krit.club logo

Physics - Mass, weight, and density

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Mass is a measure of the quantity of matter in an object at rest relative to the observer. It is a scalar quantity measured in kilograms (kgkg).

•

Weight is a gravitational force on an object that has mass. It is a vector quantity (acting towards the center of the planet) measured in Newtons (NN).

•

Mass is an intrinsic property and does not change with location, whereas weight changes depending on the gravitational field strength (gg).

•

Gravitational field strength (gg) is the force per unit mass. On Earth, gg is approximately 9.8 N/kg9.8\,N/kg (often rounded to 10 N/kg10\,N/kg in IGCSE exams).

•

Density (ρ\rho) is defined as the mass per unit volume of a substance. Its SI unit is kg/m3kg/m^3, though g/cm3g/cm^3 is frequently used.

•

Objects with a lower density than a liquid will float, while objects with a higher density will sink.

•

The volume of an irregular solid can be found using the displacement method: the volume of the object is equal to the volume of liquid it displaces in a measuring cylinder.

📐Formulae

W=m×gW = m \times g

ρ=mV\rho = \frac{m}{V}

Vsolid=Vfinal−VinitialV_{solid} = V_{final} - V_{initial}

💡Examples

Problem 1:

An astronaut has a mass of 75 kg75\,kg on Earth. Calculate their weight on Earth (g=9.8 N/kgg = 9.8\,N/kg) and their weight on the Moon (g=1.6 N/kgg = 1.6\,N/kg).

Solution:

Weight on Earth: W=75 kg×9.8 N/kg=735 NW = 75\,kg \times 9.8\,N/kg = 735\,N. Weight on Moon: W=75 kg×1.6 N/kg=120 NW = 75\,kg \times 1.6\,N/kg = 120\,N.

Explanation:

The mass remains 75 kg75\,kg in both locations, but the weight changes because the gravitational field strength (gg) is different.

Problem 2:

A metal block has a mass of 1.2 kg1.2\,kg and a volume of 150 cm3150\,cm^3. Calculate the density of the metal in g/cm3g/cm^3.

Solution:

First, convert mass to grams: 1.2 kg=1200 g1.2\,kg = 1200\,g. Then use the formula: ρ=1200 g150 cm3=8.0 g/cm3\rho = \frac{1200\,g}{150\,cm^3} = 8.0\,g/cm^3.

Explanation:

To find density in g/cm3g/cm^3, the mass must be in grams and the volume in cm3cm^3.

Problem 3:

An irregular stone is placed in a measuring cylinder containing 40 ml40\,ml of water. The water level rises to 65 ml65\,ml. If the stone has a mass of 125 g125\,g, what is its density?

Solution:

Volume of stone: V=65 ml−40 ml=25 ml=25 cm3V = 65\,ml - 40\,ml = 25\,ml = 25\,cm^3. Density: ρ=125 g25 cm3=5.0 g/cm3\rho = \frac{125\,g}{25\,cm^3} = 5.0\,g/cm^3.

Explanation:

Since 1 ml=1 cm31\,ml = 1\,cm^3, the displacement of water gives the volume of the irregular object directly.