# Shi Qi 20 Feb 2018

## Question

A wire that obeys Hooke’s Law of length x1 when it is in equilibrium under tension T1. It’s length becomes x2 when the tension is increased to T2. What is the extra energy stored in the wire as a result of this process?

A) 1/4 ( T2 + T1 ) ( x2 - x1 )

B) 1/4 ( T2 + T1 ) ( x2 + x1 )

C) 1/2 ( T2 + T1 ) ( x2 - x1 )

D) 1/2 ( T2 + T1 ) ( x2 + x1 )

E) ( T2 - T1 ) ( x2 - x1 )

The extra energy is stored as elastic potential energy.

The graph of tension against total length is shown below. The extra elastic potential  energy is given by the shaded area.

# All the Best to Graduating Students of 2017

All the Best to Graduating Students of 2017

A heartfelt thank you graduating students of 2017!

It has been a very memorable journey tutoring all of you in preparation for your exams this year.

The examinations have been difficult and you deserve a well-needed rest!

Wishing you the best for your results!

O-Level results will be released from mid January 2018 onwards.

A-Level results will be released from early March 2018 onwards.

Daniel, GCE O Level

Gunjita, GCE A Levels

Galbraith GCE O Level

Haley, GCE O Levels

Joshua, GCE O Levels

Matthew, iGCSE

Malcolm, GCE O Levels

Shao Xian, GCE A Levels

Tze Han, GCE O Levels

Brendan, GCE A Levels

# Shi Qi 2 Oct 2017

## Question

A metal nugget floats in between some water and mercury. The densities of the metal nugget, mercury and water are 7900 kg m-3, 13600 kg m-3 and 1000 kg m-3 respectively.

What is the ratio of the volume of the nugget submerged in water to that of mercury?

A) 0.541
B) 0.826
C) 0.924
D) 1.21

Since the nugget is suspended in the liquids, the weight of the nugget is equal to the upthrust.

The nugget displaces both water and mercury, hence, the upthrust is due to the both the weight of the water displaced and the weight of the mercury displaced.

The volume of the nugget is the sum of the volume of water displaced and the volume of mercury displaced.

# Shi Qi 22 Sep 2017

## Question

Two containers of volume 4.0m3 and 6.0m3 contain an ideal gas at pressures of 100Pa and 50Pa respectively. Their temperatures are equal. They are joined by a tube of negligible volume. The gas flows from one container to the other with no change in temperature. The final pressure will be

A) 70 Pa
B) 75 Pa
C) 80 Pa
D) 150 Pa

Before the valve is opened, we first find an expression for the initial number of moles of gas in each container:

Now, the total initial number of moles of gas can be found:

After the valve is opened, the pressure in each container is now the same and the temperature remains the same.

The new expression for the final number of moles in each container is found again:

Now, the total final number of moles of gas can be found:

Since the total number of moles remains the same,

# James 30 Aug 2017 - 2

## Question

A light ray travels from a Perspex block into the air.

The critical angle of Perspex is 42.9o

What is the angle of refraction when the angle of incidence is 36.0o?

A) 23.6o

B) 24.5o

C) 52.9o

D) 59.7o

In order to calculate the angle of refraction, the refractive index of Perspex has to be found first.

Using the critical angle,

Now that the refractive index of Perspex is known, the angle of refraction can be calculated.

# James 30 Aug 2017 - 1

## Question

The diagram shows an object O viewed using 2 mirrors.

A person looks into the mirror as shown.

At which position is the image of O seen?

The object O is first reflected in top mirror to form the first image.
(Object distance is equal to Image distance)

The first image is then reflected in the bottom mirror to obtain the final image.
(Object distance is equal to Image distance)

# Brendan 4 Jun 2017 - 2

## Question

A copper bar of length L is moving to the right with a uniform speed v in a region of uniform magnetic field of flux density B, directly perpendicularly downwards into the paper in the figure below.

The ends of the rods are rigidly connected to a voltmeter which moves with the rod. What is the reading on the voltmeter?

A) zero
B) non-zero reading less than Blv
C) BLv
D) more than BLv

Method 1:

Consider wires PQ and RS. Since both wires are moving through the magnetic field, both of them will have an induced e.m.f. of BLv.

Based on Fleming’s Right Hand rule, the induced current of PQ and RS will both be upwards.

This means that the resultant current will be zero and the e.m.f. measured by the voltmeter will be zero.

Method 2:

Consider the closed loop PQRS. As the loop moves through the magnetic field, the magnetic flux linkage remains the same since the area of the loop does not change.

Since there is no change in the magnetic flux linkage, by Faraday’s law there is no induced e.m.f. and the voltmeter will read 0 V.

# Brendan 4 Jun 2017 - 1

## Question

An e.m.f. is induced in a coil placed in a changing magnetic field. The flux density B of this field varies with time as shown below.

At which value of t is the magnitude of the e.m.f. induced in the wire a maximum?

A) 1 ms
B) 2 ms
C) 3 ms
D) 4 ms

The induced e.m.f. is given by the formula:

Hence the magnitude of the induced e.m.f. is given by:

This means that the induced e.m.f. will be the greatest when the rate of change of the magnetic flux density is the greatest.

From the graph, the rate of change of the magnetic flux density is the greatest when the gradient of the graph is the greatest.

This means that the induced e.m.f. will be the greatest at time t = 0 ms and t = 4 ms.

# James 27 Feb 2017 - 2

## Question

A car travelling at a constant speed of 30 ms-1 passes a police car, which is at rest.

The police officer accelerates at a constant rate of 3.0 ms-2 and maintains this rate of acceleration until he pulls next to the speeding car. Assume that the police car starts to move at the moment the car moves past the police car.

What is the time required for the police officer to catch up with the car?

We assume that the police car catches up with the car at time t1.

The following formula for displacement can be obtained:

When the police car catches up with the car, both have travelled the same displacement.

The police car requires 20 s to catch up with the car.

# James 27 Feb 2017 - 1

## Question

A body is thrown vertically up from the ground with an initial speed u and it reaches the maximum height h at time t0.

What is the height it reaches at ½ t0?

We first define the upwards direction as positive.

When the body reaches the maximum height, the speed of the body is 0.

Knowing this, we can find expressions for h and t0. The acceleration of the body is taken to be 10 ms-2 downwards. Since we define upwards to be positive, the acceleleration will be -10 ms-2.

When the time is ½ t0,

Now that we know the speed at ½ t0, we can find the height travelled by the body.

# James 25 Feb 2017

## Question

A student flips a coin into the air. Its initial velocity is 8.0 ms-1. Taking g = 10 ms-2 and ignoring air resistance, calculate:

a) the maximum height, h, the coin reaches, [2]
b) the velocity of the coin on returning to his hand, [1]
c) the time that the coin is in the air. [2]

a) To easily solve the problem, the velocity-time graph has to sketched.

Since there is no air resistance, the acceleration will be constant at 10ms-2 downwards. This means that the graph is a straight line.

At the highest point, the velocity of the coin will be zero.

The maximum height is given by the shaded area below.

To find the area we first need to find the value of th. Since we are taking upwards to be positive, the acceleration is -10 ms-2.

b) Since the coin falls back down to the student, the distance travelled downwards must be equal to the distance travelled upwards.

This means that the two shaded areas must be the same.

By symmetry, the final velocity of the coin must be -8.0 ms-1.

c) Since the area of the shaded areas are the same, the time taken for the upward journey and the downward journey must also be the same.

Hence the total time taken is twice the time taken for the coin to reach the highest point.

# Shao Xian 19 Jan 2017

## Question

A child of mass 50 kg is on a swing which is suspended by 4.0m ropes from a rigid support. The horizontal speed of the swing as it passes through the lowest point is 3.0 ms-1.

What is the angle θ that the ropes make with the vertical when the swing is at the highest point?

We take the lowest point as the reference to measure the height of the child as he swings.

Since the lowest point is the reference level, the gravitational potential energy is 0 at the lowest point. As the child swings from the lowest point to the highest point, kinetic energy is converted to gravitational potential energy.

Using the value of h, we can find the length of the sides of the triangle.

# Best Wishes to Graduating Students of 2016

Thank you graduating students of 2016!

It has been a very enjoyable journey tutoring all of you in preparation for your exams this year.

Examinations are always tough and you deserve a well-needed rest!

All the best for your results and I'll check in with you next year when they are released!

O-Level results will be released between 11 and 13 January 2017.

A-Level results will be released between 24 and 28 Feburary 2017.

Clive, iGCSE 2016

Ashley, iGCSE 2016

David, O-Level 2016

Yusong, A-Level 2016

Ehan, A-Level 2016

Gavin, A-Level 2016

Siew Mun, A-Level 2016

Jia Yee, A-Level 2016

Kimberly, A-Level 2016

# James 29 Sep 2016

## Question

Find the potential difference across the 2.0 Ω resistor.

Find the potential difference across the open gap in the circuit.

The easiest way to find the potential difference in this case is to assign values of potential at the various points.

We first start at the sides of the cell, A and B.

We can assign any value as long as the difference between A and B is 6.0 V and A is a higher value than B.

In this case, we choose A to be 7 V and B to be 1 V.

All connecting wires are assumed to have no resistance. This means that the value of potential at all points of the same wire are of the same value.

Hence, E has a value of 7 V and C has a value of 1 V.

Since the circuit is an open circuit and no current flows, from V = I R, the potential difference across the 2.0 Ω resistor is zero.

This means that the values of the potential at both sides of the resistor are the same.

From the diagram, we can see that the difference between the potential at C and D is 0 V.

Hence, the potential difference across the 2.0 Ω resistor is zero.

From the diagram, we can see that the difference between the potential at E and D is 6 V.

Hence, the potential difference across the open gap in the circuit is 6.0 V.

# Yusong 25 Aug 2016

## Question

The figure shows a long horizontal wire PQ carrying a steady current of 50A in the direction QP. A copper wire RS of diameter 0.40mm hangs horizontally at a distance of 0.15m below wire PQ using some threads.

A current flows through RS such that there is no tension in the threads.

Identify the direction is the current and draw a free body diagram of the forces acting on RS.

If the current in RS is flowing opposite to the current in PQ, the two wires will repel each other.

Since the wire is suspended in equilibrium, the resultant force must be zero.

Fup = Fdown
T1 + T2 = FB + W

We can see that it is not possible for the tension to be zero.

If the current in RS is flowing in the same direction as the current in PQ, the two wires will repel attract other.

Since the wire is suspended, the resultant force must be zero.

Fup = Fdown
T1 + T2 + FB = W

When the tensions in both threads are zero, we have

FB = W

Hence, current in RS must be flowing in the same direction as PQ, in the direction SR.

The free body diagram when there is no tension in the threads is as shown below:

# Jia Yee 25 Aug 2016

## Question

The figure below shows a region PQRS of a uniform magnetic field directed downwards into the plane of the paper.

Electrons, all having the same speed, enter the region of the magnetic field.

a) On the figure, show the path of the electrons as they pass through the magnetic field emerging from side QR.

b) A uniform electric field is also applied in the region PQRS so that the electrons now pass undeflected through this region. On the figure, mark with an arrow labelled E, the direction of the electric field.

c) The undeflected electrons in (b) each have charge -e, mass m and speed v. State and explain the effect if any, on the particles entering the region PQRS of the same magnetic and electric fields in (b) if the particles each have

i) charge -e, mass m and speed 2v.

ii) charge +e, mass m and speed v.

a) Since the electrons are moving to the right, it is the same as a current flowing to the left. By using Flemming’s left hand rule, the magnetic force on the electrons is downwards.

According to Flemming’s left hand rule, the magnetic force will always be perpendicular to the velocity of the electrons. This means that it provides a centripetal force. Hence, the electrons will move in a circular path as shown.

Once the electrons exit the magnetic field, they will move in a straight path since there is no resultant force acting on the electrons. We ignore the effects of the weight because the acceleration of due to the weight is too small to cause any significant change in the velocity of the electrons.

b) In order for the electrons to pass through undeflected, the resultant force on the electrons must be zero. Since the magnetic force on the electrons is downwards, the electric force acting on the electrons must be of the same magnitude and acting upwards.

In order for the electric force to be acting upwards, electric field lines must be directed downwards. (Since electrons are negatively-charged, they will experience an electric force in the opposite direction to the electric field lines.)

c)  When an electron is moving with a speed of v, it passes straight through without any deflection. The magnetic force and electric forces on the electrons are given by the following formulae:

FB= Bqv
FE = qE

Since the downwards magnetic force is now greater than the upwards electric force, the resultant force is downwards and the particle will curve downwards as it passes through the region.

ii) When a particle of the opposite charge (+e) and mass (m) and same speed (2v) moves through the field, the directions of the magnetic force and electric forces are now reversed.

The electric force acts downwards and the magnetic force acts upwards on the particle.

However, both forces are still of the same magnitude and the resultant force is still zero. Hence, the particle will still pass through the region undeflected.

# David 16 Aug 2016 - 2

## Question

The figure below shows the path of ray X as it passes through a thin converging lens AB.

Draw light rays to find the position of the image formed and determine the focal length of the lens AB.

Firstly, ray X is used to find the position of the base of the image. Since ray X comes from the base of the object, it will form the base of the image. Hence, the image is located along the blue line shown below:

Next, the ray from the tip of the object that passes through the optical centre is drawn.

The intersection point of the ray and the blue line is the location of the tip of the image.

The blue line is removed and a ray parallel to the principal axis from the tip of the object is added. This ray bends at the lens and reaches the tip of the image. The focal point can be found and the focal length can now be obtained.

The focal length is 10.0 cm.

# David 16 Aug 2016 - 1

## Question

A rectangular block of wood rests on the ground as shown in the diagram below.

Which of the following is the easiest way for a force F to topple the block?

For each block we identify the pivot point and the perpendicular distance of force F from the pivot point.

We can see that the perpendicular distance for D is the greatest. Force F will produce the greatest moment for D, hence it will topple the easiest.

# Farhan 05 Aug 2016

## Question

A piece of furniture was recovered from an archaeological site and carbon dating was used to estimate its age. It was assumed that the proportion of Carbon-14 to natural carbon (Carbon-12) in living wood was 1.25 x 10-12. If the number of particles emitted from 5 g of carbon prepared from the specimen was 21 per minute, how old is the specimen?
[The half-life of Carbon-14 is 5600 years and the mass number of natural carbon is 12.]

Since the proportion of Carbon-14 is very small compared to Carbon-12, we can assume for the sake of finding the number of carbon atoms that all of the carbon in the specimen is Carbon-12.

Given that the proportion of Carbon-14 to Carbon-12, we can find the original number of Carbon-14 atoms when the piece of furniture was living wood using the following approximation. (Refer to bottom for further discussion of this approximation.)

Using the half-life, we can calculate the decay constant of Carbon-14.

Hence, we can find the initial activity when the specimen was still living wood.

The final activity of the specimen is given by

Now, we can find the age of the specimen.

## Discussion on Approximation

Based on the phrasing of the question that

“the proportion of Carbon-14 to natural carbon is k”,

However, since the k is small compared to 1,

Hence, we arrive at

# Yusong 04 Aug 2016

## Question

A simple iron-core transformer is shown below:

Suggest why thermal energy is generated in the core when the transformer is in use.