Mass and weight

1. Overview

Mass and weight are two terms often used interchangeably in daily life, but they have distinct meanings in physics. Understanding the relationship between the amount of matter in an object and the gravitational force acting upon it is fundamental to studying how objects move and interact across the universe.

Key Definitions

• Mass: A measure of the quantity of matter in an object at rest relative to the observer.
• Weight: A gravitational force on an object that has mass.
• Gravitational Field Strength ($g$): The force per unit mass exerted by a gravitational field.
• Gravitational Field: A region in which a mass experiences a force due to gravitational attraction.

Core Content

Understanding Mass

• Mass is a property of the object itself.
• It does not change regardless of where the object is located (e.g., your mass is the same on Earth, on the Moon, or floating in deep space).
• Mass is measured in kilograms (kg).

Understanding Weight

• Weight is a force, not a mass.
• Because it is a force, it is measured in Newtons (N).
• Weight acts vertically downwards towards the center of the planet or moon.
• The weight of an object depends on the gravitational field strength of the location.

Comparing Mass and Weight

• Weights (and therefore masses) can be compared using a balance.
• A beam balance compares an unknown mass against a known mass. Since gravity acts on both sides equally, it works anywhere.
• An electronic balance or spring balance measures the downward force (weight) and is usually calibrated to display the mass in kg based on Earth's gravity.

Calculating Weight To find the weight of an object, we use the formula: $$Weight = mass \times gravitational \ field \ strength$$ $$W = m \times g$$

Worked Example (Core): An apple has a mass of 0.1 kg. If the gravitational field strength on Earth is 9.8 N/kg, calculate its weight.

• $W = m \times g$
• $W = 0.1 \times 9.8$
• $W = 0.98\text{ N}$

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Extended Content (Extended Curriculum Only)

Weight as an Effect of a Gravitational Field Weight is not an intrinsic property of an object; it is an effect caused by the presence of a gravitational field. Without a gravitational field, a mass has no weight.

• A gravitational field surrounds any object with mass (like planets).
• When a second mass enters this field, the field exerts a force on it. This force is what we call weight.
• If the field strength ($g$) increases, the weight increases, even if the mass remains constant.

Worked Example (Extended): An astronaut has a mass of 70 kg. The gravitational field strength on Earth is 9.8 N/kg, and on the Moon, it is 1.6 N/kg. Calculate the astronaut's weight on both.

1. On Earth: $W = 70 \times 9.8 = 686\text{ N}$
2. On the Moon: $W = 70 \times 1.6 = 112\text{ N}$ Note: The mass remains 70 kg in both locations, but the weight changes because the field strength changes.

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Key Equations

Equation	Symbols	Units
$W = m \times g$	$W$ = Weight	Newtons (N)
	$m$ = Mass	Kilograms (kg)
	$g$ = Gravitational field strength	Newtons per kilogram (N/kg)

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Common Mistakes to Avoid

• ❌ Wrong: Assuming mass changes when you go to another planet.
• ✓ Right: Mass is constant; only weight changes depending on the gravity of the location.
• ❌ Wrong: Thinking that changing the shape or appearance of an object (like crushing a can) changes its mass.
• ✓ Right: Mass is the "quantity of matter"—changing the shape does not change the amount of atoms present.
• ❌ Wrong: Calculating weight by dividing mass by gravity ($W = m / g$).
• ✓ Right: Weight is the product of mass and gravity ($W = m \times g$). Always check that weight (in N) is a larger value than mass (in kg) when on Earth.
• ❌ Wrong: Giving weight in kg or mass in N.
• ✓ Right: Weight is a Force (Newtons); Mass is Matter (Kilograms).

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Exam Tips

1. Unit Conversion: Examiners often give mass in grams (g). You must convert this to kilograms (kg) before using the $W = mg$ formula (divide by 1000).
2. Check the $g$ value: Look closely at the front of your exam paper or the specific question. Sometimes IGCSE uses $g = 10\text{ N/kg}$ for simplicity, and sometimes it uses $g = 9.8\text{ N/kg}$.
3. The "Mass is Constant" Rule: If a multi-part question asks for the mass of an object on Earth and then its mass on the Moon, the answer is the same for both. Do not perform any calculations for mass when location changes!

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student measures the weight of a small rock using a spring balance on Earth.

(a) Define weight. [2]

(b) The spring balance reads 2.5 N. Calculate the mass of the rock. Assume the gravitational field strength on Earth is 9.8 N/kg. [2]

(c) State what would happen to the reading on the spring balance if the student took the rock and the spring balance to the Moon. [1]

Worked Solution:

(a)

1. Weight is the gravitational force acting on an object with mass. [Definition of weight]

How to earn full marks:

• Correctly state that weight is a force.
• Correctly state that it is a gravitational force.

(b)

1. Recall the equation relating weight, mass, and gravitational field strength: $W = mg$ [Recalling the correct formula]
2. Rearrange the equation to find mass: $m = \frac{W}{g}$ [Rearranging formula correctly]
3. Substitute the given values: $m = \frac{2.5 \text{ N}}{9.8 \text{ N/kg}} = 0.255 \text{ kg}$ [Substituting values and calculating mass]

How to earn full marks:

• Correctly recall and rearrange the formula relating weight, mass, and g.
• Correctly substitute the values of weight and gravitational field strength.
• State the final answer: $\boxed{0.255 \text{ kg}}$
• Include the correct unit (kg).

(c)

1. The reading on the spring balance would decrease. [Stating that weight decreases on the moon]

How to earn full marks:

• State that the reading would decrease.

Common Pitfall: Many students confuse mass and weight. Remember that mass is the amount of "stuff" in an object and stays the same everywhere, while weight is the force of gravity on that "stuff" and changes depending on the gravitational field strength.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A hot air balloon of mass 400 kg (including the basket and burners) is initially stationary on the ground. The balloon is then inflated with hot air, and it begins to rise.

(a) State that mass is a measure of the quantity of matter in an object at rest relative to the observer. [1]

(b) Before the balloon starts to move, identify the two main forces acting on the stationary balloon and state their directions. [2]

(c) As the balloon rises, it experiences an upward force (upthrust) of 4500 N and air resistance of 500 N. Calculate the magnitude of the resultant force acting on the balloon. [2]

(d) Calculate the initial acceleration of the balloon. [3]

Worked Solution:

(a)

1. Mass is a measure of the quantity of matter in an object at rest relative to the observer. [Definition of mass]

How to earn full marks:

• State the definition of mass correctly.

(b)

1. The two main forces are weight, acting downwards, and the normal contact force from the ground, acting upwards. [Identifying the two forces and their directions]

How to earn full marks:

• Identify weight acting downwards.
• Identify the normal contact force acting upwards.

(c)

1. Calculate the net upward force: $4500 \text{ N} - 500 \text{ N} = 4000 \text{ N}$ [Calculate the net upward force from the upthrust and air resistance]
2. Calculate the weight of the balloon: $W = mg = 400 \text{ kg} \times 9.8 \text{ N/kg} = 3920 \text{ N}$ [Calculating the weight of the balloon]
3. Calculate the resultant force: $4000 \text{ N} - 3920 \text{ N} = 80 \text{ N}$ [Calculating the resultant force by subtracting the weight from the net upward force]

How to earn full marks:

• Calculate the net upward force correctly.
• Calculate the weight of the balloon correctly.
• Calculate the resultant force correctly: $\boxed{80 \text{ N}}$
• State the correct unit (N).

(d)

1. Recall Newton's Second Law: $F = ma$ [Recalling Newton's Second Law]
2. Rearrange the equation to find acceleration: $a = \frac{F}{m}$ [Rearranging the formula correctly]
3. Substitute the values: $a = \frac{80 \text{ N}}{400 \text{ kg}} = 0.2 \text{ m/s}^2$ [Substituting the values and calculating the acceleration]

How to earn full marks:

• Correctly recall and rearrange Newton's Second Law.
• Correctly substitute the resultant force and mass.
• State the final answer: $\boxed{0.2 \text{ m/s}^2}$
• Include the correct unit (m/s$^2$).

Common Pitfall: When calculating the resultant force, many students forget to consider all the forces acting on the object, including weight. Make sure to draw a free-body diagram to visualize all the forces involved. Also, remember to use the resultant force in Newton's Second Law, not just any single force.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

Two students are discussing the difference between mass and weight.

(a) Define gravitational field strength. [2]

(b) Student A says "Mass and weight are the same thing, just measured in different units." Student B says "Mass is constant, but weight can change depending on location." Explain which student is correct and why. [4]

Worked Solution:

(a)

1. Gravitational field strength is defined as the gravitational force per unit mass. [Defining gravitational field strength]
2. $g = \frac{W}{m}$ is the equation that defines gravitational field strength. [Defining gravitational field strength]

How to earn full marks:

• State that gravitational field strength is force per unit mass.
• Mention that it is gravitational force.

(b)

1. Student B is correct. [Identifying the correct student]
2. Mass is a measure of the amount of matter in an object and remains constant regardless of location. [Explaining the nature of mass]
3. Weight is the force of gravity on an object and depends on the gravitational field strength, which varies from place to place (e.g., Earth vs. Moon). [Explaining the dependence of weight on gravitational field strength]
4. Therefore, weight can change while mass remains constant. [Concluding that weight changes while mass is constant]

How to earn full marks:

• Identify Student B as correct.
• Explain that mass is constant.
• Explain that weight depends on gravitational field strength.
• State that gravitational field strength varies by location.

Common Pitfall: Students often struggle to articulate the difference between mass and weight conceptually. Focus on understanding that mass is an intrinsic property of an object, while weight is an effect of gravity acting on that mass. Don't just memorize definitions; think about what they mean in different situations.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A physics student is investigating the relationship between mass and weight using various objects and a calibrated electronic balance.

(a) Describe how the student could use the electronic balance to accurately determine the weight of each object. [3]

(b) The student collects the following data:

Object	Mass (kg)	Weight (N)
A	0.10	0.98
B	0.25	2.45
C	0.40	3.92
D	0.55	5.39

Show that the gravitational field strength, $g$, is approximately 9.8 N/kg. Use the data for all four objects to justify your answer. [3]

(c) The student then takes object D (mass 0.55 kg) to the top of a tall mountain where the gravitational field strength is slightly lower, at 9.75 N/kg.

Calculate the percentage decrease in the weight of object D when taken to the top of the mountain, compared to its weight at the base of the mountain (where $g$ = 9.8 N/kg). [3]

Worked Solution:

(a)

1. Place the object on the electronic balance. [Placing the object on the balance]
2. The electronic balance directly displays the weight of the object in Newtons (N). [Reading the weight directly from the balance]
3. Ensure the balance is calibrated to zero before each measurement to eliminate systematic errors. [Calibrating the balance to ensure accuracy]

How to earn full marks:

• Describe placing the object on the balance.
• State that the balance displays weight in Newtons.
• Mention calibrating the balance.

(b)

1. Calculate $g$ for each object using $g = \frac{W}{m}$: [Recalling the relevant equation]
2. For object A: $g = \frac{0.98 \text{ N}}{0.10 \text{ kg}} = 9.8 \text{ N/kg}$ [Calculating g for object A]
3. For object B: $g = \frac{2.45 \text{ N}}{0.25 \text{ kg}} = 9.8 \text{ N/kg}$ [Calculating g for object B]
4. For object C: $g = \frac{3.92 \text{ N}}{0.40 \text{ kg}} = 9.8 \text{ N/kg}$ [Calculating g for object C]
5. For object D: $g = \frac{5.39 \text{ N}}{0.55 \text{ kg}} = 9.8 \text{ N/kg}$ [Calculating g for object D]
6. Since the calculated value of $g$ is approximately 9.8 N/kg for all four objects, this justifies that $g \approx 9.8 \text{ N/kg}$. [Concluding that the calculations show g to be around 9.8 N/kg]

How to earn full marks:

• State the formula $g = W/m$.
• Calculate g for at least three objects correctly.
• State that the values are all approximately 9.8 N/kg, justifying the statement.

(c)

1. Calculate the weight of object D at the base of the mountain: $W_1 = mg_1 = 0.55 \text{ kg} \times 9.8 \text{ N/kg} = 5.39 \text{ N}$ [Calculating the weight at the base]
2. Calculate the weight of object D at the top of the mountain: $W_2 = mg_2 = 0.55 \text{ kg} \times 9.75 \text{ N/kg} = 5.3625 \text{ N}$ [Calculating the weight at the top]
3. Calculate the decrease in weight: $5.39 \text{ N} - 5.3625 \text{ N} = 0.0275 \text{ N}$ [Calculating the difference in weight]
4. Calculate the percentage decrease: $\frac{0.0275 \text{ N}}{5.39 \text{ N}} \times 100% = 0.51%$ [Calculating the percentage decrease]

How to earn full marks:

• Calculate the weight at the base of the mountain correctly.
• Calculate the weight at the top of the mountain correctly.
• Calculate the percentage decrease correctly: $\boxed{0.51%}$
• Include the correct unit (%).

Common Pitfall: When calculating percentage decrease, make sure you divide by the original value (in this case, the weight at the base of the mountain). Also, pay close attention to units throughout the calculation to avoid errors. A common mistake is forgetting to multiply by 100% at the end to get the percentage value.