53 free, high-quality past-paper-style questions on Work, energy and power for IB Physics HL, each with a full mark scheme. Covers: Sankey diagrams; Elastic potential energy; Energy and projectile motion; and more. For when you have finished the past papers, or want to practise one topic at a time.A.3 at Standard 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.229 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-42Sankey diagramsPaper 1AEasy1 mark
The Sankey diagram, drawn to scale, shows the energy transferred each second in an electric winch. The motor drives the winch through a gearbox. The arrow for the useful output has not been labelled.
What is the efficiency of the gearbox?
Sankey diagram for the winch, drawn to scale (width of each arrow proportional to power).Show mark scheme
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Step 1Energy is conserved at the motor: the power passed to the gearbox is 500 − 100 = 400 W.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the gearbox: useful output = 400 − 40 = 360 W (the output arrow is 72 % of the width of the input arrow, as drawn).
2A-1A-45Elastic potential energyPaper 1AEasy1 mark
The plunger of a pinball machine compresses a light spring of spring constant 250 N m−1 by 6.0 cm. When released, the spring launches a ball of mass 80 g along a horizontal, frictionless track.
What is the speed of the ball as it leaves the spring?
Frictionless horizontal track (not to scale)
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Step 1All the elastic potential energy becomes kinetic energy: ½k(Δx)² = ½mv².
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All 2 steps must be completed — there is no mark for a part-answer.
3A-1A-61Energy and projectile motionPaper 1AEasy1 mark
Two identical balls are projected from the same balcony with the same speed. Ball P is projected horizontally; ball Q is projected at 40° above the horizontal. Both land on the same level ground below. Air resistance is negligible.
Which row compares the speeds of the balls on reaching the ground and the times they spend in the air?
Initial velocities only; not to scale
Speed on reaching the groundTime in the air
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Step 1Energy (A.3): ½mv² = ½mu² + mgh for both balls, with the same u and the same drop h, so the landing speeds are equal. The direction of projection does not matter.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Kinematics (A.1): the vertical motion decides the time. Q starts with an upward velocity component, so it rises first and then has further to fall; it spends longer in the air than P.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe landing speed does not depend on direction because the same kinetic energy is gained from the same loss of gravitational potential energy; and P, with zero initial vertical velocity, reaches the ground sooner.
BThe speeds are equal, but the times are not: Q’s upward vertical component delays its landing.
CCorrect: same energy transfer gives the same speed; the upward component makes Q’s flight longer.
DA longer flight does not mean a greater final speed: the total energy gained depends only on the height fallen.
5A-1A-70Conservation of mechanical energyPaper 1AEasy1 mark
A bungee jumper steps off a platform and falls vertically from rest. The elastic cord becomes taut and stretches until the jumper is momentarily at rest at the lowest point. Air resistance and energy dissipation in the cord are negligible.
Which statements about the jumper at the lowest point are correct? I. The kinetic energy of the jumper is zero. II. The elastic potential energy stored in the cord is equal to the gravitational potential energy lost by the jumper since leaving the platform. III. The resultant force on the jumper is zero.
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Step 1I: the jumper is momentarily at rest, so the kinetic energy is zero. I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: no energy is dissipated, so mechanical energy is conserved. Kinetic energy is zero at both the start and the lowest point, so all the gravitational potential energy lost is stored as elastic potential energy. II is correct.
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Step 3III: at the lowest point the cord is stretched beyond the equilibrium position, so the tension is greater than the weight. The resultant force is upwards, which is why the jumper moves back up. III is wrong.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: zero kinetic energy, and all the gravitational potential energy lost is stored in the cord; the resultant force is upwards, not zero.
BThis confuses the lowest point with the equilibrium position. At the lowest point the tension exceeds the weight.
CThis rejects I, perhaps thinking the jumper is still moving at the lowest point, and accepts III, confusing the lowest point with equilibrium.
DIII is wrong: a zero resultant force at the lowest point would leave the jumper at rest there, but the jumper accelerates back upwards.
6IBP2H-006-00-02Work done by a variable forcePaper 1AEasy1 mark
A spring that obeys Hooke's law stores elastic potential energy E when its extension is x. How much additional work is needed to increase the extension from x to 2x?
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Elastic potential energy is ½kx2. At extension 2x the stored energy is ½k(2x)2 = 4E. The additional work is therefore 4E − E = 3E. 4E is the total stored energy, not the extra energy. 2E assumes the energy is proportional to the extension. E assumes each equal step of extension needs equal work, but the force grows as the spring stretches.
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Answer C
Answer: C
Exam tip: Read whether a question asks for the total energy stored or the change in energy. On a force–extension graph, the change is the area of a trapezium, not of a triangle.
7IBP2H-006-01-01Work done by a variable forcePaper 1AEasy1 mark
An airport worker pulls a baggage cart 15 m along a level floor. The handle is held at 60° above the horizontal, and the worker applies a constant force of 40 N along the handle. What is the work done by the worker on the cart?
Handle held at 60° above the horizontal (not to scale)
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Only the component of the force along the displacement does work. W = Fs cos θ = 40 × 15 × cos 60° = 40 × 15 × 0.5 = 300 J. The value 520 J uses sin 60° by mistake, 600 J ignores the angle, and 1200 J divides by cos 60° instead of multiplying.
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Answer A
Answer: A
Common error: θ in W = Fs cos θ is the angle between the force and the displacement. Check that cos θ makes the work smaller, never larger, than Fs.
8IBP2H-006-01-02Kinetic energy and momentumPaper 1AEasy1 mark
A proton and an alpha particle have the same momentum. The mass of the alpha particle is four times the mass of the proton. What is the ratio \(\frac{\text{kinetic energy of alpha particle}}{\text{kinetic energy of proton}}\)?
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Ek = p2/2m. With equal p, Ek ∝ 1/m, so the alpha particle has 1/4 of the proton's kinetic energy. The value 1/2 uses only the speed ratio, 1 assumes equal momentum means equal energy, and 4 inverts the mass dependence.
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Answer A
Answer: A
Exam tip: When momentum is the shared quantity, use Ek = p²/2m. When speed is shared, use Ek = ½mv².
9A-1A-04Power and rate of energy transferPaper 1AMedium1 mark
A constant resultant force acts on a body that is initially at rest. At a time t after the force starts to act, the power delivered to the body by the force is P.
What is the power delivered to the body by the force at time 2t?
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Step 1A constant resultant force gives a constant acceleration a = F/m, so from rest the speed is v = at: the speed is proportional to the time.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The power delivered by a force is P = Fv. F is constant, so P ∝ v ∝ t.
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Step 3Doubling the time doubles the speed and therefore doubles the power: 2P. (Check: Ek = ½m(at)² ∝ t², and the rate of increase of Ek ∝ t.)
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the power as constant because the force is constant. The power P = Fv also depends on the speed, which keeps increasing.
BThis assumes that the kinetic energy increases at a steady rate (constant power), so that v ∝ √t. Here it is the force, not the power, that is constant.
CCorrect: P = Fv = F²t/m ∝ t, so doubling the time doubles the power.
DThis is the factor by which the kinetic energy increases (Ek ∝ t²). The power is the rate of transfer of energy, which is proportional to t.
10A-1A-15Power and rate of energy transferPaper 1AMedium1 mark
A cargo ship moves at a constant speed through calm water. The total resistive force on the ship is modelled as being proportional to the square of its speed.
The speed of the ship is doubled and is again constant. By what factor does the useful power output of the engines change, and by what factor does the useful energy output for each kilometre travelled change?
Factor for the useful power outputFactor for the useful energy per kilometre
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Step 1At constant speed the driving force equals the resistive force: F = kv². Doubling v multiplies the driving force by 4.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Power P = Fv = kv³, so the useful power output increases by a factor of 2³ = 8.
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Step 3Energy for each kilometre = work done by the driving force over 1 km = F × 1000 m ∝ v², a factor of 4. (Check: the power is 8 times larger but each kilometre takes half the time: 8 × ½ = 4.)
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the resistive force as constant, so that P = Fv only doubles and the work per kilometre is unchanged. The force here grows with v².
BThis takes the power to be proportional to the force alone (∝ v²), forgetting the factor v in P = Fv. The energy per kilometre is right.
CThe power factor is right, but the energy per kilometre has been scaled like the power, ignoring that at double speed each kilometre takes only half the time: 8 × ½ = 4.
DCorrect: P ∝ Fv ∝ v³ gives 8; the work per kilometre ∝ F ∝ v² gives 4.
12A-1A-43Work done by a constant forcePaper 1AMedium1 mark
A box of mass m is pulled a distance d up a rough slope inclined at θ to the horizontal. The pulling force F is parallel to the slope and the box moves at constant speed.
What is the work done on the box by the frictional force?
Rough slope; the box moves at constant speed
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Step 1Constant speed means zero resultant force along the slope: F = mg sin θ + Ff, so the friction force is Ff = F − mg sin θ, directed down the slope.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Friction acts opposite to the displacement (cos 180° = −1), so Wf = −(F − mg sin θ)d.
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Step 3Check: the work done by F, by the weight (−mgd sin θ) and by friction add to zero, as the kinetic energy does not change.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: friction = F − mg sin θ down the slope, and it does negative work.
BThis assumes friction alone balances the pull, forgetting that the component of the weight down the slope also opposes the motion.
CThis adds the weight component to the pull instead of letting it share the job of balancing it.
DThe magnitude is right but the sign is wrong: friction acts against the displacement, so it does negative work on the box.
13A-1A-44Work–energy principlePaper 1AMedium1 mark
An electric delivery van of mass 2400 kg speeds up from 12 m s−1 to 18 m s−1 while travelling 150 m along a road. In this distance the road rises through a vertical height of 6.0 m.
What is the average resultant force on the van over the 150 m?
Not to scale
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Step 1The work done by the resultant force equals the change in kinetic energy of the van: Fress = ΔEk.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2ΔEk = ½ × 2400 × (18² − 12²) = 2.16 × 105 J.
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Step 3Fres = 2.16 × 105/150 = 1440 N ≈ 1.4 kN. The rise of 6.0 m does not enter: the weight component along the slope is already part of the resultant.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses (Δv)² instead of Δ(v²): ½ × 2400 × 6² /150 = 290 N.
BCorrect: ½ × 2400 × (18² − 12²)/150 = 1.4 kN.
CThis adds the gain in gravitational potential energy (2400 × 9.81 × 6.0 = 1.41 × 105 J) to ΔEk and divides by 150 m: (3.57 × 105)/150 = 2.4 kN. That is the average driving force minus resistive forces, not the resultant force.
DThis omits the ½ in the kinetic energy: 2400 × (18² − 12²)/150 = 2.9 kN.
14A-1A-46Energy density of fuelsPaper 1AMedium1 mark
A standby diesel generator delivers a constant electrical power of 5.0 kW. The energy density of diesel (energy released per unit mass when it is burned) is 45.6 MJ kg−1, and the generator converts 32 % of this energy into electrical energy.
What mass of diesel does the generator burn in one hour?
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Step 1Electrical energy delivered in one hour = 5.0 × 103 × 3600 = 1.8 × 107 J.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy that must be released by the fuel = useful output/η = 1.8 × 107/0.32 = 5.6 × 107 J.
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Step 3Mass = energy released/energy density = 5.6 × 107/45.6 × 106 = 1.2 kg.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis multiplies by the efficiency instead of dividing: 1.8 × 107 × 0.32/45.6 × 106 = 0.13 kg — less fuel than would be needed even at 100 % efficiency.
BThis ignores the efficiency: 1.8 × 107/45.6 × 106 = 0.39 kg, which assumes every joule of chemical energy becomes electrical energy.
CThis divides by the wasted fraction (1 − 0.32 = 0.68): 1.8 × 107/(0.68 × 45.6 × 106) = 0.58 kg.
15A-1A-69Work done by a variable forcePaper 1AMedium1 mark
A child pulls a sledge along level snow with a rope that is kept at 30° above the horizontal. The graph shows how the tension T in the rope varies with the horizontal displacement s of the sledge.
What is the work done by the tension on the sledge during the first 10 m?
Tension in the rope against horizontal displacement of the sledge (drawn to scale).Show mark scheme
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Step 1The area under the graph is the work that a force T would do if it acted along the displacement: ½ × 4.0 × 80 + 6.0 × 80 = 160 + 480 = 640 J.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Only the component of the tension along the displacement does work: W = Fs cos θ, with θ = 30°.
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Step 3W = 640 × cos 30° = 554 J ≈ 550 J.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses sin 30°, the vertical component, which is perpendicular to the displacement and does no work: 640 × sin 30° = 320 J.
BCorrect: area 640 J multiplied by cos 30°.
CThis is the area under the graph, ignoring the angle of the rope: the full tension does not act along the displacement.
DThis treats the tension as 80 N for the whole 10 m, ignoring the triangle at the start: 80 × 10 × cos 30° = 690 J.
16IBP2H-006-00-03Energy density of fuelsPaper 1AMedium1 mark
Fuel X has a specific energy of 45 MJ kg−1 and a density of 750 kg m−3. Fuel Y has a specific energy of 120 MJ kg−1 and a density of 70 kg m−3. Which statement is correct?
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Per kilogram, Y releases more energy: 120 MJ against 45 MJ. Energy density is specific energy × density. For X this is 45×106 × 750 ≈ 3.4×1010 J m−3. For Y it is 120×106 × 70 ≈ 8.4×109 J m−3. So Y is better per unit mass but X is about four times better per unit volume. Option A assumes a high specific energy always means a high energy density. Option B reverses the comparison. Option D invents a constant product.
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Answer C
Answer: C
Common error: Specific energy is measured in J kg⁻¹ and energy density in J m⁻³. Multiply specific energy by density to convert from one to the other.
17IBP2H-006-00-04Sankey diagramsPaper 1AMedium1 mark
An irrigation system has an electric motor of efficiency 80% that drives a water pump of efficiency 60%. The electrical input to the motor is 2.5 kW. A Sankey diagram is drawn for the whole system. What is the total power shown by all of the waste arrows?
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The motor's output is 0.80 × 2.5 = 2.0 kW, so it wastes 0.50 kW. The pump's useful output is 0.60 × 2.0 = 1.2 kW, so it wastes 0.80 kW. The total waste is 0.50 + 0.80 = 1.3 kW. Equivalently, the overall efficiency is 0.80 × 0.60 = 48%, so the waste is 52% of 2.5 kW = 1.3 kW. 0.50 kW counts only the motor loss, 0.80 kW counts only the pump loss, and 1.2 kW is the useful output.
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Answer D
Answer: D
Exam tip: Efficiencies of stages in series multiply. The waste arrows on a Sankey diagram must add up to the input minus the useful output.
18IBP2H-006-00-05Energy and projectile motionPaper 1AMedium1 mark
Three identical stones are thrown from the top of a sea cliff with the same speed. Stone P is thrown at 60° above the horizontal, stone Q horizontally and stone R at 30° below the horizontal. Air resistance is negligible. Which stone hits the sea with the greatest speed?
Initial velocities only; not to scale
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Each stone starts with the same kinetic energy ½mu2. Each falls through the same vertical height h to the sea. Gravity is the only force doing work, and its work mgh depends only on the vertical displacement. So every stone arrives with kinetic energy ½mu2 + mgh and therefore the same speed. The launch angle changes the time of flight, the landing point and the direction of impact, but not the impact speed. Choosing P or R confuses a longer flight time or a downward start with greater energy.
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Answer D
Answer: D
Exam tip: If a projectile question asks only for a speed, an energy method avoids resolving components altogether.
19IBP2H-006-00-06Elastic potential energyPaper 1AMedium1 mark
A block of mass m is released from rest at a height h above the top of a vertical spring of spring constant k. The block lands on the spring and compresses it by a maximum distance x. Air resistance is negligible and the spring's mass can be ignored. Which equation is correct?
Block released above a vertical spring; x is the maximum compression
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At maximum compression the block is momentarily at rest. Its total drop is h + x, so all of the gravitational potential energy lost, mg(h + x), is stored as elastic potential energy ½kx2. Option A leaves out the extra fall x while the spring is compressed. Option C drops the factor ½ (it uses force × extension instead of the area under the force–extension graph). Option D treats the extra drop as an energy gain when it is an energy loss.
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Answer B
Answer: B
Common error: Measure the gravitational potential energy change down to the lowest point the object actually reaches, not to where it first touches the spring.
20IBP2H-006-01-03Work done by a variable forcePaper 1AMedium1 mark
In a crash test, the force exerted on a seat-belt strap rises uniformly from 0 to 200 N as the strap extends by the first 0.10 m. The force then stays constant at 200 N while the strap extends by a further 0.05 m. What is the total work done on the strap?
Force on the strap against extension
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The work done equals the area under the force–extension relationship. The rising section gives a triangle of area ½ × 200 × 0.10 = 10 J. The constant section gives a rectangle of area 200 × 0.05 = 10 J. The total is 20 J. The value 10 J counts only the first stage, 15 J treats the whole extension as linear, and 30 J assumes the maximum force acts throughout.
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Answer C
Answer: C
Exam tip: For a variable force, split the area under the F–x description into simple shapes and add them.
21IBP2H-006-01-04Sankey diagramsPaper 1AMedium1 mark
An electric motor raises a goods lift of total mass 500 kg vertically at a constant speed of 2.0 m s−1. The efficiency of the motor is 70%. What is the electrical input power to the motor?
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At constant speed, the useful output power is mgv = 500 × 9.8 × 2.0 = 9800 W. The input power is the useful power divided by the efficiency: 9800 / 0.70 = 14 000 W = 14 kW. The value 6.9 kW multiplies by the efficiency instead of dividing, 9.8 kW ignores the efficiency, and 4.2 kW is only the power wasted.
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Answer D
Answer: D
Common error: The input power must always be larger than the useful output power. If your answer is smaller, you have multiplied by the efficiency instead of dividing.
22IBP2H-006-01-05Kinetic energy and momentumPaper 1AMedium1 mark
A supply capsule of mass m approaches a space station of mass 3m. In the station's initial rest frame, the capsule moves at speed u. The capsule docks and the two move off together. What fraction of the initial kinetic energy is transferred to other forms during docking?
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Momentum is conserved: mu = 4m v, so v = u/4. The final kinetic energy is ½(4m)(u/4)2 = mu2/8. The initial kinetic energy is mu2/2. The fraction remaining is 1/4, so the fraction transferred is 3/4. The answer 0 confuses conservation of momentum with conservation of kinetic energy. The answer 1/4 is the fraction remaining, not the fraction lost. The answer 1/2 holds only for equal masses.
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Answer D
Answer: D
Exam tip: In a perfectly inelastic collision with one body initially at rest, the fraction of kinetic energy lost is M/(m + M), where M is the mass of the stationary body.
23IBP2H-006-01-06Work done by a constant forcePaper 1AMedium1 mark
A shuffleboard disc is released at 2.0 m s−1 and slides 4.0 m before stopping. The friction force is constant. The disc is released again on the same surface at 3.0 m s−1. How far does it slide?
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By the work–energy principle, the work done against friction equals the loss of kinetic energy: Fd = ½mv2, so d ∝ v2. The new distance is d = 4.0 × (3.0/2.0)2 = 9.0 m. The value 4.0 m assumes the stopping distance depends only on friction, 4.9 m assumes d ∝ √v, and 6.0 m assumes d ∝ v.
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Answer D
Answer: D
Common error: Increasing the speed by 50% does not increase the stopping distance by 50%. Kinetic energy, and so the work needed to remove it, scales with v².
24IBP2S-005-03-09Conservation of mechanical energyPaper 1AHard1 mark
In a simple seismometer, a mass m hangs from a light spring of spring constant k. The mass is attached when the spring is at its natural length and is then released from rest. Air resistance is negligible. What is the elastic potential energy stored in the spring when the mass reaches its lowest point?
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At the lowest point the speed is zero, so all the gravitational energy lost has become elastic energy: \(mgx = \frac{1}{2}kx^2\). This gives x = 2mg/k, which is twice the equilibrium extension. The elastic energy is then \(E_p = mgx = \frac{2m^2g^2}{k}\). Option A is the energy stored at the equilibrium position, which is what you would get if the mass were lowered slowly. Option D wrongly squares the factor of 2 a second time.
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Answer C
Answer: C
Common error: A released mass overshoots the equilibrium position. Use energy conservation to find the extension; do not use mg = kx.
25IBP2S-005-03-10Work done by a variable forcePaper 1AHard1 mark
A force-meter spring of spring constant k stores energy E when a force F is applied. The spring is cut into two equal halves. The same force F is applied to one half. What energy does the half-spring store?
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Under force F the whole spring extends by x = F/k. Each half carries the full tension F and extends by x/2, so a half-spring has spring constant 2k. For a fixed force, \(E_p = \frac{1}{2}Fx = \frac{F^2}{2k}\). Doubling k halves the stored energy, so the half-spring stores E/2. Choosing 2E assumes a stiffer spring always stores more energy. Choosing E/4 squares the factor of 2. Choosing E assumes that the same force always gives the same stored energy, ignoring the smaller extension.
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Answer B
Answer: B
Exam tip: Decide first whether force or extension is held constant. Use F²/2k for fixed force and ½kx² for fixed extension.
26IBP2H-006-00-07Power and rate of energy transferPaper 1AHard1 mark
On a level motorway, the air resistance on a car is proportional to the square of its speed. All other resistive forces are negligible. The car's steady speed increases from 20 m s−1 to 30 m s−1. By what factor does the work done against air resistance for each kilometre travelled increase?
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The work done per kilometre is W = F × 1000 m, where F ∝ v2. The distance is fixed, so W scales with F. The factor is (30/20)2 = 2.25. 3.375 = 1.53 is the factor for the power, P = Fv. However, the car covers each kilometre in less time, so the energy per kilometre rises only by the force ratio. 1.5 assumes the force is proportional to speed, or constant work per unit speed. 1.25 is the fractional increase (2.25 − 1), not the factor.
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Answer C
Answer: C
Exam tip: Decide whether the question asks for energy per distance (proportional to force) or energy per time (power, proportional to force × speed).
27IBP2H-006-00-08Energy transfer in a wind turbinePaper 1AHard1 mark
The maximum power available to a wind turbine is proportional to the rate at which the wind brings kinetic energy through the area swept by the blades. A turbine delivers power P. It is replaced by a turbine with blades twice as long, in a place where the wind speed is half as great. Assume the efficiency is unchanged. What power does the new turbine deliver?
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The mass of air reaching the blades each second is ρAv, so the power is ½(ρAv)v2 = ½ρAv3, with A = πr2. Doubling r multiplies A by 4. Halving v multiplies v3 by 1/8. The new power is 4 × 1/8 × P = P/2. Answer P comes from using only the v2 in the kinetic energy and forgetting that the mass flow rate also depends on v. P/4 treats area as proportional to blade length. 2P uses v instead of v3.
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Answer B
Answer: B
Common error: The v³ dependence comes from two separate factors: v² from the kinetic energy of each kilogram of air, and v from the mass of air arriving per second.
28IBP2H-006-01-07Kinetic energy and momentumPaper 1AHard1 mark
Two spacecraft modules of masses 1200 kg and 3600 kg are at rest together in deep space. A light, compressed spring between them is released and pushes them apart. All of the spring's elastic potential energy is transferred to the kinetic energy of the two modules. What fraction of the spring's elastic potential energy becomes kinetic energy of the 1200 kg module?
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The total momentum stays zero, so both modules have momenta of equal magnitude p. Since Ek = p2/2m, the 1200 kg module has 3 times the kinetic energy of the 3600 kg module. Its share is therefore 3/(3 + 1) = 3/4. The answer 1/4 is the share of the heavier module. The answer 1/2 assumes the energy is shared equally. The answer 9/10 uses the square of the speed ratio (9 : 1) but ignores the masses.
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Answer C
Answer: C
Exam tip: When two bodies push apart from rest, the lighter one always takes the larger share of the kinetic energy. The kinetic energies are in the inverse ratio of the masses.
29IBP2H-006-01-08Power and rate of energy transferPaper 1AHard1 mark
A car starts from rest on a level road. Its engine delivers a constant output power. Resistive forces are negligible. The car takes time t to reach speed v. How long does it take, from rest, to reach speed 2v?
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With constant power and no losses, the energy supplied is Pt = ½mv2, so t ∝ v2. Doubling the speed therefore takes 4t. The answer 2t assumes a constant driving force (uniform acceleration), but at constant power the force falls as the speed rises. The answers √2 t and 8t use the wrong power of v.
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Answer C
Answer: C
Common error: Constant power is not the same as constant force. Since P = Fv, the driving force decreases as the speed increases.
30A-1B-05Energy transfers in collisionsPaper 1BMedium7 marks
A rubber ball of mass 45 g is dropped from rest from a height of 2.00 m above a concrete floor. A video camera with a vertical scale behind the ball records the maximum height hn that the ball reaches after the nth bounce. Each height has an uncertainty of ±0.01 m. Air resistance is negligible.
A student suggests that the ball loses the same fraction of its kinetic energy at every bounce. The graph shows ln(hn/m) against n.
bounce number n
0
1
2
3
4
5
hn / m
2.00
1.26
0.81
0.53
0.34
0.21
ln(hn/m)
0.69
0.23
−0.21
−0.63
−1.08
−1.56
Graph drawn to scale
(a)
(i)
Test the student's suggestion using at least three of the measured heights.
(2)
(b)
(i)
Determine, using the full range of the graph, the fraction of the kinetic energy retained at each bounce.
(2)
(ii)
Predict the number of the first bounce after which the ball fails to reach a height of 0.030 m.
(1)
(c)
(i)
The specific heat capacity of the rubber is 1.7 × 103 J kg−1 K−1. Assume that half of the kinetic energy lost at the first bounce becomes internal energy of the ball. Estimate the temperature rise of the ball caused by this bounce.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
If the same fraction of energy is lost each time, hn+1/hn is constant; calculates at least three ratios, e.g. 0.63, 0.64, 0.65, 0.64
✓ 1
A test using two heights only scores [1 max].
The ratios are constant (≈ 0.64) within the uncertainty of the data (±0.01 m is ±3 % of 0.34 m), so the suggestion is supported
✓ 1
Conclusion must be consistent with the ratios calculated.
Part (b)(i)
Gradient = (−1.56 − 0.69)/5 = −0.45
✓ 1
Accept −0.43 to −0.47. Using only the first bounce scores [1 max].
ln hn = ln h0 + n ln k, so k = e−0.45 = 0.64
✓ 1
Accept 0.63–0.65. Allow ECF from the gradient. Must be consistent with (a)(i).
Part (b)(ii)
2.00 × 0.64n < 0.030 gives n > ln(0.015)/ln 0.64 = 9.4, so the 10th bounce
✓ 1
Allow ECF from (b)(i). Accept 10 for any k between 0.63 and 0.65.
Part (c)(i)
Energy lost = (1 − 0.64) × 0.045 × 9.81 × 2.00 = 0.32 J
✓ 1
Allow ECF from (b)(i). Accept mg(h0 − h1) = 0.33 J.
ΔT = ½ × 0.32/(0.045 × 1.7 × 103) = 2.1 × 10−3 K
✓ 1
Accept 2.0–2.2 × 10−3 K.
Answers: (a)(i) ratio ≈ 0.64; supported · (b)(i) 0.64 · (b)(ii) 10th bounce · (c)(i) 2.1 × 10−3 K (the remaining parts are explanations — see the table above)
31A-1B-16Elastic potential energyPaper 1BMedium7 marks
A ball of mass 25.0 g rests on a light vertical spring of spring constant 320 N m−1. The spring is compressed by a distance x, measured with a ruler to ±0.1 cm, and released. The maximum height h reached by the ball above its release position is found from a video recording.
If all the elastic potential energy is transferred to gravitational potential energy of the ball, ½kx² = mgh, so h = (k/2mg)x².
x / cm
2.0
2.5
3.0
3.5
4.0
4.5
h / m
0.23
0.37
0.53
0.72
0.93
1.19
Graph drawn to scale. The point for x = 4.5 cm has not been plotted.
(a)
(i)
Calculate x² for x = 4.5 cm, in m², together with its absolute uncertainty.
(2)
(ii)
Determine the gradient of the graph of h against x², using the full range of the data. Give its unit.
(2)
(iii)
Determine the fraction of the stored elastic potential energy that is transferred to gravitational potential energy of the ball.
(2)
(iv)
Suggest one reason why your answer to (a)(iii) is less than 1.
(1)
Show mark scheme
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Mark
Notes
Part (a)(i)
x² = (0.045)² = 20.3 × 10−4 m²
✓ 1
Accept 20.25 × 10−4.
Fractional uncertainty = 2 × 0.1/4.5 = 4.4 %, so ±0.9 × 10−4 m²
✓ 1
Absolute uncertainty to 1 s.f.
Part (a)(ii)
Line through the origin and (20.3 × 10−4, 1.19): gradient = 1.19/20.3 × 10−4 = 586
Air resistance on the rising ball; OR the spring is not light, so some energy becomes kinetic and gravitational potential energy of the spring; OR energy dissipated as internal energy in the spring (and sound)
✓ 1
Allow ECF from (a)(iii). Any one physically justified reason.
Answers: (a)(i) (20.3 ± 0.9) × 10−4 m² · (a)(ii) 586 m−1 · (a)(iii) 0.90 (the remaining parts are explanations — see the table above)
32A-1B-22Efficiency and energy transferPaper 1BMedium6 marks
A small electric winch, powered from a 6.00 V supply, lifts a hook carrying a load of mass M through a height of 0.800 m at a steady speed. For each load the current I is read from an ammeter (±0.01 A) and the time t for the lift is measured with a stopwatch (±0.2 s). The electrical energy supplied is E = VIt; the uncertainty in the supply voltage is negligible.
The graph shows E against M.
M / kg
I / A
t / s
E / J
0.100
0.34
2.3
4.69
0.200
0.38
2.8
6.38
0.300
0.42
3.3
8.32
0.400
0.47
3.5
9.87
0.500
0.51
3.8
11.63
0.600
0.55
4.0
Graph drawn to scale. The point for M = 0.600 kg has not been plotted.
(a)
(i)
Calculate E for M = 0.600 kg together with its absolute uncertainty.
(2)
(ii)
The gradient of the graph is the extra electrical energy needed for each extra kilogram lifted. Determine, using the full range of the data, the efficiency with which this extra energy is transferred to gravitational potential energy of the load.
(2)
(iii)
Determine the intercept of the line of best fit on the E axis and suggest what it represents. Hence explain why the overall efficiency of lifting the 0.600 kg load is smaller than your answer to (a)(ii).
33A-1B-26Energy density of fuelsPaper 1BMedium7 marks
A spirit burner containing ethanol heats 0.250 kg of water in a thin metal can. The mass Δm of ethanol burned is found by weighing the burner before and after heating on a balance that reads to ±0.01 g, so each value of Δm is ±0.02 g. The temperature rise ΔT of the water is ±0.4 K. The specific heat capacity of water is 4180 J kg−1 K−1.
Only a fraction η of the energy released by the burning ethanol is transferred to the water, so ΔT = (ηEd/mwcw)Δm, where Ed = 29.7 MJ kg−1 is the energy density of ethanol. The graph shows ΔT against Δm.
Δm / g
0.40
0.80
1.20
1.60
2.00
2.40
ΔT / K
3.7
7.3
10.9
14.7
18.1
21.7
Graph drawn to scale: line of best fit (solid), lines of maximum and minimum gradient (dashed).
(a)
(i)
Determine the gradient of the line of best fit. Give its unit.
(2)
(ii)
The lines of maximum and minimum gradient pass through the ends of the error bars of the first and last points. Determine the absolute uncertainty in the gradient.
(2)
(b)
(i)
Determine η, together with its absolute uncertainty.
(2)
(ii)
Suggest one change to the apparatus that would increase the value of η.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient using the full range, e.g. (21.7 − 3.7)/(2.40 − 0.40) = 9.0 K g−1
✓ 1
Accept 8.8–9.2 K g−1.
Unit K g−1 (or 9.0 × 103 K kg−1)
✓ 1
Accept °C g−1.
Part (a)(ii)
Maximum gradient (22.1 − 3.3)/2.00 = 9.4 K g−1; minimum gradient (21.3 − 4.1)/2.00 = 8.6 K g−1
✓ 1
Accept values read from the dashed lines within ±0.1 K g−1.
Uncertainty = (9.4 − 8.6)/2 = ±0.4 K g−1
✓ 1
Allow ECF from (a)(i) for the comparison. Accept the larger of (max − best) and (best − min).
Part (b)(i)
Energy reaching the water per kilogram of ethanol = gradient × mwcw = 9.0 × 103 × 0.250 × 4180 = 9.40 × 106 J kg−1, so η = 9.40/29.7 = 0.32
✓ 1
Allow ECF from (a)(i). Gradient must be converted to K kg−1; a power-of-ten error scores 0 for this mark. Accept η = 0.31–0.33.
Fractional uncertainty of η = that of the gradient, 0.4/9.0 = 4.4 %, so η = 0.32 ± 0.01
✓ 1
Allow ECF from (a)(ii). Accept 32 % ± 1 % (or ± 1.4 %). The uncertainties in mw, cw and Ed are negligible.
Part (b)(ii)
Reduce the energy transferred to the surroundings: e.g. surround the flame and can with a draught shield; put a lid on the can and insulate its sides; place the can closer to the flame; use a can with a blackened base
✓ 1
Do not accept "use more accurate instruments" or "repeat the experiment": these do not change η.
Answers: (a)(i) 9.0 K g−1 · (a)(ii) ±0.4 K g−1 · (b)(i) η = 0.32 ± 0.01 (the remaining parts are explanations — see the table above)
A team tests a new runner coating on a sledge of mass 12.0 kg. The sledge is released from rest on a uniform snow slope inclined at 20.0° to the horizontal, from different heights h above a pair of timing gates at the bottom of the slope, which measure its speed v. A constant frictional force f acts along the slope; air resistance is negligible.
Applying the work–energy principle gives v² = 2h(g − f/(m sin 20.0°)). The graph shows v² against h with the line of best fit. The specific latent heat of fusion of ice is 3.34 × 105 J kg−1.
h / m
v / m s−1
v² / m² s−2
0.50
2.41
5.8
1.00
3.62
13.1
1.50
4.53
20.5
2.00
5.22
27.2
2.50
5.84
34.1
3.00
6.46
41.7
Graph drawn to scale
(a)
(i)
Determine the gradient of the line of best fit. Give its unit.
(2)
(ii)
Hence determine f.
(2)
(iii)
For the run from h = 3.00 m, half of the work done against friction melts snow at 0 °C under the runners. Estimate the mass of snow melted.
(2)
(iv)
The line meets the h axis at about 0.08 m. State and explain whether this affects your answer to (a)(ii).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Large triangle on the line, e.g. from (0.083, 0) to (3.00, 41.6): 41.6/2.92 = 14.2
✓ 1
Accept 13.8–14.6.
Unit m s−2
✓ 1
Part (a)(ii)
Gradient = 2(g − f/(m sin 20.0°)), so f = m sin 20.0° (g − gradient/2) = 12.0 × 0.342 × (9.81 − 7.12)
✓ 1
Allow ECF from (a)(i).
f = 11 N
✓ 1
Accept 10–12 N.
Part (a)(iii)
Work done against friction = f × h/sin 20.0° = 11 × 3.00/0.342 = 97 J
✓ 1
Allow ECF from (a)(ii).
Mass = ½ × 97/3.34 × 105 = 1.4 × 10−4 kg
✓ 1
Accept 1.3–1.6 × 10−4 kg. Using the vertical height instead of the distance along the slope scores 0 for the first mark.
Part (a)(iv)
No: a constant offset in h (e.g. the gates measure v about 8 cm above the level from which h was measured) shifts the line sideways without changing its gradient, and f depends only on the gradient
✓ 1
Allow ECF from (a)(ii): the answer must refer to the method actually used.
Answers: (a)(i) 14.2 m s−2 · (a)(ii) 11 N · (a)(iii) 1.4 × 10−4 kg (the remaining parts are explanations — see the table above)
35A-2-26Elastic potential energyPaper 2Easy12 marks
A child bounces on a pogo stick. The combined mass of the child and the stick is 32 kg. The graph shows how the force needed to compress the spring of the stick varies with its compression. At the lowest point of each bounce the spring is compressed by 0.12 m and the child and stick are momentarily at rest.
Force F needed to compress the spring against the compression (drawn to scale).
(a)
(i)
Use the graph to determine the spring constant of the spring.
(1)
(ii)
Determine the elastic potential energy stored at the lowest point.
(2)
(iii)
Show that, if no energy were dissipated, the centre of mass of the child and stick would rise about 0.28 m from the lowest point to the highest point.
(2)
(b)
(i)
The centre of mass actually rises 0.22 m. Calculate the efficiency of the transfer from elastic to gravitational potential energy.
(2)
(ii)
The child makes 2.1 bounces per second at a steady height. Calculate the average power the child must supply.
(2)
(c)
(i)
The child stands still on the stick. Calculate the compression of the spring.
(1)
(ii)
The child then bounces gently without the stick leaving the ground. Calculate the frequency of the resulting simple harmonic motion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
k = gradient = 1440/0.12 = 1.2 × 104 N m−1
✓ 1
Accept 1.1–1.3 × 104 N m−1.
Part (a)(ii)
Area under the graph (triangle) = ½ × 0.12 × 1440
✓ 1
= 86 J
✓ 1
Allow ECF from (a)(i) if ½kx² is used. Accept 86.4 J.
Energy dissipated per bounce = 86.4 − 69.1 = 17.3 J
✓ 1
Allow ECF from (b)(i).
P = 17.3 × 2.1 = 36 W
✓ 1
Accept 35–37 W.
Part (c)(i)
x = mg/k = 32 × 9.81/1.2 × 104 = 0.026 m
✓ 1
Allow ECF from (a)(i).
Part (c)(ii)
f = (1/2π)√(k/m)
✓ 1
The oscillation is about the equilibrium position of (c)(i).
f = (1/2π)√(1.2 × 104/32) = 3.1 Hz
✓ 1
Allow ECF from (a)(i).
Answers: (a)(i) 1.2 × 104 N m−1 · (a)(ii) 86 J · (b)(i) 0.80 · (b)(ii) 36 W · (c)(i) 0.026 m · (c)(ii) 3.1 Hz (the remaining parts are explanations — see the table above)
36IBP2H-006-00-09Elastic potential energyPaper 2Easy4 marks
An archer draws a bowstring back 0.50 m. The force needed increases uniformly from zero to 180 N over this distance. The arrow has a mass of 0.025 kg.
Force needed to draw the bowstring against distance drawn
(a)
Calculate the elastic potential energy stored in the bow when it is fully drawn.
(2)
(b)
When the string is released, 80% of the stored energy is transferred to the arrow as kinetic energy. Calculate the speed of the arrow as it leaves the bow.
(2)
Show mark scheme
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Notes
Part (a)
Energy = area under the force–distance graph = ½Fx
✓ 1
= ½ × 180 × 0.50 = 45 J
✓ 1
Part (b)
Kinetic energy of arrow = 0.80 × 45 = 36 J
✓ 1
v = √(2 × 36 / 0.025) = 54 m s−1
✓ 1
ECF from (a)
—
Exam tip: When a force increases uniformly from zero, the work done is the average force (half the maximum) multiplied by the distance.
37IBP2H-006-01-09Work done by a variable forcePaper 2Easy5 marks
A kite-surfer is pulled across a lake by a kite. The tension in the kite lines is a constant 600 N, directed at 35° above the horizontal. The surfer moves 80 m in a straight horizontal line in 16 s at constant speed.
Side view (not to scale)
(a)
Calculate the work done on the surfer by the tension in the kite lines.
(2)
(b)
State why the vertical component of the tension does no work on the surfer.
(1)
(c)
State the work done on the surfer by the water and air resistance during the 80 m.
(1)
(d)
Calculate the average power delivered to the surfer by the kite.
(1)
Show mark scheme
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Mark
Notes
Part (a)
Uses W = Fs cos θ: 600 × 80 × cos 35°
✓ 1
W = 3.9 × 104 J (3.93 × 104 J)
✓ 1
Part (b)
The vertical component is perpendicular to the (horizontal) displacement, so there is no displacement in its direction
✓ 1
Part (c)
−3.9 × 104 J, because the speed is constant so the net work is zero (ECF from (a))
✓ 1
Part (d)
P = W/t = 3.93 × 104 / 16 = 2.5 × 103 W. Also accept 2.4 × 103 W from the rounded work, or P = Fv cos θ = 600 × 5.0 × cos 35°. ECF from (a)
✓ 1
Exam tip: At constant velocity the net work is zero. The work done by the driving force and the work done by resistance are equal in size and opposite in sign. Carry unrounded values into later parts to avoid rounding drift.
38A-2-24Efficiency and energy transferPaper 2Medium13 marks
A disused mine shaft is converted into a gravity energy store. When electricity is plentiful, electric winches raise heavy blocks of total mass 5.0 × 105 kg through a height of 250 m. When electricity is needed, the blocks are lowered at a constant speed of 1.2 m s−1 and the winches act as generators. The winch motors have an efficiency of 80 % and the generators an efficiency of 90 %.
(a)
(i)
Show that the energy stored when all the blocks are raised through 250 m is about 1.2 × 109 J.
(1)
(ii)
A lithium-ion battery stores 0.60 MJ kg−1. Compare the energy stored per kilogram of blocks with this value.
(2)
(b)
(i)
Calculate the electrical power output while the blocks descend.
(2)
(ii)
Calculate the time for which the store can deliver this power.
(1)
(c)
(i)
Determine the electrical energy needed to raise all the blocks and the overall efficiency of one complete store-and-return cycle.
(2)
(ii)
The generators deliver their output at 11 kV through cables of total resistance 0.80 Ω. Determine the fraction of the output power that is dissipated in the cables.
(3)
(iii)
Outline a Sankey diagram for one complete cycle that starts with 100 MJ of electrical energy supplied to the motors.
(2)
Show mark scheme
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Notes
Part (a)(i)
ΔEp = 5.0 × 105 × 9.81 × 250 = 1.23 × 109 J
✓ 1
Must see 1.23 × 109 or full substitution.
Part (a)(ii)
Energy per kilogram = gΔh = 9.81 × 250 = 2452 J kg−1
✓ 1
Allow ECF from (a)(i).
The battery stores about 0.60 × 106/2452 ≈ 240 times more per kilogram
39IBP2H-006-00-10Energy density of fuelsPaper 2Medium7 marks
The table gives data for three fuels.
Fuel | Specific energy / MJ kg−1 | Density / kg m−3 Petrol | 46 | 740 Ethanol | 30 | 789 Liquid hydrogen | 142 | 71
A car engine has an efficiency of 25%. On a motorway the car travels at a steady 25 m s−1 with a useful mechanical power output of 15 kW.
(a)
Calculate the energy density of ethanol.
(1)
(b)
The car has a 50 L tank (1 L = 1.0×10−3 m3) filled with ethanol. Determine the motorway range of the car.
(3)
(c)
Outline why liquid hydrogen is used as a fuel for launch rockets despite its low density.
(1)
(d)
A car with the same engine efficiency runs on liquid hydrogen. Determine the volume of liquid hydrogen needed to give the same range as in (b).
(2)
Show mark scheme
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Notes
Part (a)
30×106 × 789 = 2.37×1010 ≈ 2.4×1010 J m−3
✓ 1
Part (b)
Useful energy per km = 15×103 × (1000 / 25) = 6.0×105 J
✓ 1
Fuel energy needed per km = 6.0×105 / 0.25 = 2.4×106 J
✓ 1
Energy in tank = 0.050 × 2.37×1010 = 1.18×109 J, so range = 1.18×109 / 2.4×106 ≈ 490 km
✓ 1
Accept equivalent methods, e.g. useful energy = 0.25 × tank energy and time = useful energy / 15 kW, then distance = 25 m s−1 × time. Accept 480–500 km. ECF from (a).
—
Part (c)
It has by far the highest specific energy, so the least fuel mass has to be carried and accelerated for each joule released
✓ 1
Part (d)
Energy density of hydrogen = 142×106 × 71 = 1.01×1010 J m−3
✓ 1
The same fuel energy, 1.18×109 J, is needed, so V = 1.18×109 / 1.01×1010 = 0.117 m3 ≈ 117 L
✓ 1
Accept 0.12 m3 or 120 L. Accept the ratio method: 50 L × (2.37×1010 / 1.01×1010). ECF from (b).
—
Exam tip: Work out the input energy needed per kilometre first. Every range or tank-size calculation then becomes a single division.
40IBP2H-006-00-11Power and rate of energy transferPaper 2Medium7 marks
An electric bicycle has a battery that supplies 360 W to a motor controller. The controller wastes 5.0% of its input power as internal energy. The motor converts 85% of the power it receives into mechanical power. Friction in the chain and gears then wastes a further 12 W before the power reaches the rear wheel.
(a)
Calculate the mechanical power delivered to the rear wheel by the motor.
(2)
(b)
Describe the main features of a Sankey diagram for this system.
(2)
(c)
The rider pedals, adding a further 120 W at the wheel. The rider and bicycle have a combined mass of 95 kg. They climb a road that rises 1.0 m vertically for every 20 m travelled along the road, against a constant resistive force of 22 N. Determine the steady speed of the bicycle up the hill.
(3)
Show mark scheme
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Notes
Part (a)
Power into motor = 0.95 × 360 = 342 W
✓ 1
Power at wheel = 0.85 × 342 − 12 = 279 W
✓ 1
Part (b)
A single input arrow representing 360 W, with all arrow widths proportional to power
✓ 1
Three waste arrows branching away, representing 18 W, 51 W and 12 W, and a useful output arrow of 279 W continuing straight on
✓ 1
Part (c)
Total power at wheel = 279 + 120 = 399 W
✓ 1
Driving force needed = mg × (1/20) + 22 = 95 × 9.8 × 0.050 + 22 = 68.6 N
✓ 1
v = P / F = 399 / 68.6 = 5.8 m s−1
✓ 1
ECF from (a)
—
Common error: A percentage efficiency applies to the power entering that stage, not to the original battery output. Apply each loss in sequence.
41IBP2H-006-00-12Energy and projectile motionPaper 2Medium6 marks
A ski jumper of mass 62 kg starts from rest at the top of an in-run. The take-off point is 38 m vertically below the start. On the in-run, 4.5 kJ of energy is transferred to internal energy by friction and air resistance. The jumper leaves the take-off at 10° above the horizontal and lands at a point 45 m vertically below the take-off. Ignore air resistance during the flight.
Not to scale
(a)
Show that the take-off speed is about 24 m s−1.
(2)
(b)
Calculate the speed of the jumper on landing.
(2)
(c)
Explain why the answer to (b) does not depend on the take-off angle of 10°.
(2)
Show mark scheme
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Notes
Part (a)
Kinetic energy at take-off = 62 × 9.8 × 38 − 4500 = 1.86×104 J
✓ 1
v = √(2 × 1.86×104 / 62) = 24.49 m s−1, which is about 24 m s−1
v = 38.5 m s−1 (accept 38 m s−1; using 24 m s−1 gives 38.2 m s−1, also accepted)
✓ 1
Allow error carried forward from (a).
—
Part (c)
The work done by gravity during the flight depends only on the vertical displacement of 45 m, not on the path taken
✓ 1
The take-off speed is fixed by the in-run energy whatever the angle, and kinetic energy depends only on speed, not direction, so the landing kinetic energy and speed are the same for any angle
✓ 1
Exam tip: In a 'show that' question, give your answer to one more significant figure than the value stated, to prove you have not simply worked back from it.
42IBP2H-006-00-13Elastic potential energyPaper 2Medium7 marks
A pole vaulter of mass 70 kg sprints along the runway at 9.5 m s−1. During the run-up her centre of mass is 1.0 m above the ground.
(a)
Calculate the greatest height her centre of mass could reach if all of her kinetic energy were converted into gravitational potential energy.
(2)
(b)
In the vault, her centre of mass actually rises to 5.9 m, where she still has a horizontal speed of 1.0 m s−1. Determine the minimum work done by her muscles during the vault.
(3)
(c)
Outline the role of the pole in the energy transfers during the vault.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Rise h = v2 / 2g = 9.52 / (2 × 9.8) = 4.6 m
✓ 1
Maximum height = 4.6 + 1.0 = 5.6 m
✓ 1
Part (b)
Energy at top = 70 × 9.8 × 4.9 + ½ × 70 × 1.02 = 3361 + 35 = 3.40×103 J, measured from the run-up height
✓ 1
Initial kinetic energy = ½ × 70 × 9.52 = 3.16×103 J
✓ 1
Work done by muscles ≥ 3.40×103 − 3.16×103 ≈ 240 J
✓ 1
Part (c)
As the pole bends, kinetic energy is stored as elastic potential energy in the pole
✓ 1
As the pole straightens, this energy is returned as gravitational potential energy of the vaulter; any energy dissipated in the pole must be made up by work from her muscles
✓ 1
Common error: Measure height changes of the centre of mass from its starting height of 1.0 m, not from the ground.
43IBP2H-006-01-10Efficiency and energy transferPaper 2Medium8 marks
A sports scientist compresses the foam midsole of a running shoe in a materials tester. The table gives the compressive force F for each compression x.
x / mm: 0, 2.0, 4.0, 6.0, 8.0 F / N: 0, 200, 500, 900, 1400
Compressive force on the midsole against compression
(a)
Outline how the data show that the midsole does not obey Hooke's law.
(1)
(b)
By treating the force–compression relationship as straight-line segments between adjacent data points, estimate the work done in compressing the midsole by 8.0 mm.
(3)
(c)
The force increases with compression at an increasing rate. Explain whether your estimate in (b) is too large or too small.
(2)
(d)
When the load is removed, the midsole returns 3.7 J of energy to the runner. Calculate the percentage of the work done that is returned, and state what happens to the remainder.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Equal increases in x produce increasingly larger increases in F (or F/x is not constant), so F is not proportional to x
✓ 1
Part (b)
Recognises that the work done is the area under the F–x relationship, found as the sum of trapezium areas
44IBP2H-006-01-12Work–energy principlePaper 2Medium7 marks
A tram of mass 3.6 × 104 kg uses regenerative braking on a level track. While slowing from 15 m s−1 to 5.0 m s−1, the system stores 45% of the lost kinetic energy in supercapacitors. The tram later uses this stored energy to accelerate from 5.0 m s−1 back to 15 m s−1. The motor converts stored energy to kinetic energy with an efficiency of 80%.
(a)
Calculate the kinetic energy lost by the tram during braking.
(2)
(b)
Calculate the energy stored in the supercapacitors.
(1)
(c)
The kinetic energy needed to re-accelerate the tram from 5.0 m s−1 to 15 m s−1 equals the kinetic energy lost in (a). Determine the percentage of this kinetic energy that the stored energy can supply.
(2)
(d)
Outline two reasons why the energy stored is less than the kinetic energy lost.
First valid reason, e.g. work is done against air resistance and rolling resistance during braking
✓ 1
Second, different valid reason, e.g. resistive (Joule) heating in the generator windings and circuits; OR friction in the bearings and gearing; OR energy dissipated in friction brakes if they are also used
✓ 1
Exam tip: For a chain of energy transfers, the overall efficiency is the product of the efficiencies of each stage.
45A-2-04Power and rate of energy transferPaper 2Hard13 marks
A cyclist and bicycle have a combined mass of 80 kg. The resistive force on them is modelled as F = kv², where v is the speed and k is a constant. On a horizontal road the cyclist reaches a steady speed of 12.0 m s−1 while delivering a power of 300 W to the bicycle.
(a)
(i)
Show that k is about 0.17 in SI units.
(2)
(ii)
Determine the fundamental SI unit of k.
(1)
(b)
(i)
The cyclist stops pedalling and freewheels down a straight road inclined at 5.0° to the horizontal. Determine the terminal speed.
(2)
(ii)
The cyclist climbs the same road at a steady speed, again delivering 300 W. Determine this speed.
(3)
(c)
(i)
Calculate the mechanical power needed to ride up this road at 12.0 m s−1.
(1)
(ii)
An electrically assisted bicycle lets the cyclist ride up this road at 12.0 m s−1 while still delivering 300 W. The rest is supplied by a motor of efficiency 80 % powered by a 36 V battery. Calculate the current from the battery.
(2)
(iii)
The battery stores 0.50 kW h of energy. Determine the distance the cyclist can climb at 12.0 m s−1 before the battery is empty.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
At steady speed the driving force equals the resistive force: F = P/v = 300/12.0 = 25 N
✓ 1
k = 25/12.0² = 0.174
✓ 1
Must see 0.174 or full substitution.
Part (a)(ii)
k = F/v²: kg m s−2/(m² s−2) = kg m−1
✓ 1
Answer in fundamental units required; "N s² m−2" alone scores 0.
Part (b)(i)
At terminal speed mg sin 5.0° = kv²: 80 × 9.81 × sin 5.0° = 68.4 N
✓ 1
v = √(68.4/0.174) = 19.8 m s−1
✓ 1
Allow ECF from (a)(i). Accept 19.6–20.2 m s−1.
Part (b)(ii)
Driving force = mg sin 5.0° + kv², so 300 = v(68.4 + 0.174v²)
✓ 1
Allow ECF from (a)(i) and (b)(i).
First estimate ignoring drag: v ≈ 300/68.4 = 4.39 m s−1; drag at this speed is only about 3.3 N
✓ 1
Or any valid numerical/iterative solution.
v = 4.2 m s−1
✓ 1
Accept 4.1–4.3 m s−1. 4.4 m s−1 (drag ignored) scores [2 max].
Part (c)(i)
P = (68.4 + 25) × 12.0 = 1121 W
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 1.1 kW.
Part (c)(ii)
Motor output = 1121 − 300 = 821 W; electrical input = 821/0.80 = 1026 W
✓ 1
Allow ECF from (c)(i).
I = P/V = 1026/36 = 28.5 A
✓ 1
Accept 28–29 A. 23 A (efficiency ignored) scores [1 max].
Part (c)(iii)
0.50 kW h = 0.50 × 3.6 × 106 = 1.8 × 106 J; time = 1.8 × 106/1026 = 1754 s
✓ 1
Allow ECF from (c)(ii).
Distance = 12.0 × 1754 = 21 km
✓ 1
Accept 20–22 km.
Answers: (a)(ii) kg m−1 · (b)(i) 19.8 m s−1 · (b)(ii) 4.2 m s−1 · (c)(i) 1121 W · (c)(ii) 28.5 A · (c)(iii) 21 km (the remaining parts are explanations — see the table above)
A lorry of mass 3.2 × 104 kg descends a long hill, keeping a constant speed by using its brakes. Its brakes then fail, and the driver steers into an escape lane. The lane slopes upwards at 8.0° to the horizontal and is filled with deep, loose gravel. The lorry enters the lane at 28 m s−1 and stops after travelling 110 m along it. Assume that the resistive force of the gravel is constant.
Side view of the escape lane (angle exaggerated)
(a)
(i)
Before the brakes failed, the lorry descended a vertical height of 150 m at constant speed. 40 % of the energy dissipated by braking was absorbed by the brake drums, of total mass 120 kg and specific heat capacity 450 J kg−1 K−1. Estimate the rise in temperature of the brake drums.
(2)
(ii)
State one assumption made in your estimate, and suggest why it explains the brake failure.
(1)
(b)
(i)
Calculate the kinetic energy of the lorry as it enters the lane and the gravitational potential energy it gains before it stops.
(2)
(ii)
Determine the resistive force exerted by the gravel.
(2)
(iii)
Show that the stopping distance on an upward slope of angle θ is u²/[2(g sin θ + F/m)], and hence explain why a fully loaded lorry stops in a longer distance than an empty one, if F is the same for both.
(2)
(c)
(i)
Calculate the rate at which the gravel dissipates energy as the lorry enters the lane.
(1)
(ii)
The same gravel bed is laid on level ground. Determine the stopping distance from 28 m s−1, and explain why escape lanes are built sloping upwards.
(2)
Show mark scheme
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Notes
Part (a)(i)
Energy to drums = 0.40 × 3.2 × 104 × 9.81 × 150 = 1.9 × 107 J
✓ 1
At constant speed all the gravitational potential energy lost is dissipated.
ΔT = 1.9 × 107/(120 × 450) = 349 K
✓ 1
Accept 340–360 K.
Part (a)(ii)
No energy is lost from the drums to the air during the descent (or the 40 % is spread uniformly); a rise of several hundred kelvin so overheats the brakes that friction falls (brake fade)
✓ 1
Allow ECF from (a)(i). Any valid assumption with a link to failure.
Work done by the gravel = loss of mechanical energy = 1.25 × 107 − 4.8 × 106 = 7.7 × 106 J
✓ 1
Allow ECF from (b)(i).
F = 7.7 × 106/110 = 7.0 × 104 N
✓ 1
Accept 6.9–7.1 × 104 N.
Part (b)(iii)
Work–energy: ½mu² = (mg sin θ + F)s, so s = u²/[2(g sin θ + F/m)]
✓ 1
Symbolic working required.
Larger m makes F/m smaller, so the denominator is smaller and s is larger (the slope term g sin θ is the same for both)
✓ 1
Part (c)(i)
P = Fv = 7.0 × 104 × 28 = 2.0 × 106 W
✓ 1
Allow ECF from (b)(ii).
Part (c)(ii)
All the kinetic energy is dissipated by the gravel: 1.25 × 107/7.0 × 104 = 178 m
✓ 1
Allow ECF from (b)(ii). Accept 175–180 m.
On the slope about 40 % of the energy becomes gravitational potential energy, so a shorter lane is needed; the gravel stops the stored energy from returning the lorry down the slope
✓ 1
Answers: (a)(i) 349 K · (b)(i) 1.25 × 107 J; 4.8 × 106 J · (b)(ii) 7.0 × 104 N · (c)(i) 2.0 × 106 W · (c)(ii) 178 m (the remaining parts are explanations — see the table above)
47A-2-40Energy transfer in a wind turbinePaper 2Hard20 marks
A horizontal-axis wind turbine has three blades that sweep a circle of diameter 90 m. Its hub is 80 m above the ground. The graph shows the electrical power output of the turbine for different wind speeds. The density of air is 1.2 kg m−3.
Electrical output power P of the turbine against wind speed (drawn to scale).
(a)
(i)
Show that the kinetic energy carried per second by wind of speed v through an area A perpendicular to the wind is ½ρAv³.
(2)
(ii)
Show that ½ρAv³ has the fundamental SI units of power.
(1)
(iii)
Determine the overall efficiency of the turbine when the wind speed is 10 m s−1.
(2)
(b)
(i)
At a wind speed of 10.0 m s−1 the air passes through the blades at an average speed of 7.0 m s−1 and leaves well behind the turbine at 4.0 m s−1. Determine the force exerted by the air on the rotor.
(3)
(ii)
Calculate the torque of this force about the base of the tower.
(1)
(iii)
Show that, in this model, the power removed from the wind is about 2.2 MW, and suggest why it is greater than the electrical output read from the graph.
(2)
(c)
(i)
At a wind speed of 10 m s−1 the rotor turns at 16 revolutions per minute. Calculate the speed of the tips of the blades.
(1)
(ii)
Each blade has a mass of 6.5 × 103 kg and its centre of mass is 15 m from the axis of rotation. Determine the magnitude of the resultant force that the hub exerts on one blade. Ignore the weight of the blade.
(2)
(iii)
A damaged blade is repaired so that the centre of mass of the rotor no longer lies on its axis of rotation. Explain why the tower then vibrates when the rotor turns.
(2)
(d)
(i)
Assume that the overall efficiency of the turbine stays at the value found in (a)(iii). Determine the wind speed at which the turbine would reach its rated output of 2.0 MW, and comment on your answer with reference to the graph.
(2)
(ii)
Explain why no turbine can transfer all of the kinetic energy of the wind passing through it.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
Mass of air passing per second = ρ × volume per second = ρAv
✓ 1
Symbolic working required.
Kinetic energy per second = ½(ρAv)v² = ½ρAv³
✓ 1
Part (a)(ii)
kg m−3 × m² × m³ s−3 = kg m² s−3 = (kg m s−2)(m s−1) = N m s−1 = W
✓ 1
Fundamental units kg m² s−3 must be seen.
Part (a)(iii)
Power in the wind = ½ × 1.2 × π × 45² × 10³ = 3.82 × 106 W
✓ 1
Allow ECF from (a)(i). Using 90 m as the radius scores 0 for this mark.
Output read from the graph ≈ 1.14 MW; efficiency = 1.14/3.82 = 0.30
✓ 1
Accept 1.1–1.2 MW and 0.28–0.32.
Part (b)(i)
Mass flow = ρA × 7.0 = 1.2 × 6362 × 7.0 = 5.34 × 104 kg s−1
✓ 1
Allow ECF from (a)(iii) area.
Rate of change of momentum of the air = 5.34 × 104 × (10.0 − 4.0)
✓ 1
The speed 7.0 m s−1 sets the mass flow; the change of speed is 6.0 m s−1.
F = 3.2 × 105 N
✓ 1
Accept 3.1–3.3 × 105 N.
Part (b)(ii)
τ = 3.2 × 105 × 80 = 2.6 × 107 N m
✓ 1
Allow ECF from (b)(i).
Part (b)(iii)
½ × 5.34 × 104 × (10.0² − 4.0²) = 2.24 × 106 W
✓ 1
Allow ECF from (b)(i) mass flow.
Energy is dissipated between the air and the grid: drag and turbulence at the blades, friction in the bearings and gearbox, and heating in the generator
✓ 1
Any one valid loss.
Part (c)(i)
ω = 16 × 2π/60 = 1.68 rad s−1; v = ωr = 1.676 × 45 = 75 m s−1
✓ 1
Using 90 m as the radius (151 m s−1) scores 0.
Part (c)(ii)
The force is the centripetal force on the blade, whose mass may be treated as concentrated at its centre of mass: a = ω²r = 1.676² × 15 = 42.1 m s−2
✓ 1
Allow ECF from (c)(i) for ω. Using 45 m scores 0 for this mark.
F = 6.5 × 103 × 42.1 = 2.7 × 105 N (about four times the weight of the blade)
✓ 1
Accept 2.7–2.8 × 105 N.
Part (c)(iii)
The centre of mass of the rotor now moves in a circle about the axis, so a resultant (centripetal) force must act on the rotor; it is provided by the shaft and bearings
✓ 1
Do not accept "the rotor is unbalanced" without reference to a force.
The direction of this force turns with the rotor, so (Newton’s third law) the rotor exerts on the tower a force whose direction changes once every revolution: a periodic force that makes the tower vibrate
✓ 1
Accept a reference to a periodic driving force at the rotation frequency (possible resonance).
Part (d)(i)
Power in the wind needed = 2.0 × 106/0.30 = 6.7 × 106 W
✓ 1
Allow ECF from (a)(iii).
v = (2P/ρA)1/3 = (2 × 6.7 × 106/(1.2 × 6362))1/3 = 12 m s−1, the speed at which the graph reaches 2.0 MW
✓ 1
Accept 11.8–12.3 m s−1. The comment must refer to the graph.
Part (d)(ii)
To remove all the kinetic energy the air would have to leave the blades at zero speed
✓ 1
Allow ECF from (a)(i) and (b)(iii).
Air that has stopped would block the air arriving behind it, so the mass flow through the rotor would fall to zero; some kinetic energy must remain to carry the air away
✓ 1
Do not accept "because of friction".
Answers: (a)(iii) 0.30 · (b)(i) 3.2 × 105 N · (b)(ii) 2.6 × 107 N m · (c)(i) 75 m s−1 · (c)(ii) 2.7 × 105 N · (d)(i) 12 m s−1(the remaining parts are explanations — see the table above)
48IBP2S-005-03-15Conservation of mechanical energyPaper 2Hard8 marks
A bungee jumper of mass 70 kg falls from rest from the point where the cord is attached. The unstretched cord is 20 m long and obeys Hooke's law, with spring constant 140 N m−1. Treat the jumper as a point mass. Air resistance and the mass of the cord are negligible.
(a)
Calculate the speed of the jumper at the instant the cord becomes taut.
(2)
(b)
Determine the maximum extension of the cord.
(3)
(c)
Calculate the maximum elastic potential energy stored in the cord.
(1)
(d)
Determine the extension of the cord when the jumper's speed is greatest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Use ½mv² = mgh, so v = √(2 × 9.8 × 20)
✓ 1
v ≈ 19.8 m s−1
✓ 1
Part (b)
At the lowest point Ek = 0, so GPE lost = EPE gained: mg(20 + x) = ½kx²
Speed is greatest when the resultant force is zero, i.e. when kx = mg
✓ 1
x = (70 × 9.8) / 140 = 4.9 m
✓ 1
Common error: The gravitational energy lost includes the 20 m free fall as well as the extension x. Using mgx alone underestimates the maximum extension.
49IBP2H-006-00-15Sankey diagramsPaper 2Hard7 marks
In a grain store, grain falls vertically onto a conveyor belt at a steady rate of 40 kg s−1. The belt moves horizontally at a constant 1.5 m s−1. Each grain arrives with negligible horizontal velocity and quickly reaches the speed of the belt.
(a)
Calculate the rate at which the grain gains kinetic energy.
(1)
(b)
Determine the extra power the belt motor must deliver to keep the belt moving at constant speed once the grain is arriving.
(2)
(c)
Explain why the answers to (a) and (b) are different.
(2)
(d)
Further along, the belt carries the grain at the same speed and rate up an incline to a hopper 6.0 m higher. The motor drives the whole belt. It converts 70% of its electrical input power into mechanical power delivered to the belt. Ignore all other energy losses in the belt system. Determine the electrical input power to the motor.
(2)
Show mark scheme
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Mark
Notes
Part (a)
½ × (Δm/Δt) × v2 = ½ × 40 × 1.52 = 45 W
✓ 1
Part (b)
Force = rate of change of momentum = (Δm/Δt) × v = 40 × 1.5 = 60 N
✓ 1
P = Fv = 60 × 1.5 = 90 W
✓ 1
Part (c)
While each grain is being accelerated it slides relative to the belt, and kinetic friction does work that is dissipated as internal energy (thermal energy)
✓ 1
The energy dissipated equals the kinetic energy gained, so only half of the 90 W (45 W) becomes kinetic energy of the grain
✓ 1
Part (d)
Mechanical output power of the motor = 90 + 40 × 9.8 × 6.0 = 90 + 2352 ≈ 2.44×103 W
✓ 1
Input power = 2.44×103 / 0.70 ≈ 3.5×103 W
✓ 1
Allow the value of (b) carried forward. Using g = 9.81 m s−2 gives the same answer.
—
Exam tip: When mass is added to a moving system at a steady rate, find the force from the rate of change of momentum and the power from P = Fv. Do not assume all of that power becomes kinetic energy.
State, with a reason, whether the tendon obeys Hooke's law over the whole range of the data.
(1)
(b)
Estimate the elastic potential energy stored in the tendon at an extension of 8.0 mm.
(3)
(c)
Determine the stiffness of the tendon in its linear region.
(2)
(d)
While running, the tendon is stretched to 8.0 mm 85 times per minute. It returns 93% of the stored energy during recoil. Calculate the average rate at which the tendon returns energy to the runner.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
No: force is not proportional to extension, e.g. F/x rises from 0.20 to 0.40 kN mm−1, or the force increases for equal 2.0 mm steps are 0.4, 0.8, 1.0 and 1.0 kN, so the graph is not a straight line through the origin
✓ 1
Part (b)
Energy stored = area under the force–extension graph from 0 to 8.0 mm
✓ 1
Valid method for the area of the curved graph, with correct units handled (kN × mm = J). For example, trapezium rule over 2.0 mm strips: 0.4 + 1.6 + 3.4 + 5.4 J; any other valid area method is equally acceptable
✓ 1
Energy ≈ 11 J (accept any answer from 10 J to 11 J, e.g. 10.8 J)
✓ 1
Part (c)
Gradient of the linear region (4.0 mm to 8.0 mm) = (3.2 − 1.2)×103 N / (4.0×10−3 m)
✓ 1
= 5.0×105 N m−1
✓ 1
Part (d)
Energy returned per minute = 0.93 × 10.8 × 85 ≈ 854 J
✓ 1
Average power = 854 / 60 ≈ 14 W (allow ECF from (b))
✓ 1
Common error: ½Fx gives the stored energy only when the force–extension graph is a straight line through the origin. For curved data, estimate the area under the graph instead.
51IBP2H-006-01-14Energy transfers in collisionsPaper 2Hard8 marks
In a nuclear reactor, fast neutrons are slowed down by elastic collisions with carbon-12 nuclei in a graphite moderator. Assume each collision is head-on and that the carbon nucleus is initially at rest. Take the mass of a carbon-12 nucleus as 12 times the mass of a neutron.
(a)
State what is meant by an elastic collision.
(1)
(b)
A particle of mass m moving at speed u collides head-on and elastically with a stationary particle of mass M. Show that the final velocity of the first particle is v = (m − M)u/(m + M).
(3)
(c)
Calculate the fraction of its kinetic energy that a neutron keeps after one such collision with a carbon-12 nucleus.
(2)
(d)
Estimate the number of collisions needed to reduce a neutron's kinetic energy from 2.0 MeV to 0.025 eV.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A collision in which the total kinetic energy is conserved, as well as momentum
✓ 1
Part (b)
Momentum: mu = mv + MV, so m(u − v) = MV; kinetic energy: ½mu2 = ½mv2 + ½MV2, so m(u2 − v2) = MV2
✓ 1
Dividing the energy equation by the momentum equation gives u + v = V
✓ 1
Substituting for V: m(u − v) = M(u + v), which rearranges to v = (m − M)u/(m + M)
Sets up (121/169)n = 0.025 / (2.0 × 106) = 1.25 × 10−8, or an equivalent equation using the fraction from (c)
✓ 1
Solves for n, e.g. n = ln(8.0 × 107) / ln(169/121) ≈ 54.5, giving 55 collisions; accept 56 from using the rounded value 0.72 (n ≈ 55.4); allow ECF from (c)
✓ 1
Exam tip: When the same fraction of energy is kept in every collision, the energy falls geometrically. Take logarithms of both sides to find the number of collisions, and round up to a whole number.
52IBP2H-006-01-15Power and rate of energy transferPaper 2Hard7 marks
An electric car travels on a level road. The total resistive force on it is F = 220 + 0.40v2, where F is in N and v is the speed in m s−1. The battery stores 2.16 × 108 J. The efficiency of the transfer from battery to useful work at the wheels is 85%.
(a)
Calculate the resistive force at a constant speed of 25 m s−1.
(1)
(b)
Calculate the output power needed at the wheels to keep the car moving at 25 m s−1.
(1)
(c)
Determine the range of the car at a constant 25 m s−1.
(2)
(d)
Determine the percentage increase in range if the car instead travels at a constant 20 m s−1.
(2)
(e)
The journey at 20 m s−1 takes longer. Explain why the car still uses less energy per kilometre.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = 220 + 0.40 × 252 = 470 N
✓ 1
Part (b)
P = Fv = 470 × 25 ≈ 1.2 × 104 W (ECF from (a))
✓ 1
Part (c)
Useful energy = 0.85 × 2.16 × 108 = 1.84 × 108 J
✓ 1
Range = 1.84 × 108 / 470 ≈ 3.9 × 105 m (ECF from (a))
✓ 1
Part (d)
F at 20 m s−1 = 220 + 0.40 × 400 = 380 N, giving a range of 4.8 × 105 m
53IBP2H-006-01-16Work–energy principlePaper 2Hard7 marks
A gymnast of mass 52 kg lets go of a bar and falls from rest. Her feet fall 1.8 m before they first touch the top surface of a landing mat. The mat behaves like a spring that obeys Hooke's law and is compressed by a maximum of 0.25 m. Assume air resistance is negligible and that the gymnast stays upright and rigid during the landing.
Gymnast released from rest above the mat (not to scale)
(a)
Determine the effective spring constant of the mat.
(3)
(b)
Calculate the maximum force exerted by the mat on the gymnast, and express it as a multiple of her weight.
(2)
(c)
Explain how bending her knees as she lands reduces the maximum force on her body.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Energy conservation, including the extra fall during compression: mg(h + x) = ½kx2
✓ 1
Gravitational potential energy lost = 52 × 9.8 × 2.05 ≈ 1.04 × 103 J
✓ 1
k = 2 × 1.04 × 103 / 0.252 ≈ 3.3 × 104 N m−1
✓ 1
Part (b)
F = kx = 3.34 × 104 × 0.25 ≈ 8.4 × 103 N (allow ECF from (a))
✓ 1
F / mg = 8.4 × 103 / 510 ≈ 16 (16.4)
✓ 1
Part (c)
The kinetic energy to be removed is about the same, but it is now removed over a larger stopping distance (mat compression plus knee bend)
✓ 1
Since work done = force × distance, the average force, and so the peak force, is smaller
✓ 1
Common error: Include the gravitational potential energy lost during the compression itself. The gymnast falls h + x in total, not just h.
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