OxfordAQA International GCSE Physics: Resultant forces and Newton's laws (9203)

This guide follows section 3.1.3, Resultant forces (statements a to h), of the 9203 OxfordAQA International GCSE Physics specification, which applies to exams from May/June 2018. The qualification is untiered, so everything here is for all candidates. Worked examples run through every statement. For short recall notes use the revision notes, and for questions with answers use the practice questions.

Related pages: the OxfordAQA International GCSE Physics hub, the printable checklist, the broader overview of forces and the free diagnostics if you want to find your weak spots first.

What this unit covers

Spec	What you must be able to do
3.1.3 a	State Newton’s Third Law and name the pair of forces in an interaction
3.1.3 b	Replace several forces by one resultant; find the resultant of opposite or parallel forces in a line; find the resultant of two coplanar forces by scale drawing
3.1.3 c	Know that a non-zero resultant force causes acceleration
3.1.3 d	Define acceleration; know that changing direction at constant speed is acceleration; know deceleration is negative acceleration; use a = Δv / t
3.1.3 e	Find acceleration from the gradient of a velocity-time graph
3.1.3 f	Use the area under a velocity-time graph to find how far an object travels
3.1.3 g	State Newton’s First Law: zero resultant means constant velocity or rest
3.1.3 h	State Newton’s Second Law and use F = ma

Newton’s Third Law (3.1.3 a)

If object A pushes or pulls object B, then B pushes or pulls A with a force of the same size in the opposite direction. The two forces form a pair. Four features identify a genuine pair:

• they are the same type of force (both gravitational, both contact, both magnetic and so on)
• they are equal in size
• they point in opposite directions
• they act on different objects

The last point matters most. Because the two forces act on different objects, they never cancel each other out in a resultant-force calculation for one object.

Example: a mug rests on a desk. The Earth pulls the mug down (its weight). The pair of that force is the mug pulling the Earth up, with the same size, also gravitational. The desk pushing up on the mug is a contact force, and its pair is the mug pushing down on the desk. Weight and the desk’s push do not make a Third Law pair, because both act on the mug and they are different types of force. They happen to balance because the mug is at rest.

Resultant force (3.1.3 b)

A number of forces acting on an object can be replaced by one force that has the same effect on its motion. This is the resultant force.

Forces in a straight line

Choose one direction as positive, add forces that point that way and subtract those that point the other way.

Worked example 1. A sledge is pulled forwards by a 450 N force. Friction of 180 N and a 70 N headwind both act backwards. Find the resultant.

Forwards positive:  450 - 180 - 70 = 200 N

The resultant is 200 N forwards. If the answer had come out negative, the resultant would point backwards.

Parallel forces in the same direction simply add: two people pushing a car with 300 N and 250 N give 550 N.

Two coplanar forces by scale drawing

Coplanar forces act in the same plane, for example two ropes pulling on a buoy on the water surface. When they are not in a line, you find the resultant by drawing.

Method:

1. Choose a scale that makes the drawing large, such as 1 cm to 1 N.
2. Draw the first force as a line of the correct length and direction.
3. From the end of that line, draw the second force at the correct angle, so the arrows run head to tail.
4. Draw the line from the start of the first arrow to the end of the second. This is the resultant.
5. Measure its length and convert back using the scale. Measure its angle with a protractor.

Worked example 2. Two forces of 8 N and 5 N act on an object. The angle between their directions is 60°. Find the resultant by scale drawing.

Use 1 cm to 1 N. Draw an 8.0 cm line. At its end, draw a 5.0 cm line so that it points 60° away from the direction of the first line. Join the start to the finish. The closing line measures about 11.4 cm, so the resultant is about 11 N. A protractor shows about 22° from the 8 N force. Calculation by trigonometry gives 11.4 N and 22.4°, so your drawing should agree within the accuracy of your ruler and protractor.

Check: the resultant must be less than 13 N (8 + 5, the largest possible) and more than 3 N (8 - 5, the smallest possible).

A special case is two forces at right angles. For 12 N and 9 N the drawing gives a 15 N resultant, at about 37° from the 12 N force.

A resultant force causes acceleration (3.1.3 c)

Any non-zero resultant force makes the object accelerate. It may speed up, slow down or change direction. The force changes the velocity rather than just keeping the object moving.

Acceleration (3.1.3 d)

Acceleration measures how quickly velocity changes.

a = Δv / t
acceleration (m/s²) = change in velocity (m/s) / time taken (s)

Here Δv is final velocity minus initial velocity. It is the change in velocity, so a sign matters.

Worked example 3. A car increases its velocity from 8 m/s to 26 m/s in 9 s. Find its acceleration.

Δv = 26 - 8 = 18 m/s
a = 18 / 9 = 2.0 m/s²

Deceleration is negative acceleration. Worked example 4: a train slows from 40 m/s to rest in 25 s.

Δv = 0 - 40 = -40 m/s
a = -40 / 25 = -1.6 m/s²

The train decelerates at 1.6 m/s². You may quote the answer as “-1.6 m/s²” or as a deceleration of 1.6 m/s², but do not mix the two ideas.

Changing direction is accelerating

Velocity has direction as well as size. A car going round a roundabout at a steady 12 m/s keeps changing direction. Its velocity changes, so it accelerates although its speed is unchanged. A resultant force must therefore be acting on it. For an object moving in a circle, that resultant force is directed towards the centre of the circle.

Velocity-time graphs (3.1.3 e and f)

Reading a velocity-time graph:

• gradient = acceleration (m/s² when the axes are m/s and s)
• flat section = constant velocity, so zero acceleration
• straight slope = steady acceleration
• downward slope = deceleration
• area under the line, down to the time axis = how far it has moved

Worked example 5. A cyclist’s velocity rises steadily from 0 to 12 m/s in the first 6 s, stays at 12 m/s for 8 s, then falls steadily to 0 in the last 6 s. Find the acceleration of each stage, the whole distance and the average speed.

Stage 1 gradient: 12 / 6   =  2.0 m/s²
Stage 2 gradient: 0        =  0 m/s²
Stage 3 gradient: -12 / 6  = -2.0 m/s²

Stage 1 area (triangle):   1/2 x 6 x 12 = 36 m
Stage 2 area (rectangle):  8 x 12       = 96 m
Stage 3 area (triangle):   1/2 x 6 x 12 = 36 m
Total distance = 36 + 96 + 36 = 168 m

Average speed = 168 / 20 = 8.4 m/s

For a trapezium, use (sum of parallel sides) / 2 x width. Where a graph is curved, count squares: work out the distance each square represents (width in seconds multiplied by height in m/s), count whole squares and combine part squares, then multiply.

Take care with the axes. A gradient on a distance-time graph is speed, but a gradient on a velocity-time graph is acceleration.

Newton’s First Law (3.1.3 g)

If the resultant force on an object is zero:

• a moving object continues to move at the same velocity (same speed and same direction)
• a stationary object stays at rest

Forces may still act. They are balanced. A car driving at a steady 20 m/s along a level road has a driving force equal to the resistive forces, so the resultant is zero and the velocity does not change. The reverse is also true: an object with constant velocity must have zero resultant force.

Explaining first-law situations: when a bus brakes, nothing slows an unrestrained passenger, so the body keeps its old velocity until a seat belt or seat pushes on it. Say that the velocity stays the same until a resultant force acts, not that a force “keeps it going”.

Newton’s Second Law (3.1.3 h)

If the resultant force on an object is not zero, the object accelerates in the direction of the resultant force. The relationship is

F = ma
resultant force (N) = mass (kg) x acceleration (m/s²)

Rearranged, this gives a = F / m or m = F / a. The newton is defined so that 1 N of resultant force produces 1 m/s² in a 1 kg object. Doubling the resultant force doubles the acceleration. Doubling the mass halves it.

Worked example 6. A van of mass 1250 kg has a driving force of 3400 N against resistive forces totalling 900 N. Find its acceleration.

Resultant = 3400 - 900 = 2500 N
a = F / m = 2500 / 1250 = 2.0 m/s²

Worked example 7. A resultant force acts on a ball of mass 0.45 kg and gives it an acceleration of 20 m/s². Find the force, then find the mass of an object that accelerates at 1.5 m/s² under a 36 N resultant force.

F = ma = 0.45 x 20 = 9.0 N
m = F / a = 36 / 1.5 = 24 kg

Worked example 8 (combining ideas). A cyclist and bicycle have a total mass of 85 kg. The cyclist’s forward force is 140 N and the resistive force is 60 N. How long to reach 6.0 m/s from rest, and how far is travelled?

Resultant = 140 - 60 = 80 N
a = 80 / 85 = 0.941 m/s²
t = Δv / a = 6.0 / 0.941 = 6.4 s
distance = area under v-t graph = 1/2 x 6.4 x 6.0 = 19 m

Because the resultant is constant, the acceleration is constant and the velocity-time graph is a straight line from the origin.

Common errors

• Pairing weight with the desk’s push. Both act on the same object, so they are not a Third Law pair.
• Saying a Third Law pair “cancels”. The forces act on different objects.
• Putting the driving force into F = ma when the resultant is needed.
• Calling a zero resultant “no forces”. Forces can act and balance.
• Forgetting that circular motion at steady speed involves acceleration.
• Reading the gradient of a distance-time graph as acceleration.
• Using the total time rather than the time for one stage when finding the area or gradient of a stage.
• Leaving the sign out of Δv, or giving a deceleration as a positive acceleration in a calculation.
• Measuring the angle of a resultant from the wrong force in a scale drawing, or not stating the reference line.

Next steps

Next, condense this with the revision notes and test it with the practice questions. For motion graphs earlier in the specification see the forces practice set, and use the diagnostics to check your progress.

Official syllabus

Source document: OxfordAQA International GCSE Physics (9203), version 5.2, exams May/June 2018 onwards. Scope here: section 3.1.3, a to h.