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P10.2 Weight and Terminal Velocity

GCSE Combined Science · AQA · Chapter P10: Force and Motion

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: mass and weight, W = m g, falling freely, air resistance 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 calculate weight using W = m g
  • Describe the motion of an object falling freely under gravity (acceleration about 9.8 m/s²)
  • Describe how the air resistance on an object depends on its speed and its area
  • 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.
  • 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 resistive forces on it balance its weight.

Mass and weight

  • 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)

weight = mass × gravitational field strength W = m g

W in newtons, N; m in kilograms, kg; g in newtons per kilogram, N/kg. Earth 9.8 N/kg; Moon 1.6 N/kg; Mars 3.7 N/kg

Worked example: a lander on Mars

A lander weighs 3430 N on the Earth. Calculate its weight on Mars.
Mass first: m = W ÷ g = 3430 N ÷ 9.8 N/kg = 350 kg (the mass does not change)
Weight on Mars: W = 350 kg × 3.7 N/kg = 1295 N

Answer: about 1300 N

Exam tip: Weight is a force, so its unit is always the newton. Weight acts through the object’s centre of mass and is directly proportional to mass (W ∝ m).

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. Dropped together from 1.5 m, a steel ball and a wooden ball of the same size land at almost the same moment, because air resistance is tiny compared with their weights. A dropped leaf drifts down slowly because the air resistance on it soon becomes as big as its weight.

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. The air resistance on an object is larger when:

  • its speed is higher
  • its area facing the direction of motion is larger (it is less streamlined).
  • 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 25 g ball bearing falls through oil. At one moment the drag on it is 0.080 N. Calculate its acceleration at that moment.
m = 0.025 kg; weight = 0.025 × 9.8 = 0.245 N
resultant force = 0.245 − 0.080 = 0.165 N downwards; a = F ÷ m = 0.165 ÷ 0.025 = 6.6 m/s²

Answer: 6.6 m/s² downwards

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 about 54 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 6 m/s), safe for landing.
  • F: landing – the ground stops the skydiver.

Worked example: just after the parachute opens

A skydiver of mass 70 kg has 1800 N of air resistance on him just after his parachute opens.
weight = 70 × 9.8 = 686 N; resultant force = 1800 − 686 = 1114 N upwards
a = 1114 ÷ 70 = 15.9 m/s² upwards

Answer: he decelerates at about 16 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. He slows down but is still moving downwards.
  • Thinking heavier objects always fall faster. With no air resistance, every object accelerates at g.

Revision questions

Try these, then check your answers below each one.

Question 1. Calculate the weight on the Earth of a dog of mass 12 kg. (g = 9.8 N/kg)

Answer: 117.6 N

Working: W = 12 × 9.8 = 117.6 N.

Question 2. What is the unit of weight? (Choose from: newton (N), kilogram (kg), newton per kilogram (N/kg), metre per second squared (m/s²).)

Answer: newton (N)

Working: Weight is the force of gravity on an object, so it is measured in newtons.

Question 3. The graph shows the velocity of a skydiver. During which part does she first fall at terminal velocity? (Choose from: A, B, C, D, E.)

Answer: C

Working: C is the first flat section, where air resistance equals weight and the velocity is constant. E is the second, lower terminal velocity after the parachute opens.

Related lessons

Continue in the full course

This lesson is section P10.2 (AGCSC3902) of our free GCSE Combined Science 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 Combined Science course.

 

 

 

 

 

 

 

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