AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2024: Question 2

✅ Correct answer

• As the spring decompresses, its elastic potential energy decreases.
• That energy is transferred to the gymnast, so the gymnast’s kinetic energy increases as they leave the floor.
• Alternatively: the gymnast’s gravitational potential energy increases as they rise.
• Or: the spring exerts a force on the gymnast, so work is done on the gymnast.

💡 Key knowledge

• Elastic potential energy is stored when a spring is stretched or compressed.
• When the spring returns to its natural length, energy is released.
• That released energy can become kinetic energy and then gravitational potential energy.

🧠 Exam technique

• Use clear “energy transfer” language.
• State what decreases and what increases.
• Link the spring’s action to the gymnast’s motion.

❌ Common errors

• Saying only “the spring pushes them up” without explaining energy transfer.
• Writing that elastic potential energy increases during decompression.
• Mixing up kinetic energy and gravitational potential energy.

📐 Calculation

Given: k = 8500 N/m, e = 1.2 cm = 0.012 m

1. Convert the compression into metres: 1.2 cm = 0.012 m
2. Use the equation: Eₑ = 0.5 × k × e²
3. Substitute values: Eₑ = 0.5 × 8500 × 0.012²
4. Calculate: 0.012² = 0.000144
5. Eₑ = 0.5 × 8500 × 0.000144 = 0.612 J

Answer: 0.612 J (accept 0.61 J )

💡 Key knowledge

• Use the elastic potential energy equation: Eₑ = 0.5ke²
• Compression must be in metres, not centimetres.
• The unit for energy is joule, J .

🧠 Exam technique

• Always show substitution before calculation.
• Include units at the end.
• If you round, keep it sensible: 0.612 J → 0.61 J is fine.

❌ Common errors

• Using 1.2 instead of 0.012.
• Forgetting to square the compression.
• Writing the answer in newtons instead of joules.

✅ Correct answer

• Measure the original length of the spring with a metre rule.
• Measure the compressed length of the spring with the same ruler.
• Calculate compression using: compression = original length − compressed length

Alternative method: measure the weight of the mass and use e = F / k .

💡 Key knowledge

• Compression is a change in length.
• To get a valid reading, compare the spring before and after loading.
• In practical work, a metre rule is the expected measuring tool.

🧠 Exam technique

• Give both the measurement and the calculation.
• Use “original minus compressed” in the right order.

❌ Common errors

• Measuring only one length.
• Subtracting in the wrong order.
• Not mentioning a ruler/metre rule.

✅ Correct answer

• Masses could fall off the spring.
• Doing it on the floor means they are less likely to cause injury or damage.

💡 Key knowledge

• This is a safety answer.
• GCSE practical questions often want a specific hazard and a specific reason it is safer.

🧠 Exam technique

• Include the hazard first, then the safety reason.
• “Less likely to cause injury” is enough if linked to falling masses.

❌ Common errors

• Saying “it is easier” without a safety reason.
• Referring to the table falling instead of the masses.
• Not making the danger clear.

📐 Calculation

1. Read the top value: 330 N
2. Read the bottom value: 80 N
3. Find the difference: Δy = 330 − 80 = 250 N

Answer: 250 N

🧠 Exam technique

• Use the axis values exactly as shown on the graph.
• Always include the unit.

❌ Common errors

• Reading from the wrong grid line.
• Forgetting the unit N.
• Using x-axis values for Δy.

📐 Calculation

1. Read the right-hand x-value: 0.052 m
2. Read the left-hand x-value: 0.010 m
3. Find the difference: Δx = 0.052 − 0.010 = 0.042 m

Answer: 0.042 m

🧠 Exam technique

• Compression is measured in metres here.
• Check you are using the plotted line values, not random grid squares.

❌ Common errors

• Using 0.042 as 4.2 cm without showing the conversion.
• Writing 0.52 or 0.0052 by misreading the graph.
• Leaving off the unit.

📐 Calculation

1. Use the gradient formula for a force–compression graph: k = ΔF / Δx
2. Substitute the values: k = 250 / 0.042
3. Calculate: k = 5952.38...
4. Round to 3 significant figures: 5950 N/m

Answer: 5.95 × 10³ N/m or 5950 N/m

💡 Key knowledge

• The gradient of a force–compression graph is the spring constant.
• Spring constant unit: N/m
• Use values from your answers to 02.5 and 02.6 if needed.

🧠 Exam technique

• Show the formula, substitution, and final rounded answer.
• Give the answer to 3 significant figures if asked.
• Keep the unit attached to the final answer.

❌ Common calculation traps

• Using the reciprocal: Δx / ΔF instead of ΔF / Δx .
• Forgetting the unit is N/m .
• Rounding too early.
• Using the wrong values from the graph.

💡 Top responses

• Used precise energy language: “elastic potential energy decreases” and “kinetic energy increases”.
• Showed graph readings clearly and used the correct gradient idea.
• Included units every time.

❌ Where marks were lost

• Vague statements like “it helps him jump” without explaining how.
• Incorrect unit conversions from cm to m.
• Graph answers without subtraction shown.

🧠 Quick recall checklist

• Energy changes: Eₑ → KE → GPE
• Equation: Eₑ = 0.5ke²
• Gradient = spring constant
• Units: J, N, m, N/m