Problem: Materials #4

A cube of ice is floating in a salt solution of density 1.25 times the density of pure water in a beaker. When the ice melts, the level of the solution in the vessel
A. rises
B. falls
C. remains unchanged
D. falls at first and then rises to the same height as before

Solution

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The answer is A.

A non-mathematical explanation would be that since the ice is floating, its mass would be the same as the mass of salt water displaced by it, as per Archimedes’ principle. Now consider the ice melted where the resulting water gradually “fills in” the space in the solution once occupied by the ice. The mass of the resulting water would still be the same as the mass of salt water displaced earlier, but the resulting water will occupy a larger space compared to the salt water displaced, because its density is lower (density is mass per unit volume). Hence the solution level rises.

A mathematical proof is as follows:

Mass of displaced salt solution = Mass of resulting water


This shows that the resulting water occupies more space compared to the displaced salt water in the ratio 5:4.

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Question source: Malaysia National Physics Competition 2007 (Secondary Level)

Problem: Materials #3

Within a certain type of star called a neutron star, the material at the centre has a mass density of 1.0 × 10¹⁸ kg m^-3. If a small sphere of this material of radius 1.0 × 10^-5 m were somehow transported to the surface of the earth, what would be the weight of this sphere? (Assume that g = 10 m s^-2)
A. 1000 N
B. 4200 N
C. 4.2 × 10⁴ N
D. 7.0 × 10⁴ N
E. 3.1 × 10⁹ N

Solution

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The answer is C.

From the information given, we need to first find the volume of this sphere, then find its mass and lastly calculate its weight. Therefore, we need the following formulas:


Weight, W = mg

where r is the radius of the sphere.


Step 1- Finding the volume of the sphere:


Step 2 - Finding the mass of the sphere:
The second formula can be rearranged to m = Vρ, so

m	= (4.189 × 10-15 m3)(1.0 × 1018 kg m-3) = 4.189 × 103 kg

Step 3 - Finding the weight of the sphere:

W	= (4.189 × 103 kg)(10 m s-2) = 4.189 × 104 kg m s-2 = 4.189 × 104 N

Rounding off, we have W = 4.2 × 10⁴ N.

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Question source: Malaysia National Physics Competition 2007 (Secondary Level)

Problem: Kinematics #4

A train covers half the distance of its journey with a speed 20 m s^-1 and the other half with a speed of 40 m s^-1. The average speed of the train during the whole journey is
A. 25 m s^-1
B. 27 m s^-1
C. 30 m s^-1
D. 32 m s^-1
E. 35 m s^-1

Solution

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The answer is B.

Let the total distance travelled by the train be d and the time taken for it to complete the first half and second half of the journey be t₁ and t₂ respectively. The distance travelled each half-journey is d/2 and it is the product of their respective travelling velocity and time taken.

For the first half-journey,


For the second half-journey,


Combining the above two equations, we get



Now we have the relationship between t₁ and t₂, we can eliminate one of them from our calculations. For convenience, let us state the d in terms of t₂ and find the average speed, which is the total distance travelled divided by the total time taken.



Rounding off to the nearest integer, we have 27 m s^-1.

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Question source: Malaysia National Physics Competition 2007 (Secondary Level)

The Concept of… Force

Force is one of the most fundamental concepts of physics. The concept of force has formed part of statics and dynamics since ancient times. The modern concept of force is commonly explained in terms of Newton's three laws of motion set forth in his Principia Mathematica (1687).


What is force?

Force can be described as a push or pull. A force upon an object is always a result of interaction with another object. When we say an object is acted by a force, there must be another object applying a force to it.

Force is also a vector, that is, it has magnitude and direction. We can fully describe a force by saying how strong it is (its magnitude) and in which direction it is acting. The SI Unit for force is newton (named after none other than Isaac Newton) and its symbol is N.


Types of Force

Forces can be categorised into two broad categories:

• Contact forces are those types of forces which result when two interacting objects are perceived to be physically interacting with each other. Examples of contact forces include frictional forces, tensional forces and applied forces.
• Action-at-a-distance forces are those types of forces which result even when the two interacting objects are not in physical contact with each other, yet are able to exert a push or pull despite their physical separation. Gravitational forces, magnetic forces and electrical forces are examples of this type of force.

The Effects of Force

Basically, a net force that acts on an object will produce an acceleration. For example, by applying a force, a stationary object can be set in motion, whereas a moving object can be made to move faster or slower and/or change its direction of motion (since acceleration also involves direction). In short, a force can alter the state of motion of an object; this is described in Newton’s first law of motion. The relationship between force and acceleration is further described in Newton’s second law of motion.

More often than not, a force that acts on an object does not cause the whole object to accelerate, but rather a part of it. Therefore a force can also cause the rotation and deformation of an object.

Photo by max's picks

When a force causes acceleration, it can transfer energy in the process. To illustrate this point, imagine a moving billard ball colliding with an 8-ball, as shown in the photo above. The billard ball will exert a force on the 8-ball so that the 8-ball accelerates and its velocity increases gradually. This increase in velocity in turn results in an increase in its kinetic energy. For this interaction to obey the law of conservation of energy, the object supplying the force must lose energy, therefore the application of force in this case causes energy to be transferred.

However, the presence of force does not necessary result in the transfer of energy. Now imagine a ball hitting a wall and bouncing back (assume the wall does not absorb any energy from the ball). Although acted by a force, the wall does not move and the ball continues to move with the same speed in the opposite direction. The energy of the ball remains the same.


What Causes Force?

It is mentioned above that an object acting by a force undergoes acceleration. Conversely, an object decelerating exerts a force on another object. This apparent symmetry can be explained using Newton's third law of motion and is the cause of contact forces.

For action-at-a-distance forces to act on an object, the object merely has to be within the influence of a field, like gravitational field, magnetic field and electrical field. Depending on the properties of the object, it may react towards more than one field. The exact reason why a field exerts a force is unknown, but the fact is that this is a phenomenon of our Universe.

An object that exerts a force on another object has the potential of transferring work to it, as mentioned above. Therefore, it is safe to say that for an object to exert a force on another object, it must first have energy (note that mass is also energy – concentrated energy – in agreement with the concept of mass-energy equivalence).

References

• "force." Encyclopædia Britannica. 2007. Encyclopædia Britannica Online. 11 Aug. 2007 <http://www.britannica.com/eb/article-9034834/force>
• "Force," Microsoft® Encarta® Online Encyclopedia 2007. 11 Aug. 2007. <http://encarta.msn.com/encyclopedia_761555718/Force.html>

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

The Relationship between Variables (Part I)

One of the most important mathematical operations in physics is finding the relationship between variables. Through the study of these relationships, we can know how a change in one variable affects another variable, thus enabling us to make predictions and conclusions easily.

For example, an automobile maker will want to know how the mass of a car affects its acceleration, so that they can design cars with optimum performance.


Direct Proportion

Suppose that in an experiment four sets of readings are obtained for the quantities of x, y, p and q as in the tables below.

x	1	2	3	4
y	1	2	3	4

Table 1

p	1	2	3	4
q	3	6	9	12

Table 2


In Table 1, we see that when x is doubled, y doubles; when is x trebled, y trebles; when x is halved, y halves and so on.

Similarly in Table 2, when p is doubled, q doubles; when p is trebled, q trebles; when p is halved, q halves and so on.

Now, two graphs are plotted using the values in Table 1 and Table 2, as shown below:

Since both graphs are similar in the sense that they are straight line graphs passing through the origin, the relationship of x with y and p with q are somewhat similar. We say that both pairs of variables are directly proportional to each other or varies directly with each other. In symbols,

x ∝ y and p ∝ q

But there is a difference in gradient (slope) between the two graphs, as illustrated above. When we want to define the relationship between the two variables exactly, we have to take into account this difference. Because they are straight line graphs passing through the origin,

where x and y are the x and y values of any point on the graphs.

The constant obtained is called the constant of proportionality or variation and is given a symbol, such as k, so that the relation between y and x can be summed up by the equation


Therefore, we can conclude that the relationship between x and y in Table 1 is
y = x
(k = 1)

The relationship between p and q in Table 2 is
p = 3q
(k = 3)

Notes:

1. The constant of proportionality defines how y increases with x. Every increase of 1 unit of x will result in k increase in y.
2. In practice, because of inevitable experimental errors, the readings seldom show the relation so clearly as here.
3. Assuming that there are points (x₁, y₁), (x₂, y₂), (x₃, y₃), … and x ∝ y, we can say

or y₁ = kx₁, y₂ = kx₂, y₃ = kx₃…

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

• Force and mass
• Work done and force
• Speed of wave and its frequency

>> Continue to Part II

Problem: Kinematics #2

The figure above shows two objects, P and Q moving with velocities 30 m s^-1 and 20 m s^-1 respectively towards each other on a straight line.
How long, after that instant, will P and Q meet?
A. 100.0 s
B. 83.3 s
C. 20.0 s
D. 13.3 s

Solution

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The answer is C.

Let the distance travelled by P and Q before they meet each other be d₁ and d₂ respectively. Here, we can form an equation:

d1 + d2	= 1000 m
d1	= 1000 m - d2	- (1)

Since we know the speeds of P and Q, their distance travelled is the product of their speed and time taken, t. t is also the time when P and Q meet.

d₁ = (30 m s^-1)t
d₂ = (20 m s^-1)t

Substituting in equation (1),

d1	= 1000 m - d2
(30 m s-1)t 50t m s-1 ∴ t	= 1000 m - (20 m s-1)t = 1000 m = 20 s

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Question source: Malaysia National Physics Competition Panel

Problem: Materials #2

The density of an object is ¾ of the density of water. If the object is floating on the water surface, then the ratio between the volume of object above the water surface and the volume of object below the water surface is
A. 1:4
B. 1:3
C. 3:4
D. 4:3

Solution

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The answer is B.

This is a problem that requires the application of Archimedes’ principle.

We know that the density of the object is ¾ of the density of water, so the weight of the object, W is given by


Where
V_obj – Volume of the object
ρ_water – Density of water
g – The acceleration due to gravity

Because the object is floating freely,
Upthrust = Weight of the object

But upthrust depends on the volume of object submerged. Let r be the ratio between the volume of object submerged and the total volume of the object, so that rV_obj is the volume of object submerged.




We can conclude that ¼ of the object is above the water surface and ¾ of the object is below the water surface.

Therefore, the ratio between the volume of object above the water surface and the volume of object below the water surface is 1:3.

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Question source: Malaysia National Physics Competition Panel

Problem: Materials #1

An object has a weight of 4 N in air, 3 N in water and 2.8 N in a salt solution. If the density of water is 1000 kg m^-3, the density of the salt solution is
A. 830 kg m^-3
B. 1070 kg m^-3
C. 1200 kg m^-3
D. 1430 kg m^-3

Solution

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The answer is C.

This is a problem that requires the application of Archimedes’ principle.

The object has a weight of 4 N in air. Because the upthrust on the object when it is in air is negligible, we can say that the weight of the object is 4 N.

When the object is (completely immersed) in water, its weight is 3 N, so it experiences an apparent loss of weight of 1 N due to the upthrust acting on the object.

Upthrust = Apparent loss of weight
and
Upthrust = Vρ_waterg

Where
V – volume of the object
ρ_water – density of water
g – the acceleration due to gravity

Now the object is immersed in a salt solution with density ρ_salt and its weight becomes 2.8 N. It experiences an apparent loss of weight of 1.2 N, therefore
Upthrust = Vρ_saltg = 1.2 N - (2)

Because the volume of object remains the same, we substitute equation 1 into equation 2.

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Question source: Malaysia National Physics Competition Panel

Archimedes' Principle Explained

Archimedes' principle is a law that explains buoyancy or upthrust. It states that

When a body is completely or partially immersed in a fluid it experiences an upthrust, or an apparent loss in weight, which is equal to the weight of fluid displaced.

According to a tale, Archimedes discovered this law while taking a bath. After making this discovery, he is said to have leapt out of his bathtub and ran through the streets of Syracuse naked shouting "Eureka!".

An object experiences upthrust due to the fact that the pressure exerted by a fluid on the lower surface of a body being greater than that on the top surface, since pressure increses with depth. Pressure, p is given by p = hρg, where:
h is the height of the fluid column
ρ (rho) is the density of the fluid
g is the acceleration due to gravity


Let us confirm this principle theoretically. On the figure on the left, a solid block is immersed completely in a fluid with density ρ. The difference in the force exerted, d on the top and bottom surfaces with area a is due to the difference in pressure, given by

d = h₂aρg – h₁aρg = (h₂ – h₁)aρg

But (h₂– h₁) is the height of the wooden block. So, (h₂ – h₁)a is the volume of the solid block, V.

d = Vρg
∴ Upthrust = Vρg

In any situation, the volume of fluid displaced (or the volume of the object submerged) is considered to calculate upthrust, because (h₂ – h₁) is the height of the solid block only when it is completely immersed. Furthermore, the pressure difference of the fluid acts only on the immersed part of an object.

Now, moving back to Vρg. Since V is the volume of fluid displaced, then the product of V, ρ and g is the weight of the fluid displaced. So, we can say that

Upthrust = Weight of the fluid displaced

Compare this conclusion with the statement above summarising Archimedes' principle. Are they the same? Well, not totally. The “apparent loss in weight” was not mentioned in my explanation.

In the figure on the left, there are arrows on the top and bottom of the solid block. The downward arrow represent the weight of the block pulling it downwards and the upward arrow represent the upthrust pushing it upwards. If one were to measure the weight of the solid block when it is immersed in the fluid, he will find that the weight of the block is less than that in air. There is a so-called “apparent loss in weight”, because the buoyant force has supported some of the block’s weight.

Weight in air – Upthrust	=	Weight in fluid
Upthrust	=	Weight in air - Weight in fluid

∴ Upthrust = Apparent loss in weight


Objects Floating Freely

When an object is floating freely (i.e. neither sinking nor moving vertically upwards), then the upthrust must be fully supporting the object’s weight. We can say

Upthrust on body = Weight of floating body

By Archimedes’ principle,

Upthrust on body = Weight of fluid displaced

Therefore,

Weight of floating body = Weight of fluid displaced

This result, sometimes called the “principle of floatation”, is a special case of Archimedes’ principle and can be stated:

A floating body displaces its own weight of fluid.

If a body cannot do this, even when completely immersed, it sinks.


See Also

• Problem: Materials #1 - a sample problem involving Archimedes' principle.
• Materials - for related articles and problems.

Problem: Kinematics #1

A stone is released from a height of 20 m and allowed to fall in a straight line towards the ground. Ignoring air resistance, calculate
a) the time taken for the stone to reach the ground.
b) the velocity of the stone just before it touches the ground
(Assume that g = 10 m s^-2)

Solution

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The answers for (a) and (b) are 2 seconds and 20 m s^-1 respectively.

This is a problem utilising the equations for uniform acceleration. First, we list out the available information:
u = 0
a = g = 10 m s^-2
s = 20 m

For (a):
We want to find the value of t and we know the values of s, a and u, so we utilise the equation s = ut + ½at²


For (b):
We want to find the value of v and we know the values of s, a and u, so we utilise the equation v² = u² + 2as


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The Equations for Uniform Acceleration

There are four equations for uniform acceleration (also known as the kinematic equations) which are used to describe the motion of an object:

1.
2.
3.
4.

Where:
u – initial velocity
v – final velocity
a – acceleration
t – time
s – displacement


You may be wondering – why there are four equations instead of one? Notice that each equation has one variable missing. For example, the first equation doesn’t have the variable displacement, s, in it. So, if you were to calculate something that involves displacement, equation 1 is not your choice. Basically, if you know any three of u, v, a, t and s, the others can be found using one or more of the above equations. But take note that you can use these equations only when the acceleration of an object is constant throughout its motion.

It’ll be useful to memorize these formulas if you’re going to solve problems involving motion, but do you know how to derive them? Knowing how to derive these formulas is useful because you will understand how these equations originated and this knowledge acts as a backup in case you forgot any one of them – you just have to derive it and they just appear out of nowhere!


Deriving the Equations

Suppose the velocity of a body increases at a consistent rate from u to v in time t, the body is said to be accelerating uniformly and uniform acceleration a is given by


Since the velocity is increasing steadily, the average velocity is the mean of the initial and final velocities:


If s is the displacement of the body in time t, then the average velocity is equal displacement/time or s/t, so we can say


By substituting equation 1 into equation 2, we have


By rearranging equation 1, we know that t = (v-u)/a. We substitute this value of t into equation 2 and we have


There’re other ways of deriving these equations. Can you find them?


See Also

• Problem: Kinematics #1 - a sample problem that requires the usage of these equations.

Problem: Waves #1

The frequency of oscillation of a simple pendulum is f₁. When the length of the pendulum l is increased to 2l, the frequency of oscillation of the simple pendulum is f₂. The value of the ratio f₁/f₂ is


Solution

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The answer is D.

The formula for the periodic time, T of a simple pendulum is given by

and we know that


Therefore, the formula for f of a simple pendulum is given by


Now we know that*

[* refer to this article]

Moving on,



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Question source: Malaysia National Physics Competition Panel