IB Sports, Exercise And Health Science SL Paper 2 Practice Paper (Structured Questions, 10 Questions, 100 Marks)

IB Sports, Exercise and Health Science SL Paper 2 Practice Paper (Structured Questions, 10 Questions, 100 Marks)

2026-11-02

IB SEHS SL

IB SEHS SL Paper 2 structured questions: energy systems, VO2max and Karvonen zones, Fick equation, levers, fatigue, heat, body fat, skill learning, ACWR, nutrition.

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Paper details　Level: IB　|　Subject: Sports, Exercise and Health Science SL　|　Time: 75 minutes　|　Marks: 100　|　Questions: 10

Section A Energy systems, cardiorespiratory and biomechanics (Q1–Q4, 40 marks)

Show the formula and each calculation step; give units with every numerical answer.

#	Question	Answer and marking notes
1	Q1 An athlete of 70 kg has fat-free tissue equal to about 40% of body mass. Each kilogram of fat-free tissue holds 0.25 mol of usable ATP-PC and 1 mol of ATP releases about 50 kJ; the phosphagen system lasts about 10 s. (a) Calculate the total moles of ATP and the total energy available from the ATP-PC system. (b) Calculate the maximum average power (kW) of that system. (c) Another athlete obtains energy from glycolysis of 90 g of glucose (net 2 mol ATP per mole of glucose, molar mass 180 g/mol); calculate the energy and the ratio of ATP-PC energy to glycolytic energy. (d) If a 60 s sprint costs 800 kJ in total, calculate the percentage supplied by ATP-PC.	(a) Fat-free tissue = 70 x 0.40 = 28 kg; ATP = 28 x 0.25 = 7.0 mol; energy = 7.0 x 50 = 350 kJ. (3 marks: mass, moles, energy) (b) Power = 350 kJ / 10 s = 35 kW. (2 marks) (c) Glucose 90 / 180 = 0.5 mol, giving 1.0 mol ATP, i.e. 50 kJ; ratio 350 / 50 = 7. (3 marks) (d) 350 / 800 x 100 = 43.75%. (2 marks) Marking notes: no credit for saying the system is fast without calculation; the power unit must be kW or kJ/s. (2 marks)
2	Q2 A 20-year-old athlete of 70 kg has a VO2max of 55 mL/kg/min and a resting heart rate of 60 bpm. (a) Convert this to an absolute VO2max (L/min) and to METs (1 MET = 3.5 mL/kg/min). (b) Taking maximum heart rate as 220 minus age, use the Karvonen method to find the heart rates at 70% and 85% intensity. (c) If 1 L of oxygen yields about 5 kcal, calculate the energy cost per minute at maximum intensity, and explain why training at that intensity cannot be sustained for long.	(a) 55 x 70 / 1000 = 3.85 L/min; 55 / 3.5 = 15.7 METs. (3 marks) (b) HRmax = 220 – 20 = 200 bpm; at 70%: 60 + 0.70 x (200 – 60) = 158 bpm; at 85%: 60 + 0.85 x 140 = 179 bpm. (3 marks) (c) 3.85 x 5 = 19.25 kcal/min. The intensity is close to maximal aerobic power, so lactate accumulates quickly and glycogen is depleted within minutes. (4 marks) Marking notes: Karvonen must use heart-rate reserve; using 0.7 x 200 directly loses marks.
3	Q3 The same athlete has a resting stroke volume of 70 mL, resting heart rate of 70 bpm and arteriovenous oxygen difference of 50 mL/L. At maximal exercise stroke volume is 120 mL, heart rate 190 bpm and the arteriovenous difference is 160 mL/L. (a) Calculate cardiac output (L/min) at rest and at maximum. (b) Use the Fick equation to find oxygen uptake in both states and express each in mL/kg/min. (c) Calculate the percentage increase in stroke volume and state which factor mainly raises cardiac output in a trained athlete.	(a) Rest: 70 x 70 = 4,900 mL/min = 4.9 L/min; maximum: 120 x 190 = 22,800 mL/min = 22.8 L/min. (3 marks) (b) Fick: oxygen uptake = cardiac output x arteriovenous difference. Rest 4.9 x 50 = 245 mL/min = 3.5 mL/kg/min; maximum 22.8 x 160 = 3,648 mL/min = 52.1 mL/kg/min. (3 marks) (c) (120 – 70) / 70 x 100 = 71.4%. In trained athletes the rise comes mainly from stroke volume (larger ventricles, stronger contraction), while maximum heart rate falls slightly. (4 marks)
4	Q4 (a) In a biceps curl the muscle applies 300 N at a moment arm of 0.04 m, while an 80 N load acts at 0.30 m from the elbow. Calculate both moments and decide whether the load can be lifted. (b) What force must the muscle produce to lift that load, and what is the mechanical advantage (effort arm / load arm)? (c) An athlete throws a javelin at 12 m/s at 45 degrees (g = 9.81 m/s², ignore air resistance); find the horizontal range, maximum height and time of flight.	(a) Muscle moment = 300 x 0.04 = 12 N m; load moment = 80 x 0.30 = 24 N m; the moment is insufficient, so the load cannot be lifted. (3 marks) (b) Required force = 24 / 0.04 = 600 N; mechanical advantage = 0.04 / 0.30 = 0.133. This is a third-class lever: mechanically disadvantaged but fast with a large range of movement. (2 marks) (c) Range = 12 squared x sin 90 / 9.81 = 14.7 m; maximum height = (12 x sin 45) squared / (2 x 9.81) = 3.67 m; flight time = 2 x 12 x sin 45 / 9.81 = 1.73 s. (3 marks) Marking notes: moment unit must be N m; range and height must be in m.

Section B Muscle, thermoregulation and body composition (Q5–Q7, 30 marks)

Explain each answer with the underlying physiological mechanism; naming alone is not enough.

#	Question	Answer and marking notes
5	Q5 A rower performs 60 s of maximal effort and peak force falls from 620 N to 480 N. (a) Calculate the percentage drop in force and give two peripheral causes of fatigue. (b) Compare type I and type IIx fibres in contraction speed, fatigue resistance and main energy system (one point each). (c) Explain why the lactate threshold (about 4 mmol/L) predicts endurance performance better than VO2max, and why lactate is not the only cause of fatigue.	(a) (620 – 480) / 620 x 100 = 22.6%. (2 marks) Peripheral causes: accumulation of inorganic phosphate and hydrogen ions inhibiting myosin cross-bridge cycling; glycogen depletion; impaired calcium release and reuptake (any two, 1 mark). (b) Type I: slow contraction, high fatigue resistance, mainly aerobic; type IIx: fast contraction, easily fatigued, mainly phosphagen and glycolysis. (3 marks) (c) The lactate threshold reflects the ability to clear lactate at an intensity that can still be sustained and maps directly onto race pace, whereas VO2max is only a ceiling that may not be usable. Lactate rises with hydrogen ions, lowering pH and inhibiting phosphofructokinase and calcium sensitivity, but inorganic phosphate, ADP and glycogen depletion also limit contraction, so fatigue cannot be blamed on lactate alone. (4 marks) Marking notes: circular statements such as lactate makes the muscle acidic earn no credit.
6	Q6 A 65 kg player competes for 90 minutes at 32 degrees C and 70% relative humidity. Body mass falls from 65.0 kg to 63.8 kg while only 0.4 L of fluid is drunk. (a) Find the net mass loss (kg) and its percentage of body mass; if endurance falls by about 2% per 1% mass lost, estimate the performance loss. (b) Give two reasons why heat loss is less effective in humid conditions and suggest two practical cooling or hydration measures. (c) If power output is 240 W at 20% mechanical efficiency, find the heat production rate; taking 2,430 kJ evaporated per litre of sweat, how much sweat per hour is needed for thermal balance?	(a) Loss = 1.2 kg, net loss = 1.2 – 0.4 = 0.8 kg; 0.8 / 65 x 100 = 1.23%; performance loss about 1.23 x 2 = 2.5%. (3 marks) (b) High humidity reduces the vapour pressure gradient so sweat evaporates poorly and evaporative cooling fails; skin blood flow competes with working muscle for blood, lowering oxygen delivery. Measures: small drinks with electrolytes every 15-20 minutes, ice towels or cooling vests, shorter warm-up, shaded breaks. (3 marks) (c) Metabolic rate = 240 / 0.20 = 1,200 W, so heat = 1,200 – 240 = 960 W = 0.96 kJ/s; per hour 0.96 x 3,600 = 3,456 kJ; / 2,430 = 1.42 L/h. (4 marks) Marking notes: sweat volume and net loss must be distinguished; drink more water alone earns no credit.
7	Q7 A 20-year-old male athlete is 72 kg and 1.78 m tall with a three-site skinfold sum of 48 mm. Use D = 1.10938 – 0.0008267S + 0.0000016S squared – 0.0002574A (S in mm, A in years) and the Siri equation body fat % = 495 / D – 450. (a) Calculate BMI and body fat percentage. (b) Find fat mass and fat-free mass. (c) If the target is 10% body fat with fat-free mass unchanged, find the target body mass and the fat to be lost; explain why BMI alone misjudges athletes.	(a) BMI = 72 / 1.78 squared = 22.7 kg/m2; D = 1.10938 – 0.0396816 + 0.0036864 – 0.005148 = 1.0682; body fat = 495 / 1.0682 – 450 = 13.4%. (4 marks) (b) Fat mass = 0.1338 x 72 = 9.6 kg; fat-free mass = 72 – 9.6 = 62.4 kg. (2 marks) (c) Target mass = 62.4 / 0.90 = 69.3 kg, so 72 – 69.3 = 2.7 kg of fat must be lost. (2 marks) BMI uses only mass and height, so muscular athletes score high while their fat is normal; skinfolds and girths show composition, so athletes should be judged on body composition, not BMI. (2 marks)

Section C Skill learning, injury and nutrition (Q8–Q10, 30 marks)

Support each answer with data or training principles and a sport-specific example.

#	Question	Answer and marking notes
8	Q8 A beginner practises basketball free throws in five blocks of 10 attempts and scores 4, 5, 6, 7 and 8. (a) Calculate the success rate of each block and the percentage-point and relative improvement from block one to block five. (b) Describe the three Fitts and Posner stages of skill learning and identify the stage this player is most likely in. (c) A retention test one week later gives 70%; find the gap from block five, and compare the roles of knowledge of results and knowledge of performance.	(a) 40%, 50%, 60%, 70%, 80%; improvement 80 – 40 = 40 percentage points; relative improvement 40 / 40 x 100 = 100%. (3 marks) (b) Cognitive stage: heavy verbal guidance, stiff movements, many errors; associative stage: coordination improves, errors reduce, self-correction begins; autonomous stage: movement is automatic and tactics can be processed at the same time. This player is still refining technique and needs cues, so the associative stage is most likely. (4 marks) (c) Gap = 80 – 70 = 10 percentage points; part of the gain may be day-to-day form rather than learning, and only retention tests reveal real learning. Knowledge of results (scored or not) gives outcome information; knowledge of performance (release angle, follow-through) gives movement information. Beginners need both, but knowledge of performance matters more for technique. (3 marks)
9	Q9 A runner totals 3,200 AU of training load (RPE x minutes) in the last 7 days, against a 28-day average of 2,400 AU per week; one session is RPE 7 for 90 minutes. (a) Calculate that session load and the acute:chronic workload ratio (ACWR). (b) Given that 0.8-1.3 is the safer ACWR band, find the upper limit of chronic load at an acute load of 3,200 AU and comment on this runner. (c) Explain the mechanisms of two common running injuries and list three objective criteria for return to competition after rehabilitation.	(a) Session load = 7 x 90 = 630 AU; ACWR = 3,200 / 2,400 = 1.33. (3 marks) (b) Chronic upper limit = 3,200 / 1.3 = 2,462 AU. At 2,400 AU the ratio of 1.33 sits just above the safe band, a yellow flag for a rapid load spike: injury risk rises, so acute load should be held or trimmed and low-intensity recovery sessions added. (4 marks) (c) Medial tibial stress syndrome: repeated impact and plantar flexion load the posteromedial tibia, irritating periosteum and flexor attachments; iliotibial band syndrome: the band slides over the lateral femoral epicondyle during knee flexion and extension, with friction and weak hip abductors raising pressure. Return criteria: pain 0-2/10 and absent while running, strength at least 90% of the uninjured side, and sport-specific tasks such as single-leg hops and cutting performed pain-free and symmetrically. (3 marks)
10	Q10 A 70 kg athlete carbohydrate-loads with a target of 8 g/kg carbohydrate and 1.6 g/kg protein per day and 3,000 kcal in total (4 kcal/g for carbohydrate and protein, 9 kcal/g for fat). (a) Calculate the grams and energy supplied by carbohydrate and protein. (b) Find the grams of fat and the percentage of energy from each macronutrient (they must total 100%). (c) Explain the physiological role of a high-carbohydrate meal 3 hours before competition, and why large amounts of fat or fibre are unwise then.	(a) Carbohydrate = 8 x 70 = 560 g, giving 2,240 kcal; protein = 1.6 x 70 = 112 g, giving 448 kcal. (4 marks) (b) Fat energy = 3,000 – 2,240 – 448 = 312 kcal, i.e. 34.7 g; percentages 74.7%, 14.9% and 10.4%, totalling 100%. (3 marks) (c) A high-carbohydrate meal restores liver and muscle glycogen and stabilises blood glucose, so fuel supply is adequate early in the game. High-fat food empties from the stomach slowly and can cause bloating and nausea; high fibre also slows emptying and increases gut motility, which may cause gastrointestinal upset during play. (3 marks) Marking notes: percentages must total 100%; eating carbohydrate is good without calculation earns no credit.

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