OxfordAQA International GCSE Physics: Terminal velocity, centre of mass, moments and levers (9203)

This guide covers the OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards. It covers sections 3.1.6 (Forces and terminal velocity), 3.1.7 (Centre of mass) and 3.1.8 (Moments and levers). The qualification is not tiered.

Use it with the revision notes and the practice questions. The wider topic is in Forces and their effects, and the course pages are the OxfordAQA Physics hub and the printable checklist. If you are unsure where you stand, take a free diagnostic.

What this unit covers

Section	What you must be able to do
3.1.6 a	Explain that drag on an object in a fluid grows as its speed grows
3.1.6 b	Describe how a falling object reaches terminal velocity
3.1.6 c	Explain what a parachute does; draw and interpret velocity-time graphs with the forces
3.1.6 d	Explain how streamlining reduces drag, using a shark and a car
3.1.7 a-c	Define centre of mass; find it for an irregular lamina; link it to stability
3.1.8 a	Calculate a moment using M = F × d
3.1.8 b	Use the principle of moments to find a force or a distance
3.1.8 c	Explain toppling for vehicles and balancing toys
3.1.8 d	Explain how a simple lever acts as a force multiplier

Drag and terminal velocity (3.1.6 a-b)

A fluid is a liquid or a gas. An object moving through a fluid pushes fluid out of its way, and the fluid pushes back. This frictional force is drag. It acts opposite to the direction of motion. The faster the object moves, the greater the drag.

Take a skydiver leaving a plane. Weight acts down and drag acts up.

1. At the start the speed is zero, so drag is zero. The resultant force is the whole weight and the acceleration is at its largest.
2. As speed rises, drag rises. The resultant force (weight minus drag) gets smaller, so the acceleration gets smaller. She is still speeding up, but less quickly.
3. Eventually drag equals weight. The resultant force is zero, so there is no acceleration. The speed stays constant: the terminal velocity.

Use resultant force = mass × acceleration, with gravitational field strength 10 N/kg.

Worked example 1. A skydiver of mass 80 kg falls from rest. Find the acceleration (a) at the instant of leaving, and (b) when drag is 320 N. (c) State the drag at terminal velocity.

• Weight = 80 × 10 = 800 N.
• (a) Drag is 0, so resultant = 800 N and a = 800 ÷ 80 = 10 m/s².
• (b) Resultant = 800 − 320 = 480 N, so a = 480 ÷ 80 = 6.0 m/s².
• (c) Resultant is zero, so drag = weight = 800 N.

Parachutes and velocity-time graphs (3.1.6 c)

A parachute has a very large area, so at any given speed it produces far more drag than a body alone. The parachutist reaches terminal velocity at a much lower speed, which allows a safe landing.

Worked example 2. The 80 kg skydiver opens a parachute while falling fast. At that instant drag is 2400 N. Describe what happens next.

• Resultant = 2400 − 800 = 1600 N upwards, so a = 1600 ÷ 80 = 20 m/s² upwards: the skydiver slows sharply.
• As speed falls, drag falls. When drag is back to 800 N the forces balance and a new, much lower, terminal velocity is reached.

Reading a velocity-time graph

• The gradient is the acceleration.
• A horizontal line means constant velocity and zero resultant force.
• A curve that flattens means acceleration is falling because drag is rising.

For a skydiver who opens a parachute the graph has four parts:

Stage	Shape of line	Forces
Falling, parachute closed	Steep, then curving over	Weight bigger than drag, resultant shrinking
First terminal velocity	Horizontal, high	Weight = drag
Parachute opens	Line drops steeply	Drag much bigger than weight
Second terminal velocity	Horizontal, low	Weight = drag again

Worked example 3. A skydiver of mass 75 kg has weight 750 N. Her speed is recorded:

Time (s)	0	2	4	6	8	10	12
Speed (m/s)	0	14	23	28	30	30	30

(a) Find the acceleration between 0 and 2 s. (b) Find the average drag between 6 and 8 s. (c) State the terminal velocity and the drag at 11 s.

• (a) a = (14 − 0) ÷ 2 = 7.0 m/s².
• (b) a = (30 − 28) ÷ 2 = 1.0 m/s². Resultant = 75 × 1.0 = 75 N. Drag = 750 − 75 = 675 N (an average over the interval).
• (c) The speed stops changing from 8 s, so the terminal velocity is 30 m/s. Drag equals weight, 750 N.

Streamlining (3.1.6 d)

Streamlining shapes an object so fluid flows smoothly past it. Less drag means a higher maximum speed.

• Shark. The body is smooth and tapered at both ends, and the skin is smooth. Water flows past with little disturbance, so the drag is small. This is an adaptation that lets the shark swim fast with less energy.
• Car. The body is smooth and curved, with a sloping bonnet and windscreen and a low front. Air flows over it more smoothly, so drag is smaller. A roof box or open window breaks up the flow and increases drag. This is a design feature.

In an answer, say that the fluid flows smoothly, that drag is reduced, and what follows (higher maximum speed, or less energy used).

Centre of mass (3.1.7)

The centre of mass is the point at which the mass of an object may be thought to be concentrated. The weight can be treated as acting at this point. The centre of mass can lie outside the material: for a ring it is at the middle of the hole.

• A symmetrical object has its centre of mass on the axis of symmetry. If there are two axes, it is where they cross.
• A freely suspended object comes to rest with its centre of mass directly below the point of suspension.

Finding the centre of mass of an irregular thin lamina

1. Make three small holes near the edge of the card.
2. Hang the card from a pin held in a clamp through one hole, so it can swing freely.
3. Hang a plumb line (a thread with a weight) from the same pin. When everything is still, mark two points on the card along the thread.
4. Remove the card and draw a line through the two marks.
5. Repeat for the other holes. The lines cross at the centre of mass.

Why it works: a freely hanging card has its centre of mass directly below the pin, so the centre of mass lies on the vertical line. It must lie on every such line. The third line is a check. Keep the pin loose so friction does not hold the card in the wrong position.

Stability

The position of the centre of mass affects stability. An object is more stable with a low centre of mass and a wide base. A tall, narrow object with a high centre of mass tips easily.

Moments (3.1.8 a)

The turning effect of a force is called the moment.

moment (newton-metre, N m) = force (newton, N) × perpendicular distance (metre, m)

M = F × d

The distance d is measured at right angles to the force, from the pivot to the force’s line of action. A longer spanner, or a push at a door handle rather than near the hinge, gives a bigger moment for the same force. If a force acts at an angle, use the perpendicular distance from the pivot to its line of action, not the length of the handle.

Worked example 4. A force of 150 N is applied at 0.28 m from a nut, at right angles to the spanner. (a) Find the moment. (b) A longer spanner lets you apply the same moment at 0.35 m. What force is needed?

• (a) M = 150 × 0.28 = 42 N m.
• (b) F = M ÷ d = 42 ÷ 0.35 = 120 N.

The principle of moments (3.1.8 b)

If an object is not turning, the total clockwise moment about any pivot equals the total anticlockwise moment about the same pivot.

Method:

1. Choose the pivot and mark each force with a moment about it.
2. Sort the forces into clockwise and anticlockwise.
3. Write: sum of clockwise moments = sum of anticlockwise moments.
4. Substitute and solve.

Worked example 5. A seesaw balances with a child of weight 300 N sitting 1.8 m from the pivot. An adult sits on the other side, 1.2 m from the pivot. Find the adult’s weight and the upward force of the pivot.

• Anticlockwise moment = 300 × 1.8 = 540 N m.
• Clockwise moment = F × 1.2. Set them equal: F = 540 ÷ 1.2 = 450 N.
• The pivot supports both weights: 300 + 450 = 750 N upwards.

Worked example 6. A uniform metre rule weighs 1.2 N. It balances on a pivot at the 35 cm mark with a weight W hung at the 5 cm mark. Find W.

• The uniform rule’s weight acts at the 50 cm mark, 0.15 m from the pivot, clockwise.
• Clockwise: 1.2 × 0.15 = 0.18 N m.
• W is 0.30 m from the pivot, anticlockwise: W × 0.30.
• W = 0.18 ÷ 0.30 = 0.60 N.

Toppling (3.1.8 c)

The weight of an object acts through its centre of mass. While the vertical line of action of the weight stays inside the base, the weight gives a moment that returns the object upright. If it falls outside the base, there is a resultant moment about the tipping edge and the object topples.

Worked example 7. A box weighs 200 N and is 0.50 m wide, with its centre of mass at the middle. A horizontal push acts 0.80 m above the floor. Find the smallest push that starts to tip it about the lower edge on the far side.

• The weight acts 0.25 m from the tipping edge: restoring moment = 200 × 0.25 = 50 N m.
• Push moment about the same edge = F × 0.80.
• It begins to tip when F × 0.80 = 50, so F = 62.5 N.

Vehicles: a bus or lorry with a heavy load stacked high has a high centre of mass, so a smaller tilt (for example on a sloping road or in a sharp turn) brings the line of action outside the wheels. Balancing toys: the toy has weights on long arms so the centre of mass is below the pivot. When it is nudged, the weight gives a moment that swings it back.

Levers (3.1.8 d)

A lever is a rigid bar that turns about a pivot. A simple lever can be a force multiplier: a small force (the effort) far from the pivot overcomes a large force (the load) close to the pivot. Crowbars, wheelbarrows and nutcrackers work this way.

Worked example 8. A crowbar is pushed with 80 N at 1.20 m from the pivot. The load acts 0.15 m from the pivot. Find the load that can be lifted.

• Load × 0.15 = 80 × 1.20 = 96 N m.
• Load = 96 ÷ 0.15 = 640 N.

The force is multiplied by 1.20 ÷ 0.15 = 8. In return, the effort moves eight times as far as the load.

Common errors

• Saying a skydiver at terminal velocity has “no forces”. The forces are balanced, not absent.
• Mixing centimetres and metres in one calculation.
• Saying a tall object topples because it is heavy, not because the line of action of the weight is outside the base.

Official syllabus

This guide follows the OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Check the current version on the OxfordAQA website.