OxfordAQA International GCSE Physics: Terminal velocity, centre of mass, moments and levers (9203) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.1.6 to 3.1.8 of the OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards. Use gravitational field strength 10 N/kg where a value is needed. Learn the content first from the study guide and the revision notes. Course pages: the OxfordAQA Physics hub and the checklist. More forces practice is in Forces and Their Effects practice.

Questions

1. State what is meant by terminal velocity, and give the size of the resultant force on an object that is moving at its terminal velocity. [2]

2. A skydiver of mass 62 kg is falling.

(a) Calculate the skydiver’s weight. [1] (b) At one instant the drag is 310 N. Calculate the resultant force and the acceleration. [3] (c) State the drag when the skydiver reaches terminal velocity. [1]

3. Explain why opening a parachute reduces the terminal velocity of a falling person. [3]

4. A skydiver jumps from a plane, falls for some time, opens a parachute and then lands. Describe the velocity-time graph for the fall up to landing, and state how drag compares with weight in each stage. [6]

5. A model load of mass 0.050 kg falls through still air attached to a small parachute. Ignore upthrust. Its speed is measured:

Time (s)	0	0.1	0.2	0.3	0.4	0.5
Speed (m/s)	0	0.8	1.4	1.7	1.8	1.8

(a) State the terminal velocity. [1] (b) Calculate the weight of the load and state the drag at terminal velocity. [2] (c) Calculate the average drag between 0.2 s and 0.3 s. [3]

6. Explain how the shape of a shark and the shape of a car each reduce drag. [4]

7. Describe how you would find the centre of mass of a thin piece of card with an irregular shape. [5]

8. (a) State where the weight of an object can be considered to act. [1] (b) A van is carrying heavy boxes. Explain why it is more likely to topple on a sloping road when the boxes are on the roof than when they are on the floor. [3]

9. A door is pushed with a force of 60 N at right angles to the door, 0.90 m from the hinges.

(a) Calculate the moment of the force about the hinges. [2] (b) The same moment is needed by a force acting 0.15 m from the hinges. Calculate that force. [2]

10. A uniform plank 3.0 m long is balanced on a pivot at its centre. A load of 200 N is placed 1.2 m to the left of the pivot and a load of 150 N is placed 0.50 m to the left of the pivot. A force F pushes down on the right, 1.4 m from the pivot, to keep the plank level.

(a) Explain why the weight of the plank can be ignored. [1] (b) Calculate F. [3] (c) Calculate the upward force of the pivot on the plank. [2]

11. A wheelbarrow carries a load of weight 360 N, which acts 0.60 m from the axle of the wheel. The hands lift 1.5 m from the axle.

(a) Calculate the upward force needed from the hands to hold the barrow level. [3] (b) Explain why the wheelbarrow is called a force multiplier, and state what is given up in exchange. [2]

12. A cabinet weighs 600 N and is 0.80 m wide. Its centre of mass is at its middle. It stands on a floor and is pushed horizontally, 1.5 m above the floor.

(a) Explain why the cabinet topples once its weight acts outside its base. [2] (b) Calculate the smallest push that makes it start to tip about its lower edge. [4]

Answers

1. Terminal velocity is the constant velocity reached by a falling object when drag has increased until it equals the driving force (weight) [1]. The resultant force is zero [1]. [2] Examiner insight: answers that say the object has “no forces” on it tend to lose credit; the forces are present but balanced.

2. (a) W = 62 × 10 = 620 N [1] (b) Resultant = 620 − 310 = 310 N downwards [1]; a = 310 ÷ 62 [1] = 5.0 m/s² [1] (c) Drag equals weight = 620 N [1] Examiner insight: write the equation and the substitution, so that your method is clear even if you slip on the final number.

3. The parachute has a large area, so at any given speed the drag is much greater [1]. The drag balances the weight at a lower speed [1]. The weight is unchanged, so the terminal velocity is lower [1]. [3] Examiner insight: “reduces gravity” or “reduces weight” is a common wrong statement; link the larger area to larger drag at a given speed.

4. Falling with the parachute closed: the line is a curve that gets less steep as drag grows, with drag less than weight [1]. The line then levels at a constant high speed, drag equal to weight [1]. After the parachute opens: speed falls steeply, drag greater than weight [1]. The gradient then reduces as drag falls with speed [1]. The line levels again at a much lower speed [1] with drag again equal to weight [1]. [6] Examiner insight: name the forces at each stage, not just the shape of the line; the shape and the force comparison are different points.

5. (a) 1.8 m/s [1] (b) W = 0.050 × 10 = 0.50 N [1]; drag at terminal velocity = 0.50 N [1] (c) a = (1.7 − 1.4) ÷ 0.1 = 3.0 m/s² [1]; resultant = 0.050 × 3.0 = 0.15 N [1]; drag = 0.50 − 0.15 = 0.35 N [1] Examiner insight: the drag is the weight minus the resultant, not the resultant on its own; showing the resultant first makes the logic clear.

6. Shark: a smooth, tapered body [1] lets water flow smoothly past, giving little drag [1]. Car: a smooth, curved body with a sloping front [1] lets air flow smoothly, so drag is reduced (higher top speed or less fuel used) [1]. [4] Examiner insight: a mark for the shape and a mark for its effect are separate; saying only “it is streamlined” repeats the question.

7. Make several small holes near the edge of the card [1]. Hang the card from a pin through one hole so it swings freely [1]. Hang a plumb line from the same pin and mark where it lies on the card [1]. Draw a line through the marks [1]. Repeat with other holes; the lines cross at the centre of mass [1]. [5] Examiner insight: the plumb line and the repeat from a second hole are the details that are easy to leave out; mention both.

8. (a) At the centre of mass [1] (b) Boxes on the roof raise the centre of mass [1]. On a slope, the line of action of the weight reaches the edge of the wheels at a smaller tilt [1]. Once the line is outside the base there is a resultant moment that topples the van [1]. [3] Examiner insight: explain through the position of the line of action of the weight relative to the base; “it is heavier at the top” does not explain the toppling.

9. (a) M = F × d [1]; 60 × 0.90 = 54 N m [1] (b) F = 54 ÷ 0.15 [1] = 360 N [1] Examiner insight: the unit N m is part of the answer; N or N/m after a moment is an error.

10. (a) Its weight acts at the centre, which is the pivot, so it has no moment [1] (b) Anticlockwise moments = 200 × 1.2 + 150 × 0.50 = 240 + 75 = 315 N m [1]; clockwise moment = F × 1.4 [1]; F = 315 ÷ 1.4 = 225 N [1] (c) The plank is in equilibrium so the upward force equals the downward forces [1]: 200 + 150 + 225 = 575 N [1] Examiner insight: state which moments are clockwise and which anticlockwise before equating them; part (c) is often forgotten because it is not a moment calculation.

11. (a) Moment of the load about the axle = 360 × 0.60 = 216 N m [1]; 216 = F × 1.5 [1]; F = 144 N [1] (b) The effort of 144 N is smaller than the 360 N load [1]; the hands move further than the load does [1]. [2] Examiner insight: the pivot of a wheelbarrow is the axle of the wheel; choosing the wrong pivot leads to the wrong distances.

12. (a) The weight gives a moment about the tipping edge [1]; when its line of action is outside the base this moment turns the cabinet further over instead of back [1]. [2] (b) Distance of weight from the tipping edge = 0.40 m [1]; restoring moment = 600 × 0.40 = 240 N m [1]; push moment = F × 1.5 [1]; F = 240 ÷ 1.5 = 160 N [1] Examiner insight: take both moments about the same edge; a pivot taken at the centre of the cabinet misses the point of the question.

Where marks are usually lost

• Saying a skydiver at terminal velocity has no forces, or has stopped falling.
• Writing “reduces gravity” for a parachute.
• Describing a graph shape without saying what drag and weight are doing.
• Taking drag as the resultant force instead of weight minus resultant.
• Forgetting the plumb line, or hanging a lamina from only one hole.
• Explaining toppling by weight alone, not by where the line of action falls.
• Using a distance that is not perpendicular to the force, or mixing cm and m.
• Leaving out the pivot force in a balanced-plank question.
• Choosing the wrong pivot when the object tips about an edge or an axle.

Next steps

• Revision notes for this unit
• Study guide for this unit
• OxfordAQA International GCSE Physics hub
• Printable checklist
• Take any of the free 10-minute diagnostics
• Book a free trial class

Official syllabus

These questions follow the OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards, published by OxfordAQA.