S5 PE Practice Paper (Set 2): Heart Rate Zones, BMI, Torque, Nutrition, Training Load And Hydration

S5 PE Practice Paper (Set 2): Heart Rate Zones, BMI, Torque, Nutrition, Training Load and Hydration

2026-11-15

S5 PE

S5 PE paper (Set 2) with answers: heart rate zones for a 16-year-old, BMI and energy balance, torque and lever advantage, nutrient grams and hydration.

Paper details　S5 PE practice paper (Set 2) with full answers covering heart rate zones, BMI, torque, nutrient grams, training load and hydration for Hong Kong secondary students.

Heart Rate Monitoring and Training Intensity

The data below come from a resting test of a sixteen-year-old athlete: age 16 years, resting heart rate 58 beats per minute. Training intensity is set by the heart rate reserve method.

#	Question	Answer and marking notes
1	(1) Calculate the maximum heart rate by subtracting age from 220 and state the unit.	Answer: Maximum heart rate is 204 beats per minute (220 – 16 = 204).
2	(2) Use the heart rate reserve method to find the target heart rate at 60 per cent intensity; give a whole number.	Answer: 58 + 0.60 x (204 – 58) = 58 + 87.6 = 145.6, about 146 beats per minute.
3	(3) Use the same method to find the target heart rate at 80 per cent intensity; give a whole number.	Answer: 58 + 0.80 x (204 – 58) = 58 + 116.8 = 174.8, about 175 beats per minute.
4	(4) State the difference between the two intensity zones and explain how this difference is used when planning training.	Answer: The difference is 29.2 beats per minute (175 – 146). Coaches use the 60 per cent zone for aerobic base work and the 80 per cent zone for interval work, and track whether the athlete can hold the higher zone steadily.

Body Mass Index and Energy Balance

A student weighs 62.5 kg and is 1.68 m tall. World Health Organization adult BMI bands are: below 18.5 underweight, 18.5 to 24.9 normal, 25.0 to 29.9 overweight.

#	Question	Answer and marking notes
5	(1) Calculate the student body mass index; give one decimal place.	Answer: 62.5 / (1.68 x 1.68) = 62.5 / 2.8224 = 22.1.
6	(2) Use the bands above to judge the body mass status of the student.	Answer: A BMI of 22.1 lies between 18.5 and 24.9, so the student is in the normal band.
7	(3) To bring the BMI down to 21.0, what body mass should be reached and how much mass must be lost?	Answer: 21.0 x 2.8224 = 59.3 kg, so 3.2 kg must be lost (62.5 – 59.3).
8	(4) With a daily energy deficit of 500 kcal and 7700 kcal per kilogram of fat, how many days are needed?	Answer: 3.2 x 7700 = 24640 kcal; 24640 / 500 = 49.28, so about 49 days.

Torque and Lever Systems

An athlete lifts a load with the forearm: resistance 250 N, resistance arm 0.40 m, effort arm 0.20 m, and the system is in equilibrium.

#	Question	Answer and marking notes
9	(1) Calculate the torque produced by the resistance about the pivot and state the unit.	Answer: 250 x 0.40 = 100.0 newton metres.
10	(2) Calculate the effort needed to keep the system in equilibrium.	Answer: 100.0 / 0.20 = 500 N.
11	(3) Calculate the mechanical advantage of this lever.	Answer: Resistance arm divided by effort arm = 0.40 / 0.20 = 2.0.
12	(4) Identify the class of this lever and give one human movement of the same class.	Answer: It is a second class lever, with the resistance between the pivot and the effort. A human example is rising on the toes, where the ankle joint is the pivot and the calf muscles lift body weight.

Energy Systems and Sports Nutrition

A 70 kg athlete runs for 30 minutes at 12 kcal per minute; the energy sources are 60 per cent carbohydrate, 25 per cent fat and 15 per cent protein.

#	Question	Answer and marking notes
13	(1) Calculate the total energy expenditure of the session.	Answer: 12 x 30 = 360 kcal.
14	(2) Calculate the energy from carbohydrate and the equivalent grams (4 kcal per gram).	Answer: 360 x 0.60 = 216 kcal; 216 / 4 = 54 g.
15	(3) Calculate the energy from fat and the equivalent grams (9 kcal per gram).	Answer: 360 x 0.25 = 90 kcal; 90 / 9 = 10 g.
16	(4) Calculate the energy from protein and the equivalent grams, then explain why carbohydrate is replaced after long sessions.	Answer: 360 x 0.15 = 54 kcal; 54 / 4 = 13.5 g. Replacing carbohydrate after long sessions restores muscle glycogen, speeds recovery and limits the drop in performance in the next session.

Progressive Overload and Training Volume

An athlete has a one repetition maximum of 80 kg. The coach plans four strength sessions a week, starting at 70 per cent of the maximum and adding 2.5 per cent each week.

#	Question	Answer and marking notes
17	(1) Calculate the starting training load.	Answer: 80 x 0.70 = 56 kg.
18	(2) With 3 sets of 8 repetitions per session, calculate the training volume in kilograms.	Answer: 56 x 8 x 3 = 1344 kg.
19	(3) Adding 2.5 per cent each week, what is the training load in week four?	Answer: 56 x 1.025 to the fourth power = 61.8 kg, to one decimal place.
20	(4) Explain why the weekly increase in load should not be too large.	Answer: A large jump makes tendons and joints adapt more slowly than muscle, raising the risk of overuse injury. Small steps allow volume to build steadily while keeping movement quality.

Hydration and Injury Prevention

A 62.5 kg athlete trains football for 90 minutes. Sweat rate is 1.2 litres per hour and 0.9 litre of sports drink is consumed during the session.

#	Question	Answer and marking notes
21	(1) Calculate the total sweat loss of the session.	Answer: 1.2 x 1.5 = 1.8 litres.
22	(2) Calculate the net fluid deficit after the session.	Answer: 1.8 – 0.9 = 0.9 litre.
23	(3) Calculate the body mass loss as a percentage; give two decimal places.	Answer: 0.9 / 62.5 x 100 = 1.44 per cent.
24	(4) Give two practical suggestions on post-session rehydration and heat illness prevention.	Answer: Replace each litre lost with 1.2 to 1.5 litres of electrolyte drink in small doses, cool down in shade, watch for dizziness and nausea, and suspend high intensity work if body mass loss exceeds two per cent.

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