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Motion, Forces and Energy - Mass, weight and density

Grade 12IGCSEPhysics

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

🔑Concepts

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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).

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Weight is a gravitational force on an object that has mass. Unlike mass, weight is a vector quantity measured in Newtons (NN) and its direction is towards the center of the planet.

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The gravitational field strength (gg) is the force per unit mass. On Earth, g≈9.8 m/s2g \approx 9.8\,m/s^2 (often taken as 10 m/s210\,m/s^2 for IGCSE calculations).

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Density is defined as the mass per unit volume of a substance. It determines whether an object will float or sink in a given fluid.

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To find the density of an irregular solid, the displacement method is used: the volume of the object is equal to the volume of the liquid it displaces in a measuring cylinder or displacement (Eureka) can.

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Objects with a density less than the density of a fluid (ρobject<ρfluid\rho_{object} < \rho_{fluid}) will float, while those with a higher density will sink.

📐Formulae

W=m⋅gW = m \cdot g

ρ=mV\rho = \frac{m}{V}

Vdisplaced=Vfinal−VinitialV_{displaced} = V_{final} - V_{initial}

1 g/cm3=1000 kg/m31\,g/cm^3 = 1000\,kg/m^3

💡Examples

Problem 1:

An astronaut has a mass of 75 kg75\,kg on Earth. If the gravitational field strength on the Moon is 1.6 m/s21.6\,m/s^2, calculate the astronaut's weight on the Moon.

Solution:

W=m⋅gmoonW = m \cdot g_{moon} W=75 kg⋅1.6 m/s2W = 75\,kg \cdot 1.6\,m/s^2 W=120 NW = 120\,N

Explanation:

Mass is an intrinsic property and does not change regardless of location. Therefore, the mass remains 75 kg75\,kg on the Moon. Weight is the product of mass and the local gravitational field strength.

Problem 2:

A metal cylinder has a mass of 135 g135\,g. When it is immersed in a measuring cylinder containing 50 cm350\,cm^3 of water, the water level rises to 65 cm365\,cm^3. Calculate the density of the metal in g/cm3g/cm^3.

Solution:

V=Vfinal−Vinitial=65 cm3−50 cm3=15 cm3V = V_{final} - V_{initial} = 65\,cm^3 - 50\,cm^3 = 15\,cm^3 ρ=mV\rho = \frac{m}{V} ρ=135 g15 cm3=9.0 g/cm3\rho = \frac{135\,g}{15\,cm^3} = 9.0\,g/cm^3

Explanation:

First, find the volume of the cylinder using the displacement of water. Then, divide the mass by the calculated volume to find the density.

Mass, weight and density - Revision Notes & Key Formulas | IGCSE Grade 12 Physics