Question 1
Fig. 1.1 shows the speed–time graph for a bus journey.
(a) (i)Using the information in Fig. 1.1, determine the maximum speed of the bus during the journey. (Sub-topic – 1.2)
▶️Answer/Explanation
The maximum speed of the bus during the journey is 18 m/s.
(ii) Using the information in Fig. 1.1, determine the speed of the bus at time = 65s. On Fig. 1.1, show how you obtained this information. (Sub-topic – 1.2)
▶️Answer/Explanation
The speed of the bus at time = 65s is 12 m/s. To obtain this, draw a vertical line from 65s on the time axis to the speed-time curve and then a horizontal line to the speed axis.
(b) Describe how the speed of the bus changes between time = 60s and time = 80s. (Sub-topic – 1.2)
▶️Answer/Explanation
Between time = 60s and time = 80s, the speed of the bus decreases until it comes to a stop (speed = 0 m/s) at around 75s. After that, the bus remains stationary.
(c) Determine the distance travelled by the bus between time = 0 and time = 10s. (Sub-topic – 1.2)
▶️Answer/Explanation
The distance travelled by the bus between time = 0 and time = 10s is 90 meters. This is calculated by finding the area under the speed-time graph, which is a triangle with a base of 10s and a height of 18 m/s. The area is given by:
\[ \text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 18 = 90 \, \text{m} \]
(d) Fig. 1.2 shows the speed–time graph for another bus journey.
The driver sees a hazard ahead and applies the brakes at time = 40 s.
The bus reduces its speed from 14.0 m/s to 6.0 m/s in a time of 5.0 s.
On Fig. 1.2, draw the speed–time graph for the bus as it reduces its speed. (Sub-topic – 1.2)
▶️Answer/Explanation
To draw the speed-time graph for the bus as it reduces its speed:
- Start the line at (40s, 14 m/s).
- Draw a straight line decreasing to (45s, 6 m/s).
This line represents the deceleration of the bus over the 5-second period.
Question 2
A farmer uses a rope to lift a barrel of fruit from the ground to a platform, as shown in Fig. 2.1. (Sub-topic – 1.7.1)
(a) The farmer lifts the barrel of fruit at a constant speed.
(i) State the energy store of the barrel of fruit that increases as the barrel rises.
▶️Answer/Explanation
Answer: Gravitational potential energy.
Explanation: As the barrel is lifted, its height increases, which results in an increase in its gravitational potential energy. This is because gravitational potential energy is directly related to the height of an object above a reference point.
(ii) The weight of the barrel of fruit is 140 N. Show that the work done on the barrel of fruit in lifting it from the ground to the platform is approximately 450 J.
▶️Answer/Explanation
Answer: Work done = Force × Distance = 140 N × 3.2 m = 448 J ≈ 450 J.
Explanation: Work done is calculated using the formula \( W = F \times d \), where \( F \) is the force (weight of the barrel) and \( d \) is the distance moved (height lifted). Here, the weight of the barrel is 140 N, and the height lifted is 3.2 m. Therefore, the work done is \( 140 \times 3.2 = 448 \) J, which is approximately 450 J.
(b) The farmer wants to make the process faster. He buys a machine to lift the barrels of fruit. (Sub-topic – 1.7.3)
(i) The output power of the machine is 75 W. The work done in lifting a barrel of fruit onto the platform is 450 J. Calculate the time taken for the machine to lift a barrel of fruit onto the platform.
▶️Answer/Explanation
Answer: Time = Work done / Power = 450 J / 75 W = 6 s.
Explanation: Power is defined as the rate at which work is done, so \( P = \frac{W}{t} \). Rearranging this formula gives \( t = \frac{W}{P} \). Here, the work done is 450 J, and the power is 75 W. Therefore, the time taken is \( \frac{450}{75} = 6 \) seconds.
(ii) The machine uses an electric motor. The farmer installs some wind turbines to supply electrical power for the farm. Suggest one environmental reason for using wind turbines rather than using a diesel (fossil fuel) generator.
▶️Answer/Explanation
Answer: Wind turbines do not produce greenhouse gas emissions, unlike diesel generators which emit carbon dioxide and other pollutants.
Explanation: Wind turbines generate electricity using wind energy, which is a renewable source of energy. Unlike diesel generators, which burn fossil fuels and release greenhouse gases, wind turbines do not contribute to air pollution or global warming. This makes them a more environmentally friendly option.
Question 3
The mass of a glass bottle is 0.18 kg.
(a) Calculate the weight of the bottle. (Sub-topic – 1.3)
▶️Answer/Explanation
Answer: 1.8 N
Explanation: Weight is calculated using the formula \( W = m \times g \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (9.8 m/s²).
\( W = 0.18 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.764 \, \text{N} \).
Rounding to two significant figures, the weight is 1.8 N.
(b) The bottle contains 2.7 kg of cooking oil. The density of the cooking oil is 0.92 g/cm³. Calculate the volume of the cooking oil. (Sub-topic – 1.4)
▶️Answer/Explanation
Answer: 2900 cm³
Explanation: Volume is calculated using the formula \( V = \frac{m}{\rho} \), where \( m \) is the mass and \( \rho \) is the density.
First, convert the mass from kg to grams: \( 2.7 \, \text{kg} = 2700 \, \text{g} \).
\( V = \frac{2700 \, \text{g}}{0.92 \, \text{g/cm}^3} = 2934.78 \, \text{cm}^3 \).
Rounding to two significant figures, the volume is 2900 cm³.
(c) A cookery student pours some cooking oil into a glass bowl containing water, as shown in Fig. 3.1.
The student accidentally drops a plastic spoon and a metal spoon into the bowl. The densities of the spoons and liquids are shown in Table 3.1. (Sub-topic – 1.4)
On Fig. 3.1, label a suggested position for each spoon after each has fallen into the bowl.
Use the letter P to label the position of the plastic spoon and the letter M to label the position of the metal spoon.
▶️Answer/Explanation
Answer: P on top liquid surface, M on bottom of bowl
Explanation: The plastic spoon has a density of 0.76 g/cm³, which is less than the density of water (1.0 g/cm³) and cooking oil (0.92 g/cm³), so it will float on the surface of the top liquid (cooking oil). The metal spoon has a density of 8.7 g/cm³, which is much higher than both water and cooking oil, so it will sink to the bottom of the bowl.
Question 4
(a) Fig. 4.1 shows a pan on a hotplate. The hotplate heats the pan and the water. (Sub-topic – 2.3.1, 2.3.3)
Use your ideas about thermal energy transfer to explain why the pan has:
(i) a wooden handle
(ii) shiny sides
(iii) a copper base.
▶️Answer/Explanation
Solution:
(i) The wooden handle is used because wood is a poor conductor of heat. This reduces the amount of thermal energy transferred to the handle, making it safe to touch even when the pan is hot.
(ii) The shiny sides of the pan are designed to reflect infrared radiation, which reduces the amount of heat lost to the surroundings by radiation. This helps to keep the pan and its contents hot.
(iii) The copper base is used because copper is an excellent conductor of heat. This allows for efficient transfer of thermal energy from the hotplate to the pan and its contents, ensuring even heating.
(b) Fig. 4.2 shows a heater for warming a room. When there is hot water in the heater, thermal energy transfers from the water to the room. (Sub-topic – 2.3.2)
Explain how thermal energy from the heater warms the entire room. Use your ideas about the density of air. You may draw on Fig. 4.2.
▶️Answer/Explanation
Solution:
The heater warms the air around it, causing the air particles to gain kinetic energy and move further apart. This makes the air less dense, so it rises. As the warm air rises, cooler air from the room moves in to take its place, creating a convection current. This process continues, circulating warm air throughout the room and ensuring that the entire room is heated evenly. The convection current is driven by the difference in air density caused by the temperature variation.
Question 5 (Sub-topic – 3.3)
(a) Fig. 5.1 shows regions of the electromagnetic spectrum in order of increasing wavelength. Two of the regions are unlabelled.
(i) Complete Fig. 5.1 by writing the name of each unlabelled region in the correct box.
(ii) State two properties that are the same for all regions of the electromagnetic spectrum.
▶️Answer/Explanation
(i) The unlabelled regions are ultraviolet (left-hand box) and microwaves (right-hand box).
(ii) Two properties that are the same for all regions of the electromagnetic spectrum are:
- They are all transverse waves.
- They all travel at the same speed in a vacuum (3.0 × 108 m/s).
(b) (i) State one use for infrared radiation.
(ii) State one harmful effect of excessive exposure to infrared radiation.
▶️Answer/Explanation
(i) One use for infrared radiation is in thermal imaging or remote controllers for televisions.
(ii) One harmful effect of excessive exposure to infrared radiation is skin burns.
Question 6
(a) A student shines a ray of red light into a rectangular glass block, as shown in Fig. 6.1. (Sub-topic – 3.2.2)
(i) Draw the normal at the point where the ray of red light enters the glass block.
▶️Answer/Explanation
Answer: The normal should be drawn perpendicular to the surface of the glass block at the point where the ray of light enters.
(ii) On Fig. 6.1, label each ray using words from the list
diffracted diffused dispersed incident reflected refracted
▶️Answer/Explanation
Answer: The ray entering the glass block should be labeled as “incident,” the ray reflecting off the surface should be labeled as “reflected,” and the ray passing through the glass block should be labeled as “refracted.”
(b) Fig. 6.2 and Fig. 6.3 each show two parallel rays of light travelling through air towards a lens. (Sub-topic – 3.2.3)
For each lens, draw the path of the two rays as they pass through the lens and back into the air.
▶️Answer/Explanation
Answer:
- For Fig. 6.2 (converging lens), the rays should converge to a focal point after passing through the lens.
- For Fig. 6.3 (diverging lens), the rays should diverge as if they are coming from a focal point on the same side as the incoming rays.
(c) State the seven colours of visible light. Give the colours in order of frequency. (Sub-topic – 3.2.4)
▶️Answer/Explanation
Answer: The seven colours of visible light in order of increasing frequency are: Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV).
Question 7
Two identical resistors, \( R_1 \) and \( R_2 \), are connected to a 24 V battery, as shown in Fig. 7.1.
The value of each resistor is 50 \(\Omega\). (Sub-topic – 4.3.2)
(a) Calculate the combined resistance of \( R_1 \) and \( R_2 \) when they are connected as shown in Fig. 7.1.
▶️Answer/Explanation
The resistors \( R_1 \) and \( R_2 \) are connected in series. The combined resistance \( R_{\text{total}} \) in a series connection is given by: \[ R_{\text{total}} = R_1 + R_2 \] Given \( R_1 = 50 \Omega \) and \( R_2 = 50 \Omega \): \[ R_{\text{total}} = 50 \Omega + 50 \Omega = 100 \Omega \] Therefore, the combined resistance is 100 \(\Omega\).
(b) Show that the current in the circuit is approximately 0.25 A. (Sub-topic – 4.2.2)
▶️Answer/Explanation
The current \( I \) in the circuit can be calculated using Ohm’s Law: \[ I = \frac{V}{R} \] where \( V = 24 \) V and \( R = 100 \Omega \) (from part (a)): \[ I = \frac{24 \text{ V}}{100 \Omega} = 0.24 \text{ A} \] This is approximately 0.25 A.
(c) Determine the potential difference (p.d.) across \( R_1 \). (Sub-topic – 4.2.3)
▶️Answer/Explanation
The potential difference across \( R_1 \) can be calculated using Ohm’s Law: \[ V_1 = I \times R_1 \] where \( I = 0.24 \) A (from part (b)) and \( R_1 = 50 \Omega \): \[ V_1 = 0.24 \text{ A} \times 50 \Omega = 12 \text{ V} \] Therefore, the potential difference across \( R_1 \) is 12 V.
(d) Calculate the power transferred in \( R_1 \). (Sub-topic – 4.2.5)
▶️Answer/Explanation
The power \( P \) transferred in \( R_1 \) can be calculated using the formula: \[ P = I^2 \times R_1 \] where \( I = 0.24 \) A (from part (b)) and \( R_1 = 50 \Omega \): \[ P = (0.24 \text{ A})^2 \times 50 \Omega = 0.0576 \times 50 = 2.88 \text{ W} \] Alternatively, using \( P = V \times I \): \[ P = 12 \text{ V} \times 0.24 \text{ A} = 2.88 \text{ W} \] Therefore, the power transferred in \( R_1 \) is 2.88 W.
(e) A student connects \( R_1 \), \( R_2 \) and the battery to make a different circuit. The resistors \( R_1 \) and \( R_2 \) are connected so their combined resistance is as small as possible.
Draw a circuit diagram to show how \( R_1 \) and \( R_2 \) are connected to the battery. (Sub-topic – 4.3.2)
▶️Answer/Explanation
To achieve the smallest possible combined resistance, \( R_1 \) and \( R_2 \) should be connected in parallel. The circuit diagram would show both resistors connected directly across the battery terminals.
Question 8
(a) Fig. 8.1 shows an arrangement for transmitting electricity generated by a power station. (Sub-topic – 4.5.6)
The step-up transformer has 500 turns on the primary coil.
Calculate the number of turns on the secondary coil of the step-up transformer. Use the information given in Fig. 8.1.
▶️Answer/Explanation
Solution:
To calculate the number of turns on the secondary coil, we use the transformer equation:
\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]
Where:
- \( V_p \) = primary voltage = 25 kV
- \( V_s \) = secondary voltage = 360 kV
- \( N_p \) = number of turns on the primary coil = 500
- \( N_s \) = number of turns on the secondary coil (to be calculated)
Rearranging the equation to solve for \( N_s \):
\[ N_s = N_p \times \frac{V_s}{V_p} = 500 \times \frac{360}{25} = 7200 \]
Therefore, the number of turns on the secondary coil is 7200.
(b) State two benefits of using high voltages for transmitting electricity. (Sub-topic – 4.5.6)
▶️Answer/Explanation
Solution:
Two benefits of using high voltages for transmitting electricity are:
- Reduced power loss: High voltages reduce the current in the transmission lines, which in turn reduces the power loss due to resistance (\( P = I^2R \)).
- Efficiency: High voltage transmission is more efficient as it allows for the transfer of electrical energy over long distances with minimal energy loss.
Question 9
(a) Describe what is meant by alternating current (a.c.). (Sub-topic – 4.2.2)
▶️Answer/Explanation
Alternating current (a.c.) is an electric current that periodically reverses direction. In an a.c. circuit, the flow of electric charge changes direction back and forth at regular intervals. This is in contrast to direct current (d.c.), which flows in only one direction. The frequency of the alternating current is the number of complete cycles per second, measured in hertz (Hz).
(b) A teacher demonstrates how a loudspeaker works by using the equipment shown in Fig. 9.1.
There is an alternating current in the coil. The paper cone and coil vibrate as shown in Fig. 9.1.
(i) Explain why the paper cone vibrates. Use your ideas about magnetism. (Sub-topic – 4.5.3)
▶️Answer/Explanation
The paper cone vibrates due to the interaction between the magnetic field of the permanent magnet and the magnetic field produced by the alternating current in the coil. When the alternating current flows through the coil, it creates a varying magnetic field. This varying magnetic field interacts with the fixed magnetic field of the permanent magnet, causing the coil to move back and forth. Since the coil is attached to the paper cone, the movement of the coil causes the paper cone to vibrate, producing sound waves.
(ii) When the paper cone vibrates, the teacher hears a sound. Suggest a value for the frequency of the alternating current. Include the unit. (Sub-topic – 4.5.3)
▶️Answer/Explanation
A typical value for the frequency of the alternating current in a loudspeaker is 1 kHz (1000 Hz). This frequency is within the audible range for humans and is commonly used in audio applications to produce clear and audible sound.
Question 10
(a) Fig. 10.1 represents all the particles in a lithium atom. (Sub-topic – 5.1.1)
(i) State the proton number (atomic number) of the lithium atom in Fig. 10.1.
▶️Answer/Explanation
Answer: 3
Explanation: The proton number (atomic number) of lithium is 3, as it has 3 protons in its nucleus.
(ii) Determine the nucleon number (mass number) of the lithium atom in Fig. 10.1.
▶️Answer/Explanation
Answer: 7
Explanation: The nucleon number (mass number) is the sum of protons and neutrons in the nucleus. Lithium has 3 protons and 4 neutrons, so the nucleon number is 7.
(iii) Describe how a lithium atom changes to form a positive ion.
▶️Answer/Explanation
Answer: A lithium atom loses an electron to form a positive ion.
Explanation: Lithium has 3 electrons. When it loses one electron, it becomes a positively charged ion (Li⁺) because it now has more protons than electrons.
(b) The half-life of iodine-131 is 8 days. A sample contains 80 mg of iodine-131. Calculate the time taken to decay until 10 mg of iodine-131 remain in the sample. (Sub-topic – 5.2.4)
▶️Answer/Explanation
Answer: 24 days
Explanation: The half-life of iodine-131 is 8 days. Starting with 80 mg, after each half-life, the amount of iodine-131 is halved:
- After 8 days: 80 mg → 40 mg
- After 16 days: 40 mg → 20 mg
- After 24 days: 20 mg → 10 mg
Therefore, it takes 24 days for the sample to decay to 10 mg.
Question 11
Fig. 11.1 represents the Earth in orbit around the Sun. (Sub-topic – 6.1.1)
(a) (i) State the name of the force that keeps the Earth in orbit around the Sun.
▶️Answer/Explanation
The force that keeps the Earth in orbit around the Sun is gravitational force.
(ii) State the time taken by the Earth to complete one orbit of the Sun. Include the unit.
▶️Answer/Explanation
The time taken by the Earth to complete one orbit of the Sun is 365.25 days.
(iii) State the time taken by the Earth to rotate once on its axis. Include the unit.
▶️Answer/Explanation
The time taken by the Earth to rotate once on its axis is 24 hours.
(iv) The Sun consists mainly of two gases. State the names of the two gases.
▶️Answer/Explanation
The Sun consists mainly of hydrogen and helium.
(b) (i) Most of the radiation from the Sun consists of visible light and two other regions of the electromagnetic spectrum. State the name of one of the other two regions. (Sub-topic – 6.2.1)
▶️Answer/Explanation
One of the other two regions of the electromagnetic spectrum is infrared or ultraviolet.
(ii) The speed of visible light is \( 3.0 \times 10^8 \, \text{m/s} \). Calculate the time taken for visible light to travel from the Sun to the Earth when the Earth is in the position shown in Fig. 11.1.
▶️Answer/Explanation
To calculate the time taken for light to travel from the Sun to the Earth, we use the formula:
\[ \text{time} = \frac{\text{distance}}{\text{speed}} \]
Given:
- Distance from the Sun to the Earth = \( 1.5 \times 10^{11} \, \text{m} \)
- Speed of light = \( 3.0 \times 10^8 \, \text{m/s} \)
Substituting the values:
\[ \text{time} = \frac{1.5 \times 10^{11} \, \text{m}}{3.0 \times 10^8 \, \text{m/s}} = 500 \, \text{s} \]
Therefore, the time taken for visible light to travel from the Sun to the Earth is 500 seconds.