OxfordAQA International GCSE Physics: Forces, Hooke's law, vectors and distance-time motion (9203)

This guide is aligned to the OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards. It covers sections 3.1.1 (Forces and their interactions) and 3.1.2 (Motion). The qualification is untiered. After reading, use the revision notes and the practice questions.

Related pages: course hub, printable checklist, Forces and their effects overview, forces revision notes and the free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do
3.1.1 a	Sort forces into contact and non-contact (field) forces
3.1.1 b	Describe friction and air resistance, including heating
3.1.1 c	Describe the pair of forces when two objects interact, as vectors
3.1.1 d	Tell scalars from vectors and draw a vector as an arrow
3.1.1 e	Use W = m × g
3.1.1 f	Describe stretching, bending and compressing; elastic and inelastic distortion
3.1.1 g	Explain that a stretched spring stores elastic potential energy
3.1.1 h	Use F = k × e, the limit of proportionality, and the required practical
3.1.2 a-d	Draw and read distance-time graphs; speed from gradient; v = s/t; average speed

Contact and non-contact forces (3.1.1 a)

Contact forces need the two objects to touch. Examples: friction, air resistance, tension in a rope, and the normal contact force (the push of a surface on whatever rests on it, at right angles to the surface).

Non-contact (field) forces act across a gap. Examples: gravity, electrostatic force between charged objects, and the magnetic force.

Friction and air resistance (3.1.1 b)

Friction is a force between two surfaces that impedes motion. It acts opposite to the direction an object moves (or tries to move), and it may result in heating. Rub your hands together and they warm up. Brake pads pressed on a wheel rim grow hot, and a meteor glows because of friction with the air.

Air resistance is a form of friction. It is the frictional force on an object moving through air. It increases as the object goes faster, and a streamlined shape reduces it. A crouching cyclist presents a smaller area to the air, so there is less air resistance.

Force pairs as vectors (3.1.1 c)

Forces never occur alone. When object A pushes or pulls object B, B pushes or pulls A back. The two forces have equal magnitude and point in opposite directions. Each force is a vector, so each has a size and a direction.

The two forces of a pair act on different objects. If you push a wall with 30 N to the right, the wall pushes your hand with 30 N to the left. The 30 N on the wall and the 30 N on your hand do not cancel, because they act on different things. Section 3.1.3 states this as Newton’s third law.

Scalars and vectors (3.1.1 d)

A scalar has magnitude (size) only. A vector has magnitude and direction.

Scalars	Vectors
distance	displacement
speed	velocity
time	acceleration
	force
	momentum

Distance is how far you travel along your path. Displacement is how far the finish is from the start in a straight line, with its direction stated.

Drawing a vector. Choose a scale, draw an arrow that many units long, and point it in the direction of the vector. The length shows the magnitude and the head shows the direction.

Worked example 1. A force of 15 N acts to the right on a box. Use a scale of 1 cm to 5 N.

• Length = 15 ÷ 5 = 3.0 cm.
• Draw a horizontal arrow 3.0 cm long with the head on the right.

Weight (3.1.1 e)

Weight is the pull of gravity on an object. Its size depends on the gravitational field strength, g, wherever the object happens to be.

weight (N) = mass (kg) × gravitational field strength (N/kg)
W = m × g

You do not need to remember g. It is given in the question, so use the value supplied. Mass is a scalar measured in kg and does not change from place to place. Weight is a vector (it points towards the centre of the planet) and does change.

Worked example 2. A sample has a mass of 3.5 kg. Find its weight where g = 3.7 N/kg.

W = m × g = 3.5 × 3.7 = 12.95 N

Weight = 13 N (2 significant figures).

Worked example 3. A tool weighs 20 N where g = 1.6 N/kg. Its mass is W ÷ g = 20 ÷ 1.6 = 12.5 kg, and it would keep that mass anywhere.

Changing shape: elastic and inelastic (3.1.1 f)

When more than one force acts on an object, it can change shape by stretching, bending or compressing. A single force just moves the object; to deform it, forces must act on it from different sides, such as hands pulling the two ends of a spring.

• Elastic distortion: the object goes back to its first shape once the forces stop acting.
• Inelastic distortion: the object stays out of shape afterwards. A paper clip bent open is an example.

To test which kind you have, measure the length, load the object, unload it and measure again. A longer length means inelastic distortion.

Elastic potential energy (3.1.1 g)

Stretching a spring needs a force acting through a distance, so work is done on the spring. For an elastic object that work is stored as elastic potential energy. Released, the spring returns to its original length and transfers the energy, for example as kinetic energy of a toy dart.

The calculation of the energy stored belongs to the energy section of the specification. For the equation and examples see the energy revision notes.

Hooke’s law (3.1.1 h)

For an object behaving elastically, the extension is directly proportional to the force applied, provided the limit of proportionality is not exceeded.

F = k × e
F: force (N)    k: spring constant (N/m)    e: extension (m)

The extension is the new length minus the original length. It is not the new length itself. The spring constant k is a constant for a given spring and tells you how stiff it is: a large k means a stiff spring.

On a force-extension graph, the plot is a straight line starting at the origin, and stays straight until the limit of proportionality. Its gradient is k. Beyond that point the line curves, and doubling the force produces more than double the extension.

Worked example 4. A spring has an original length of 12.0 cm. A force of 4.0 N gives an extension of 8.0 cm.

e = 8.0 cm = 0.080 m
k = F ÷ e = 4.0 ÷ 0.080 = 50 N/m

The spring constant is 50 N/m.

Now find the extension for 3.5 N, assuming the limit has not been exceeded.

e = F ÷ k = 3.5 ÷ 50 = 0.070 m = 7.0 cm

Under a 2.5 N load the extension is 0.050 m = 5.0 cm, so the length is 12.0 + 5.0 = 17.0 cm.

Convert centimetres to metres before using k in N/m, and never put the new length into the equation.

Required practical: force and extension of a spring

Aim: investigate the relationship between force and extension for a spring.

Method

1. Clamp the spring vertically from a stand. Fix a metre rule beside it, vertical and close to the spring, and attach a pointer (or use the bottom of the spring) to read against the rule.
2. Read the original length with no load.
3. Add a 100 g mass. Its weight is the force on the spring (W = m × g, with g supplied). Wait until the spring is still and read the new length with your eye level with the pointer.
4. Add masses one at a time, reading the length each time. Then remove them in the same steps and read again, to see whether the spring comes back to its starting length.
5. Calculate extension = length − original length for each load.

Variables. Independent: force (the weight added). Dependent: extension. Control: the same spring, the same rule, the same way of reading.

Results table

Force (N)	Extension (cm)
0	0
1.0	2.0
2.0	4.1
3.0	6.0
4.0	8.0
5.0	10.1
6.0	13.0
7.0	19.5

Graph. Plot force on the y-axis and extension on the x-axis, and draw a best-fit straight line through the first points and the origin. The points up to 5.0 N lie on a line, so the gradient is about 5.0 ÷ 0.101 = 49.5, or 50 N/m. At 6.0 N the point sits below the line, and at 7.0 N it is well below it. The limit of proportionality is at about 5.0 N. Force divided by extension gives about 50 N/m up to 5.0 N, then 46 N/m at 6.0 N and 36 N/m at 7.0 N.

Improving the practical. Read at eye level to avoid parallax, check for a zero error, use smaller mass steps near the limit and repeat readings to find a mean. Wear eye protection, secure the stand and keep feet clear of falling masses. Do not load far past the limit, or the spring is permanently stretched.

Distance-time graphs (3.1.2 a-b)

If an object moves in a straight line, its distance from a certain point can be plotted against time. Time goes on the x-axis.

• A horizontal line: the object is stationary.
• A straight sloping line: constant speed. A steeper line means a faster speed.
• A curve that steepens means speeding up; one that flattens means slowing down.

Speed is the gradient:

speed = change in distance ÷ change in time

For a curve, draw a tangent at the time you want and find the gradient of the tangent.

Velocity and average speed (3.1.2 c-d)

The velocity of an object is its speed in a given direction:

v = s ÷ t
v: velocity (m/s)    s: displacement (m)    t: time (s)

The same equation gives the average speed of an object whose motion is non-uniform, using the total distance and the total time.

Worked example 5. A cyclist travels 50 m in the first 10 s, is stationary for the next 15 s, then travels a further 90 m in 20 s.

• Speed, 0 to 10 s: 50 ÷ 10 = 5.0 m/s.
• Speed, 25 s to 45 s: 90 ÷ 20 = 4.5 m/s.
• Total distance = 50 + 90 = 140 m. Total time = 10 + 15 + 20 = 45 s.
• Average speed = 140 ÷ 45 = 3.1 m/s (2 significant figures).

The rest counts as time. Averaging the three speeds gives 3.2 m/s, which is the wrong method.

Worked example 6. A walker goes 60 m east, then 20 m west, taking 40 s in all.

• Distance = 60 + 20 = 80 m, so average speed = 80 ÷ 40 = 2.0 m/s.
• Displacement = 60 − 20 = 40 m east, so average velocity = 40 ÷ 40 = 1.0 m/s east.

Common errors

• Using the new length, not the extension, in F = k × e.
• Taking the average of the speeds instead of total distance ÷ total time.
• Reading a flat section of a distance-time graph as steady movement. It means the object is not moving.
• Calling a distance a vector. Distance is a scalar; displacement is the vector.

Next steps

Attempt the practice questions, then check progress on the course checklist and the free diagnostics.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, version 5.2, for exams May/June 2018 onwards, published by Oxford International AQA Examinations. Sections 3.1.1 and 3.1.2, and the required practical on force and extension for a spring.