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10.2 Weight and Terminal Velocity

GCSE Physics · AQA · Chapter 10: Force and Motion

A free GCSE Physics lesson for AQA (8463) from OBA School: mass and weight, W = m g, falling freely, drag and terminal velocity. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Explain the difference between mass and weight and give the unit of each
  • Calculate weight, mass or gravitational field strength using W = m g
  • Describe the motion of an object falling freely under gravity
  • Explain, using forces, why an object falling through a fluid reaches a terminal velocity
  • Interpret the velocity–time graph of a skydiver before and after the parachute opens

Key words

  • Mass: The quantity of matter in an object, in kilograms (kg). It is the same everywhere.
  • Weight: The force acting on an object due to gravity, in newtons (N).
  • Gravitational field strength: The force of gravity on each kilogram, g, in N/kg. About 9.8 N/kg at the Earth’s surface.
  • Newtonmeter: A calibrated spring balance used to measure forces such as weight.
  • Centre of mass: The single point where the weight of an object can be taken to act.
  • Free fall: Falling with weight as the only force acting.
  • Drag: The resistive force on an object moving through a fluid (a liquid or a gas). In air it is called air resistance.
  • Terminal velocity: The constant, maximum velocity reached by a falling object when the drag on it equals its weight.

Mass and weight

In everyday speech people mix up mass and weight, but in physics they are different quantities.

  • What it is; Mass the amount of matter in an object; Weight the force on an object due to gravity
  • Unit; Mass kilogram, kg; Weight newton, N
  • Scalar or vector?; Mass scalar (size only); Weight vector (acts downwards, towards the centre of the Earth)
  • Does it change?; Mass no – the same on the Earth, the Moon or in space; Weight yes – it depends on the gravitational field strength
  • Measured with; Mass a balance; Weight a newtonmeter (calibrated spring balance)

The weight of an object acts at a single point called its centre of mass. On a diagram, draw the weight arrow from this point.

Calculating weight: W = m g

weight = mass × gravitational field strength W = m g

W = weight in newtons, N; m = mass in kilograms, kg; g = gravitational field strength in newtons per kilogram, N/kg

Rearranged: m = W ÷ g and g = W ÷ m

Values of g: Earth 9.8 N/kg; Moon 1.6 N/kg; Mars 3.7 N/kg

Worked example: a student on the Earth

Calculate the weight of a student of mass 62 kg.
W = m g = 62 kg × 9.8 N/kg = 607.6 N

Answer: W = 610 N (2 significant figures)

Worked example: a rover on Mars

A rover weighs 1960 N on the Earth. Calculate its weight on Mars.
Mass first: m = W ÷ g = 1960 N ÷ 9.8 N/kg = 200 kg (the mass does not change)
Weight on Mars: W = 200 kg × 3.7 N/kg = 740 N

Answer: 740 N

Exam tip: Weight is a force, so its unit is always the newton. A question that asks for a weight “in kg” is really asking for the mass.

Falling freely

If the only force on a falling object is its weight, the object is in free fall. Its acceleration is:

a = F ÷ m = (m × g) ÷ m = g, about 9.8 m/s² near the Earth’s surface

The mass cancels, so all objects in free fall have the same acceleration. In a vacuum tube, a coin and a feather dropped together land together. In air, the feather is slowed much more by air resistance.

So g has two meanings with the same value: gravitational field strength (9.8 N/kg) and the acceleration of free fall (9.8 m/s²).

Falling through a fluid: terminal velocity

A fluid (liquid or gas) exerts a drag force on an object moving through it. Drag acts against the motion and increases as the speed increases.

  • Just released: the speed is zero, so there is no drag. Resultant force = weight, so the acceleration is g.
  • Speeding up: drag increases, so the resultant force (weight − drag) decreases. The object still speeds up, but its acceleration gets smaller.
  • Terminal velocity: drag = weight, so the resultant force is zero and the acceleration is zero. The object keeps falling at a constant velocity.

Worked example: a ball bearing in oil

A 20 g ball bearing falls through oil. At one moment the drag on it is 0.050 N. Calculate its acceleration at that moment.
m = 0.020 kg; weight = 0.020 × 9.8 = 0.196 N
resultant force = 0.196 − 0.050 = 0.146 N downwards
a = F ÷ m = 0.146 ÷ 0.020 = 7.3 m/s²

Answer: 7.3 m/s² downwards

Key fact: An object with a larger weight, or a more streamlined shape, needs more speed before the drag balances its weight, so it has a higher terminal velocity.

The skydiver

  • A: just after jumping, air resistance is small; acceleration is about 9.8 m/s² (steepest gradient).
  • B: air resistance increases with speed; the resultant force falls, so the gradient (acceleration) decreases.
  • C: air resistance = weight; first terminal velocity (here 50 m/s).
  • D: the parachute opens. The large area makes air resistance much greater than the weight, so the resultant force is upwards and the skydiver decelerates (still moving down).
  • E: as speed falls, air resistance falls until it equals the weight again: a new, lower terminal velocity (about 5 m/s), safe for landing.
  • F: landing – the ground stops the skydiver.

Worked example: just after the parachute opens

A skydiver of mass 80 kg has 2000 N of air resistance on her just after her parachute opens.
weight = 80 × 9.8 = 784 N; resultant force = 2000 − 784 = 1216 N upwards
a = 1216 ÷ 80 = 15.2 m/s² upwards

Answer: she decelerates at about 15 m/s²

Common mistakes

  • Giving weight in kilograms. Weight is a force, measured in newtons.
  • Saying there are no forces on an object at terminal velocity. The forces are balanced, not absent.
  • Saying the skydiver moves upwards when the parachute opens. She slows down but is still moving downwards.
  • Thinking heavier objects fall faster in a vacuum. With no air resistance, every object accelerates at g.
  • Using g = 10 N/kg when a question gives 9.8 N/kg. Always use the value in the question.

Revision questions

Try these, then check your answers below each one.

Question 1. Calculate the weight on the Earth of a dog of mass 7.5 kg. (g = 9.8 N/kg)

Answer: 73.5 N

Working: W = m g = 7.5 kg × 9.8 N/kg = 73.5 N.

Question 2. What is the unit of weight? (Choose from: newton (N), kilogram (kg), newton per kilogram (N/kg), joule (J).)

Answer: newton (N)

Working: Weight is the force on an object due to gravity, so it is measured in newtons. Mass is measured in kilograms.

Question 3. An astronaut has a mass of 85 kg. Calculate her weight on the Moon, where g = 1.6 N/kg.

Answer: 136 N

Working: W = 85 kg × 1.6 N/kg = 136 N.

Related lessons

Continue in the full course

This lesson is section 10.2 (AGCPH1002) of our free GCSE Physics course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Physics course.

 

 

 

 

 

 

 

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