Problem: Heat #1

The air in a room consists principally of nitrogen and oxygen. The molecular weight of nitrogen (N₂) is 28 g/mole and the molecular weight of oxygen (O₂) is 32 g/mole. The air is in thermal equilibrium. Which statement is true?

A.	The average speeds of the nitrogen and oxygen molecules are equal.
B.	The average kinetic energies of the nitrogen and oxygen molecules are equal.
C.	The oxygen molecules have average speeds that are greater than for the nitrogen molecules.
D.	The nitrogen molecules have average kinetic energies that are greater than for the oxygen molecules.
E.	None of the above statements are true.

Solution

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The answer is B. When it is said that the air is in thermal equilibrium, all the constituents of the air - including nitrogen and oxygen gases - are having the same temperature. Because temperature is the measure of the average kinetic energy of the atoms and molecules of a substance, the average kinetic energies of the nitrogen and oxygen molecules must be equal.

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Question source: Saskatchewan High School Physics Scholarship Competition 2006

Problem: Dynamics #1

A stone thrown vertically upwards from the Earth’s surface travels until a height of h and falls back to its original position. If air resistance can be ignored, which graph below best represents the change of force, F which acts on the stone with its distance travelled, s?

A.		C.
B.		D.

Solution

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The answer is D. Because the stone is thrown upwards, the gravitational field strength of Earth at the stone’s original position and at the apex of the upward motion of the stone can be assumed to be the same. Therefore, the gravitational force of Earth acting on the stone, which is the weight of the stone, remains constant. All other forces acting on it are negligible.

From the point of view of kinematics, what actually changes is the stone’s velocity and speed. The shape of the velocity – time graph of the stone will look exactly like the one in A, while the shape of the speed – time graph of the stone will look exactly like the one in C.

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Question source: Malaysia National Physics Competition Panel

Problem: Measurements #1

x_1, y₁ and x_2, y₂ are two sets of values that are related according to the following equation:

The points (x_1, y₁) and (x_2, y₂) lie on the graph

A.		C.
B.		D.

Solution

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The answer is C.

The above equation can be rearranged in this way:



When xy = k, that means y is inversely proportional to x. The graph y versus x will be like the one in C.

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Question source: Malaysia National Physics Competition Panel

The Relationship between Variables (Part II)

This article is a continuation from Part I, which discusses about direct proportion.

Inverse Proportion

Two sets of readings for the quantities x and y are given in the table below.

x	3	4	6	12
y	4	3	2	1

Here, when x is doubled, y is halved; when x is trebled, y is one-third its previous value and so on.

We say that y is inversely proportional to x, or y varies inversely as x, that is

The graph of y versus x is as shown:

Notice that the product of x and y is the same for every point (12 in this case). Hence this product is known as the constant of proportionality or variation and it is given a symbol such as k. Therefore, we can write the exact relationship between x and y as

Assuming that there are points (x₁, y₁), (x₂, y₂), (x₃, y₃), …, we can say

xy = x₁y₁ = x₂y₂ = x₃y₃ … = k

Notes:

1. When a variable is proportional to two or more variables, we can represent this relationship as a joint variation. For example,
   When
   y ∝ x and y ∝ z
   therefore
   y ∝ xz or y = kxz
   
   When
   
   therefore
   

2. In every proportional relationship, the converse is also true. For example,

If y ∝ x, then x ∝ y.

Examples of physical quantities that vary inversely with each other are:

• Pressure and volume
• Frequency and periodic time
• Gravitational force and the distance squared

To be continued…

Problem: Kinematics #3

The figure above shows a displacement versus time squared (t²) graph for the motion of an object.
Which of the following motion can be represented by this graph?
A. A ball that is travelling at terminal velocity.
B. A ball that is falling freely from a stationary position.
C. A ball that bounces back from the floor.
D. A ball that is travelling on a rough surface.

Solution

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The answer is C.

The graph shows that s is directly proportional to t² or
s ∝ t²
∴ s = kt²

Without modifying the points on the graph, we can manipulate this relationship by dividing both sides by t to become:


But, s/t is the velocity of the object. So,
v = kt

Using the relationship between v and t, a graph can be drawn with similarities with the above graph because the gradient, k remains the same and it too will pass through the origin.

From here, we know that the velocity of the object increases uniformly from zero (acceleration is constant). An object free-falling from a stationary position will experience this kind of motion.

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Question source: Malaysia National Physics Competition Panel