Answers – 1.3 – Mass and Weight

IGCSE Physics  |  Practice Test — Answers & Worked Solutions

Section A — Recall
Questions 1–10
1.

Write down the definition of mass.

Answer

Mass is the amount of matter in an object, measured relative to the observer when the object is at rest.

Mass = the quantity of matter in an object at rest relative to the observer
Revision: Mass is measured in kilograms (kg). It does not change depending on where the object is — the mass is the same on Earth, the Moon, or anywhere in space.
2.

Write down the definition of weight.

Answer

Weight is the gravitational force acting on an object that has mass.

Weight = the gravitational force on an object  (unit: newton, N)
Revision: Weight is a force, so it is measured in newtons (N). Unlike mass, weight changes depending on location because the strength of gravity is different in different places.
3.

State the unit of mass and the unit of weight.

Answer
Mass → kilogram (kg)   |   Weight → newton (N)
Revision: A common exam mistake is to give weight in kg. Remember: weight is a force, so its unit is always newtons (N), not kilograms.
4.

Write down the equation for gravitational field strength $g$. Define each symbol and state its unit.

Answer
$$g = \frac{W}{m}$$
  • $g$ = gravitational field strength — unit: N/kg
  • $W$ = weight — unit: N
  • $m$ = mass — unit: kg
$g = \dfrac{W}{m}$, which rearranges to $W = m \times g$
Revision: You need to be able to rearrange this formula. To find mass: $m = W \div g$. To find weight: $W = m \times g$.
5.

State the approximate value and unit of gravitational field strength at the surface of the Earth.

Answer
$g \approx 9.8\ \text{N/kg}$
Revision: This means every 1 kg of mass has a gravitational force of approximately 9.8 N acting on it at Earth’s surface. This value changes on other planets — it is about 1.6 N/kg on the Moon and 3.7 N/kg on Mars.
6.

Apart from being a measure of force per unit mass, what else does $g$ represent in terms of motion?

Answer

$g$ is also equal to the acceleration of free fall.

$g \approx 9.8\ \text{m/s}^2$ (acceleration of free fall near Earth’s surface)
Revision: The value 9.8 N/kg and 9.8 m/s² are the same quantity described in two ways. As a field strength it tells you the force per kg. As an acceleration it tells you how quickly a freely falling object speeds up.
7.

An object is taken from Earth to the Moon. What happens to its mass? Explain your answer.

Answer

The mass does not change.

Mass stays the same — it is the amount of matter in the object, which does not depend on location.
Revision: Mass is a property of the object itself, not of its surroundings. It does not matter where in the universe the object is — the amount of matter inside it is always the same.
8.

An object is taken from Earth to the Moon. What happens to its weight? Explain your answer.

Answer

The weight decreases.

Weight decreases because the Moon’s gravitational field strength (1.6 N/kg) is much less than Earth’s (9.8 N/kg).
Revision: Weight = $m \times g$. The mass stays the same, but $g$ is smaller on the Moon, so the weight is smaller. Weight depends on where the object is; mass does not.
9.

Name one piece of equipment used to compare the masses of two objects.

Answer
A beam balance
Revision: A beam balance compares masses by balancing the gravitational forces on both sides. Because both sides experience the same $g$, the balance gives the correct result anywhere — even on the Moon or another planet.
10.

Define what is meant by a gravitational field. Supplement

Answer

A gravitational field is a region of space where a mass experiences a force due to gravity.

A gravitational field is a region of space in which a mass experiences a gravitational force.
Revision: Any object with mass creates a gravitational field around it. The strength of the field decreases as you move further away from the object. On Earth’s surface, $g \approx 9.8\ \text{N/kg}$.
Section B — Application
Questions 11–20
11.

A bag has a mass of 4 kg. Calculate its weight on Earth. ($g = 9.8\ \text{N/kg}$)

Answer
  1. Formula: $$W = m \times g$$
  2. Given: $m = 4\ \text{kg}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$W = 4 \times 9.8$$
  4. Arithmetic: $4 \times 9.8 = 4 \times (10 – 0.2) = 40 – 0.8 = 39.2$
  5. Answer: $$W = 39.2\ \text{N}$$
$W = 39.2\ \text{N}$
Revision: Always include the unit (N) in your answer. Weight is a force, not a mass.
12.

An object weighs 19.6 N on Earth ($g = 9.8\ \text{N/kg}$). Calculate its mass.

Answer
  1. Formula: $$g = \frac{W}{m} \implies m = \frac{W}{g}$$
  2. Given: $W = 19.6\ \text{N}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$m = \frac{19.6}{9.8}$$
  4. Check: $9.8 \times 2 = 19.6$  ✓
  5. Answer: $$m = 2\ \text{kg}$$
$m = 2\ \text{kg}$
Revision: To find mass, rearrange $W = m \times g$ to get $m = W \div g$.
13.

A rock weighs 98 N on Earth ($g = 9.8\ \text{N/kg}$). Calculate the mass of the rock.

Answer
  1. Formula: $$m = \frac{W}{g}$$
  2. Given: $W = 98\ \text{N}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$m = \frac{98}{9.8}$$
  4. Check: $9.8 \times 10 = 98$  ✓
  5. Answer: $$m = 10\ \text{kg}$$
$m = 10\ \text{kg}$
14.

An astronaut has a mass of 60 kg. Calculate their weight on the Moon, where $g = 1.6\ \text{N/kg}$.

Answer
  1. Formula: $$W = m \times g$$
  2. Given: $m = 60\ \text{kg}$, $g = 1.6\ \text{N/kg}$
  3. Substitute: $$W = 60 \times 1.6$$
  4. Arithmetic: $60 \times 1.6 = 60 \times 1 + 60 \times 0.6 = 60 + 36 = 96$
  5. Answer: $$W = 96\ \text{N}$$
$W = 96\ \text{N}$
Revision: The astronaut’s mass is still 60 kg on the Moon — only the weight changes because $g$ is different.
15.

A probe has a mass of 3 kg. Calculate its weight on Mars, where $g = 3.7\ \text{N/kg}$.

Answer
  1. Formula: $$W = m \times g$$
  2. Given: $m = 3\ \text{kg}$, $g = 3.7\ \text{N/kg}$
  3. Substitute: $$W = 3 \times 3.7$$
  4. Arithmetic: $3 \times 3.7 = 3 \times 3 + 3 \times 0.7 = 9 + 2.1 = 11.1$
  5. Answer: $$W = 11.1\ \text{N}$$
$W = 11.1\ \text{N}$
16.

An object weighs 44.1 N on Earth ($g = 9.8\ \text{N/kg}$).
(a) Calculate the mass of the object.  (b) Calculate the weight of the object on the Moon ($g = 1.6\ \text{N/kg}$).

Answer
(a) Mass of the object
  1. Formula: $$m = \frac{W}{g}$$
  2. Given: $W = 44.1\ \text{N}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$m = \frac{44.1}{9.8}$$
  4. Check: $9.8 \times 4 = 39.2$; $9.8 \times 0.5 = 4.9$; $39.2 + 4.9 = 44.1$  ✓   so $m = 4.5$
  5. Answer: $$m = 4.5\ \text{kg}$$
$m = 4.5\ \text{kg}$
(b) Weight on the Moon
  1. Formula: $$W = m \times g$$
  2. Given: $m = 4.5\ \text{kg}$, $g = 1.6\ \text{N/kg}$
  3. Substitute: $$W = 4.5 \times 1.6$$
  4. Arithmetic: $4 \times 1.6 = 6.4$; $0.5 \times 1.6 = 0.8$; $6.4 + 0.8 = 7.2$
  5. Answer: $$W = 7.2\ \text{N}$$
$W = 7.2\ \text{N}$
17.

Two objects are placed on a beam balance. Object A has a mass of 500 g. Object B also has a mass of 500 g.
(a) What does the balance show on Earth?  (b) The same experiment is repeated on the Moon. What does the balance show? Explain why.

Answer
(a) On Earth

The balance is level (balanced). Both objects have equal mass, so equal gravitational force acts on each side.

The balance is balanced — both sides carry the same mass (500 g).
(b) On the Moon

The balance is still balanced.

A beam balance compares masses by comparing the gravitational force on each side. On the Moon, $g$ is smaller — but it is the same $g$ acting on both sides. The ratio of forces is unchanged, so the balance still reads level.

Still balanced — a beam balance measures mass, which does not change with location.
Revision: A beam balance gives the correct mass reading anywhere in the universe. A spring balance (which measures force directly) would give a different reading on the Moon because the gravitational force is smaller.
18.

A space probe on the Moon has a weight of 24 N ($g_\text{Moon} = 1.6\ \text{N/kg}$).
(a) Calculate the mass of the probe.  (b) Calculate the weight of the probe on Earth ($g = 9.8\ \text{N/kg}$).

Answer
(a) Mass of the probe
  1. Formula: $$m = \frac{W}{g}$$
  2. Given: $W = 24\ \text{N}$, $g = 1.6\ \text{N/kg}$
  3. Substitute: $$m = \frac{24}{1.6}$$
  4. Arithmetic: $1.6 \times 10 = 16$; $1.6 \times 5 = 8$; $16 + 8 = 24$  ✓   so $m = 15$
  5. Answer: $$m = 15\ \text{kg}$$
$m = 15\ \text{kg}$
(b) Weight on Earth
  1. Formula: $$W = m \times g$$
  2. Given: $m = 15\ \text{kg}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$W = 15 \times 9.8$$
  4. Arithmetic: $15 \times 10 = 150$; $15 \times 0.2 = 3$; $150 – 3 = 147$
  5. Answer: $$W = 147\ \text{N}$$
$W = 147\ \text{N}$
19.

Calculate the weight of a 25 kg object on a planet where $g = 12\ \text{N/kg}$.

Answer
  1. Formula: $$W = m \times g$$
  2. Given: $m = 25\ \text{kg}$, $g = 12\ \text{N/kg}$
  3. Substitute: $$W = 25 \times 12$$
  4. Arithmetic: $25 \times 12 = 25 \times 10 + 25 \times 2 = 250 + 50 = 300$
  5. Answer: $$W = 300\ \text{N}$$
$W = 300\ \text{N}$
20.

A student’s weight on Earth is 637 N ($g = 9.8\ \text{N/kg}$). Calculate their mass.

Answer
  1. Formula: $$m = \frac{W}{g}$$
  2. Given: $W = 637\ \text{N}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$m = \frac{637}{9.8}$$
  4. Arithmetic: $9.8 \times 60 = 588$; $9.8 \times 5 = 49$; $588 + 49 = 637$  ✓   so $m = 65$
  5. Answer: $$m = 65\ \text{kg}$$
$m = 65\ \text{kg}$
Section C — Challenge
Questions 21–25
21.

An astronaut has a mass of 80 kg.
(a) Calculate the astronaut’s weight on Earth ($g = 9.8\ \text{N/kg}$).  (b) Calculate the astronaut’s weight on the Moon ($g = 1.6\ \text{N/kg}$).  (c) The astronaut is weighed using a beam balance on the Moon. Will the reading be different from on Earth? Explain your answer.

Answer
(a) Weight on Earth
  1. Formula: $$W = m \times g$$
  2. Given: $m = 80\ \text{kg}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$W = 80 \times 9.8$$
  4. Arithmetic: $80 \times 10 = 800$; $80 \times 0.2 = 16$; $800 – 16 = 784$
  5. Answer: $$W = 784\ \text{N}$$
$W = 784\ \text{N}$
(b) Weight on the Moon
  1. Formula: $$W = m \times g$$
  2. Given: $m = 80\ \text{kg}$, $g = 1.6\ \text{N/kg}$
  3. Substitute: $$W = 80 \times 1.6$$
  4. Arithmetic: $80 \times 1 = 80$; $80 \times 0.6 = 48$; $80 + 48 = 128$
  5. Answer: $$W = 128\ \text{N}$$
$W = 128\ \text{N}$
(c) Beam balance reading on the Moon

The beam balance will show the same reading — 80 kg.

A beam balance works by comparing the gravitational force on both sides. On the Moon, $g$ is smaller, but it acts equally on both sides. The ratio of forces is the same, so the balance is still level at 80 kg.

Beam balance reads 80 kg — mass is unchanged, and a beam balance measures mass, not weight.
Revision: This is a key distinction. A spring balance would give a lower reading on the Moon (because it measures force). A beam balance gives the same reading anywhere because it compares masses.
22.

A student makes the following statement: “My mass is 50 kg on Earth. So on the Moon my mass must be less, because the Moon’s gravity is weaker.”
(a) Identify and explain the scientific error.  (b) Calculate the student’s weight on Earth ($g = 9.8\ \text{N/kg}$).  (c) Calculate the student’s weight on the Moon ($g = 1.6\ \text{N/kg}$).

Answer
(a) The error

The student is confusing mass with weight. Mass does not change with location — it is the amount of matter in the object. Only weight changes, because weight depends on the gravitational field strength ($g$), which is different on the Moon.

Error: mass does not change with location. The student’s mass is still 50 kg on the Moon. Their weight would be less.
(b) Weight on Earth
  1. Formula: $$W = m \times g$$
  2. Given: $m = 50\ \text{kg}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$W = 50 \times 9.8$$
  4. Arithmetic: $50 \times 10 = 500$; $50 \times 0.2 = 10$; $500 – 10 = 490$
  5. Answer: $$W = 490\ \text{N}$$
$W = 490\ \text{N}$
(c) Weight on the Moon
  1. Formula: $$W = m \times g$$
  2. Given: $m = 50\ \text{kg}$, $g = 1.6\ \text{N/kg}$
  3. Arithmetic: $50 \times 1.6 = 80$
  4. Answer: $$W = 80\ \text{N}$$
$W = 80\ \text{N}$
Revision: The student’s mass is 50 kg everywhere. On Earth their weight is 490 N; on the Moon it drops to 80 N. This is one of the most common exam mistakes — always be clear about which quantity you are discussing.
23.

An object has a weight of 78.4 N on Earth ($g = 9.8\ \text{N/kg}$).
(a) Calculate the mass of the object.  (b) Calculate the weight on a planet where $g = 6.2\ \text{N/kg}$.  (c) Explain, using the concept of gravitational field strength, why the weight is different. Supplement

Answer
(a) Mass
  1. Formula: $$m = \frac{W}{g}$$
  2. Given: $W = 78.4\ \text{N}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$m = \frac{78.4}{9.8}$$
  4. Check: $9.8 \times 8 = 78.4$  ✓
  5. Answer: $$m = 8\ \text{kg}$$
$m = 8\ \text{kg}$
(b) Weight on the other planet
  1. Formula: $$W = m \times g$$
  2. Given: $m = 8\ \text{kg}$, $g = 6.2\ \text{N/kg}$
  3. Arithmetic: $8 \times 6 = 48$; $8 \times 0.2 = 1.6$; $48 + 1.6 = 49.6$
  4. Answer: $$W = 49.6\ \text{N}$$
$W = 49.6\ \text{N}$
(c) Explanation using gravitational field strength Supplement

Weight is the force that a gravitational field exerts on a mass. The stronger the gravitational field, the greater the force on the same mass.

This planet has $g = 6.2\ \text{N/kg}$, which is less than Earth’s $g = 9.8\ \text{N/kg}$. Its gravitational field is weaker, so it exerts a smaller force on the object. Since $W = m \times g$ and $g$ is smaller, the weight is smaller.

Smaller $g$ → weaker gravitational field → smaller force on the mass → smaller weight.
Revision: The mass (8 kg) is the same on both planets. Only the weight changes because $g$ is different. This is the core idea of Supplement point 5.
24.

Supplement
(a) State what a gravitational field is and describe how it affects a mass placed within it.  (b) Explain how weight is the effect of a gravitational field on a mass.  (c) An object of mass 2 kg is taken to deep space where $g \approx 0\ \text{N/kg}$. State its weight and whether its mass has changed. Explain.

Answer
(a) Gravitational field

A gravitational field is a region of space where a mass experiences a gravitational force. Any object with mass creates a gravitational field around it. A mass placed inside this field will feel a pull towards the source of gravity.

A gravitational field is a region where a mass experiences a force due to gravity.
(b) Weight as the effect of a gravitational field

When a mass is placed inside a gravitational field, the field exerts a force on it. This force is what we call weight. The stronger the gravitational field (higher $g$), the greater the force on the mass, and therefore the greater the weight.

Weight is the force a gravitational field exerts on a mass: $W = m \times g$.
(c) Object in deep space

Weight: In deep space, $g \approx 0\ \text{N/kg}$, so:

$$W = m \times g = 2 \times 0 = 0\ \text{N}$$

Mass: The mass is still 2 kg. Mass is the amount of matter in the object — it does not depend on gravity. Only weight changes.

Weight = 0 N in deep space. Mass remains 2 kg — mass never changes with location.
Revision: An object in deep space is weightless (weight = 0 N) but it still has the same mass. This is why astronauts float in space — there is negligible gravitational field acting on them, so no weight force pulls them down.
25.

A new planet is discovered with $g = 4.9\ \text{N/kg}$. A probe of mass 200 kg is sent to this planet. Supplement
(a) Calculate the weight of the probe on the new planet.  (b) Calculate the weight of the probe on Earth ($g = 9.8\ \text{N/kg}$).  (c) A scientist uses a beam balance to measure a rock on this planet. It reads 5 kg. What would the rock’s mass be on Earth? Explain.

Answer
(a) Weight on the new planet
  1. Formula: $$W = m \times g$$
  2. Given: $m = 200\ \text{kg}$, $g = 4.9\ \text{N/kg}$
  3. Substitute: $$W = 200 \times 4.9$$
  4. Arithmetic: $200 \times 5 = 1000$; $200 \times 0.1 = 20$; $1000 – 20 = 980$
  5. Answer: $$W = 980\ \text{N}$$
$W = 980\ \text{N}$
(b) Weight on Earth
  1. Formula: $$W = m \times g$$
  2. Given: $m = 200\ \text{kg}$, $g = 9.8\ \text{N/kg}$
  3. Substitute: $$W = 200 \times 9.8$$
  4. Arithmetic: $200 \times 10 = 2000$; $200 \times 0.2 = 40$; $2000 – 40 = 1960$
  5. Answer: $$W = 1960\ \text{N}$$
$W = 1960\ \text{N}$
(c) Rock’s mass on Earth

The rock’s mass on Earth is still 5 kg.

A beam balance compares masses. It gives the correct mass reading regardless of the gravitational field strength, because both sides of the balance experience the same $g$. Mass does not change with location — so a rock that has a mass of 5 kg on the new planet also has a mass of 5 kg on Earth.

Mass = 5 kg on Earth — mass is constant everywhere; only weight changes with location.
Revision: The weight of this rock would be different on Earth ($W = 5 \times 9.8 = 49\ \text{N}$) compared to the new planet ($W = 5 \times 4.9 = 24.5\ \text{N}$). But the mass is always 5 kg.

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