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Wonders of Physics

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N2004 P1 Q15 – Pressure P1V1 = P2V2

Water if depth 10 m exerts a pressure equal to atmospheric pressure. An air bubble rises to the surface of a lake which is 20 m deep. When the bubble reaches the surface, its volume is 6 cm3.



 What was the volume of the air bubble at the bottom of the lake?


A) 2 cm3          B) 3 cm3         C) 12 cm3      D) 18 cm3



Solutions: Option A
Since the air bubble is enclosed, PV at A is equal to PV at B.

                  PAVA = PBVB

(Patm + Pwater) VA = (Patm) VB

        (10 + 20) VA = 10 x 6

                      VA = 60 / 30

                           = 2 cm3


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N2005/P2/A1 – Tension in Spring

The figure below shows a 0.40 kg mass hanging at rest from a spring.


(a) State what is meant by the mass of an object.

Mass is the amount of matter in the body. SI unit is kg.

(b) (i) On the figure above, draw an arrow showing the ling of action and the direction for each of the two forces that act on the mass. Write the name of the force next to each arrow.


(ii) The gravitational field strength is 10 N/kg. Calculate the values of the two forces you have drawn in (i).

W = mg = 0.40 x 10 = 4 N
Weight = Tension = 4 N (Newton’s 1st law, net force = 0N)

(c) The mass is pulled downwards and then released. Explain, in terms of any changes in the forces acting on the mass, why the mass accelerates upwards.

As the spring is pulled downwards, there is a gain in elastic potential energy. The tension in the spring increases compared to previously.
Since Tension > Weight, there is an resultant force acting upwards, hence there is an acceleration upwards.

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N2004P1Q4 – Dynamics + Kinematics

Trolley X and trolley Y are joined by a stretched spring. Trolley X has twice the mass of trolley Y.


When the trolleys are released, the acceleration of X is   to the right.  What is the initial acceleration of trolley Y to the left?Solutions:Key concept here is Newton’s 3rd Law. Action equals reaction. Both trolleys are connected by the spring, hence the tension force experienced by each trolley is equal and opposite.

Fx    =    Fy
ma       =    ma
2m x 2    =    m x a
a        =   4 m/s2

Another important concept to know. In Fig A, a box of m is pulled with a force of 40 N. Given that the frictional force is 10 N. The resultant force is 30N. Box is accelerating.


In Fig B, if another identical box m is added on top of the existing box (no change to the floor), the total mass is 2m, the frictional force now will double = 20 N. Hence resultant force now will be 20N. The acceleration of the box will be lower than previously.

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N2007P1Q17 – Thermal Properties

Four bars, all exactly the same size, are each placed with one end in boiling water.The times taken for the temperature of the other end to increase by 2 oC are measured.

Material of bar Time for 2oC rise / s
Aluminium 10
Copper 5
Cork 800
Styrofoam 1200

To make a large metal tank with the least heat loss, which materials should be used for the tank and its insulation?

Tank Insulation
A Aluminium Cork
B Aluminium Styrofoam
C Copper Cork
D Copper Styrofoam

Solutions: Option B

This question can be a bit tricky. It depends on which approach you view the question. Simply using conduction will be a much easier way to get the answer. The hint to use conduction is from the first paragraph. Time taken for temp of the other end to increase by 2 oC is about conduction.

To build such a tank, metal (aluminium or copper) has to be used and insulation on the external wall (cork or styrofoam).

To contain boiling water with least heat loss, both tank and insulation have to be good insulator (poor conductor) to reduce heat lost to surrounding. Hence tank should be aluminium and insulation should be styrofoam (option B)

On the other hand, if you approach the question in term of heat capacity, it will be a bit tedious and you do not have the values.Firstly, if aluminium takes a longer time to rise by 2 oC, it has a higher specific heat capacity. But you cannot merely look at specific heat capacity. You have to look at heat capacity as the mass of the container is important.

Copper: Density = 8940 kg/m3 and specific heat capacity = 400 J/kg.K

Aluminium: Density = 2700 kg/m3 specific heat capacity = 900J/kg.K

Assumptions: Volume of copper and aluminium are the same (same shape of container) say 0.02 m3, Initial temperature of metal is 30oC

Considering the amount of thermal energy gained from water as metal temperature reaches 80oC

Heat lost by water = heat gain by metal


Mass of copper = 8940 x 0.02 = 178.8 kg

Heat lost be water = heat gained by copper = mcθ = 178.8 x 400 x (80-30) = 3576000 J


Mass of Aluminium = 2700 x 0.02 = 54 kg

Heat lost be water = heat gained by aluminium = mcθ = 54 x 900 x (80-30) = 2430000J

From the calculation, it is obvious that the aluminium gains lesser thermal energy from the water, hence water will remain warmer compared to when using copper. Hence using this approach, it is still Option B. But important to know that you have to consider heat capacity [C], not specific heat capacity (c). [ C = mc]

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N2007P1Q27 – Current Electricity – Brightness of bulbs

X and Y are lamps with filaments made of the same material. The filament of lamp X is thicker and shorter than that of lamp Y. Each lamp is connected to the mains and switched on. Which is the brighter lamp and which lamp has the larger resistance?

Brighter lamp Larger resistance

Solutions: Option B

Using the formula, R = ρl/A     (ρ: resistivity (Ωm), l: length (m) and A: cross-sectional area (m2) Since both are of the same material, resistivity ρ is the same. X is thicker (cross-sectional area) and shorter, X will have lower resistance than Y.

Hence Y has larger resistance. Both are then connected to mains.

You can assume they are connected individually as shown below. Potential difference across each bulb is the same as the mains.


Since brightness depends on power, using P = IV, bulb X is brighter.


N2005P1Q24 Sound – Reflection of sound from 2 walls

A man who is standing at a point X between two parallel walls (as shown in the diagram) fires a starting pistol.


He hears the first echo after 0.6s and another one after 0.8s. How long after firing the pistol will he hear the next echo?


View video tutorial to visualise the distance travelled.

Time taken to hear next echo = 0.6 + 0.8 = 1.4s

The distance travelled by sound (left and right) from X is the same during 3rd and 4th echos.

Hence 3rd and 4th echos actually coincide. And time taken is 0.6 + 0.8 = 1.4s.