IB MYP Sciences – Forces Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers forces for IB MYP Sciences. It is aligned to the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014, which lists forces among the topics explored in the MYP sciences on-screen examinations. It suits MYP years 4 and 5; there is no SL/HL split. MYP has no prescribed content list – schools design their own units – and this set covers a topic the brief names.

Each question is labelled with the criterion it mainly trains. Real MYP work is judged against criterion level descriptors (levels 1–8), so the [1] points below are a revision aid, not IB marks.

Learn the content first with the forces study guide. For investigation method, see investigation skills exam preparation.

Use g = 9.8 N/kg on Earth unless a question says otherwise. Give answers to 3 significant figures where they are not exact.

Questions

1. (Criterion A) Classify each of these forces as a contact force or a non-contact force: friction, magnetic force, tension, weight. [2]

2. (Criterion A) A rock sample has a mass of 5.0 kg. Take g on Mars as 3.7 N/kg.

(a) Calculate the weight of the rock on Earth. [1] (b) Calculate the weight of the rock on Mars. [1] (c) State the mass of the rock on Mars. [1]

3. (Criterion A) A skydiver of weight 750 N is falling at a steady speed. The air resistance on her is 750 N. Explain, using Newton’s first law, why her speed is not changing, and name this speed. [3]

4. (Criterion A) A model rocket of mass 2.0 kg is launched straight up. The engine gives an upward thrust of 35 N. Ignore air resistance. Calculate the rocket’s acceleration at launch. [4]

5. (Criterion A) A bus starts from rest and accelerates steadily to 15 m/s in 10 s. It travels at 15 m/s for 20 s, then slows steadily to rest in 6 s.

(a) Calculate the acceleration during the first 10 s. [2] (b) Use the area under the velocity-time graph to find the total distance travelled. [3] (c) Calculate the average speed for the whole journey. [1]

6. (Criterion A) A tractor has a weight of 48 000 N. Each of its four tyres touches the ground over an area of 0.30 m².

(a) Calculate the pressure the tractor exerts on the ground. [2] (b) A farmer fits wider tyres, giving a total contact area of 1.6 m². Calculate the new pressure and explain why this helps on soft, wet fields. [2]

7. (Criterion A) A gardener lifts the handles of a wheelbarrow. The wheel axle is the pivot. The load of 600 N acts 0.40 m from the axle. The gardener lifts at right angles to the handles, 1.5 m from the axle. Ignore the weight of the barrow.

(a) Calculate the upward force the gardener must apply to just lift the load. [3] (b) Explain why longer handles would make lifting easier. [1]

8. (Criterion B) Plan an investigation into how the area of a parachute canopy affects the terminal velocity of a small falling package. Include a hypothesis with a reason, the variables, a method, the range and number of readings, and one safety point. [7]

9. (Criterion C) A student pulled a 0.80 kg trolley along a bench with different forces and measured its acceleration. These results are fictional.

Force / N	Trial 1 a / m/s²	Trial 2 a / m/s²	Trial 3 a / m/s²
0.40	0.42	0.44	0.43
0.80	0.93	0.91	0.95
1.20	1.41	1.45	1.08
1.60	1.90	1.92	1.94
2.00	2.40	2.43	2.37

The means for 0.40, 0.80, 1.60 and 2.00 N are 0.43, 0.93, 1.92 and 2.40 m/s².

(a) Identify the anomalous result and calculate the mean acceleration for 1.20 N. [2] (b) Using the mean values at 0.40 N and 2.00 N, calculate the gradient of the acceleration-force graph. [2] (c) Use your gradient to estimate the mass of the trolley, and compare it with 0.80 kg. [2] (d) The line of best fit does not pass through the origin. It cuts the force axis at about 0.05 N. Explain this and suggest one improvement to the method. [2]

10. (Criterion A) A swimmer pushes water backwards with her hands and moves forwards. Use Newton’s third law to explain how she moves. [3]

11. (Criterion D) A 70 kg passenger is travelling at 14 m/s when the car crashes. Without a seat belt, airbag and crumple zone, the passenger is stopped in 0.10 s by hitting the dashboard. With all three, the stopping time is 0.35 s.

Calculate the average force on the passenger in each case, then evaluate these safety features, including one limitation and one economic or ethical factor, and reach a conclusion. [10]

Answers

1. Friction and tension are contact forces [1]. Magnetic force and weight are non-contact forces [1]. [2] Examiner insight: One mark covers each pair, so one misplaced force loses a whole mark.

2. (a) W = mg = 5.0 × 9.8 = 49 N [1] (b) W = 5.0 × 3.7 = 18.5 N [1] (c) 5.0 kg – mass does not change with location [1] Examiner insight: Mass in newtons, or weight in kilograms, earns nothing even with the right number.

3. The upward air resistance equals her 750 N weight, so the resultant force is zero [1]. By Newton’s first law, with no resultant force her velocity stays constant, so she neither speeds up nor slows down [1]. This speed is her terminal velocity [1]. Examiner insight: “No forces act on her” is wrong and scores nothing; name a zero resultant force and link it to constant velocity.

4. Weight = 2.0 × 9.8 = 19.6 N [1] Resultant force = 35 − 19.6 = 15.4 N upwards [1] a = F / m = 15.4 / 2.0 [1] a = 7.7 m/s² upwards [1] Examiner insight: Using 35 N as F gives 17.5 m/s², which loses the resultant-force mark and the final mark, though the F = ma method mark can still be earned.

5. (a) a = (v − u) / t = (15 − 0) / 10 [1] = 1.5 m/s² [1] (b) Accelerating: ½ × 10 × 15 = 75 m; steady: 15 × 20 = 300 m [1]; slowing: ½ × 6 × 15 = 45 m [1]; total = 420 m [1] (c) Average speed = 420 / 36 = 11.7 m/s [1] Examiner insight: A wrong total from (b) can still earn the (c) mark if the division is shown.

6. (a) Total area = 4 × 0.30 = 1.2 m² [1]; p = 48 000 / 1.2 = 40 000 Pa [1] (b) p = 48 000 / 1.6 = 30 000 Pa [1]. The weight is spread over a larger area, so lower pressure and less sinking [1]. Examiner insight: Dividing by 0.30 m² instead of the total area gives 160 000 Pa and loses the first mark; always check how many contact points share the weight.

7. (a) Moment of load = 600 × 0.40 = 240 N m [1]. Principle of moments: F × 1.5 = 240 [1]. F = 160 N [1] (b) A larger distance from the pivot means a smaller force is needed to make the same 240 N m moment [1]. Examiner insight: The answer “160 N” with no moment equation shown loses the method marks, so write clockwise = anticlockwise before solving.

8. Hypothesis: as canopy area increases, terminal velocity decreases [1], because a larger area gives more air resistance, so drag equals weight at a lower speed [1]. Variables: independent – canopy area; dependent – terminal velocity [1]. Controls: same package mass, same canopy material and shape, same drop height, same string length [1]. Method: drop from a fixed height, use two light gates (or video against a metre rule) near the bottom, and find speed = distance / time where speed has stopped changing [1]. Range and repeats: at least five areas (for example 100 to 900 cm²), three drops each, then find the mean [1]. Safety: drop from a stable platform, not a chair, with the area below kept clear [1]. Examiner insight: A hypothesis whose reason is not physics (drag balancing weight) stays in the lower bands.

9. (a) The anomaly is 1.08 m/s² (trial 3 at 1.20 N) [1]. Mean of the other two = (1.41 + 1.45) / 2 = 1.43 m/s² [1] (b) Gradient = (2.40 − 0.43) / (2.00 − 0.40) = 1.97 / 1.60 [1] = 1.23 m/s² per N (kg⁻¹) [1] (c) Since a = F / m, gradient = 1/m, so m = 1 / 1.23125 = 0.812 kg [1]. This is about 1.5% above 0.80 kg, so the results agree well with Newton’s second law [1]. (d) About 0.05 N of the pulling force is used to overcome friction, so the resultant force is less than the applied force [1]. Improvement: tilt the bench until the trolley rolls at constant speed with no pull, to compensate for friction [1]. Examiner insight: Including the anomaly in the mean (giving 1.31 m/s²) loses the mark; say which value you left out and why.

10. The swimmer’s hands exert a backward force on the water [1]. By Newton’s third law, the water exerts an equal force on the swimmer in the opposite direction, forwards [1]. The forces act on different objects, so they do not cancel; the forward force moves her [1]. Examiner insight: Saying the pair “cancels out” loses the final mark.

11. Without safety features: a = 14 / 0.10 = 140 m/s² [1]; F = 70 × 140 = 9800 N [1]. With safety features: a = 14 / 0.35 = 40 m/s² [1]; F = 70 × 40 = 2800 N [1]. Explanation: a longer stopping time means a smaller deceleration, so by F = ma a force 3.5 times smaller [1]. Benefit: a smaller force, spread over belt and airbag, means fewer serious injuries [1]. Limitation: the features work only if used properly – a seat belt that is not worn gives no protection [1]. Limitation: airbags inflate with great force, so a rear-facing child seat must not be placed in front of an active airbag [1]. Economic or ethical factor: the features add cost, which may put safer cars out of reach of some buyers, but requiring them by law protects everyone [1]. Conclusion: the large cut in force means the benefits outweigh the costs, provided belts are worn and children seated correctly [1]. Examiner insight: Calculation alone stays low in Criterion D; higher bands need benefits, limitations, a factor beyond science and a reasoned judgement.

Where marks are usually lost

• Using the thrust, not the resultant force, in F = ma.
• Forgetting to multiply by the number of tyres when finding the area.
• Adding heights of a velocity-time graph instead of areas.
• Taking average speed as the mean of two speeds.
• Including an anomalous value in a mean without comment.
• Describing a third-law pair as acting on the same object.
• A hypothesis with no physics in its reason.
• A Criterion D answer with calculations but no evaluation.

Next steps

• Forces revision notes
• Forces study guide
• Criteria in practice questions
• Visit the IB MYP Sciences course hub.
• Print the IB MYP Sciences checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014. The brief lists forces among the on-screen examination topics and names the four criteria used in these labels.