Mass and weight

Ask someone "how much do you weigh?" and they answer in kilograms — but a kilogram is a unit of mass, not weight. This mix-up is one of the most common in all of school physics. Mass is how much matter a body contains; weight is the pull of gravity on that matter. One is a scalar that never changes, wherever you carry the object. The other is a force, tied to the local value of $g$ , that genuinely differs from place to place. On the ICSE Class 9 syllabus, inside the chapter Laws of Motion, you need to state this distinction precisely, use $W = mg$ in calculations, and work confidently with gravitational units of force such as the kilogram-force. This article builds that understanding with one derivation, one fully worked numerical, and the traps examiners love to set.

The core insight

Mass is the fixed quantity of matter in a body (a scalar, in kg), while weight is the gravitational force on that body, $W = mg$ (a vector, in N), so weight changes with location even though mass never does.

The core idea

Mass is matter; weight is a pull. Every object is made of a certain amount of matter, and that amount is its mass, $m$ . Mass is a scalar quantity: it has size but no direction, is measured in kilograms (kg) in SI units, and stays exactly the same no matter where the object is taken. A 5 kg block has a mass of 5 kg on Earth, on the Moon, or floating in empty space, because the matter in it has not changed.

Weight is different in kind, not just in size. Weight, $W$ , is the force with which gravity pulls a body toward the centre of the Earth (or of whichever astronomical body it is near). Because it is a force, weight is a vector, always directed straight down toward the centre of that body, and its SI unit is the newton (N), not the kilogram. Its size is fixed by how strongly gravity acts there, measured by the local acceleration due to gravity, $g$ :

W = mg

where: $W$ is weight in newtons, $m$ is mass in kilograms, and $g$ is the local acceleration due to gravity in m/s². On the surface of the Earth you take $g = 9.8\ \text{m/s}^2$ throughout this syllabus; you are not required to work out how $g$ itself changes with altitude or depth, only to compare mass and weight using a given value of $g$ .

Why weight travels, mass does not. Since $g$ is not the same everywhere — roughly one-sixth as large on the Moon as on Earth — the weight of a fixed mass genuinely changes when moved, even though the mass in $W = mg$ never does. The table below packs this into one glance.

	Mass	Weight
Measures	quantity of matter	gravitational force on the body
Type	scalar	vector (force)
SI unit	kilogram (kg)	newton (N)
Changes with location?	no — constant everywhere	yes — depends on local $g$
Measured by	beam balance	spring balance

A beam balance compares masses; it does not weigh things. It balances one unknown mass against known standard masses on the other pan, using gravity equally on both sides, so it reads the same in every location (Earth or Moon alike). A spring balance, by contrast, hangs the object from a spring and reads how far gravity stretches it — since the stretch depends on the actual pull of gravity at that spot, a spring balance genuinely measures weight, and its reading would differ on the Moon.

Figure

A beam balance shown level with equal masses on both pans labelled measures mass and stays the same everywhere, next to a spring balance stretched by a hanging weight labelled measures weight and changes with g

Gravitational units of force. Because weight is felt so directly, force is often stated in gravitational units instead of newtons. The kilogram-force (kgf) is defined as the weight of a 1 kg mass at a place where $g = 9.8\ \text{m/s}^2$ , and the gram-force (gf) as the weight of a 1 g mass:

1\ \text{kgf} = 9.8\ \text{N}, \qquad 1\ \text{gf} = 9.8\times10^{-3}\ \text{N} = 980\ \text{dyne}

where: the numbers come directly from $W = mg$ with $m = 1\ \text{kg}$ (or 1 g) and $g = 9.8\ \text{m/s}^2$ . A convenient consequence is that the numerical value of an object's weight in kgf equals the numerical value of its mass in kg — a 45 kg boy weighs "45" whether you call it 45 kgf or, after multiplying by 9.8, 441 N. That agreement is a feature of the unit's definition; mass and weight remain conceptually distinct quantities.

Derivation AWeight as the gravitational force

W = mg

The goal is to show, from Newton's second law alone, why the force pulling any mass toward the Earth is proportional to that mass with the local value of $g$ as the constant of proportionality — the single fact that makes the entire mass-versus-weight comparison, and every gravitational-unit conversion in this topic, provable rather than merely stated.

Step 1. Recall Newton's second law: a resultant force $F$ acting on a body of mass $m$ produces an acceleration $a$ given by $F = ma$ . This holds for any force and any cause of acceleration whatsoever — it is not specific to gravity.

Step 2. Apply it to a body falling freely near the Earth's surface, acted on only by gravity. Experiment shows every such body, regardless of its own mass, accelerates downward at the same rate, called the acceleration due to gravity, $g \approx 9.8\ \text{m/s}^2$ at the surface. So the acceleration in Step 1 is specifically $a = g$ for this case.

Step 3. Substitute $a = g$ into $F = ma$ . The force producing this particular acceleration — the pull of the Earth's gravity on the body — is therefore $F = mg$ . This force is what we name the body's weight, $W$ , giving $W = mg$ .

Step 4. Read off the consequence that drives the whole topic. Because $m$ is fixed for a given body, $W$ scales directly with $g$ : take the same body somewhere $g$ is smaller (the Moon) and $W$ shrinks in exact proportion, while $m$ itself never changes. Mass is the unchanging cause; weight is the location-dependent effect. Consistent with $g = 9.8\ \text{m/s}^2$ used throughout: a 1 kg mass has weight $1 \times 9.8 = 9.8\ \text{N}$ , which is exactly the definition of 1 kgf used above.

Worked example

A boy has a mass of 45 kg. Take $g = 9.8\ \text{m/s}^2$ . Find (a) his weight in newtons, (b) his weight in kilogram-force, and (c) his weight on the Moon, where $g_{\text{Moon}} = 1.63\ \text{m/s}^2$ , along with his mass there.

Step 1. Apply $W = mg$ on Earth. $W = 45 \times 9.8 = 441\ \text{N}$ . This is his weight in SI units.

Step 2. Convert to gravitational units. Since $1\ \text{kgf} = 9.8\ \text{N}$ , $W = \dfrac{441}{9.8}\ \text{kgf} = 45\ \text{kgf}$ — numerically the same as his mass in kg, exactly as the definition of kgf guarantees.

Step 3. Find the mass and weight on the Moon. Mass does not depend on location, so his mass on the Moon is still $m = 45\ \text{kg}$ . His weight there is $W_{\text{Moon}} = 45 \times 1.63 = 73.35\ \text{N}$ , roughly one-sixth of his weight on Earth, because the Moon's $g$ is roughly one-sixth of Earth's.

Step 4. State the answer. His weight is 441 N (= 45 kgf) on Earth and 73.35 N on the Moon, while his mass stays 45 kg in both places — the clearest possible illustration that mass is fixed and weight is not.

Competitive stretch

If instead you are told an astronaut's weight on the Moon is 98 N, her mass follows by rearranging $W = mg$ : $m = W/g = 98/1.63 = 60.1\ \text{kg}$ . Because mass is location-independent, her weight back on Earth would be $W = mg = 60.1 \times 9.8 = 589\ \text{N}$ — always work through the mass, never scale weights directly by a ratio of distances or guesswork.

Common traps

Quoting weight in kilograms — kg is a unit of mass; weight must be in newtons, or in gravitational units (kgf, gf) if the question asks for those.
Treating weight as constant like mass — weight depends on $g$ and genuinely differs from place to place, even though this syllabus does not ask you to compute how $g$ itself varies with altitude or depth.
Believing a beam balance measures weight — it compares two masses against each other and gives the same reading anywhere; only a spring balance, which responds to the local pull of gravity, measures weight.
Forgetting weight is a vector — it always points toward the centre of the attracting body, unlike mass, which is a scalar with magnitude only and no direction.

Ask the tutor: Want more practice problems comparing mass and weight across Earth, the Moon, and gravitational units — ask the AI tutor to generate a fresh set.

If a beam balance always reads the same everywhere, does that mean it is actually measuring mass and not weight

Yes, exactly. A beam balance compares an unknown mass against known standard masses using two pans, and it balances when the gravitational pull on both sides is equal — since gravity acts equally on both pans regardless of how strong or weak $g$ is at that location, the comparison, and hence the reading, comes out the same everywhere. That is precisely why it measures the location-independent quantity, mass, and not weight.

Why does the numerical value of weight in kgf always equal the numerical value of mass in kg

Because that is how the kilogram-force is defined: 1 kgf is defined to be the weight of exactly 1 kg of mass at $g = 9.8\ \text{m/s}^2$ . So for any mass $m$ in kg, its weight in kgf is just $m$ again, while its weight in newtons is $mg = 9.8m$ . The two numbers matching is a built-in feature of the unit's definition, not a coincidence of physics — but the units kg (mass) and kgf (force) still name different kinds of quantity.

Why doesn't this topic ask me to calculate how g changes with height or depth

That needs the inverse-square law of gravitation applied to varying distance from Earth's centre, which goes beyond ICSE Class 9 Laws of Motion. You are expected only to know $g$ differs from place to place (Earth versus the Moon, say) and to use $W = mg$ with whatever value of $g$ a question gives you — not to derive how $g$ itself changes with altitude or depth.

Mass and weight

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads