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AQA PhysicsForces

Forces and motion

Calculate speed, acceleration and resultant force.

Start here

The key idea

A resultant force changes an object's velocity. Balanced forces do not cause acceleration.

Equation to know

resultant force = mass x acceleration

Forces And Motion
object4 N10 Nresultant force = 6 N to the right

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Scalars and vectors

A scalar quantity has only a magnitude (size), such as distance, speed, mass, energy and temperature.A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force and momentum.Vectors are usually drawn as arrows whose length represents the magnitude and whose direction shows the direction of the quantity.Forces are vectors, so direction matters when adding them.

2

Resultant force and Newton's laws

The resultant force is the single force that has the same effect as all the forces acting on an object.Newton's first law states that an object stays at rest or moves at constant velocity unless acted on by a resultant force.Newton's second law gives resultant force = mass x acceleration (F = m x a).Newton's third law states that when two objects interact they exert equal and opposite forces on each other.

3

Acceleration and the equations of motion

Acceleration = change in velocity / time taken (a = (v - u) / t), measured in m/s2.For uniform acceleration you can also use v2 = u2 + 2 x a x s, where s is distance.The gradient of a velocity-time graph gives acceleration, and the area under it gives the distance travelled.A negative acceleration (deceleration) means the object is slowing down.

4

Terminal velocity

When a falling object speeds up, the air resistance acting upwards increases.Terminal velocity is reached when the air resistance balances the weight, so the resultant force is zero and acceleration becomes zero.The object then falls at a constant (maximum) velocity.Opening a parachute increases air resistance, briefly making the resultant force upward, so the object decelerates to a new, lower terminal velocity.

Key terms

Definitions to learn

Resultant force

The single force that has the same effect as all the forces acting on an object.

Weight

The force on an object due to gravity, given by weight = mass x gravitational field strength.

Acceleration

The rate of change of velocity, measured in m/s2.

Terminal velocity

The constant maximum velocity reached when air resistance balances weight.

Inertia

The tendency of an object to stay at rest or in uniform motion (related to its mass).

Worked example

A 1200 kg car accelerates at 2.5 m/s2. Calculate the resultant force.

1

Use F = ma.

2

Substitute the values.

Final answer

3000 N

Exam habit

State F = ma then identify the resultant force direction.For terminal velocity, always explain that the resultant force is zero — not that forces are removed.Include units: newtons, kg, m/s².

Watch out

Do not confuse mass in kilograms with weight in newtons.

Examiner tips

How to score full marks

  • 1Mass is in kg and weight is in N — never confuse them; use weight = m x g with g = 9.8 N/kg.
  • 2Always quote a unit in your final answer; many calculation marks are lost for missing units.
  • 3When a question says constant velocity or steady speed, write that the resultant force is zero.
Quick-fire quiz

Test yourself

Pick an answer — you'll see instantly if it's right.

A 900 kg car accelerates at 3 m/s². What is the resultant force?

An object moves at constant velocity. What must be true about the forces acting on it?

A skydiver reaches terminal velocity. Which statement is correct?

A car goes from rest to 20 m/s in 8 s. What is its acceleration?

Which quantity is a vector?

Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A cyclist increases speed from 4 m/s to 10 m/s in 3 seconds. Find the acceleration.
Mark scheme
  1. 1.Use a = change in velocity / time.
2 m/s2
2A force of 18 N accelerates a trolley at 1.5 m/s2. Find its mass.
Mark scheme
  1. 1.Rearrange m = F / a.
12 kg
3Explain why a skydiver eventually reaches terminal velocity.
Mark scheme
  1. 1.Compare weight and drag.
As speed rises, drag increases until it equals weight. The resultant force becomes zero, so acceleration stops.
4State the difference between a scalar and a vector quantity, giving one example of each.[2 marks]
Mark scheme
  1. 1.Define scalar.
  2. 2.Define vector.
  3. 3.Give examples.
A scalar has only magnitude e.g. speed/mass (1); a vector has magnitude and direction e.g. velocity/force (1).
5Calculate the weight of a 60 kg student. (g = 9.8 N/kg)[2 marks]
Mark scheme
  1. 1.Use weight = m x g.
  2. 2.Substitute values.
W = m x g = 60 × 9.8 (1) = 588 N (1).
6A 1200 kg car accelerates at 2.5 m/s2. Calculate the resultant force.[2 marks]
Mark scheme
  1. 1.Use F = m x a.
  2. 2.Substitute 1200 × 2.5.
F = m x a = 1200 × 2.5 (1) = 3000 N (1).
7A car travelling at 8 m/s accelerates uniformly to 20 m/s in 6 s. Calculate the acceleration.[3 marks]
Mark scheme
  1. 1.Use a = (v - u) / t.
  2. 2.Substitute.
  3. 3.Evaluate.
a = (v - u) / t = (20 - 8) / 6 (1) = 12 / 6 (1) = 2 m/s2 (1).
8A skydiver jumps from a plane. Explain, in terms of forces, how she reaches terminal velocity and what happens to her motion when she opens her parachute.[5 marks]
Mark scheme
  1. 1.Initially weight greater than air resistance.
  2. 2.Air resistance increases with speed.
  3. 3.Forces balance at terminal velocity.
  4. 4.Parachute increases air resistance.
  5. 5.New lower terminal velocity.
At first weight is greater than air resistance so she accelerates (1). As speed increases, air resistance increases (1). When air resistance equals weight the resultant force is zero, so she falls at constant (terminal) velocity (1). Opening the parachute increases air resistance so it is now greater than weight, giving an upward resultant force and she decelerates (1). Air resistance then falls until forces balance again at a new, lower terminal velocity (1).
9A car of mass 900 kg brakes from 30 m/s to rest in 6.0 s. Calculate the braking force, assuming it is constant.[3 marks]
Mark scheme
  1. 1.Find acceleration using a = (v - u) / t.
  2. 2.Use F = m x a (deceleration, so force is negative/opposing).
  3. 3.State the magnitude of the braking force.
a = (0 - 30) / 6.0 = -5.0 m/s2 (1); F = m x a = 900 × 5.0 (1) = 4500 N (1)
10Use the equation v2 = u2 + 2as to find the distance a ball travels while accelerating from rest at 3.0 m/s2 to a final speed of 12 m/s.[3 marks]
Mark scheme
  1. 1.u = 0, v = 12, a = 3.0.
  2. 2.Rearrange v2 = u2 + 2as to s = (v2 - u2) / (2a).
  3. 3.Substitute and evaluate.
s = (v2 - u2) / (2a) = (122 - 0) / (2 × 3.0) (1) = 144 / 6 (1) = 24 m (1)
11Explain, using Newton's first and second laws, why a heavier lorry travelling at the same speed as a car takes longer to stop when the same braking force is applied.[3 marks]
Mark scheme
  1. 1.Newton's first law: an object won't decelerate without a resultant force.
  2. 2.Newton's second law: a = F / m.
  3. 3.Same force, larger mass means smaller deceleration.
  4. 4.Smaller deceleration over same speed change means longer stopping time.
Newton's first law tells us the lorry will continue at constant velocity until a resultant force acts; applying brakes provides that force (1); Newton's second law, a = F / m, shows that for the same braking force, a larger mass (the lorry) produces a smaller deceleration (1); a smaller deceleration means it takes longer to change the velocity from the initial speed to zero, so the stopping distance and time are greater (1)
12A velocity-time graph shows a straight line from (0, 0) to (5, 20), then a horizontal line from (5, 20) to (10, 20), then a straight line from (10, 20) to (14, 0). Calculate the acceleration in the first phase, the distance travelled in the second phase, and the deceleration in the third phase.[3 marks]
Mark scheme
  1. 1.Acceleration phase 1: gradient = (20 - 0) / (5 - 0).
  2. 2.Distance phase 2: area = 20 × (10 - 5).
  3. 3.Deceleration phase 3: gradient = (0 - 20) / (14 - 10); take magnitude.
Acceleration = 20 / 5 = 4.0 m/s2 (1); distance in phase 2 = 20 × 5 = 100 m (1); deceleration = (0 - 20) / (14 - 10) = -20 / 4 = -5.0 m/s2, magnitude 5.0 m/s2 (1)
13Explain the difference between thinking distance and braking distance, and give two factors that affect each one.[2 marks]
Mark scheme
  1. 1.Thinking distance: time before brakes applied; linked to reaction time.
  2. 2.Braking distance: distance after brakes applied; linked to deceleration.
  3. 3.Two factors for thinking distance.
  4. 4.Two factors for braking distance.
Thinking distance is the distance the vehicle travels between the driver perceiving a hazard and applying the brakes; it depends on reaction time, which is increased by tiredness, alcohol, distractions or drugs (1); braking distance is the distance travelled after the brakes are applied until the vehicle stops; it increases at higher speeds (v2 relationship), and is greater with worn brakes, bald tyres, or a wet/icy road surface (1) — award 1 per section with two valid factors each, max 2
14A rocket of mass 4000 kg is launched vertically. Its engines provide an upward thrust of 60 000 N. Taking g = 9.8 N/kg, calculate the resultant force on the rocket at launch and its initial acceleration.[3 marks]
Mark scheme
  1. 1.Weight = m x g = 4000 × 9.8.
  2. 2.Resultant force = thrust - weight (upward positive).
  3. 3.Acceleration = resultant force / mass.
Weight = 4000 × 9.8 = 39200 N (1); resultant force = 60000 - 39200 = 20800 N upward (1); a = F / m = 20800 / 4000 = 5.2 m/s2 (1)
15Explain, using Newton's third law, what happens when a person pushes against a wall. Identify the action-reaction pair clearly, state the direction of each force, and explain why the person may move but the wall does not.[3 marks]
Mark scheme
  1. 1.Person pushes wall with force in one direction.
  2. 2.Wall exerts equal and opposite force on person (Newton's third law).
  3. 3.Forces act on different objects — this is key.
  4. 4.Person accelerates if unbalanced; wall does not move because forces on it are balanced and it is fixed.
When the person pushes the wall, they exert a force on the wall directed towards it (action force); by Newton's third law, the wall exerts an equal and opposite force on the person directed away from the wall (reaction force) (1); these two forces are equal in magnitude and opposite in direction, but they act on different objects — one on the wall, one on the person (1); the person may accelerate away because the reaction force from the wall is the resultant force on them (assuming low friction on their feet); the wall does not move because it is fixed to the ground and the external forces on the wall are balanced by the ground's reaction (1)
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