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)

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

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

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

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