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Years 9–10 Pūtaiao / Science

Physics • Years 9–10 • Practice bank

Physics Problem Bank: Forces and Motion

Use these questions for retrieval practice, worked examples, homework, small-group coaching or extension. The emphasis is on showing a physical model, choosing the right relationship, carrying units, and explaining what the answer means.

Best for

Years 9–10 forces and motion, revision, mixed-ability workshops and substitute-ready practice.

Kaiako use

Select a short set rather than assigning everything. Ask students to annotate diagrams and units before calculating.

Ākonga use

Write the relationship, substitute values with units, calculate, then explain the physical meaning in one sentence.

Convention

Unless your kaiako specifies otherwise, use g = 9.8 N/kg near Earth’s surface and give sensible significant figures.

Relationship bank

Speed: v = d/t   •   Acceleration: a = Δv/t   •   Net force: F = ma   •   Weight: W = mg   •   Momentum: p = mv

Rule: Do not choose a formula just because its letters appear in the question. First decide what physical change is happening.

A. Forces and free-body reasoning

1. Pushing a trolley

Net forceExplain + calculate

A 12 kg trolley is pushed forward with 42 N. Friction acts backward with 18 N.

  1. Calculate the net force.
  2. Calculate the trolley’s acceleration.
  3. Explain why using 42 N directly in F = ma would be wrong.

2. Constant speed

A cyclist travels along a flat road at constant speed. The forward driving force is 95 N.

  1. What is the net horizontal force?
  2. What total resistive force acts backward?
  3. Explain your answer using Newton’s first law.

3. Waka moving through calm water

A paddling crew produces an average forward thrust of 620 N while drag is 470 N. The loaded waka and crew have a total mass of 500 kg.

  1. Find the net horizontal force.
  2. Find the acceleration.
  3. Predict what happens to the acceleration if drag rises while thrust stays the same.

B. Mass, weight and acceleration

4. Mass is not weight

A field pack has a mass of 7.5 kg.

  1. Calculate its weight on Earth using g = 9.8 N/kg.
  2. State the mass if the pack is taken to the Moon.
  3. Explain why the mass stays the same even though its weight changes.

5. Reading the units

Three students give the following answers for the weight of a 60 kg person: A: 588 kg, B: 588 N, C: 6.12 N.

  1. Which answer is physically plausible?
  2. Identify the unit error in answer A.
  3. Show the calculation that supports your choice.

C. Motion and graph thinking

6. Sprint data

A runner increases speed from 2.0 m/s to 8.0 m/s in 3.0 s.

  1. Calculate the change in velocity.
  2. Calculate the average acceleration.
  3. What additional information would you need to calculate the net force on the runner?

7. Describe before calculating

A velocity–time graph is horizontal at 6 m/s from 0–4 s, rises evenly to 12 m/s from 4–7 s, then is horizontal at 12 m/s from 7–10 s.

  1. Describe the motion in each interval.
  2. Calculate the acceleration from 4–7 s.
  3. Which intervals have zero acceleration?

D. Momentum and collision reasoning

8. Same speed, different momentum

A 0.15 kg ball and a 1.2 kg medicine ball each move at 4.0 m/s.

  1. Calculate the momentum of each.
  2. Which is harder to stop? Use momentum in your explanation.

9. Change in momentum

A 0.20 kg ball moving east at 10 m/s is brought to rest.

  1. Calculate its initial momentum.
  2. Calculate the change in momentum, taking east as positive.
  3. Explain why the sign of the change matters.

E. Challenge: diagnose the reasoning

10. Three claims

For each statement, write agree or disagree, then justify it with physics.

  1. “If an object is moving, there must be a net force in the direction of motion.”
  2. “A heavier object always falls faster because it has more weight.”
  3. “If forces are balanced, an object can still be moving.”

Kaiako answer key

  1. Net 24 N forward; a = 24/12 = 2.0 m/s². The applied force is not the net force because friction opposes motion.
  2. Net force 0 N; resistive force 95 N backward. Constant velocity means zero acceleration, so forces balance.
  3. Net 150 N forward; a = 150/500 = 0.30 m/s². More drag with unchanged thrust reduces net force and therefore acceleration.
  4. W = 7.5 × 9.8 = 73.5 N; mass remains 7.5 kg. Mass measures matter; weight is gravitational force.
  5. B; weight is measured in newtons. 60 × 9.8 = 588 N.
  6. Δv = 6.0 m/s; a = 6/3 = 2.0 m/s². Need the runner’s mass for F = ma.
  7. 0–4 s constant 6 m/s; 4–7 s uniformly accelerating; 7–10 s constant 12 m/s. a = (12−6)/3 = 2.0 m/s². Zero acceleration in 0–4 s and 7–10 s.
  8. Ball: 0.15×4 = 0.60 kg·m/s. Medicine ball: 1.2×4 = 4.8 kg·m/s. The medicine ball has greater momentum at the same speed.
  9. Initial p = 0.20×10 = +2.0 kg·m/s; final 0; Δp = −2.0 kg·m/s. The negative sign shows the change is opposite the chosen positive direction.
  10. 1 disagree: constant-velocity motion needs no net force. 2 disagree: in the same gravitational field the acceleration due to gravity is the same when air resistance is neglected. 3 agree: balanced forces mean zero acceleration, not necessarily zero velocity.

Teaching note: Accept equivalent reasoning. The quality target is model → relationship → units → physical interpretation, not answer-only arithmetic.