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What is the difference between mass and weight, and why does a falling object reach a steady speed?

Mass and weight, the equation W = mg, free fall, air resistance and how a falling object reaches terminal velocity.

A CCEA GCSE Physics answer on the difference between mass and weight, the equation W = mg, free fall and how air resistance leads a falling object to reach a constant terminal velocity.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

CCEA wants you to distinguish mass from weight, use the equation W = mg, describe free fall, and explain how air resistance causes a falling object to reach a constant terminal velocity. The terminal-velocity explanation is a classic extended-answer question.

The answer

Mass and weight

So a 10 kg10\ \text{kg} bag has a weight of 98 N98\ \text{N} on Earth but only about 16 N16\ \text{N} on the Moon, while its mass stays 10 kg10\ \text{kg}.

Free fall

When an object falls with only gravity acting (no air resistance), it is in free fall. Its acceleration equals gg, about 9.8 m/s29.8\ \text{m/s}^2, the same for all masses, because a heavier object has a larger weight but also a larger mass to accelerate.

Terminal velocity

When an object falls through air, air resistance (drag) acts upward and increases with speed.

The stages are:

  1. At the start, only weight acts, so the resultant force is large and the object accelerates.
  2. As speed rises, air resistance rises, so the resultant force and acceleration fall.
  3. When air resistance equals weight, the resultant force is zero and speed is constant: terminal velocity.

Worked example: a falling raindrop

Examples in context

Example 1. A skydiver opening a parachute
At terminal velocity the diver falls steadily. Opening the parachute hugely increases air resistance, so resistance now exceeds weight, the resultant force is upward and the diver decelerates to a new, slower terminal velocity safe for landing.
Example 2. A feather and a hammer
In a vacuum (no air resistance) both fall with the same acceleration gg and land together, as demonstrated on the Moon, because free-fall acceleration does not depend on mass.
Example 3. A shuttlecock
A badminton shuttlecock has a very low weight but a large surface area, so air resistance grows quickly and matches its weight at a low speed. It therefore reaches terminal velocity almost at once and drops slowly, which is why it falls so much more gently than a smooth ball of the same mass.

A velocity-time graph for a falling object shows the link clearly: the line starts steep (large acceleration near gg), curves as it levels off (acceleration falling as air resistance grows), and finally becomes horizontal once terminal velocity is reached and the resultant force is zero.

Try this

Q1. State the difference between mass and weight. [2 marks]

  • Cue. Mass is the amount of matter (kg), the same everywhere; weight is the force of gravity (N) and depends on gg.

Q2. Find the weight of a 12 kg12\ \text{kg} object on Earth, g=9.8 N/kgg = 9.8\ \text{N/kg}. [2 marks]

  • Cue. W=mg=12×9.8=118 NW = mg = 12 \times 9.8 = 118\ \text{N} (about 120 N120\ \text{N}).

Q3. State the condition for an object to fall at terminal velocity. [1 mark]

  • Cue. Air resistance equals weight, so the resultant force is zero.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style3 marksCalculate the weight of a 65 kg student on Earth, where the gravitational field strength is 9.8 N/kg, and state how the student's mass and weight would change on the Moon.
Show worked answer →

Weight on Earth:

W=mg=65×9.8=637 N.W = mg = 65 \times 9.8 = 637\ \text{N}.

On the Moon the mass stays 65 kg (mass does not change with location), but the weight is smaller because the Moon's gravitational field strength is about one sixth of Earth's, so the weight would be roughly 106 N.

Markers reward W=mgW = mg, the value 637 N, mass unchanged, and weight smaller on the Moon.

CCEA style5 marksDescribe and explain, in terms of forces, how the speed of a skydiver changes from the moment they jump until they reach terminal velocity, before the parachute opens.
Show worked answer →

At the start the only significant force is weight, so the resultant force is large and downward and the skydiver accelerates.

As speed increases, air resistance increases (it depends on speed). The resultant force gets smaller, so the acceleration decreases.

Eventually air resistance equals weight. The resultant force is zero, so by Newton's first law the skydiver falls at a constant maximum speed, the terminal velocity.

Markers reward: weight only at first so large acceleration; air resistance rises with speed; resultant force and acceleration fall; at terminal velocity air resistance equals weight, resultant force zero, constant speed.

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