Question 1
Fig. 1.1 shows a straight section of a river where the water is flowing from right to left at a speed of 0.54 m / s.
A swimmer starts at point P and swims at a constant speed of 0.72 m/s relative to the water and at right angles to the current.
(a) (i) Determine, relative to the river bank, both the magnitude and direction of the swimmer’s velocity. (Sub-topic – 1.2)
▶️Answer/Explanation
Magnitude of velocity: 0.90 m/s
Direction of velocity: 53° to the river bank
Explanation:
The swimmer’s velocity relative to the river bank can be found using vector addition. The swimmer’s velocity relative to the water (0.72 m/s) and the water’s velocity (0.54 m/s) are perpendicular to each other. Using Pythagoras’ theorem:
\[ v = \sqrt{(0.72)^2 + (0.54)^2} = \sqrt{0.5184 + 0.2916} = \sqrt{0.81} = 0.90 \, \text{m/s} \]
The direction can be found using trigonometry:
\[ \tan \theta = \frac{0.72}{0.54} \Rightarrow \theta = \tan^{-1}\left(\frac{0.72}{0.54}\right) \approx 53^\circ \]
(ii) After 1.5 minutes, the swimmer reaches point Q. Calculate the distance between P and Q. (Sub-topic – 1.2)
▶️Answer/Explanation
Distance: 81 m
Explanation:
First, convert the time from minutes to seconds:
\[ 1.5 \, \text{minutes} = 90 \, \text{seconds} \]
Using the formula for distance:
\[ \text{Distance} = \text{Velocity} \times \text{Time} = 0.90 \, \text{m/s} \times 90 \, \text{s} = 81 \, \text{m} \]
(b) When the swimmer is crossing the river, his actions produce a constant forward force on his body. Explain why he moves at a constant speed. (Sub-topic – 1.5)
▶️Answer/Explanation
Explanation:
The swimmer moves at a constant speed because the forward force he applies is balanced by the backward force of water resistance (friction). When the resultant force is zero, there is no acceleration, and the swimmer maintains a constant speed according to Newton’s first law of motion.
Question 2
Fig. 2.1 shows a motorcyclist accelerating along a straight horizontal section of track.
The motorcyclist and motorcycle have a combined mass of 240 kg.
(a) On the straight horizontal section of the track, the motorcyclist accelerates from rest at
(i) The motorcyclist reaches the end of the straight section of track in 5.3 s.
Calculate the speed of the motorcyclist at the end of the straight section. (Sub-topic – 1.2)
▶️Answer/Explanation
Solution:
Given:
Acceleration, \( a = 7.2 \, \text{m/s}^2 \)
Time, \( t = 5.3 \, \text{s} \)
Initial speed, \( u = 0 \, \text{m/s} \) (since the motorcyclist starts from rest)
Using the equation of motion:
\( v = u + at \)
\( v = 0 + (7.2 \times 5.3) \)
\( v = 38.16 \, \text{m/s} \)
Therefore, the speed of the motorcyclist at the end of the straight section is 38 m/s.
(ii) Calculate the resultant force on the motorcyclist and motorcycle on the straight section of track. (Sub-topic – 1.5)
▶️Answer/Explanation
Solution:
Given:
Mass, \( m = 240 \, \text{kg} \)
Acceleration, \( a = 7.2 \, \text{m/s}^2 \)
Using Newton’s second law:
\( F = ma \)
\( F = 240 \times 7.2 \)
\( F = 1728 \, \text{N} \)
Therefore, the resultant force on the motorcyclist and motorcycle is 1700 N.
(b) At the end of the straight section, the track remains horizontal but bends to the right, as shown in Fig. 2.1. When the motorcyclist reaches the bend, she travels around the bend in a circular path at a constant speed.
(i) Velocity is a vector quantity. State how a vector quantity differs from a scalar quantity. (Sub-topic – 1.1)
▶️Answer/Explanation
Solution:
A vector quantity has both magnitude and direction, whereas a scalar quantity has only magnitude. For example, velocity is a vector quantity because it includes both speed (magnitude) and direction, while speed is a scalar quantity as it only includes magnitude.
(ii) Describe what happens to the velocity of the motorcyclist as she travels around the bend at constant speed. (Sub-topic – 1.2)
▶️Answer/Explanation
Solution:
As the motorcyclist travels around the bend at constant speed, the direction of her velocity changes continuously. Even though the speed remains constant, the change in direction means that the velocity (which is a vector quantity) is changing. This change in velocity indicates that the motorcyclist is accelerating, even though her speed is constant.
(iii) Explain why there must be a resultant force on the motorcyclist as she travels around the bend. (Sub-topic – 1.5)
▶️Answer/Explanation
Solution:
There must be a resultant force on the motorcyclist as she travels around the bend because her velocity is changing direction. According to Newton’s first law of motion, an object will continue in a straight line at constant speed unless acted upon by a resultant force. Since the motorcyclist is changing direction, a resultant force (centripetal force) must be acting on her to cause this change in direction. This force is directed towards the center of the circular path and is necessary to keep the motorcyclist moving in a curved path.
Question 3
(a) A rubber balloon is inflated with helium and sealed so that no helium escapes. The balloon is positioned immediately below the ceiling in a room. Heaters are switched on and the temperature of the air in the room increases. (Sub-topic – 2.1.2)
When the heaters are first switched on, the temperature of the air immediately below the ceiling increases more quickly than the temperature of the air in the rest of the room. Explain why this happens.
▶️Answer/Explanation
Explanation:
When the heaters are switched on, the air near the heaters gets heated first. Hot air is less dense than cold air, so it rises to the ceiling. As a result, the air immediately below the ceiling gets heated more quickly than the air in the rest of the room. This is because the hot air rises and accumulates near the ceiling, causing the temperature to increase more rapidly in that area.
(b) The temperature of the helium in the balloon increases and as the rubber stretches, the volume occupied by the helium increases.
(i) State what happens to the motion of the helium particles as the temperature increases. (Sub-topic – 2.1.2)
▶️Answer/Explanation
Explanation:
As the temperature increases, the kinetic energy of the helium particles also increases. This means that the particles move faster and collide with each other and the walls of the balloon more frequently and with greater force.
(ii) As the rubber stretches and the volume of the helium increases, the pressure of the helium remains constant. Explain, in terms of the particles of helium, how the pressure of the helium remains constant.
(Sub-topic – 2.1.2)
▶️Answer/Explanation
Explanation:
As the temperature increases, the helium particles move faster and collide with the walls of the balloon more forcefully, which would normally increase the pressure. However, as the rubber stretches, the volume of the balloon increases. The increase in volume means that the particles have more space to move around, reducing the frequency of collisions with the walls. The two effects (increased kinetic energy and increased volume) balance each other out, resulting in a constant pressure.
Question 4
A student investigates the efficiency of a filament lamp. Fig. 4.1 shows the filament lamp with its glass bulb immersed in water in a beaker. (Sub-topic – 1.7.3)
The reading on the thermometer in the water is 19.0°C. Only the glass of the lamp is in contact with the water and the electrical connections are completely insulated. The lamp is switched on. At the end of the experiment, the temperature of the water is 21.5°C.
(a) The mass of the water in the beaker is 600 g and the specific heat capacity of water is 4200 J/(kg°C).
(i) Show that the increase in the internal energy of the water is 6300 J.
(ii) In the experiment, the lamp is switched on for 500 s. The power supplied to the filament lamp is 13 W. The useful energy from the lamp is transferred as light. The energy that increases the temperature of the water is wasted energy.
Determine the maximum possible efficiency of the filament lamp.
▶️Answer/Explanation
(i) Calculation of the increase in internal energy:
The increase in internal energy of the water can be calculated using the formula:
\[ \Delta E = mc\Delta\theta \]
Where:
\( m = 600 \, \text{g} = 0.6 \, \text{kg} \) (mass of water),
\( c = 4200 \, \text{J/(kg°C)} \) (specific heat capacity of water),
\( \Delta\theta = 21.5°C – 19.0°C = 2.5°C \) (temperature change).
Substituting the values:
\[ \Delta E = 0.6 \times 4200 \times 2.5 = 6300 \, \text{J} \]
Therefore, the increase in internal energy of the water is 6300 J.
(ii) Calculation of the maximum possible efficiency:
The total energy supplied to the lamp is:
\[ E_{\text{total}} = P \times t = 13 \, \text{W} \times 500 \, \text{s} = 6500 \, \text{J} \]
The useful energy output is the energy transferred as light, which is the total energy minus the wasted energy:
\[ E_{\text{useful}} = 6500 \, \text{J} – 6300 \, \text{J} = 200 \, \text{J} \]
The efficiency of the lamp is given by:
\[ \text{Efficiency} = \left( \frac{E_{\text{useful}}}{E_{\text{total}}}} \right) \times 100\% = \left( \frac{200}{6500} \right) \times 100\% \approx 3.1\% \]
Therefore, the maximum possible efficiency of the filament lamp is 3.1%.
(b) The efficiency of the lamp is less than the value determined in (a)(ii). Suggest one reason for this. (Sub-topic – 1.7.3)
▶️Answer/Explanation
One possible reason for the efficiency being less than the calculated value is that some of the energy from the lamp is lost to the surroundings, such as through heat transfer to the air or the beaker, rather than being entirely used to increase the temperature of the water. Additionally, some energy may be lost as light that escapes the beaker and is not absorbed by the water.
Question 5
Fig. 5.1 shows a block ABCD made of glass that has a refractive index of 1.5. The block has one curved side AB and three straight sides, BC, CD and DA.
There are right angles at C and D. The curved side AB is one quarter of the circumference of a circle that has its centre at point P.
A ray of monochromatic light enters the block through the curved side AB and strikes side BC at P. Some light emerges into the air and some is reflected.
(a) State what is meant by monochromatic. (Sub-topic – 3.2.2)
▶️Answer/Explanation
Monochromatic light refers to light that consists of a single frequency or wavelength. It is light of a single color, such as red or blue, without any mixture of other colors.
(b) Explain why the ray of light does not change direction when it enters the block through side AB. (Sub-topic – 3.2.2)
▶️Answer/Explanation
The ray of light does not change direction when it enters the block through side AB because it is incident along the normal (perpendicular) to the surface. When light enters a medium along the normal, the angle of incidence is 0°, and thus, the angle of refraction is also 0°. Therefore, the light continues in the same direction without bending.
(c) Show that the critical angle \( c \) for glass of refractive index 1.5 is \( 42^\circ \). (Sub-topic – 3.2.2)
▶️Answer/Explanation
The critical angle \( c \) can be calculated using the formula:
\[ \sin c = \frac{1}{n} \]
where \( n \) is the refractive index of the glass. Given \( n = 1.5 \):
\[ \sin c = \frac{1}{1.5} = 0.6667 \]
Taking the inverse sine of 0.6667:
\[ c = \sin^{-1}(0.6667) \approx 42^\circ \]
Thus, the critical angle \( c \) for glass of refractive index 1.5 is \( 42^\circ \).
(d)(i) State and explain what happens to the light that strikes P when angle \( \theta \) increases to \( 45^\circ \). (Sub-topic – 3.2.2)
▶️Answer/Explanation
When angle \( \theta \) increases to \( 45^\circ \), which is greater than the critical angle \( c \) (42°), the light undergoes total internal reflection. This means that all the light is reflected back into the glass block and no light is refracted out into the air. Total internal reflection occurs because the angle of incidence exceeds the critical angle, causing the light to be completely reflected within the denser medium (glass).
(ii) When \( \theta = 45^\circ \), the reflected light strikes side CD. Describe what happens when this reflected light strikes side CD. (Sub-topic – 3.2.2)
▶️Answer/Explanation
When the reflected light strikes side CD, it will again undergo total internal reflection if the angle of incidence at side CD is greater than the critical angle. Since the light is already traveling within the glass block and the angle of incidence at side CD is likely to be greater than the critical angle, the light will be reflected back into the glass block. This process can continue, with the light being reflected back and forth within the glass block until it eventually exits the block at an angle less than the critical angle or is absorbed by the material.
Question 6
A mobile phone (cell phone) network uses microwaves of frequency 1.9 × 109 Hz to transmit and receive signals.
The speed of microwaves in air is 3.0 × 108 m / s
(a) Calculate the wavelength of these microwaves in air. (Sub-topic – 3.3)
▶️Answer/Explanation
Answer: 0.16 m
Explanation: The wavelength (\(\lambda\)) can be calculated using the formula: \[ \lambda = \frac{v}{f} \] where \(v\) is the speed of the wave and \(f\) is the frequency. Substituting the given values: \[ \lambda = \frac{3.0 \times 10^8 \, \text{m/s}}{1.9 \times 10^9 \, \text{Hz}} = 0.16 \, \text{m} \]
(b) State two reasons why microwaves are used for mobile phone (cell phone) signals. (Sub-topic – 3.3)
▶️Answer/Explanation
Answer: 1. Microwaves only need short aerials/antennas.
2. Microwaves can penetrate some walls.
Explanation: Microwaves are suitable for mobile phone signals because they require shorter antennas, making them practical for portable devices. Additionally, their ability to penetrate walls allows for better signal reception indoors.
(c)(i) Describe, with the aid of a diagram, how a digital signal differs from an analogue signal. (Sub-topic – 3.3)
▶️Answer/Explanation
Answer: A digital signal consists of discrete values, typically represented as high (1) and low (0), whereas an analogue signal varies continuously over a range of values.
Explanation: A digital signal is represented by a series of binary values (0s and 1s), making it less susceptible to noise and distortion. In contrast, an analogue signal is a continuous wave that can take any value within a range, making it more prone to interference.
(ii) State two advantages of using digital signals rather than analogue signals. (Sub-topic – 3.3)
▶️Answer/Explanation
Answer: 1. Faster data transmission rate.
2. Noise can be easily removed from the signal.
Explanation: Digital signals allow for faster data transmission and can be easily regenerated, reducing the impact of noise and distortion over long distances. This makes digital signals more reliable for communication.
Question 7
Fig. 7.1 shows a circuit that contains a battery, a switch, a voltmeter and three 40Ω resistors, \( R_1 \), \( R_2 \) and \( R_3 \). (Sub-topic code: 4.3.3)
The switch is open and resistors \( R_1 \) and \( R_2 \) form a potential divider.
(a) Describe what is meant by a potential divider.
▶️Answer/Explanation
A potential divider is a circuit that splits or shares the electromotive force (e.m.f.) or voltage of a power source between two or more resistors or components connected in series. The voltage is divided in proportion to the resistances of the resistors. In this case, resistors \( R_1 \) and \( R_2 \) form a potential divider, sharing the voltage from the battery.
(b) The reading on the voltmeter is 7.5V. (Sub-topic – 4.3.2)
(i) Calculate the electromotive force (e.m.f.) of the battery.
▶️Answer/Explanation
The e.m.f. of the battery is 15V. This is because the voltmeter reads the voltage across \( R_2 \), which is half of the total e.m.f. when \( R_1 \) and \( R_2 \) are equal (both 40Ω). Therefore, the total e.m.f. is twice the reading on the voltmeter, which is 7.5V × 2 = 15V.
(ii) The switch is closed.
Calculate the resistance of the complete circuit.
▶️Answer/Explanation
When the switch is closed, resistors \( R_2 \) and \( R_3 \) are in parallel. The combined resistance of \( R_2 \) and \( R_3 \) is calculated as follows: \[ \frac{1}{R_{23}} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20} \] Therefore, \( R_{23} = 20Ω \). The total resistance of the circuit is the sum of \( R_1 \) and \( R_{23} \): \[ R_{\text{total}} = R_1 + R_{23} = 40Ω + 20Ω = 60Ω \] So, the resistance of the complete circuit is 60Ω.
(c) Calculate the reading on the voltmeter when the switch is closed. (Sub-topic – 4.3.2)
▶️Answer/Explanation
When the switch is closed, the total resistance of the circuit is 60Ω, and the e.m.f. of the battery is 15V. The current in the circuit is: \[ I = \frac{V}{R} = \frac{15V}{60Ω} = 0.25A \] The voltage across \( R_1 \) is: \[ V_1 = I \times R_1 = 0.25A \times 40Ω = 10V \] Therefore, the voltage across \( R_2 \) (which is the reading on the voltmeter) is: \[ V_2 = V_{\text{total}} – V_1 = 15V – 10V = 5V \] So, the reading on the voltmeter when the switch is closed is 5V.
Question 8
The electricity supplied to a town is transmitted using a high-voltage cable. A transformer in the town has a soft-iron core. (Sub-topic – 4.5.6)
(a) Explain the principle of operation of a simple iron-cored transformer.
▶️Answer/Explanation
A simple iron-cored transformer works on the principle of electromagnetic induction. When an alternating current (AC) flows through the primary coil, it generates a changing magnetic field in the iron core. This changing magnetic field induces an electromotive force (emf) in the secondary coil, which is wound around the same core. The iron core helps to concentrate the magnetic field and ensures efficient transfer of energy from the primary to the secondary coil. The voltage induced in the secondary coil depends on the ratio of the number of turns in the primary coil to the number of turns in the secondary coil.
(b) The transformer steps the supply voltage down from 220 000 V to 33 000 V. (Sub-topic – 4.5.6)
(i) There are 450 turns on the secondary coil. Calculate the number of turns on the primary coil.
(ii) The electrical power transferred to the transformer by the high-voltage cable is 77 MW. Calculate the current in the primary coil.
▶️Answer/Explanation
(i) The number of turns on the primary coil can be calculated using the transformer equation: \[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] Where:
- \( V_p \) = Primary voltage = 220 000 V
- \( V_s \) = Secondary voltage = 33 000 V
- \( N_p \) = Number of turns on the primary coil
- \( N_s \) = Number of turns on the secondary coil = 450
Rearranging the equation to solve for \( N_p \): \[ N_p = N_s \times \frac{V_p}{V_s} = 450 \times \frac{220\,000}{33\,000} = 450 \times 6.6667 = 3000 \] Therefore, the number of turns on the primary coil is 3000. (ii) The current in the primary coil can be calculated using the power equation: \[ P = V_p \times I_p \] Where:
- \( P \) = Power = 77 MW = 77 × 106 W
- \( V_p \) = Primary voltage = 220 000 V
- \( I_p \) = Current in the primary coil
Rearranging the equation to solve for \( I_p \): \[ I_p = \frac{P}{V_p} = \frac{77 \times 10^6}{220\,000} = 350 \, \text{A} \] Therefore, the current in the primary coil is 350 A.
Question 9 (Sub-topic – 5.2.4)
Fig. 9.1 represents all the particles in a neutral atom of a radioactive isotope X1.
(a) Determine the number of neutrons in this atom and explain how the answer is obtained.
▶️Answer/Explanation
Number of neutrons = 7
Explanation: The number of neutrons can be determined by subtracting the number of protons (which is equal to the number of electrons in a neutral atom) from the total number of nucleons (protons + neutrons). In this case, the atom has 5 protons and 12 nucleons, so the number of neutrons is 12 – 5 = 7.
(b) The isotope X1 is a beta emitter that decays to the stable isotope X2.
(i) Describe how a nucleus of X2 differs from a nucleus of X1.
▶️Answer/Explanation
X2 has one more proton and one fewer neutron than X1. Specifically, X2 has 6 protons and 6 neutrons, whereas X1 has 5 protons and 7 neutrons.
(ii) Suggest why isotope X2 is stable whereas X1 is not stable.
▶️Answer/Explanation
Isotope X2 is stable because it has a balanced number of protons and neutrons, which results in a more stable nuclear configuration. X1, on the other hand, has an excess of neutrons, making it unstable and prone to radioactive decay.
(c) The half-life of X1 is approximately 20 ms.
(i) Define the term half-life.
▶️Answer/Explanation
Half-life is the time taken for half of the radioactive nuclei in a sample to decay.
(ii) Suggest one reason why isotopes with very short half-lives are especially hazardous.
▶️Answer/Explanation
Isotopes with very short half-lives are hazardous because they decay rapidly, releasing a large amount of radiation in a short period of time, which can cause significant damage to living tissue.
Question 10
Pluto is a dwarf planet. Fig. 10.1 shows the direction of motion of Pluto as it follows its elliptical orbit around the Sun. (Sub-topic – 6.1.1)
(a) Point X is the point in the orbit closest to the Sun and point Y is the point furthest away.
The orbital speed of Pluto varies as it orbits the Sun.
(i) Describe how the speed of Pluto varies as it moves from X to Y and then back to X.
(ii) Explain, in terms of energy transfers, why the speed of Pluto varies in this way.
▶️Answer/Explanation
(i) The speed of Pluto decreases as it moves from X to Y and then increases as it moves from Y back to X.
(ii) As Pluto moves from X to Y, its gravitational potential energy increases while its kinetic energy decreases, causing it to slow down. Conversely, as Pluto moves from Y back to X, its gravitational potential energy decreases while its kinetic energy increases, causing it to speed up. This variation in speed is due to the conservation of energy, where the total energy (kinetic + potential) remains constant throughout the orbit.
(b) The average temperature on the surface of Pluto is 43 K. (Sub-topic – 2.2.2)
(i) Convert this temperature to a value in degrees Celsius (°C).
(ii) Pluto has a white surface, as shown in Fig. 10.2. As Pluto rotates, the white surface alternately faces towards and away from the Sun.
Explain how this affects the temperature of Pluto as it rotates on its own axis.
▶️Answer/Explanation
(i) To convert from Kelvin to Celsius, use the formula: \[ T(°C) = T(K) – 273.15 \] \[ T(°C) = 43 – 273.15 = -230.15 °C \] So, the temperature on the surface of Pluto is approximately -230 °C.
(ii) The white surface of Pluto is a poor absorber and good reflector of infrared radiation. When the white surface faces the Sun, it reflects most of the solar radiation, leading to a smaller increase in temperature. When the white surface faces away from the Sun, it emits less infrared radiation, resulting in a smaller decrease in temperature. This leads to less variation in temperature as Pluto rotates compared to a darker surface, which would absorb and emit more radiation, causing greater temperature fluctuations.