 1
t = 5.5 × 104/7.2 = 7.6 × 103 s «2.1 h»
✓ 1
Award [1] max for 6.1 × 103 s «80 % omitted». Allow ECF from (c).
Part (f)
chemical energy = 20.4 × 7.64 × 103/0.25 = 6.2 × 105 J
✓ 1
ALTERNATIVE: 5.5 × 104/(0.25 × 0.442 × 0.80). Allow ECF from (b) and (e).
m = 6.2 × 105/1.6 × 107 = 0.039 kg «39 g»
✓ 1
Award [2] for CNA.
Answers: (a)(i) 13.6 J · (b) 20.4 W · (c) 0.44 · (e) 7.6 × 103 s · (f) 0.039 kg (the remaining parts are explanations — see the table above)
104A-2-90Elastic potential energyPaper 2Easy12 marks
The foam midsole of a running shoe is tested in a machine that compresses it and then lets it return to its original thickness, as happens during one step of a run. The machine records the force F on the midsole and its compression x. Figure 1 shows the results. During loading (straight line) the midsole obeys Hooke’s law. The area under the unloading curve represents the energy returned by the midsole as it recovers its thickness.
Figure 1 — graph drawn to scale
(a)
Determine the spring constant of the midsole during loading.
(1)
(b)
Show that the elastic potential energy stored in the midsole at maximum compression is about 8 J.
(1)
(c)
The manufacturer claims that the midsole returns 85 % of the energy stored in it. Determine whether Figure 1 supports this claim.
(3)
(d)
State what is represented by the area between the loading line and the unloading curve.
(1)
(e)
During a run, one midsole is compressed 650 times per kilometre. Calculate the energy dissipated in the midsole during a 3.0 km run.
(2)
(f)
The midsole has a mass of 0.080 kg and its specific heat capacity is 1.7 × 103 J kg−1 K−1. Calculate the rise in temperature of the midsole if none of the energy in (e) leaves it, and suggest why the actual rise in temperature is much smaller.
(2)
(g)
A heavier runner exerts a maximum force of 2200 N on the same midsole. Assuming that the loading line remains straight, deduce the elastic potential energy stored at maximum compression.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
k = 1800/9.0 × 10−3 = 2.0 × 105 N m−1
✓ 1
Accept any correct pair of values read from the line. Unit required.
Part (b)
½ × 2.0 × 105 × (9.0 × 10−3)² = 8.10 «J»
✓ 1
Accept ½ × 1800 × 9.0 × 10−3. Must see full substitution OR answer to 3 s.f.
Part (c)
area under the unloading curve found by counting squares «≈ 126 small squares, each 0.5 × 10−3 m × 100 N = 0.05 J»
✓ 1
Accept the trapezium rule or other valid method.
energy returned ≈ 6.3 J
✓ 1
Accept 6.0–6.6 J.
6.3/8.1 = 0.78 «78 %», which is less than 85 %, so the graph does not support the claim
✓ 1
MP3 needs a comparison with 85 %. Allow ECF from MP2.
Part (d)
the energy dissipated «transferred to internal energy of the foam» in one compression and recovery
✓ 1
Accept “energy lost as thermal energy per step”.
Part (e)
energy dissipated per compression = 8.1 − 6.3 = 1.8 J
✓ 1
Allow ECF from (c).
1950 × 1.8 = 3.5 × 103 J
✓ 1
Award [2] for CNA.
Part (f)
ΔT = 3.5 × 103/(0.080 × 1.7 × 103) = 26 K
✓ 1
Allow ECF from (e).
during the run «which lasts many minutes» energy is transferred from the midsole to the surrounding air, the foot and the ground «by conduction and convection»
✓ 1
OWTTE.
Part (g)
x = 2200/2.0 × 105 = 0.011 m OR EH = F²/2k
✓ 1
EH = ½ × 2200 × 0.011 = 12 J
✓ 1
Award [1] max for 9.9 J «energy taken as proportional to the force».
Answers: (a) 2.0 × 105 N m−1 · (b) 8.1 J · (c) 6.3 J, 78 % · (e) 3.5 × 103 J · (f) 26 K · (g) 12 J (the remaining parts are explanations — see the table above)
105A-2-97Conservation of energyPaper 2Easy10 marks
A school team builds a model trebuchet (Figure 1). A counterweight of mass 15.0 kg on the short end of a pivoted arm falls through a vertical height of 0.90 m. The long end of the arm swings a sling that releases a projectile of mass 0.200 kg. A video shows that the projectile leaves the sling at 14.0 m s−1, at a point 1.6 m higher than its starting position.
Figure 1 — Diagram NOT accurately drawn
(a)
Calculate the loss of gravitational potential energy of the counterweight.
(1)
(b)
Determine the efficiency of the trebuchet. Take the useful output to be the gain in mechanical energy of the projectile.
(3)
(c)
Suggest two reasons why the efficiency is much less than 100 %.
(2)
(d)
A team member claims that doubling the mass of the counterweight will double the launch speed. Assuming that the efficiency and the release point are unchanged, deduce whether this claim is correct.
(3)
(e)
State one change to the design of the trebuchet that would increase its efficiency.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ΔEp = 15.0 × 9.81 × 0.90 = 132 J
✓ 1
Part (b)
kinetic energy of the projectile = ½ × 0.200 × 14.0² = 19.6 J
✓ 1
gain in Ep of the projectile = 0.200 × 9.81 × 1.6 = 3.14 J
✓ 1
η = (19.6 + 3.14)/132 = 0.17 «17 %»
✓ 1
Award [2] max for 0.15 «gain in Ep of the projectile omitted». Allow ECF from (a).
Part (c)
at release the counterweight and arm are still moving, so they keep kinetic energy that is not transferred to the projectile
✓ 1
work is done against friction at the pivot «and air resistance on the arm and sling», transferring energy to internal energy; energy also goes to sound and vibration of the frame
✓ 1
Any two distinct reasons, one mark each.
Part (d)
the loss of Ep of the counterweight doubles, so the useful output doubles to 45.5 J
✓ 1
kinetic energy of the projectile = 45.5 − 3.14 = 42.3 J, so v = √(2 × 42.3/0.200) = 20.6 m s−1
✓ 1
Allow ECF from (b).
20.6 m s−1 is much less than 28 m s−1, so the claim is incorrect: Ek ∝ v², so doubling the energy increases the speed by a factor of only about √2
✓ 1
Accept an argument based on v ∝ √Ek without the Ep correction for [2] max.
Part (e)
use a lighter arm «so less kinetic energy is left in the arm» OR lubricate the pivot / use a low-friction bearing OR adjust the sling so that release happens when the counterweight is at its lowest point
✓ 1
Any one.
Answers: (a) 132 J · (b) 0.17 · (d) 20.6 m s−1(the remaining parts are explanations — see the table above)
106IBP2S-005-01-11Sankey diagramsPaper 2Easy7 marks
An emergency radio contains a small generator that is turned by a hand crank. A person turning the crank does work at a rate of 8.0 W. The generator delivers 5.2 W of electrical power. The rest of the input power is transferred to the surroundings as thermal energy and sound.
(a)
State the principle of conservation of energy.
(1)
(b)
Calculate the efficiency of the generator.
(2)
(c)
Describe a Sankey diagram, drawn to scale, for the generator.
(2)
(d)
The radio's rechargeable battery stores 90 % of the electrical energy it receives from the generator. Calculate the time for which the crank must be turned to store 1.5 kJ in the battery.
(2)
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Notes
Part (a)
Energy cannot be created or destroyed, only transferred or transformed from one form to another, so the total energy of an isolated (closed) system stays constant.
✓ 1
Part (b)
Efficiency = useful output power ÷ input power = 5.2 ÷ 8.0
✓ 1
= 0.65 or 65 %
✓ 1
Part (c)
A single input arrow labelled 'work done by hand, 8.0 W', with a width representing 8.0 units, splits into two arrows.
✓ 1
One arrow of width 5.2 units continues straight on and is labelled electrical. The other arrow, of width 2.8 units, bends away and is labelled thermal energy/sound to the surroundings.
✓ 1
Part (d)
Rate of energy storage = 0.90 × 5.2 = 4.68 W
✓ 1
t = 1500 ÷ 4.68 ≈ 320 s (3.2×102 s)
✓ 1
Exam tip: In a Sankey description, give the width of every arrow and check that the output widths add up to the input width.
108IBP2S-005-03-11Elastic potential energyPaper 2Easy5 marks
A physiotherapy patient strengthens a knee by stretching an elastic resistance band. The band obeys Hooke's law and has spring constant 350 N m−1. In each repetition the band is stretched slowly by 0.40 m.
(a)
Calculate the force exerted on the band at an extension of 0.40 m.
(1)
(b)
Calculate the elastic potential energy stored in the band at an extension of 0.40 m.
(2)
(c)
The patient completes 12 repetitions in 30 s. Calculate the mean power developed by the patient in stretching the band.
(2)
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Notes
Part (a)
F = kx = 350 × 0.40 = 140 N
✓ 1
Part (b)
Ep = ½kx² = ½ × 350 × 0.40²
✓ 1
= 28 J
✓ 1
Part (c)
Total work done = 12 × 28 = 336 J (ECF from (b))
✓ 1
P = 336 / 30 ≈ 11 W
✓ 1
Common error: Power uses the total work for all repetitions divided by the total time, not the energy of a single stretch.
Outline whether the string obeys Hooke's law over this range.
(2)
(b)
Determine the spring constant of the string.
(2)
(c)
Calculate the elastic potential energy stored in the string at an extension of 8.0 mm.
(2)
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Notes
Part (a)
Force increases by equal steps of 3.0 N for each 2.0 mm, so F/x is constant
✓ 1
Force is proportional to extension, so the string obeys Hooke's law
✓ 1
Part (b)
k = F/x = 12.0 / 8.0×10−3
✓ 1
= 1.5×103 N m−1
✓ 1
Part (c)
Ep = ½kx² = ½ × 1500 × (8.0×10−3)², or ½Fx = ½ × 12.0 × 8.0×10−3 (ECF from (b))
✓ 1
= 0.048 J (4.8×10−2 J)
✓ 1
Common error: Convert millimetres to metres before calculating. Leaving the extension in mm makes k 1000 times too small and the energy wrong by a large factor.
110IBP2S-005-02-11Friction & inclined planesPaper 2Easy7 marks
In a curling match, a stone of mass 19.0 kg is released on horizontal ice with a speed of 2.40 m s−1. It slides in a straight line for 32.0 m before coming to rest. Assume the frictional force on the stone is constant.
(a)
Calculate the initial kinetic energy of the stone.
(2)
(b)
State the work done by friction on the stone as it comes to rest.
(1)
(c)
Calculate the magnitude of the frictional force.
(2)
(d)
Determine the coefficient of dynamic friction between the stone and the ice.
(2)
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Notes
Part (a)
Ek = ½ × 19.0 × 2.402
✓ 1
= 54.7 J
✓ 1
Part (b)
−54.7 J (accept 54.7 J of work done against friction) . Allow error carried forward from the previous part.
✓ 1
Part (c)
F = W/s = 54.7/32.0
✓ 1
= 1.71 N . Allow error carried forward from the previous part.
✓ 1
Part (d)
Normal force = mg = 19.0 × 9.81 = 186 N
✓ 1
μ = 1.71/186 = 9.2 × 10−3 . Allow error carried forward from the previous part; accept g = 9.8 m s−2.
✓ 1
Exam tip: The mass cancels in the final answer, giving μ = v²/(2gs). Use this as a quick check on your numerical working.
111IBP2S-005-04-11Energy density of fuelsPaper 2Easy5 marks
A hiker uses a butane camping stove to heat 0.50 kg of water from 15 °C to 100 °C. The specific heat capacity of water is 4200 J kg−1 K−1. Butane has a specific energy of 4.9×107 J kg−1. Only 40% of the energy released by the burning butane is transferred to the water.
(a)
State what is meant by the specific energy of a fuel.
(1)
(b)
Calculate the energy transferred to the water.
(2)
(c)
Determine the mass of butane burned.
(2)
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Notes
Part (a)
Energy released per unit mass of fuel (when it is burned/used)
✓ 1
Part (b)
Q = mcΔT = 0.50 × 4200 × 85
✓ 1
= 1.8×105 J (1.785×105 J)
✓ 1
Part (c)
Energy released by fuel = 1.785×105 / 0.40 = 4.46×105 J
✓ 1
Mass = 4.46×105 / 4.9×107 = 9.1×10−3 kg (9.1 g); accept 9.2×10−3 kg if the rounded 1.8×105 J is used
✓ 1
Allow error carried forward from the previous part.
—
Common error: The temperature change is 85 K, not 100 K. Also divide by the efficiency, not multiply, because the fuel must release more energy than the water gains.
112IBP2S-005-04-12Energy density of fuelsPaper 2Easy5 marks
Aircraft designers are comparing fuels for a future airliner. The table gives data for three fuels.
Fuel | Specific energy / MJ kg−1 | Density / kg m−3 Jet fuel (kerosene) | 45.6 | 840 Liquid hydrogen | 142 | 71 Methanol | 19.9 | 792
(a)
Calculate the energy density of liquid hydrogen.
(2)
(b)
Outline, with reference to the data, one advantage of liquid hydrogen as an aircraft fuel.
(1)
(c)
Determine the ratio \(\frac{\text{volume of liquid hydrogen}}{\text{volume of jet fuel}}\) needed to store the same amount of energy.
(2)
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Notes
Part (a)
Energy density = specific energy × density = 142×106 × 71
✓ 1
= 1.01×1010 J m−3 (accept 1.0×1010 J m−3)
✓ 1
Part (b)
It has the largest specific energy (about 3 times that of jet fuel), so a smaller mass of fuel is needed for the same energy, which reduces the aircraft's mass
✓ 1
Part (c)
Energy density of jet fuel = 45.6×106 × 840 = 3.83×1010 J m−3
✓ 1
Ratio = 3.83×1010 / 1.01×1010 ≈ 3.8
✓ 1
Allow error carried forward from the energy density of liquid hydrogen calculated in the first part.
—
Exam tip: For the same energy, the required volume is inversely proportional to the energy density. The ratio of volumes is therefore the inverse ratio of energy densities, with the fuel of larger energy density on top.
113IBP2S-005-00-11Sankey diagramsPaper 2Easy4 marks
A defibrillator delivers a pulse of 200 J of electrical energy to a patient's chest in a time of 4.0 ms. All of this energy comes from a storage component inside the defibrillator, which loses no energy during the pulse.
(a)
Calculate the average power delivered during the pulse.
(2)
(b)
Before the pulse, a battery charges the storage component with the 200 J in 6.0 s. The charging process has an efficiency of 0.80. Determine the average output power of the battery during charging.
(2)
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Part (a)
Uses P = E/t = 200 / (4.0 × 10−3)
✓ 1
P = 5.0 × 104 W
✓ 1
Part (b)
Energy supplied by battery = 200 / 0.80 = 250 J
✓ 1
Power = 250 / 6.0 = 42 W (accept 41.7 W)
✓ 1
Common error: Convert milliseconds to seconds before dividing. Leaving the time as 4.0 gives an answer that is 1000 times too small. In (b), divide by the efficiency, because the battery must supply more energy than is stored.
115A-2-09Conservation of energyPaper 2Medium12 marks
A lorry of mass 1.2 × 104 kg travels down a long, straight hill at a constant speed of 15 m s−1. The road down the hill is 3.0 km long and the lorry descends a vertical height of 150 m. The average total resistive force from air resistance and rolling friction is 1.8 kN. The rest of the energy is transferred to the brakes.
The brake discs have a total mass of 60 kg and are made of steel of specific heat capacity 450 J kg−1 K−1.
Figure 1 — Sankey diagram for (c), to be completed
(a)
(i)
State the change in the kinetic energy of the lorry during the descent.
(1)
(ii)
Calculate the decrease in the gravitational potential energy of the lorry.
(1)
(b)
Show that the energy transferred to the brakes is about 12 MJ.
(2)
(c)
Complete, on Figure 1, the Sankey diagram for the energy transfers during the descent.
(2)
(d)
The brake discs are at 20 °C at the top of the hill and their maximum safe temperature is 350 °C. Assume that all the energy transferred to the brakes remains in the discs. Determine whether the discs exceed their maximum safe temperature.
(3)
(e)
Outline why, in practice, the discs reach a lower temperature than the one you found in (d).
(1)
(f)
Calculate the average rate at which energy is transferred to the brakes.
(2)
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Notes
Part (a)(i)
zero «the speed is constant»
✓ 1
Part (a)(ii)
ΔEp = 1.2 × 104 × 9.81 × 150 = 1.77 × 107 J
✓ 1
Accept 17.7 MJ or 18 MJ.
Part (b)
work done against the resistive forces = 1800 × 3000 = 5.4 × 106 J
✓ 1
energy to the brakes = 1.77 × 107 − 5.4 × 106 «= 1.23 × 107 J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
two output bands labelled: thermal energy in the brakes «12.3 MJ» AND thermal energy from work against air resistance/rolling friction «5.4 MJ»
✓ 1
band widths in proportion: brakes ≈ 69 % and resistive forces ≈ 31 % of the input width «brakes band about twice as wide»
✓ 1
Allow ECF from (b). The two output widths must add up to the input width.
Part (d)
ΔT = Q/mc = 1.23 × 107 / (60 × 450) = 454 K
✓ 1
Accept 444 K from 12 MJ.
final temperature = 20 + 454 = 474 °C
✓ 1
Allow ECF from (b).
474 °C > 350 °C so the discs exceed their maximum safe temperature
✓ 1
The decision must be consistent with the candidate’s temperature.
Part (e)
energy is transferred from the hot discs to the surrounding air «by convection and radiation» during the descent
✓ 1
Do not accept “heat is lost” without a destination or mechanism.
Part (f)
time = 3000 / 15 = 200 s
✓ 1
P = 1.23 × 107 / 200 = 6.1 × 104 W
✓ 1
Allow ECF from (b). Accept 60 kW from 12 MJ.
Answers: (a)(ii) 1.77 × 107 J · (b) 1.23 × 107 J · (d) 474 °C; exceeds · (f) 6.1 × 104 W (the remaining parts are explanations — see the table above)
116A-2-10Elastic potential energyPaper 2Medium12 marks
A railway wagon of mass 1.8 × 104 kg rolls at 1.5 m s−1 along a level track into a buffer stop. The buffer stop has two identical spring buffers side by side, and both are compressed by the same amount. Figure 1 shows how the force F needed to compress one buffer varies with its compression x. Friction is negligible.
Figure 1 — graph drawn to scale
(a)
Determine the spring constant of one buffer.
(1)
(b)
Show that the kinetic energy of the wagon is about 20 kJ.
(1)
(c)
Each buffer can be compressed by at most 0.30 m. Determine whether the buffers can stop the wagon.
(3)
(d)
Calculate the maximum deceleration of the wagon.
(2)
(e)
Suggest why the buffers are also fitted with dampers that transfer energy to internal energy.
(2)
(f)
The buffers are replaced by stiffer ones that also stop the wagon. Explain the effect on the maximum deceleration of the wagon.
(3)
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Notes
Part (a)
gradient = 90 × 103 / 0.30 = 3.0 × 105 N m−1
✓ 1
Accept any correct pair of values read from the line.
Part (b)
½ × 1.8 × 104 × 1.5² «= 2.03 × 104 J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
energy stored in the two buffers: 2 × ½kx² = 2.03 × 104 J
✓ 1
x = √(2.03 × 104 / 3.0 × 105) = 0.26 m
✓ 1
Allow ECF from (a) and (b).
0.26 m < 0.30 m so the buffers stop the wagon
✓ 1
Award [1 max] for using one buffer only «x = 0.37 m, “cannot stop it”».
Part (d)
maximum force = 2kx = 2 × 3.0 × 105 × 0.260 = 1.56 × 105 N
✓ 1
Allow ECF from (c).
a = 1.56 × 105 / 1.8 × 104 = 8.7 m s−2
✓ 1
Award [2] for CNA.
Part (e)
without dampers the elastic potential energy stored is all returned to the wagon as kinetic energy
✓ 1
so the wagon would rebound at 1.5 m s−1; dampers dissipate the energy so the wagon comes to rest
✓ 1
OWTTE.
Part (f)
the same kinetic energy is stored, ½kx² = constant, so the maximum compression is smaller
✓ 1
the maximum force kx is larger «kx = √(2kE), which increases with k»
✓ 1
so the maximum deceleration is greater
✓ 1
MP3 only scores if MP2 scores.
Answers: (a) 3.0 × 105 N m−1 · (b) 2.03 × 104 J · (c) 0.26 m; stops the wagon · (d) 8.7 m s−2(the remaining parts are explanations — see the table above)
An electric bicycle and its rider have a total mass of 90 kg. The rider does not pedal. The bicycle climbs a straight hill at a constant speed of 4.0 m s−1; the road rises 1.0 m vertically for every 20 m travelled along it. The battery supplies a current of 8.0 A to the motor at a potential difference of 36 V. The motor and drive transfer 75 % of the electrical energy they receive into work done on the bicycle; the rest becomes thermal energy.
When fully charged, the battery can deliver a total charge of 5.0 × 104 C at 36 V.
The diagram is an incomplete Sankey diagram for the energy transfers each second during the climb.
Sankey diagram, band widths drawn to scale
(a)
(i)
Calculate the electrical power supplied by the battery.
(1)
(ii)
Calculate the time for which a fully charged battery can supply this current.
(2)
(b)
(i)
Show that the rate at which the bicycle and rider gain gravitational potential energy is about 180 W.
(2)
(ii)
Determine the total resistive force acting on the bicycle and rider.
(3)
(c)
Complete the Sankey diagram, labelling each branch you add with the energy transfer it represents and its value in watts.
(2)
(d)
Calculate the overall efficiency of the bicycle, taking the gain in gravitational potential energy as the useful output.
(1)
(e)
The battery has a mass of 3.6 kg. Calculate the energy density of the battery, and suggest one reason why a battery is used in the bicycle rather than a small petrol engine, even though petrol has an energy density of about 46 MJ kg−1.
(2)
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Notes
Part (a)(i)
P = IV = 8.0 × 36 = 288 W
✓ 1
Part (a)(ii)
t = Q/I = 5.0 × 104/8.0
✓ 1
= 6.3 × 103 s
✓ 1
Accept 6250 s or 1.7 h.
Part (b)(i)
vertical speed = 4.0 × (1.0/20) = 0.20 m s−1 OR component of weight along the road = 90 × 9.81/20 = 44.1 N
✓ 1
P = 90 × 9.81 × 0.20 = 177 «W»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
power transferred to the bicycle = 0.75 × 288 = 216 W
✓ 1
power used against resistive forces = 216 − 177 = 39 W
✓ 1
F = P/v = 39/4.0 = 9.9 N
✓ 1
Accept 9.0–10 N «9.0 N follows from using 180 W». Allow ECF.
Part (c)
the 216 W band divided into two labelled branches: work done against resistive forces «≈ 39 W» AND gain in gravitational potential energy «≈ 177 W»
✓ 1
Allow ECF from (b). Accept “thermal energy / air and road” for the resistive branch.
branch widths in proportion: the resistive branch about one-fifth of the width of the 216 W band, the two new branches together equal to the 216 W band
✓ 1
By eye.
Part (d)
η = 177/288 = 0.61 «= 61 %»
✓ 1
Accept 0.63 «from 180 W». Allow ECF.
Part (e)
energy stored = QV = 5.0 × 104 × 36 = 1.8 × 106 J, so energy density = 0.50 MJ kg−1
✓ 1
any one: the electric motor transfers a much larger fraction of its input to useful work than a small petrol engine OR no exhaust gases are emitted where the bicycle is used OR it is quieter OR the battery can be recharged from renewable sources
✓ 1
Answers: (a)(i) 288 W · (a)(ii) 6.3 × 103 s · (b)(i) 177 W · (b)(ii) 9.9 N · (d) 61 % · (e) 0.50 MJ kg−1(the remaining parts are explanations — see the table above)
118A-2-26Conservation of energyPaper 2Medium13 marks
A roller-coaster car of total mass 480 kg passes point A, at the top of the first drop, at 2.0 m s−1. Graph 1 shows how the height of the track above the lowest point B varies with the distance along the track. The track from A to B is 78 m long and the track from B to the next crest C is 52 m long.
Graph 1, drawn to scaleGraph 2, axes for (b)
(a)
Assume first that no energy is dissipated.
(i)
Calculate the loss of gravitational potential energy of the car between A and B.
(1)
(ii)
Show that the speed of the car at B would be about 30 m s−1.
(1)
(b)
Sketch, on graph 2, how the kinetic energy of the car would vary with the distance along the track from A to C if no energy were dissipated.
(2)
(c)
In practice the car passes B at 27 m s−1.
(i)
Determine the energy dissipated between A and B.
(3)
(ii)
Calculate the average resistive force on the car between A and B.
(1)
(iii)
The average resistive force between B and C is the same as between A and B. Determine whether the car reaches C.
(3)
(d)
Suggest two reasons why the resistive force on the car is not constant along the track.
(2)
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Notes
Part (a)(i)
ΔEp = 480 × 9.81 × 45 = 2.12 × 105 J
✓ 1
Part (a)(ii)
½ × 480 × v² = ½ × 480 × 2.0² + 2.12 × 105 so v = 29.8 «m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f. Do not award if the kinetic energy at A is omitted «29.7 m s−1».
Part (b)
shape: kinetic energy rises from a small value at A to a maximum at B «78 m», then falls to a lower value at C, mirroring the height profile «smooth curve»
✓ 1
values: about 2.1 × 105 J at B AND about 8.1 × 104 J at C
✓ 1
Allow ± 0.1 × 105 J. Allow ECF from (a).
Part (c)(i)
total energy at A = 960 + 2.12 × 105 = 2.129 × 105 J «relative to B»
✓ 1
Ek at B = ½ × 480 × 27² = 1.75 × 105 J
✓ 1
energy dissipated = 3.8 × 104 J
✓ 1
Award [2] max for 3.7 × 104 J «kinetic energy at A omitted».
Part (c)(ii)
F = 3.79 × 104/78 = 490 N
✓ 1
Accept 470–490 N. Allow ECF.
Part (c)(iii)
energy needed to rise to C = 480 × 9.81 × 28 = 1.32 × 105 J
✓ 1
work done against the resistive force = 486 × 52 = 2.5 × 104 J
✓ 1
1.57 × 105 J < 1.75 × 105 J «the kinetic energy at B», so the car reaches C
✓ 1
MP3 requires a comparison. The car arrives with about 1.8 × 104 J. Allow ECF.
Part (d)
air resistance increases with speed «so it is largest near B»
✓ 1
the friction in the wheels depends on the normal force, which changes with the slope and curvature of the track «largest in the dip at B»
✓ 1
Accept any other valid physical reason.
Answers: (a)(i) 2.12 × 105 J · (a)(ii) 29.8 m s−1 · (c)(i) 3.8 × 104 J · (c)(ii) 490 N · (c)(iii) yes (the remaining parts are explanations — see the table above)
A car of mass 1300 kg travels along a straight, level road. Its engine delivers a constant useful output power of 45 kW. The graph shows how the total resistive force on the car varies with its speed.
Graph drawn to scale
(a)
Determine the acceleration of the car at a speed of 15 m s−1.
(3)
(b)
(i)
Draw, on the graph, a curve showing how the driving force varies with speed between 15 m s−1 and 50 m s−1.
(2)
(ii)
Hence determine the top speed of the car.
(2)
(c)
Explain why the acceleration of the car decreases as its speed increases, even though the output power is constant.
(2)
(d)
The car climbs a straight hill on which the road rises 1.0 m for every 10 m travelled along it. The output power is still 45 kW and the resistive force varies with speed as shown on the graph. Determine the top speed of the car on the hill.
(3)
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Notes
Part (a)
driving force = P/v = 4.5 × 104/15 = 3000 N
✓ 1
resistive force from the graph = 480 N
✓ 1
Accept 460–500 N.
a = (3000 − 480)/1300 = 1.9 m s−2
✓ 1
Allow ECF.
Part (b)(i)
at least three correct points, eg 3000 N at 15 m s−1, 1500 N at 30 m s−1, 900 N at 50 m s−1
✓ 1
Allow ± 100 N.
smooth curve with a negative gradient whose magnitude decreases as the speed increases «not a straight line», crossing the resistive-force curve
✓ 1
Part (b)(ii)
the top speed is where the driving force equals the resistive force «the curves cross: resultant force zero»
✓ 1
35 m s−1
✓ 1
Accept 34–36 m s−1. Allow ECF from the candidate’s curve.
Part (c)
the driving force F = P/v decreases as the speed increases
✓ 1
while the resistive force increases, so the resultant force, and hence a = F/m, decreases
✓ 1
Part (d)
component of the weight along the road = 1300 × 9.81 × 0.10 = 1280 N
✓ 1
at top speed the driving force P/v equals the resistive force + 1280 N «eg the resistive-force curve raised by 1280 N crosses the driving-force curve»
✓ 1
Accept a trial: at 22 m s−1 P/v = 2050 N and the opposing force is 1960 N.
v = 23 m s−1
✓ 1
Accept 21–24 m s−1. Allow ECF from the candidate’s curve in (b)(i).
Answers: (a) 1.9 m s−2 · (b)(ii) 35 m s−1 · (d) 23 m s−1(the remaining parts are explanations — see the table above)
A pumped-storage power station has an upper reservoir whose water surface is, on average, 320 m above the turbines. A full upper reservoir holds 7.0 × 106 m3 of usable water of density 1000 kg m−3.
When generating, water flows down to the turbines at 140 m3 s−1 and the station converts 88 % of the gravitational potential energy lost by the water into electrical energy. When pumping, 85 % of the electrical energy supplied to the pumps becomes gravitational potential energy of the water.
Figure 1 — grid for (e); each small square represents 10 MJ
(a)
Show that the gravitational potential energy released when 1.0 kg of water descends 320 m is about 3 kJ.
(1)
(b)
Calculate the gravitational potential energy stored in a full upper reservoir.
(1)
(c)
(i)
Calculate the electrical power output of the station when it is generating.
(2)
(ii)
The electrical energy is transmitted at a potential difference of 400 kV. Calculate the current.
(1)
(d)
Calculate the overall efficiency of storing electrical energy in the reservoir and later recovering it.
(2)
(e)
Draw, on Figure 1, a Sankey diagram to scale for one complete cycle of pumping and generating that starts with 100 MJ of electrical energy. Label each band.
(3)
(f)
Explain, with reference to your answer to (a), why pumped-storage stations need very large reservoirs placed at a large height.
(3)
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Notes
Part (a)
ΔEp = mgΔh = 1.0 × 9.81 × 320 = 3139 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
E = 7.0 × 109 × 9.81 × 320 = 2.2 × 1013 J
✓ 1
Accept 6.1 GWh.
Part (c)(i)
mass flow = 1.4 × 105 kg s−1, so rate of loss of Ep = 1.4 × 105 × 9.81 × 320 = 4.39 × 108 W
✓ 1
P = 0.88 × 4.39 × 108 = 3.9 × 108 W
✓ 1
Award [2] for CNA.
Part (c)(ii)
I = P/V = 3.87 × 108/4.0 × 105 = 970 A
✓ 1
Allow ECF from (c)(i).
Part (d)
efficiency = 0.85 × 0.88
✓ 1
= 0.748 «75 %»
✓ 1
Award [2] for CNA. Award [1] max for adding or averaging the efficiencies.
Part (e)
a loss band of width 15 MJ «1.5 squares» leaving at the pumping stage, labelled as thermal energy «in pumps, pipes and water»
✓ 1
a second loss band of width 10 MJ «about 1 square» leaving at the generating stage, labelled
✓ 1
output band of width 75 MJ «7.5 squares» labelled electrical energy out; widths add up to 100 MJ
✓ 1
Allow ECF from (d).
Part (f)
each kilogram of water stores only about 3.1 kJ, a very small energy per unit mass «compared with about 46 MJ kg−1 for a fuel»
✓ 1
so a very large mass «volume» of water is needed to store a useful amount of energy
✓ 1
the energy stored per kilogram is proportional to the height «gΔh», so a large height reduces the volume needed
✓ 1
Answers: (a) 3139 J · (b) 2.2 × 1013 J · (c)(i) 3.9 × 108 W · (c)(ii) 970 A · (d) 75 % (the remaining parts are explanations — see the table above)
121A-2-53Conservation of energyPaper 2Medium13 marks
An electric bus of mass 1.4 × 104 kg reaches a speed of 12 m s−1 between bus stops. When it brakes, its motor acts as a generator: this regenerative braking transfers part of the kinetic energy of the bus to its battery. On average 60 % of the kinetic energy that the bus has at the start of braking is stored in the battery. The battery has a terminal potential difference of 650 V.
(a)
Calculate the kinetic energy of the bus at 12 m s−1.
(1)
(b)
Show that the energy stored in the battery each time the bus stops is about 6 × 105 J.
(1)
(c)
Calculate the charge that flows into the battery each time the bus stops.
(2)
(d)
Explain why only 60 % of the kinetic energy is stored in the battery.
(2)
(e)
A planner proposes that the bus should reach 18 m s−1 between stops, because “more kinetic energy means more energy is recovered”. Discuss this proposal.
(4)
(f)
Without regenerative braking, the bus uses 1.3 kWh of energy per kilometre. On a 12 km route it makes 25 stops from 12 m s−1. Estimate the percentage of the energy used on the route that regenerative braking recovers. (1 kWh = 3.6 × 106 J)
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Ek = ½ × 1.4 × 104 × 12² = 1.0 × 106 J
✓ 1
Accept 1.01 × 106 J.
Part (b)
0.60 × 1.01 × 106 = 6.05 × 105 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
V = W/q so q = W/V
✓ 1
q = 6.05 × 105/650 = 930 C
✓ 1
Allow ECF from (b). Award [2] for CNA.
Part (d)
work is done against resistive forces «air resistance, rolling resistance» during braking, so part of the kinetic energy never reaches the generator
✓ 1
energy is transferred as thermal energy in the generator, the electrical circuits and the battery «resistive heating» «and in friction brakes needed at low speed»
✓ 1
Any two distinct mechanisms.
Part (e)
Ek ∝ v², so the energy recovered per stop increases by a factor of (18/12)² = 2.25
✓ 1
but the energy needed to accelerate the bus also increases by 2.25, so the energy not recovered per stop rises from 4.0 × 105 J to 9.1 × 105 J
✓ 1
air resistance is also larger at higher speed, so more work is done against it between stops
✓ 1
so the proposal increases the energy used on the route: it should be rejected
✓ 1
Award MP4 for a conclusion consistent with MP2 or MP3.
Part (f)
energy used = 12 × 1.3 × 3.6 × 106 = 5.6 × 107 J
✓ 1
energy recovered = 25 × 6.0 × 105 = 1.5 × 107 J
✓ 1
Allow ECF from (b).
percentage = 27 %
✓ 1
Accept 26–28 %.
Answers: (a) 1.0 × 106 J · (b) 6.0 × 105 J · (c) 930 C · (f) 27 % (the remaining parts are explanations — see the table above)
A small wind turbine supplies a remote research station. Its blades sweep out a circle of radius 2.5 m. The wind blows at 8.0 m s−1 perpendicular to the plane of the blades. The density of the air is 1.2 kg m−3. At this wind speed the electrical power output of the turbine is 1.8 kW.
(a)
Show that the mass of air that passes through the circle swept out by the blades each second is about 190 kg.
(1)
(b)
Calculate the kinetic energy of the air that passes through the circle each second.
(1)
(c)
The wind speed doubles. Deduce the factor by which the power available from the wind increases.
(2)
(d)
Calculate the efficiency of the turbine.
(1)
(e)
The station needs an average power of 1.2 kW. Records show that the wind blows at 8.0 m s−1 for 40 % of the time and at 4.0 m s−1 for the rest of the time. Determine whether the turbine can meet the average demand. Assume that the efficiency of the turbine does not change.
(3)
(f)
An engineer instead uses the mean wind speed of 5.6 m s−1 to estimate the average power output. Explain why this method gives an average power that is too small.
(2)
(g)
Calculate the minimum radius of the circle swept by the blades for a turbine of the same efficiency to meet the demand.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
mass per second = ρAv = 1.2 × π × 2.5² × 8.0 = 188 «kg»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
½ × 188 × 8.0² = 6.0 × 103 J «so 6.0 kW is available»
✓ 1
Allow ECF from (a).
Part (c)
the mass of air passing through the circle each second doubles «it is ρAv»
✓ 1
the kinetic energy of each kilogram is multiplied by 4 «∝ v²», so the power increases by a factor of 8 «P = ½ρAv³»
✓ 1
Part (d)
η = 1800/6.03 × 103 = 0.30 «30 %»
✓ 1
Allow ECF from (b).
Part (e)
at 4.0 m s−1 the output is 1800 × (4.0/8.0)³ = 225 W
✓ 1
Award [0] for MP1 if a proportionality to v or v² is used.
average = 0.40 × 1800 + 0.60 × 225 = 855 W
✓ 1
0.86 kW < 1.2 kW, so the turbine cannot meet the demand
✓ 1
MP3 needs a comparison. Allow ECF from MP2.
Part (f)
P ∝ v³, so the output at the mean speed «617 W» is less than the mean of the outputs
✓ 1
Accept a calculation of the output at 5.6 m s−1.
the periods at 8.0 m s−1 contribute far more energy than the low-speed periods, and averaging the speeds first loses this
✓ 1
OWTTE.
Part (g)
P ∝ r², so r = 2.5 × √(1200/855)
✓ 1
Allow ECF from (e).
r = 3.0 m
✓ 1
Accept 2.9–3.0 m.
Answers: (a) 188 kg · (b) 6.0 × 103 J · (d) 30 % · (e) 855 W: cannot meet the demand · (g) 3.0 m (the remaining parts are explanations — see the table above)
123A-2-91Kinetic energy & momentumPaper 2Medium14 marks
A crash cushion is placed in front of a concrete bridge support at the side of a motorway. When a vehicle hits the cushion head-on, the cushion is crushed and, in a simple model, exerts a constant horizontal force of 1.25 × 105 N on the vehicle until the vehicle stops. In a test, a car of mass 1200 kg hits the cushion head-on at 25.0 m s−1.
Treat each vehicle as a particle that is not itself deformed, and ignore all other horizontal forces on it.
Diagram not to scale
(a)
(i)
Calculate the momentum of the car just before the impact.
(1)
(ii)
Show that the kinetic energy of the car just before the impact is about 3.8 × 105 J.
(1)
(b)
(i)
Calculate the distance through which the cushion is crushed as the car is brought to rest.
(2)
(ii)
Calculate the time taken for the cushion to stop the car.
(1)
(iii)
Calculate the rate at which the cushion does work on the car at the instant the impact begins.
(1)
(c)
In a second test, a van of mass 2400 kg hits an identical cushion with the same momentum as the car in (a)(i). The cushion again exerts a constant force of 1.25 × 105 N.
(i)
Use Ek = p2/2m to show that the kinetic energy of the van is half that of the car.
(1)
(ii)
Deduce how the time taken to stop the van, and the distance through which its cushion is crushed, compare with those for the car.
(3)
(d)
The cushion can be crushed through at most 3.6 m. Determine whether it can stop the car of mass 1200 kg if the car hits it at 30.0 m s−1.
(2)
(e)
Another design of cushion behaves like a spring that obeys Hooke’s law. It is designed to stop the car of (a), from 25.0 m s−1, in the same crush distance as in (b)(i). Explain why the constant-force design is safer for the people in the car.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = 1200 × 25.0 = 3.0 × 104 kg m s−1
✓ 1
Unit required.
Part (a)(ii)
½ × 1200 × 25.02 = 3.75 × 105 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
work done by the cushion «Fs» = loss of kinetic energy of the car
✓ 1
s = 3.75 × 105/1.25 × 105 = 3.00 m
✓ 1
Award [2] for CNA. Allow ECF from (a)(ii).
Part (b)(ii)
t = Δp/F = 3.0 × 104/1.25 × 105 = 0.24 s
✓ 1
ALTERNATIVE: t = s/average speed = 3.00/12.5. Allow ECF from (a)(i) or (b)(i).
Part (b)(iii)
P = Fv = 1.25 × 105 × 25.0 = 3.1 × 106 W
✓ 1
Accept 3.12 × 106 W.
Part (c)(i)
Ek = (3.0 × 104)2/(2 × 2400) = 1.88 × 105 J, which is half of 3.75 × 105 J «same p, twice the mass»
✓ 1
Must see the substitution OR the argument that Ek ∝ 1/m when p is the same.
Part (c)(ii)
time: the same change of momentum and the same force, so the same impulse FΔt and the same time «0.24 s»
✓ 1
distance: the work done by the cushion equals the kinetic energy, so the crush distance s = Ek/F is proportional to the kinetic energy
✓ 1
so the van’s cushion is crushed half as far «1.5 m»
✓ 1
Allow ECF from (b)(i).
Part (d)
Ek = ½ × 1200 × 30.02 = 5.4 × 105 J AND greatest work the cushion can do = 1.25 × 105 × 3.6 = 4.5 × 105 J
✓ 1
ALTERNATIVE: distance needed = 5.4 × 105/1.25 × 105 = 4.3 m, OR greatest speed that can be stopped = 27.4 m s−1.
5.4 × 105 J > 4.5 × 105 J «4.3 m > 3.6 m», so no, the car is not stopped by the cushion
✓ 1
MP2 needs a comparison. Do not award MP2 without a calculation.
Part (e)
the same energy must be transferred over the same distance, so the area under the force–distance graph is the same; for the Hooke’s-law cushion the force rises from zero to a maximum of twice the constant force
✓ 1
Accept a sketch of the two force–distance graphs with equal areas.
so the greatest deceleration of the car and of its passengers «and the forces on them» is twice as large OR the spring-like cushion stores elastic energy and pushes the car back «rebound»
✓ 1
OWTTE.
Answers: (a)(i) 3.0 × 104 kg m s−1 · (a)(ii) 3.75 × 105 J · (b)(i) 3.00 m · (b)(ii) 0.24 s · (b)(iii) 3.1 × 106 W · (c)(i) 1.88 × 105 J · (c)(ii) same time, 0.24 s; half the distance, 1.5 m · (d) 5.4 × 105 J > 4.5 × 105 J: not stopped (the remaining parts are explanations — see the table above)
Some aircraft carriers use an electromagnetic launch system to accelerate aircraft along a straight, horizontal track. In a test, the system accelerates a wheeled test vehicle of mass 2.4 × 104 kg from rest along a track of length 92 m. The vehicle leaves the end of the track at 72 m s−1. Assume that the force exerted on the vehicle by the launch system is constant and that resistive forces are negligible.
Figure 1 — axes for (c)(iii)
(a)
Calculate the kinetic energy of the vehicle as it leaves the track.
(1)
(b)
Show that the force exerted on the vehicle by the launch system is about 6.8 × 105 N.
(1)
(c)
(i)
Determine the average power delivered to the vehicle during the launch.
(2)
(ii)
Calculate the power delivered to the vehicle just before it leaves the track.
(1)
(iii)
Sketch, on Figure 1, a graph to show how the power delivered to the vehicle varies with time during the launch.
(2)
(d)
Between launches, energy taken from the ship’s electrical supply is stored in the launch system. Of the energy taken from the store, 62 % is transferred to kinetic energy of the vehicle.
(i)
The store is recharged in 45 s. Calculate the minimum average power that the ship’s supply must provide.
(2)
(ii)
Suggest why the energy is stored first, rather than taken directly from the ship’s supply during the launch.
(1)
(e)
The same force acts on a test vehicle of half the mass along the same track. Deduce the speed at which this vehicle leaves the track.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Ek = ½ × 2.4 × 104 × 72² = 6.2 × 107 J
✓ 1
Part (b)
work done by the resultant force = gain in kinetic energy: F = 6.22 × 107/92 = 6.76 × 105 «N»
✓ 1
Must see full substitution OR answer to 3 s.f. ALTERNATIVE: a = v²/2s and F = ma.
Part (c)(i)
time = 2s/v = 2 × 92/72 = 2.56 s
✓ 1
Accept t = v/a.
P = 6.22 × 107/2.56 = 2.4 × 107 W
✓ 1
Award [2] for CNA. Allow ECF from (a).
Part (c)(ii)
P = Fv = 6.76 × 105 × 72 = 4.9 × 107 W
✓ 1
Allow ECF from (b).
Part (c)(iii)
straight line through the origin «P = Fv with F constant and v increasing uniformly»
✓ 1
Do not accept a curve or a horizontal line.
line ending at about 49 MW at 2.6 s, and stopping there
✓ 1
Allow ECF from (c)(i) and (c)(ii).
Part (d)(i)
energy taken from the store = 6.22 × 107/0.62 = 1.0 × 108 J
✓ 1
Allow ECF from (a).
P = 1.0 × 108/45 = 2.2 × 106 W «2.2 MW»
✓ 1
Award [2] for CNA.
Part (d)(ii)
during the launch the store must deliver a power of up to about 79 MW «49/0.62» for about 2.6 s, far more than the ship’s supply can provide; storing allows a small power to be taken over a long time
✓ 1
OWTTE.
Part (e)
the same work Fs is done, so the gain in kinetic energy is the same
✓ 1
½(m/2)v² = ½m × 72², so v = 72 × √2 = 102 m s−1
✓ 1
Award [1] max for 144 m s−1.
Answers: (a) 6.2 × 107 J · (b) 6.76 × 105 N · (c)(i) 2.4 × 107 W · (c)(ii) 4.9 × 107 W · (d)(i) 2.2 × 106 W · (e) 102 m s−1(the remaining parts are explanations — see the table above)
125A-2-93Work done by a forcePaper 2Medium13 marks
An athlete puts a shot of mass 4.00 kg. In a simple model of the final push, the shot is accelerated from rest along a straight line 1.60 m long, at 38° above the horizontal. It leaves the athlete’s hand at a speed of 13.0 m s−1 in this direction, 2.10 m above level ground (Figure 1). Air resistance is negligible.
Figure 1 — Diagram NOT accurately drawn
(a)
(i)
Calculate the kinetic energy of the shot as it leaves the hand.
(1)
(ii)
Determine the average force exerted by the athlete on the shot along the line of the push.
(3)
(b)
(i)
Show that the speed of the shot at the highest point of its flight is about 10 m s−1.
(1)
(ii)
Use energy considerations to determine the maximum height of the shot above the ground.
(3)
(c)
Calculate the speed of the shot just before it lands.
(2)
(d)
Outline why your answer to (c) does not depend on the angle at which the shot is released.
(1)
(e)
The athlete increases the release speed by 5.0 %, with the same length and angle of push. Determine the percentage increase in the work done by the athlete on the shot.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ek = ½ × 4.00 × 13.0² = 338 J
✓ 1
Part (a)(ii)
gain in gravitational potential energy = 4.00 × 9.81 × 1.60 sin 38° = 38.7 J
✓ 1
work done by the athlete = gain in mechanical energy = 338 + 38.7 = 377 J
✓ 1
F = 377/1.60 = 235 N
✓ 1
Award [2] max for 211 N «gain in Ep omitted». Allow ECF.
Part (b)(i)
the horizontal component of velocity is constant: 13.0 cos 38° = 10.2 «m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
kinetic energy at the highest point = ½ × 4.00 × 10.2² = 210 J «not zero»
✓ 1
Δh = (338 − 210)/(4.00 × 9.81) = 3.26 m
✓ 1
maximum height = 2.10 + 3.26 = 5.4 m
✓ 1
Award [1] max for 10.7 m «kinetic energy at the top taken as zero». Allow ECF from (b)(i). Accept 5.3–5.4 m.
Part (c)
½mv² = 338 + 4.00 × 9.81 × 2.10 «= 420 J»
✓ 1
v = 14.5 m s−1
✓ 1
Award [2] for CNA.
Part (d)
mechanical energy is conserved and energy is a scalar: the kinetic energy at landing depends only on the kinetic energy at release and the height fallen, not on the direction of the velocity
✓ 1
OWTTE.
Part (e)
new kinetic energy = 1.05² × 338 = 373 J; the gain in Ep is unchanged
✓ 1
new work = 373 + 38.7 = 411 J, an increase of 9.2 %
✓ 1
Award [1] max for 10 % «Ep ignored» or 5 %. Allow ECF from (a).
Answers: (a)(i) 338 J · (a)(ii) 235 N · (b)(i) 10.2 m s−1 · (b)(ii) 5.4 m · (c) 14.5 m s−1 · (e) 9.2 % (the remaining parts are explanations — see the table above)
A sailplane (a glider with no engine) and its pilot have a total mass of 450 kg. The sailplane is launched by a winch on the ground, which pulls in a long cable attached to it. At the end of the launch the sailplane is 380 m above the ground and moving at 30 m s−1, and the cable is released.
Treat the cable as straight, light and inextensible.
Figure 1 — Diagram NOT accurately drawn
(a)
(i)
Calculate the gain in gravitational potential energy of the sailplane during the launch.
(1)
(ii)
The cable does 2.2 × 106 J of work on the sailplane during the launch. Determine the energy transferred by work done against air resistance during the launch.
(2)
(iii)
The launch lasts 40 s. Calculate the average power delivered to the sailplane by the cable.
(1)
(b)
Figure 1 shows the sailplane at one instant during the launch. The tension in the cable is 6.0 kN, the sailplane moves at 28 m s−1, and the angle between the cable and the velocity of the sailplane is 70°.
(i)
Calculate the rate at which the cable does work on the sailplane at this instant.
(2)
(ii)
Near the top of the launch, the cable becomes almost perpendicular to the velocity of the sailplane. Explain why the pilot releases the cable at about this point.
(2)
(c)
At the instant shown in Figure 1, the drum of the winch winds in the cable at a speed of 9.58 m s−1.
(i)
Calculate the output power of the winch at this instant.
(1)
(ii)
Suggest why your answers to (b)(i) and (c)(i) are equal.
(2)
(iii)
The electric motor of the winch takes 2.9 MJ of energy from its battery during the launch. Determine the efficiency with which this energy is transferred to mechanical energy of the sailplane.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ΔEp = 450 × 9.81 × 380 = 1.68 × 106 J
✓ 1
Part (a)(ii)
kinetic energy gained = ½ × 450 × 302 = 2.03 × 105 J
✓ 1
2.2 × 106 − 1.68 × 106 − 2.03 × 105 = 3.2 × 105 J
✓ 1
Award [1] max for 5.2 × 105 J «kinetic energy omitted». Allow ECF from (a)(i).
Part (a)(iii)
P = 2.2 × 106/40 = 5.5 × 104 W
✓ 1
Part (b)(i)
only the component of the tension along the velocity does work: P = Fv cos 70°
✓ 1
P = 6.0 × 103 × 28 × cos 70° = 5.75 × 104 W
✓ 1
Award [1] max for 1.7 × 105 W «Fv, angle ignored». Accept 5.7–5.8 × 104 W.
Part (b)(ii)
P = Fv cos θ OR W = Fs cos θ: as θ approaches 90° the cable does almost no work on the sailplane, so it no longer transfers energy to it
✓ 1
beyond this point θ > 90° and the cable would do negative work «remove energy, pull the sailplane back and down», so there is no benefit in keeping it
✓ 1
OWTTE.
Part (c)(i)
P = Fv = 6.0 × 103 × 9.58 = 5.75 × 104 W
✓ 1
Part (c)(ii)
the cable is light and does not stretch, so no energy is stored in it or transferred to it; all the work done on the cable by the winch is done by the cable on the sailplane «conservation of energy»
✓ 1
the cable is wound in at the component of the sailplane’s velocity along the cable: 28 cos 70° = 9.58 m s−1
✓ 1
Accept a calculation showing that 28 cos 70° = 9.58.
127A-2-95Energy density of fuelsPaper 2Medium10 marks
A small solar-powered helicopter of mass 1.8 kg flies on Mars, where the gravitational field strength is 3.71 N kg−1. Its rechargeable battery has a mass of 0.27 kg and stores 1.4 × 105 J when it is fully charged.
(a)
Calculate the energy density of the battery.
(1)
(b)
On one flight the helicopter climbs vertically to a height of 12 m, flies at that height and then lands at its starting point. The motors draw an electrical power of 360 W for the whole flight, which lasts 100 s.
(i)
Calculate the percentage of the energy stored in the fully charged battery that is used during the flight.
(1)
(ii)
During the climb the helicopter rises at a constant speed of 1.5 m s−1. Calculate the rate at which it gains gravitational potential energy, and express this as a percentage of the electrical power drawn by the motors.
(2)
(iii)
State the total work done on the helicopter by the gravitational force during the whole flight.
(1)
(c)
The battery is recharged by a solar panel of area 0.068 m². During the 7.0 hours of daylight in a Martian day, the mean intensity of the sunlight on the panel is 380 W m−2, and the panel converts 22 % of the incident energy into electrical energy. Each Martian day, heaters that keep the electronics warm overnight take 8.0 × 104 J from the battery. Determine whether the panel can supply enough energy each day for the heaters and one flight like the one in (b).
(4)
(d)
A petrol engine, using a fuel of energy density 4.6 × 107 J kg−1, would provide far more energy per kilogram than the battery. Suggest why such an engine cannot be used on Mars.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1.4 × 105/0.27 = 5.2 × 105 J kg−1
✓ 1
Unit required.
Part (b)(i)
energy used = 360 × 100 = 3.6 × 104 J, which is 26 % of the stored energy
✓ 1
Part (b)(ii)
rate of gain of Ep = mgv = 1.8 × 3.71 × 1.5 = 10 W
✓ 1
Award [0] for MP1 if 9.81 is used.
10.0/360 × 100 = 2.8 %
✓ 1
Allow ECF from MP1.
Part (b)(iii)
zero, because the helicopter returns to its starting height «the work done by gravity depends only on the change in height»
✓ 1
Part (c)
electrical power from the panel = 0.068 × 380 × 0.22 = 5.7 W
1.43 × 105 J > 1.16 × 105 J, so the panel supplies enough energy «with about 2.7 × 104 J to spare»
✓ 1
MP4 needs a comparison consistent with the candidate’s values.
Part (d)
the atmosphere of Mars contains almost no oxygen «it is mainly carbon dioxide», so the fuel cannot be burned; the quoted energy density assumes combustion in oxygen
✓ 1
Accept: oxygen would also have to be carried, adding much mass.
Answers: (a) 5.2 × 105 J kg−1 · (b)(i) 26 % · (b)(ii) 10 W, 2.8 % · (b)(iii) 0 · (c) 1.43 × 105 J available, 1.16 × 105 J needed (the remaining parts are explanations — see the table above)
128A-2-96Energy density of fuelsPaper 2Medium12 marks
On a dairy farm, animal manure is fed into a heated tank called a digester, where bacteria break it down and produce methane gas. The methane is burned in an engine that drives an electrical generator. The digester produces 1500 m³ of methane each day, at a density of 0.67 kg m−3. The energy density of methane is 5.5 × 107 J kg−1.
(a)
Calculate the mass of methane produced each day.
(1)
(b)
Show that the mean rate at which energy is released by burning the methane is about 640 kW.
(1)
(c)
38 % of the energy released is transferred to electrical energy. Calculate the mean electrical power output.
(1)
(d)
Heat exchangers on the engine’s cooling water and exhaust recover a further 42 % of the energy released as useful thermal energy. Calculate the overall efficiency of the system, and outline what happens to the rest of the energy released.
(2)
(e)
The farm uses a mean electrical power of 90 kW. The rest is supplied to nearby homes, each of which uses a mean power of 0.45 kW. Calculate the number of homes that can be supplied.
(2)
(f)
Each day 40 m³ of manure, of density 1.0 × 103 kg m−3 and specific heat capacity 4.0 × 103 J kg−1 K−1, enters the digester and must be warmed by 25 K. Determine whether the recovered thermal energy is enough to do this.
(3)
(g)
Methane is a greenhouse gas. Explain why burning the methane, rather than letting it escape into the atmosphere, reduces the farm’s contribution to the enhanced greenhouse effect, even though carbon dioxide is produced.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
m = ρV = 0.67 × 1500 = 1005 kg
✓ 1
Accept 1.0 × 103 kg.
Part (b)
1005 × 5.5 × 107/(24 × 3600) = 6.40 × 105 «W»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
0.38 × 6.40 × 105 = 2.4 × 105 W «243 kW»
✓ 1
Accept 243 kW from 640 kW.
Part (d)
overall efficiency = 0.38 + 0.42 = 0.80 «80 %»
✓ 1
the remaining 20 % is transferred to the surroundings as thermal energy «warm exhaust gases, hot engine and generator surfaces» and sound
✓ 1
Do not accept “energy is lost” without a destination.
Part (e)
power available for homes = 243 − 90 = 153 kW
✓ 1
Allow ECF from (c).
number of homes = 153/0.45 = 340
✓ 1
Accept 339–341. Award [1] max for 540 «farm’s use not subtracted».
Part (f)
Q = mcΔT = 4.0 × 104 × 4.0 × 103 × 25 = 4.0 × 109 J per day
✓ 1
recovered thermal energy = 0.42 × 5.53 × 1010 = 2.3 × 1010 J per day
✓ 1
ALTERNATIVE: power needed 46 kW and power recovered 269 kW.
2.3 × 1010 J > 4.0 × 109 J, so the recovered energy is enough «about 6 times what is needed»
✓ 1
MP3 needs a comparison. Allow ECF.
Part (g)
each methane molecule burned produces one carbon dioxide molecule «and water»
✓ 1
methane is a much more effective greenhouse gas than carbon dioxide «it absorbs much more infrared radiation per molecule», so replacing methane by carbon dioxide reduces the warming effect
✓ 1
Accept also: the carbon came from plants that recently absorbed CO2, so there is no net addition. OWTTE.
Answers: (a) 1005 kg · (b) 6.40 × 105 W · (c) 2.4 × 105 W · (d) 80 % · (e) 340 · (f) 4.0 × 109 J needed, 2.3 × 1010 J recovered (the remaining parts are explanations — see the table above)
129IBP2S-005-01-13Conservation of energyPaper 2Medium6 marks
Engineers test the regenerative braking system of an electric van of total mass 1500 kg. In each trial, the van brakes on a level road from an initial speed u to rest, and the energy E stored in the battery is measured.
u / m s−1 : 10, 15, 20, 25 E / kJ : 48, 101, 174, 258
(a)
Show that the kinetic energy of the van at 20 m s−1 is 300 kJ.
(1)
(b)
Determine the efficiency of energy recovery for the trials at 10 m s−1 and 25 m s−1.
(2)
(c)
Outline one reason why the efficiency of recovery decreases as the initial speed increases.
(1)
(d)
Estimate the greatest vertical height the van could climb, starting from rest, using only the energy stored in the 25 m s−1 trial. Assume there are no energy losses during the climb.
(2)
Show mark scheme
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Notes
Part (a)
Ek = ½ × 1500 × 20² = 3.0 × 105 J = 300 kJ
✓ 1
Part (b)
At 10 m s−1: Ek = 75 kJ, so efficiency = 48 ÷ 75 = 0.64 (64 %)
✓ 1
At 25 m s−1: Ek = 469 kJ, so efficiency = 258 ÷ 469 = 0.55 (55 %)
✓ 1
Part (c)
Any one valid reason, for example:
—
air resistance is much larger at high speed, so a larger fraction of the kinetic energy is dissipated to the air during braking
—
OR at high speed the friction brakes are also needed, and they transfer kinetic energy to thermal energy
—
OR at high braking power, energy losses in the generator and electronics are greater
✓ 1
Part (d)
mgh = E, so h = 258 × 103 ÷ (1500 × 9.8)
✓ 1
h = 17.6 m ≈ 18 m
✓ 1
Exam tip: Convert kJ to J before you substitute into mgh = E. An estimate to 2 significant figures is enough here.
130IBP2S-005-01-14Sankey diagramsPaper 2Medium7 marks
During a cardiac stress test, a patient pedals an exercise bike. Her muscles transfer chemical energy at a rate of 1200 W. The bike's flywheel receives 250 W of mechanical power, and the remaining power becomes thermal energy in her body.
(a)
Calculate the efficiency of the patient's muscles.
(1)
(b)
Describe a Sankey diagram for this energy transfer.
(2)
(c)
Explain why the patient's body temperature would rise during the test if her body could not lose thermal energy.
(2)
(d)
The test lasts 40 minutes. The energy density of the carbohydrate used by the muscles is 17 MJ kg−1. Determine the mass of carbohydrate used.
(2)
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Notes
Part (a)
Efficiency = 250 ÷ 1200 = 0.21 (21 %)
✓ 1
Part (b)
An input arrow representing 1200 W of chemical energy splits into a narrow straight-on arrow of 250 W, labelled mechanical/useful work
✓ 1
and a much wider arrow of 950 W (= 1200 − 250) bending away, labelled thermal energy, with arrow widths proportional to power so the 950 W arrow is about 3.8 times the width of the 250 W arrow
✓ 1
Part (c)
Thermal energy is produced in the body at 1200 − 250 = 950 W
✓ 1
If this energy is not transferred to the surroundings, for example by evaporation of sweat, the internal energy of the body increases, so its temperature rises
✓ 1
Part (d)
Energy used = 1200 × 40 × 60 = 2.88 × 106 J
✓ 1
Mass = (2.88 × 106) ÷ (17 × 106) = 0.17 kg
✓ 1
Common error: Base the energy used on the chemical input power of 1200 W, not the useful 250 W output. The muscles consume fuel for the whole input.
131IBP2S-005-01-15Elastic potential energyPaper 2Medium7 marks
A robotic lunar lander of mass 2000 kg is descending vertically at 0.80 m s−1 when its engine cuts out 1.5 m above the surface of the Moon. The gravitational field strength at the Moon's surface is 1.6 N kg−1, and there is no atmosphere. The lander has crushable legs that compress by 0.30 m as it comes to rest.
Diagram not to scale
(a)
Show that the speed of the lander just before touchdown is about 2.3 m s−1.
(2)
(b)
Explain why the energy absorbed by the crushable legs is greater than the kinetic energy of the lander at touchdown.
(2)
(c)
Determine the average force exerted by the legs on the lander during the compression.
132IBP2S-005-03-13Elastic potential energyPaper 2Medium6 marks
At a railway terminus, a carriage of mass 2.4×104 kg rolls at 0.50 m s−1 into two identical buffer springs that act side by side. Each spring has spring constant 6.0×105 N m−1 and obeys Hooke's law. Friction is negligible.
Diagram not to scale
(a)
Calculate the kinetic energy of the carriage before it reaches the buffers.
(1)
(b)
Show that the maximum compression of the buffers is about 0.07 m.
(2)
(c)
Calculate the maximum force exerted on the carriage by the buffers.
(2)
(d)
Determine the maximum deceleration of the carriage.
(1)
Show mark scheme
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Notes
Part (a)
Ek = ½ × 2.4×104 × 0.50² = 3.0×103 J
✓ 1
Part (b)
Combined spring constant = 2 × 6.0×105 = 1.2×106 N m−1, OR each spring stores 1.5×103 J
✓ 1
½ × 1.2×106 × x² = 3.0×103, so x = 0.0707 m ≈ 0.071 m
✓ 1
Part (c)
F = kx using the combined k = 1.2×106 N m−1, OR 2 × (6.0×105 × x)
✓ 1
F = 1.2×106 × 0.0707 ≈ 8.5×104 N; accept 8.4×104 N if x = 0.07 m is used (ECF from (b))
✓ 1
Part (d)
a = F/m = 8.5×104 / 2.4×104 ≈ 3.5 m s−2 (ECF from (c))
✓ 1
Exam tip: Springs side by side add their spring constants. Alternatively, split the energy equally between the two springs; both methods give the same compression.
133IBP2S-005-03-14Elastic potential energyPaper 2Medium6 marks
A small satellite (CubeSat) of mass 4.0 kg is pushed out of a deployer on a space station by a compressed spring. The spring has spring constant 250 N m−1 and is compressed by 0.12 m before release. The CubeSat leaves the deployer at 0.85 m s−1 relative to the station.
(a)
Calculate the elastic potential energy stored in the compressed spring.
(2)
(b)
Calculate the kinetic energy of the CubeSat as it leaves the deployer.
(1)
(c)
Determine the efficiency of the deployer in transferring stored elastic energy to kinetic energy of the CubeSat.
(2)
(d)
Outline one reason why the efficiency is less than 100 %.
(1)
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Notes
Part (a)
Ep = ½kx² = ½ × 250 × 0.12²
✓ 1
= 1.8 J
✓ 1
Part (b)
Ek = ½ × 4.0 × 0.85² = 1.4 J (accept 1.45 J or 1.445 J)
✓ 1
Part (c)
Efficiency = useful energy output / energy input = Ek / Ep = 1.445 / 1.8 (allow error carried forward from (a) and (b))
✓ 1
≈ 0.80, or 80 % (accept 0.78 or 78 % if the rounded value of 1.4 J is used)
✓ 1
Part (d)
Any one of the following :
✓ 1
• friction between the CubeSat and the guide rails transfers energy to thermal energy (internal energy)
—
• the spring itself gains kinetic energy or vibrates, so some energy remains in the spring or is dissipated in it
—
• energy is dissipated as internal energy in the spring or deployer structure, for example through vibration or sound in the structure
—
Exam tip: Gravity does not need to be included here: the station and the CubeSat are in free fall together, so only the spring does work on the CubeSat relative to the station. Carry unrounded values into later parts to avoid rounding errors.
134IBP2S-005-03-16Sankey diagramsPaper 2Medium6 marks
An emergency wind-up torch for disaster-relief kits stores energy in a steel spring. When fully wound, the spring stores 150 J of elastic potential energy. The spring has a mass of 0.50 kg. As the spring unwinds it drives a generator, which powers an LED. The overall efficiency from stored elastic energy to electrical energy delivered to the LED is 60 %.
(a)
Calculate the specific energy (energy per unit mass) of the spring.
(1)
(b)
The LED requires an electrical power of 0.50 W. Calculate how long the torch operates on one full winding.
(2)
(c)
A lithium battery has a specific energy of 7.2×105 J kg−1. It delivers energy to the LED with an efficiency of 90 %. Determine the mass of battery that would supply the same electrical energy to the LED as one full winding of the spring.
(2)
(d)
Outline one advantage of the spring-powered torch for use in disaster relief.
(1)
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Notes
Part (a)
150 / 0.50 = 3.0×102 J kg−1
✓ 1
Part (b)
Useful energy delivered = 0.60 × 150 = 90 J
✓ 1
t = 90 / 0.50 = 180 s
✓ 1
Part (c)
Energy required from the battery = 90 / 0.90 = 100 J (allow ECF from (b))
✓ 1
Mass = 100 / 7.2×105 ≈ 1.4×10−4 kg
✓ 1
Part (d)
Any one of the following :
✓ 1
• it can be rewound by hand many times, so no charging supply or replacement batteries are needed
—
• it does not lose stored energy through chemical self-discharge during long storage
—
Exam tip: When comparing energy stores, compare the useful energy delivered, not the stored energy. Apply each device's efficiency before calculating the mass.
135IBP2S-005-02-13Work done by a forcePaper 2Medium8 marks
A student pulls a wooden block of mass 0.500 kg up a rough adjustable ramp at constant speed. A force sensor keeps the pulling force F parallel to the ramp. The results for three ramp angles θ are shown below. Take g = 9.8 m s−2.
θ / ° : 10 | 20 | 30 F / N : 2.30 | 3.06 | 3.72
(a)
Outline why F increases as θ increases.
(2)
(b)
Show that, at θ = 30°, the coefficient of dynamic friction between the block and the ramp is about 0.3.
(3)
(c)
Calculate the work done by F when the block moves 1.20 m up the ramp at θ = 30°.
(1)
(d)
Determine the fraction of the work done by F in the previous part that is transferred to internal energy.
(2)
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Notes
Part (a)
At constant speed F = mg sin θ + friction, and the weight component down the ramp, mg sin θ, increases with θ
✓ 1
The friction force μmg cos θ decreases only slightly, so the increase in the weight component outweighs it and F increases
✓ 1
Part (b)
Weight component along the ramp = 0.500 × 9.8 × sin 30° = 2.45 N
✓ 1
Friction = 3.72 − 2.45 = 1.27 N and normal force = 0.500 × 9.8 × cos 30° = 4.24 N
✓ 1
μ = 1.27/4.24 = 0.299 ≈ 0.3
✓ 1
Part (c)
W = 3.72 × 1.20 = 4.46 J
✓ 1
Part (d)
Work done against friction = 1.27 × 1.20 = 1.52 J
✓ 1
Fraction = 1.52/4.46 = 0.34
✓ 1
Allow error carried forward from the friction force and the work done found in the earlier parts.
—
Exam tip: In a 'show that' question, give the unrounded value (0.299) before stating that it agrees with the given figure.
136IBP2S-005-04-13Sankey diagramsPaper 2Medium8 marks
An electric bicycle has a battery of mass 2.6 kg that stores 0.50 kWh of energy when fully charged. The total mass of the rider and bicycle, including the battery, is 95 kg. They travel at a constant speed of 4.0 m s−1 up a long hill. The road rises 1.0 m vertically for every 20 m travelled along it. The total resistive force on the bicycle is 18 N. Assume the motor supplies all the output power needed and that all the stored energy of the battery can be used.
Not to scale: slope exaggerated
(a)
Show that the specific energy of the battery is about 0.7 MJ kg−1.
(2)
(b)
Calculate the output power of the motor.
(3)
(c)
The motor system has an efficiency of 80%. Determine the time for which the fully charged battery can power the bicycle up the hill.
(2)
(d)
Calculate the distance travelled along the road in this time.
(1)
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Notes
Part (a)
0.50 kWh = 0.50 × 3.6×106 J = 1.8×106 J
✓ 1
1.8×106 / 2.6 = 6.9×105 J kg−1 ≈ 0.7 MJ kg−1
✓ 1
Part (b)
Vertical speed = 4.0 / 20 = 0.20 m s−1, so the rate of gain of Ep = 95 × 9.8 × 0.20 = 186 W (allow mgv sin θ with sin θ = 1/20)
✓ 1
Power against resistance = 18 × 4.0 = 72 W
✓ 1
Total output power = 258 W ≈ 2.6×102 W
✓ 1
Part (c)
Input power = 258 / 0.80 = 323 W
✓ 1
t = 1.8×106 / 323 = 5.6×103 s (about 93 min)
✓ 1
Allow error carried forward from the earlier parts.
—
Part (d)
d = 4.0 × 5.58×103 = 2.2×104 m (22 km)
✓ 1
Allow error carried forward from the time calculated in the previous part.
—
Exam tip: Only the vertical component of velocity changes gravitational potential energy: power = mg × (vertical speed).
137IBP2S-005-04-15Energy density of fuelsPaper 2Medium7 marks
During a marathon lasting 3.0 hours, a runner's body uses energy at an average rate of 1.1 kW. The energy comes from stored glycogen (a carbohydrate) and from body fat. Glycogen has a specific energy of 1.7×107 J kg−1 and fat has a specific energy of 3.8×107 J kg−1.
(a)
Calculate the total energy used by the runner's body during the race.
(1)
(b)
The runner's muscles convert 22% of the energy used by the body into useful mechanical work. Calculate the average useful mechanical power output.
(1)
(c)
The runner starts with 0.50 kg of glycogen and uses all of it. Determine the mass of fat used during the race.
(3)
(d)
Explain why it is an advantage for animals to store most of their long-term energy reserve as fat rather than carbohydrate.
(2)
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Notes
Part (a)
E = 1.1×103 × 3.0 × 3600 = 1.2×107 J (1.19×107 J)
✓ 1
Part (b)
P = 0.22 × 1.1×103 = 2.4×102 W
✓ 1
Part (c)
Energy from glycogen = 0.50 × 1.7×107 = 8.5×106 J
✓ 1
Energy from fat = 1.19×107 − 8.5×106 = 3.4×106 J
✓ 1
Mass of fat = 3.38×106 / 3.8×107 = 0.089 kg (89 g)
✓ 1
Allow 0.092 kg if the rounded total of 1.2×107 J is used. Allow error carried forward from the total energy in the first part.
—
Part (d)
Fat has a higher specific energy (more than twice that of glycogen), so a given energy is stored in a smaller mass
✓ 1
The animal therefore has less mass to carry, so it needs less energy to move
✓ 1
Exam tip: When energy comes from two sources, find the energy the first source supplies and subtract it from the total. Only then convert the remainder to a mass. Carry unrounded values through the calculation.
138IBP2S-005-00-13Power & efficiencyPaper 2Medium7 marks
A student tests a small electric motor by using it to lift loads through a vertical height of 1.50 m at constant speed. The table shows the results.
Load mass / kg: 0.20 | 0.40 | 0.60 | 0.80 Time to lift 1.50 m / s: 2.0 | 2.4 | 3.0 | 4.0 Electrical input power / W: 2.5 | 3.6 | 4.6 | 5.4
The efficiencies for the 0.20 kg, 0.40 kg and 0.80 kg loads are 0.59, 0.68 and 0.54 respectively.
(a)
Calculate the useful output power of the motor when it lifts the 0.60 kg load.
(2)
(b)
Determine the efficiency of the motor for the 0.60 kg load.
(1)
(c)
Outline how the efficiency of the motor varies with the load. Identify the load, from those tested, at which the motor is most efficient.
(2)
(d)
Suggest why the efficiency of the motor decreases for large loads.
(2)
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Notes
Part (a)
Gain in gravitational potential energy = 0.60 × 9.8 × 1.50 = 8.82 J
✓ 1
Power = 8.82 / 3.0 = 2.9 W (2.94 W)
✓ 1
Part (b)
Efficiency = 2.94 / 4.6 = 0.64 (64%)
✓ 1
Allow ECF from (a).
—
Part (c)
Efficiency first increases and then decreases as the load increases (rises to a maximum, then falls)
✓ 1
Maximum efficiency is at 0.40 kg
✓ 1
Part (d)
A larger load needs a larger current in the motor coils / larger forces in the bearings and gears
✓ 1
Thermal energy is then transferred at a greater rate, through I²R heating in the coil resistance or through friction, so a greater fraction of the input power is wasted
✓ 1
Accept any equivalent physically valid reasoning.
—
Exam tip: With table data, show the substitution for one row clearly. A correct method can still earn credit if you misread a value from the table.
0.50 m³ < 0.55 m³, so «according to the model» the tank is large enough
✓ 1
MP3 requires a comparison.
Part (c)(ii)
at such a high pressure the gas is very dense: the volume of the molecules «and the forces between them» can no longer be neglected
✓ 1
the margin «0.50 m³ against 0.55 m³» is small, so the conclusion is not reliable: the real gas may need a larger volume and the tank may be too small
✓ 1
OWTTE.
Part (d)
advantage: higher energy density so a smaller mass of fuel is needed OR no carbon dioxide is emitted where the lorry is used «only water»
✓ 1
disadvantage: must be stored at very high pressure in a heavy, bulky tank OR low energy per unit volume OR much hydrogen is produced from fossil fuels or with large energy input
✓ 1
Answers: (b)(i) 8.8 × 108 J · (b)(ii) diesel 56 kg, hydrogen 15 kg · (c)(i) 0.50 m³; large enough by the ideal-gas model (the remaining parts are explanations — see the table above)
A sledge of mass 24 kg is pulled from rest for 18 m across level snow. The rope is held at 32° above the horizontal and the tension in it is a constant 95 N. The coefficient of dynamic friction between the sledge and the snow is 0.22. Air resistance is negligible.
Diagram NOT accurately drawn
(a)
(i)
Calculate the work done on the sledge by the tension over the 18 m.
(1)
(ii)
State why the vertical component of the tension does no work on the sledge.
(1)
(b)
(i)
Show that the normal force on the sledge is about 185 N.
(2)
(ii)
Determine the speed of the sledge after it has travelled 18 m.
(3)
(c)
The rope is now held at 45° above the horizontal, with the same tension. Determine whether the sledge gains more or less kinetic energy over the 18 m than before.
(3)
(d)
Explain why there is an angle between 0° and 45° at which the kinetic energy gained by the sledge is greatest.
(3)
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Notes
Part (a)(i)
W = Fs cos θ = 95 × 18 × cos 32° = 1.45 × 103 J
✓ 1
Part (a)(ii)
it is perpendicular to the displacement «cos 90° = 0»
Must see full substitution OR answer to 4 s.f. «185.1 N».
Part (b)(ii)
Ff = 0.22 × 185 = 40.7 N
✓ 1
work done by the resultant force = 1.45 × 103 − 40.7 × 18 = 717 J = kinetic energy gained
✓ 1
Accept (80.6 − 40.7) × 18.
v = √(2 × 717/24) = 7.7 m s−1
✓ 1
Award [3] for CNA. Award [2] max for 6.6 m s−1 «friction found from mg».
Part (c)
FN = 235.4 − 95 sin 45° = 168 N, so Ff = 37.0 N
✓ 1
resultant force = 95 cos 45° − 37.0 = 30.2 N, so kinetic energy = 543 J
✓ 1
543 J < 717 J, so the sledge gains less kinetic energy
✓ 1
MP3 requires a comparison. Allow ECF from (b)(ii).
Part (d)
increasing the angle reduces the horizontal component of the tension «T cos θ», reducing the forward force
✓ 1
but it increases the vertical component, which reduces FN and therefore the frictional force
✓ 1
at small angles the reduction in friction is the larger effect «T cos θ hardly changes», at large angles the loss of horizontal component is, so the kinetic energy is greatest at an angle in between
✓ 1
Do not expect the value «about 12°». MP3 requires both effects to be compared.
Answers: (a)(i) 1.45 × 103 J · (b)(i) 185 N · (b)(ii) 7.7 m s−1 · (c) 543 J < 717 J; less (the remaining parts are explanations — see the table above)
141A-2-55Elastic potential energyPaper 2Hard15 marks
A climber of mass 70 kg is 3.0 m above the last point at which her rope is clipped to the rock face when she slips and falls vertically. She falls freely through 6.0 m before the rope becomes taut. The rope then stretches until she is momentarily at rest. While it stretches, the rope behaves as a spring of spring constant 4.0 × 103 N m−1. Air resistance and the mass of the rope are negligible.
Figure 1 — axes for (e)
(a)
Calculate the speed of the climber at the instant the rope becomes taut.
(1)
(b)
The maximum extension of the rope is x. Show that mg(6.0 + x) = ½kx², where m is the mass of the climber.
(2)
(c)
Determine x.
(2)
(d)
The rope is designed for a maximum force of 8.0 kN. Determine whether the maximum force exerted by the rope on the climber is within this limit, and calculate her maximum acceleration as a multiple of g.
(3)
(e)
Sketch, on Figure 1, a graph to show how the kinetic energy of the climber varies with the distance fallen, from the start of the fall until she is momentarily at rest.
(3)
(f)
Explain why a rope with a smaller spring constant reduces the maximum force on the climber in a fall from the same height, even though it stretches further.
(3)
(g)
State one assumption made in (b).
(1)
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Notes
Part (a)
v = √(2gh) = √(2 × 9.81 × 6.0) = 10.8 m s−1
✓ 1
Part (b)
she is at rest at the start and at the lowest point, so there is no change in kinetic energy, and she falls a total height of 6.0 + x
✓ 1
loss of gravitational potential energy mg(6.0 + x) = gain in elastic potential energy ½kx²
✓ 1
Part (c)
2.0 × 103x² − 687x − 4120 = 0
✓ 1
Allow equivalent rearrangements.
x = 1.62 m
✓ 1
Award [2] for CNA. Do not accept the negative root.
Part (d)
Fmax = kx = 4.0 × 103 × 1.62 = 6500 N
✓ 1
Allow ECF from (c).
6.5 kN < 8.0 kN, so within the limit
✓ 1
MP2 needs a comparison.
a = (kx − mg)/m = 83 m s−2 ≈ 8.4g
✓ 1
Award [0] for MP3 for kx/m «9.4g».
Part (e)
straight line through the origin to about 4.1 kJ at 6.0 m
✓ 1
curve continues to rise slightly to a maximum just beyond 6.0 m «at 6.2 m», then curves down
✓ 1
The maximum must not be at exactly 6.0 m.
Eₖ = 0 at about 7.6 m
✓ 1
Allow ECF from (c).
Part (f)
the energy the rope must store as elastic potential energy, ½kx² = mg(h + x), is mostly the fixed amount mgh «it increases only a little as x increases»
✓ 1
so a smaller k gives a larger maximum extension x
✓ 1
the maximum force kx = 2E/x is then smaller «F ∝ √k», so the climber decelerates more gently
✓ 1
OWTTE. Award MP3 for a correct numerical illustration.
Part (g)
the clip and the belay do not move or slip, so all the lost gravitational potential energy is stored in the rope «OR no energy is absorbed by the climber’s body or harness»
✓ 1
Do not accept assumptions already stated in the stem «no air resistance, light rope, rope behaves as a spring, vertical fall».
Answers: (a) 10.8 m s−1 · (c) 1.62 m · (d) 6500 N, within the limit; 8.4g(the remaining parts are explanations — see the table above)
142A-2-98Elastic potential energyPaper 2Hard12 marks
A fixed-wing survey aircraft of mass 12.0 kg is launched from a straight rail inclined at 12° to the horizontal (Figure 1). The aircraft is latched onto a carriage of mass 3.0 kg that runs along the rail. Elastic cords, which act together as a single spring of spring constant 820 N m−1 that obeys Hooke’s law for extensions up to 3.0 m, pull the carriage up the rail. The carriage is pulled back until the cords are extended by 2.40 m and is then released from rest. When the cords reach their natural length the carriage hits a buffer and stops, and the latch releases the aircraft.
A constant frictional force of 15 N acts on the carriage along the rail. The aircraft’s engine does not contribute during the launch, and air resistance is negligible.
Figure 1 — Diagram NOT accurately drawn
(a)
Show that the elastic potential energy stored in the cords before release is about 2.4 kJ.
(1)
(b)
The aircraft needs a speed of at least 16 m s−1 to fly. Determine whether it reaches this speed when it is released.
(4)
(c)
(i)
Explain why the speed of the carriage is greatest a short distance before the cords reach their natural length.
(2)
(ii)
Calculate the extension of the cords when the speed of the carriage is greatest.
(1)
(d)
A larger aircraft of mass 20.0 kg is to be launched with the same carriage and cords. Determine the extension of the cords needed for this aircraft to be released at 16 m s−1.
(3)
(e)
Suggest one reason why the actual launch speed is less than the speed predicted by this model.
(1)
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Notes
Part (a)
EH = ½ × 820 × 2.40² = 2.36 × 103 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
gain in Ep of aircraft and carriage = 15.0 × 9.81 × 2.40 sin 12° = 73 J
✓ 1
work done against friction = 15 × 2.40 = 36 J
✓ 1
½ × 15.0 × v² = 2.362 × 103 − 73 − 36 = 2.25 × 103 J, so v = 17.3 m s−1
✓ 1
Award [2] max for 19.4 m s−1 «mass of the carriage ignored throughout». Award [3] for CNA.
17.3 m s−1 > 16 m s−1, so the aircraft reaches its flying speed
✓ 1
MP4 needs a comparison. Allow ECF.
Part (c)(i)
the carriage speeds up only while the resultant force on it is up the rail, that is while the elastic force kx is greater than the component of the weight along the rail plus the friction
✓ 1
kx decreases as the cords shorten; where kx equals the opposing forces the resultant force and the acceleration are zero, so the speed is a maximum, and over the last part the resultant force is down the rail and the carriage slows slightly
✓ 1
OWTTE.
Part (c)(ii)
x = (15.0 × 9.81 × sin 12° + 15)/820 = 0.056 m
✓ 1
Accept 5.6 cm.
Part (d)
½ × 820 × x² = ½ × 23.0 × 16² + (23.0 × 9.81 × sin 12° + 15)x
✓ 1
Accept the equation in words.
410x² − 61.9x − 2944 = 0 OR a correct iteration
✓ 1
x = 2.76 m «within the 3.0 m limit»
✓ 1
Award [2] max for 2.68 m «gravity and friction ignored». Do not accept the negative root.
Part (e)
part of the elastic energy becomes kinetic energy of the cords themselves OR the cords are not perfectly elastic «energy is dissipated as internal energy in the rubber» OR air resistance acts on the aircraft and carriage
✓ 1
Any one. Do not accept “friction” alone, which is already included.
Answers: (a) 2.36 × 103 J · (b) 17.3 m s−1 · (c)(ii) 0.056 m · (d) 2.76 m (the remaining parts are explanations — see the table above)
A bulk-cargo ship travels at a constant speed of 6.5 m s−1. At this speed the total resistive force on the ship from the water and the air is 9.0 × 105 N. Of the energy released by burning fuel in the engine, 42 % becomes useful work done in propelling the ship. The energy density of the fuel is 4.1 × 107 J kg−1.
Figure 1 — Diagram NOT accurately drawn
(a)
Calculate the useful power developed by the engine.
(1)
(b)
Calculate the mass of fuel used each hour.
(2)
(c)
To save fuel, a large towing kite is flown ahead of the ship. The kite line pulls on the bow of the ship with a tension T of 2.2 × 105 N at 35° above the horizontal, in the vertical plane that contains the direction of motion (Figure 1). The speed of the ship and the resistive force are unchanged.
(i)
Calculate the work done by the kite line on the ship in one hour.
(2)
(ii)
Outline why the kite would save more fuel if its line pulled at a smaller angle to the horizontal with the same tension.
(1)
(iii)
Determine the percentage reduction in the mass of fuel used each hour.
(2)
(d)
On a voyage of 1.1 × 107 m without the kite, the operator considers “slow steaming”: reducing the speed from 6.5 m s−1 to 5.2 m s−1. The resistive force is proportional to the square of the speed. Assume that the efficiency of the engine does not change.
(i)
Show that the work done against the resistive force over the whole voyage is proportional to the square of the speed.
(2)
(ii)
Determine the mass of fuel saved on the voyage by slow steaming.
(3)
(iii)
Suggest one disadvantage of slow steaming.
(1)
Show mark scheme
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Mark
Notes
Part (a)
P = Fv = 9.0 × 105 × 6.5 = 5.85 × 106 W
✓ 1
Part (b)
power from the fuel = 5.85 × 106/0.42 = 1.39 × 107 W
✓ 1
m = 1.39 × 107 × 3600/4.1 × 107 = 1.2 × 103 kg
✓ 1
Award [1] max for 5.1 × 102 kg «efficiency omitted». Award [2] for CNA.
Part (c)(i)
distance in one hour = 6.5 × 3600 = 2.34 × 104 m
✓ 1
W = Ts cos 35° = 2.2 × 105 × 2.34 × 104 × cos 35° = 4.2 × 109 J
✓ 1
Award [1] max for 5.1 × 109 J «cos 35° omitted» or 3.0 × 109 J «sin 35° used».
Part (c)(ii)
the component of the tension along the direction of motion «T cos θ» is larger, so the line does more work on the ship for the same displacement and the engine has to do less
✓ 1
Part (c)(iii)
useful power now needed from the engine = 5.85 × 106 − 1.80 × 105 × 6.5 = 4.68 × 106 W
✓ 1
ALTERNATIVE: compare the kite’s work in one hour with the engine’s useful work in one hour.
reduction = 20 %
✓ 1
Allow ECF from (c)(i). The efficiency cancels.
Part (d)(i)
work done = Fd, and the distance d of the voyage is fixed
✓ 1
F ∝ v², so the work done ∝ v² «the power is ∝ v³, but the time taken is ∝ 1/v»
✓ 1
OWTTE.
Part (d)(ii)
work at 6.5 m s−1 = 9.0 × 105 × 1.1 × 107 = 9.9 × 1012 J
✓ 1
work at 5.2 m s−1 = 9.9 × 1012 × (5.2/6.5)² = 6.3 × 1012 J
Award [2] max for 2.8 × 105 kg «work taken as ∝ v³». Allow ECF.
Part (d)(iii)
the voyage takes longer «about 5 days more», so more ships or more crew time are needed to carry the same cargo each year, or the cargo arrives later
✓ 1
OWTTE.
Answers: (a) 5.85 × 106 W · (b) 1.2 × 103 kg · (c)(i) 4.2 × 109 J · (c)(iii) 20 % · (d)(ii) 2.1 × 105 kg (the remaining parts are explanations — see the table above)
A desert locust of mass 2.0 g jumps by suddenly straightening its hind legs. During take-off the body of the locust is accelerated from rest through a distance of 4.0 cm and leaves the ground at 3.2 m s−1. Assume that the acceleration during take-off is uniform and that the gain in gravitational potential energy during take-off is negligible.
(a)
Calculate the kinetic energy of the locust at take-off.
(1)
(b)
(i)
Determine the average resultant force on the locust during take-off.
(2)
(ii)
Express this force as a multiple of the weight of the locust.
(1)
(c)
Determine the average power delivered to the locust during take-off.
(2)
(d)
The muscles that straighten the hind legs have a total mass of 70 mg. Insect muscle can develop a power of at most about 450 W per kilogram of muscle.
(i)
Calculate the greatest power that these muscles can develop.
(1)
(ii)
Before the jump, the locust slowly uses these muscles to bend an elastic structure in each hind leg. The structures are then released suddenly. Explain how this allows the locust to reach its take-off speed.
(3)
(e)
Each of the two elastic structures stores half of the energy and behaves like a spring deformed by 1.4 mm. Assume that all the stored energy becomes kinetic energy of the locust. Calculate the spring constant of one structure.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Ek = ½ × 2.0 × 10−3 × 3.2² = 1.0 × 10−2 J
✓ 1
Award [0] if the mass is not in kg.
Part (b)(i)
work done by the resultant force = gain in kinetic energy: F × 0.040 = 1.02 × 10−2
✓ 1
ALTERNATIVE: a = v²/2s = 128 m s−2 and F = ma.
F = 0.26 N
✓ 1
Award [2] for CNA. Allow ECF from (a).
Part (b)(ii)
0.256/(2.0 × 10−3 × 9.81) = 13 «times the weight»
✓ 1
Allow ECF from (b)(i).
Part (c)
time of take-off = 2s/v = 2 × 0.040/3.2 = 0.025 s
✓ 1
P = 1.02 × 10−2/0.025 = 0.41 W
✓ 1
Allow ECF from (a). Award [2] for CNA.
Part (d)(i)
7.0 × 10−5 × 450 = 3.1 × 10−2 W
✓ 1
Part (d)(ii)
the power needed during take-off «0.41 W» is about 13 times what the muscles can develop, so the muscles cannot supply the energy in the time of take-off
✓ 1
Allow ECF.
the muscles do work slowly, storing it as elastic potential energy over a longer time «at least 1.0 × 10−2/3.1 × 10−2 = 0.33 s»
✓ 1
the stored energy is released in 25 ms: the same energy transferred in a much shorter time means a much greater power «P = ΔW/Δt»
✓ 1
OWTTE.
Part (e)
energy stored in one structure = 5.1 × 10−3 J
✓ 1
k = 2E/x² = 2 × 5.12 × 10−3/(1.4 × 10−3)² = 5.2 × 103 N m−1
✓ 1
Award [1] max for 1.0 × 104 N m−1 «the total energy used for one structure».
Answers: (a) 1.0 × 10−2 J · (b)(i) 0.26 N · (b)(ii) 13 · (c) 0.41 W · (d)(i) 3.1 × 10−2 W · (e) 5.2 × 103 N m−1(the remaining parts are explanations — see the table above)
145A-2-101Conservation of energyPaper 2Hard13 marks
A crew capsule of mass 9.0 × 103 kg returns from orbit. When it enters the upper atmosphere, 120 km above the surface of the Earth, its speed is 7.6 × 103 m s−1. Its speed has fallen to 150 m s−1 when it reaches a height of 10 km, where its parachutes open. Between these two heights the capsule travels 2.4 × 103 km along its path in 12 minutes. Assume that g = 9.81 m s−2 throughout.
Figure 1 — axes for (f)(i)
(a)
Show that the kinetic energy of the capsule at a height of 120 km is about 2.6 × 1011 J.
(1)
(b)
Calculate the decrease in the gravitational potential energy of the capsule between the two heights.
(1)
(c)
Determine the energy transferred by work done against air resistance between the two heights.
(2)
(d)
Calculate the average resistive force on the capsule along its path, and the average rate at which energy is dissipated.
(2)
(e)
About 1.0 % of the energy in (c) is transferred to the heat shield of the capsule; the rest is transferred to the surrounding air. Each kilogram of heat-shield material absorbs 5.0 × 107 J as it is heated, charred and vaporised. Calculate the mass of heat-shield material that is used up.
(2)
(f)
(i)
Figure 1 shows axes for a graph of the total mechanical energy E of the capsule (relative to the ground) against time t, from t = 0 at a height of 120 km until the parachutes open. Sketch the graph.
(2)
(ii)
Explain the shape of your graph by referring to the rate at which mechanical energy is dissipated.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
½ × 9.0 × 103 × (7.6 × 103)² = 2.60 × 1011 «J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
ΔEp = 9.0 × 103 × 9.81 × 1.10 × 105 = 9.7 × 109 J
✓ 1
Award [0] if 1.2 × 105 m is used for the change in height.
Part (c)
kinetic energy at 10 km = ½ × 9.0 × 103 × 150² = 1.0 × 108 J
Award [1] max for 2.60 × 1011 J «change in Ep omitted». Allow ECF from (b).
Part (d)
F = 2.70 × 1011/2.4 × 106 = 1.1 × 105 N
✓ 1
Allow ECF from (c).
P = 2.70 × 1011/720 = 3.7 × 108 W
✓ 1
Part (e)
energy to the heat shield = 0.010 × 2.70 × 1011 = 2.7 × 109 J
✓ 1
m = 2.7 × 109/5.0 × 107 = 54 kg
✓ 1
Allow ECF from (c). Award [2] for CNA.
Part (f)(i)
curve falling from about 2.7 × 1011 J at t = 0 to almost zero at t = 720 s
✓ 1
Allow ECF from (a) and (b).
gradient small at the start, steepest part in between, and gradient small again near the end «a reversed S-shape»
✓ 1
Do not accept a straight line.
Part (f)(ii)
the rate of dissipation is the rate of work done against air resistance, «drag force × speed», which is the magnitude of the gradient of the graph
✓ 1
near t = 0 the air is very thin, so the drag force and the rate of dissipation are small even though the speed is very large
✓ 1
near the end the speed is small, so the rate of dissipation is small even though the air is denser; in between both the density of the air and the speed are large, so the gradient is greatest
✓ 1
OWTTE. Award MP2 and MP3 only for reasons linked to drag and speed.
Answers: (a) 2.60 × 1011 J · (b) 9.7 × 109 J · (c) 2.70 × 1011 J · (d) 1.1 × 105 N, 3.7 × 108 W · (e) 54 kg (the remaining parts are explanations — see the table above)
146IBP2S-005-01-16Conservation of energyPaper 2Hard8 marks
A pole vaulter of mass 65 kg reaches 9.5 m s−1 at take-off, when her centre of mass is 1.0 m above the ground. At the top of the vault her centre of mass is 5.9 m above the ground and she is moving horizontally at 0.80 m s−1. Air resistance is negligible.
Heights are of the centre of mass; diagram not to scale
(a)
Calculate her kinetic energy at take-off.
(1)
(b)
Show that, if all this kinetic energy became gravitational potential energy, her centre of mass would rise by about 4.6 m.
(2)
(c)
Outline the energy changes that occur between take-off and the top of the vault.
(2)
(d)
Determine the minimum work done by the vaulter's muscles during the vault. Assume no energy is dissipated in the pole.
(3)
Show mark scheme
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Mark
Notes
Part (a)
Ek = ½ × 65 × 9.5² = 2.93 × 103 J
✓ 1
Part (b)
mgΔh = ½mv², so Δh = v² ÷ 2g = 9.5² ÷ (2 × 9.8)
✓ 1
Δh = 4.60 m ≈ 4.6 m
✓ 1
Part (c)
As the pole bends, kinetic energy is transferred to elastic potential energy of the pole, with some going to gravitational potential energy as she rises
✓ 1
As the pole straightens, its elastic potential energy is transferred to gravitational potential energy of the vaulter, with a small amount remaining as kinetic energy
Total energy needed at the top = 3121 + ½ × 65 × 0.80² = 3121 + 21 = 3142 J
✓ 1
Work done = 3142 − 2933 ≈ 209 J, about 2.1 × 102 J
✓ 1
Accept answers in the range 205–215 J; using g = 9.81 N kg−1 gives 212 J. Allow error carried forward from the take-off kinetic energy calculated in the first part.
—
Exam tip: Include the kinetic energy at the top of the vault. Her speed there is small but not zero, and the energy balance must account for it. Keep unrounded values until the final subtraction, because the answer is a small difference between two large energies.
147IBP2S-005-02-15Work done by a forcePaper 2Hard9 marks
A lorry of mass 1.8 × 104 kg loses its brakes on a mountain road. It enters a straight runaway ramp of loose gravel, inclined at 8.0° above the horizontal, at 28 m s−1. The gravel exerts a resistive force on the lorry along the ramp equal to 0.45 times the normal force. Assume the resistive force is constant.
Diagram not to scale; the ramp angle is exaggerated
(a)
Show that the kinetic energy of the lorry as it enters the ramp is about 7 × 106 J.
(2)
(b)
Determine the distance the lorry travels along the ramp before stopping.
(3)
(c)
Determine the percentage of the initial kinetic energy that is transferred to internal energy of the gravel and the lorry.
(2)
(d)
Explain whether the lorry remains at rest once it has stopped, assuming the maximum resistive force when stationary is also 0.45 times the normal force.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Ek = ½ × 1.8 × 104 × 282
✓ 1
= 7.06 × 106 J
✓ 1
Part (b)
Total force opposing the motion = mg(sin 8.0° + 0.45 cos 8.0°) = 2.46 × 104 + 7.87 × 104 = 1.03 × 105 N
✓ 1
Total work done against this force = Ek, so d = 7.06 × 106 / 1.03 × 105
✓ 1
d = 68 m
✓ 1
Allow error carried forward from the kinetic energy found earlier. An equivalent kinematics method using deceleration a = g(sin 8.0° + 0.45 cos 8.0°) = 5.7 m s−2 scores full marks.
—
Part (c)
Work done against the resistive force = 7.87 × 104 × 68.3 ≈ 5.4 × 106 J (accept 5.3 × 106 J to 5.4 × 106 J)
Allow error carried forward from the distance found earlier. An alternative method also scores both marks: 0.45 cos 8.0° / (sin 8.0° + 0.45 cos 8.0°) = 0.76, so 76%.
—
Part (d)
The component of the weight down the ramp (2.5 × 104 N) is smaller than the maximum resistive force available (7.9 × 104 N)
✓ 1
So the gravel can supply a force equal to the weight component, and the lorry remains at rest (equivalently, 0.45 > tan 8.0° = 0.14)
✓ 1
Exam tip: The remaining 24% of the kinetic energy becomes gravitational potential energy. Checking that the two shares add to 100% is a quick test of your answer.
148IBP2S-005-02-16Elastic potential energyPaper 2Hard9 marks
A sports-equipment company tests shoe soles using a weighted sled of mass 1.5 kg that slides along a horizontal test strip. The surface coating of the strip gets rougher along its length. The coefficient of dynamic friction μ increases uniformly from 0 at the start of the strip (x = 0) to 0.60 at the end (x = 2.0 m). The sled enters the strip at x = 0. Assume there is no friction before the strip. Take g = 9.8 m s−2.
Diagram not to scale
(a)
Outline why the work done by friction on the strip cannot be calculated as μmg multiplied by the distance travelled, using a single value of μ.
(1)
(b)
Show that the frictional force on the sled at position x is 4.41x N, where x is in metres.
(1)
(c)
Determine the work done against friction if the sled travels the full length of the strip.
(2)
(d)
Calculate the maximum entry speed for which the sled stops within the strip.
(2)
(e)
The sled enters the strip at 2.4 m s−1. Determine its stopping distance.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The frictional force is not constant because μ changes with x, so the work must be found from the area under the force–distance graph (or by using the average force)
✓ 1
Part (b)
μ = (0.60/2.0)x = 0.30x, so F = 0.30x × 1.5 × 9.8 = 4.41x N
✓ 1
Part (c)
Force at x = 2.0 m is 8.82 N; the work is the area of the triangle under the force–distance line (or average force 4.41 N × 2.0 m)
✓ 1
W = ½ × 2.0 × 8.82 = 8.8 J . ECF from part (b).
✓ 1
Part (d)
½ × 1.5 × v2 = 8.82
✓ 1
v = 3.4 m s−1 . ECF from part (c).
✓ 1
Part (e)
Initial kinetic energy = ½ × 1.5 × 2.42 = 4.32 J
✓ 1
Work against friction up to distance x = ½ × x × 4.41x = 2.205x2, so 2.205x2 = 4.32
✓ 1
x = 1.4 m . ECF from part (b). Accept the proportional method: x = 2.0 × (2.4/3.43).
✓ 1
Common error: Do not assume stopping distance is proportional to v², as it is for constant friction. Here the friction work grows as x², so the stopping distance is proportional to the entry speed. You can check your answer this way: 2.0 m × 2.4/3.43 ≈ 1.4 m.
149IBP2S-005-04-16Energy density of fuelsPaper 2Hard9 marks
A remote polar research station uses a diesel generator for electricity. Diesel has an energy density of 38 MJ L−1. The table shows the fuel consumption of the generator at different electrical output powers.
Show that the efficiency of the generator at an output of 20 kW is about 0.34.
(2)
(b)
Determine, using the data, the output power at which the generator is most efficient.
(2)
(c)
The station needs 10 kW continuously for 30 days. Two options are proposed. Option A: run the generator continuously at 10 kW. Option B: run the generator at 30 kW for part of the time, storing all its output in batteries that return 85% of the energy stored, and supply the station only from the batteries. Determine which option uses less fuel, and by how much.
(4)
(d)
Suggest why option B might still be chosen.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Input power = 5.6 × 38×106 / 3600 = 5.9×104 W
✓ 1
Efficiency = 20×103 / 5.91×104 = 0.338 ≈ 0.34
✓ 1
Part (b)
Efficiencies: 5.0 kW → 0.25; 10 kW → 0.32; 30 kW → 0.36 (at least two further values correctly calculated)
✓ 1
Most efficient at 30 kW (efficiency ≈ 0.36)
✓ 1
Part (c)
Option A: time = 720 h; fuel = 3.0 × 720 = 2.16×103 L
✓ 1
Option B: energy required from generator = (10 × 720) / 0.85 = 8.47×103 kWh
✓ 1
Running time = 8.47×103 / 30 = 282 h; fuel = 8.0 × 282 = 2.26×103 L
✓ 1
Option A uses less fuel, by about 99 L (≈ 1×102 L)
✓ 1
Part (d)
Any one: generator runs for less time so less wear/maintenance; quieter for most of the time; less exhaust near the station for much of the day; energy available if the generator fails
✓ 1
Exam tip: A higher generator efficiency does not guarantee an overall saving: include every stage (here the battery losses) before comparing options.
150IBP2S-005-00-16Work done by a forcePaper 2Hard8 marks
A cyclist and bicycle have a combined mass of 85 kg. At speed v, the air resistance is kv2, where k = 0.30 kg m−1. There is also a constant rolling resistance of 4.0 N. Take g = 9.8 m s−2.
(a)
Calculate the power output required for the cyclist to travel at a constant 8.0 m s−1 on level ground.
(2)
(b)
The cyclist climbs a slope inclined at 3.0° to the horizontal at a constant 5.0 m s−1. Determine the power output required.
(3)
(c)
The efficiency of the cyclist's muscles in converting chemical energy to useful mechanical work is 0.24. Calculate the rate at which the cyclist uses chemical energy during the climb.
(1)
(d)
An energy bar releases 1.0 MJ of chemical energy. Estimate the time for which the cyclist could climb the slope using only the energy from one bar.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Total resistive force = 0.30 × 8.02 + 4.0 = 23.2 N
✓ 1
P = Fv = 23.2 × 8.0 ≈ 190 W (186 W)
✓ 1
Part (b)
Component of weight along the slope = 85 × 9.8 × sin 3.0° = 43.6 N
✓ 1
Total force = 43.6 + 0.30 × 5.02 + 4.0 = 55.1 N
✓ 1
P = 55.1 × 5.0 ≈ 280 W (276 W)
✓ 1
Part (c)
Rate = 276 / 0.24 ≈ 1.1 × 103 W (1150 W); allow error carried forward from (b)
✓ 1
Part (d)
t = E/P = 1.0 × 106 / 1150
✓ 1
t ≈ 870 s, about 15 minutes; accept 870–910 s from rounded values, and allow error carried forward from (c)
✓ 1
Common error: On a slope, the driving force must also balance the weight component mg sin θ along the slope. Using mg or mg cos θ instead gives a power that is far too large.
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