Question 1
Calculate the charge supplied by the charger.
Suggest why, in normal use, the car needs to be recharged after travelling less than $390\text{ km}$.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.2$ — Motion (Parts $\text{(a)(i)}$)
• Topic $1.5.1$ — Effects of forces (Parts $\text{(a)(ii)}$)
• Topic $1.7.1$ — Energy (Parts $\text{(a)(iii)}$)
• Topic $4.2.2$ — Electric current (Parts $\text{(b)}$)
• Topic $1.7.3$ — Energy resources (Parts $\text{(c)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
$9.7\text{ s}$
We know the car starts from rest, so initial speed $u=0$, final speed $v=28\text{ m/s}$, and acceleration $a=2.9\text{ m/s}^2$. Using the definition of acceleration $a = \frac{\Delta v}{\Delta t}$ or $a = \frac{v-u}{t}$, we can rearrange to find the time: $t = \frac{v-u}{a} = \frac{28 – 0}{2.9} \approx 9.66\text{ s}$. Rounded to two significant figures, the time taken is $9.7\text{ s}$.
(a)(ii)
For the correct answer:
$4600\text{ N}$
According to Newton’s second law of motion, the resultant force is the product of mass and acceleration, so $F = ma$. We are given the combined mass $m = 1600\text{ kg}$ and the acceleration $a = 2.9\text{ m/s}^2$. Substituting these values gives $F = 1600 \times 2.9 = 4640\text{ N}$. To two significant figures, the force required is $4600\text{ N}$.
(a)(iii)
For the correct answer:
$630\,000\text{ J}$ / $6.3 \times 10^5\text{ J}$
The kinetic energy (KE) of a moving object is given by the equation $E_k = \frac{1}{2}mv^2$. The mass $m$ is $1600\text{ kg}$ and the speed $v$ is $28\text{ m/s}$. Putting these values into the formula gives $E_k = \frac{1}{2} \times 1600 \times (28)^2 = 800 \times 784 = 627\,200\text{ J}$. This is $6.3 \times 10^5\text{ J}$ to two significant figures.
(b)
For the correct answer:
$960\,000\text{ C}$ / $9.6 \times 10^5\text{ C}$
Electric current is the rate of flow of charge, defined by the equation $I = \frac{Q}{t}$, which we can rearrange to find the charge $Q = It$. The current is $32\text{ A}$ and the time is $8.3\text{ hours}$. First, convert the time into seconds: $8.3 \times 60 \times 60 = 29\,880\text{ s}$. Then, $Q = 32 \times 29\,880 = 956\,160\text{ C}$. This is approximately $9.6 \times 10^5\text{ C}$.
(c)
For the correct answer:
Any one explicit example of a variation from ideal conditions such as:
(repeated) acceleration / deceleration / use of brakes / varying speed / motion uphill / uneven road surface / cold weather / headwind.
The maximum range of $390\text{ km}$ is achieved under very specific, ideal, and constant conditions. In normal driving, you constantly accelerate and brake, travel at varying speeds, encounter hills and uneven roads, and face wind resistance. All these factors demand more energy from the battery than a steady, ideal journey, reducing the actual distance you can travel before needing a recharge.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.1$ — Energy (Parts $\text{(a)}$, $\text{(b)(i)}$)
• Topic $1.7.3$ — Energy resources (Parts $\text{(b)(ii)}$, $\text{(c)(i)}$, $\text{(c)(ii)}$)
• Topic $1.7.4$ — Power (Parts $\text{(b)(i)}$, $\text{(b)(ii)}$)
• Topic $4.2.5$ — Electrical energy and electrical power (Parts $\text{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
gravitational potential (energy)
Water is held at a height behind the dam. Because of its position in the Earth’s gravitational field, the energy stored is gravitational potential energy. This energy is ready to be converted into kinetic energy as the water falls.
(b)(i)
For the correct answer:
$2.0 \times 10^6\text{ J/s}$
The rate of energy transfer is the power from the falling water. The loss in gravitational potential energy per second is $\frac{\Delta E_p}{t} = \frac{mgh}{t}$. We are given the mass flow rate $\frac{m}{t} = 480\text{ kg/s}$. Using $g = 10\text{ N/kg}$ and $h = 410\text{ m}$, we get power $P = 480 \times 10 \times 410 = 1\,968\,000\text{ W}$. This is approximately $2.0 \times 10^6\text{ J/s}$.
(b)(ii)
For the correct answer:
$81\%$ OR $82\%$
Efficiency is the ratio of useful power output to total power input, multiplied by $100\%$. The useful electrical power output from the scheme is $P_{out} = IV = 270 \times 6000 = 1\,620\,000\text{ W}$. The input power from the falling water is $1\,968\,000\text{ W}$ (from part (i)). Efficiency $= \frac{1\,620\,000}{1\,968\,000} \times 100\% \approx 82.3\%$. Rounded to two significant figures, it’s either $81\%$ or $82\%$ depending on the values used.
(c)(i)
For the correct answer:
damage to habitats (for fish) / construction is expensive / droughts / flood risk if dam bursts
While hydroelectric power is renewable, it’s not without drawbacks. Building a large dam floods the valley upstream, destroying terrestrial and aquatic habitats. The construction itself is immensely expensive, and there is always a catastrophic flood risk if the dam were to fail.
(c)(ii)
For the correct answer:
biofuel / wind / geothermal / tidal / solar / wave
Any energy source that is not depleted when used and can be naturally replenished on a human timescale qualifies. Solar power, which captures energy directly from sunlight, is a prime example of another renewable source.
Question 3
Draw a circle around two other quantities in the list which are vectors.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.5.1$ — Effects of forces (Parts $\text{(a)(i)}$, $\text{(a)(ii)}$)
• Topic $1.1$ — Physical quantities and measurement techniques (Parts $\text{(a)(i)}$, $\text{(b)}$)
• Topic $1.3$ — Mass and weight (Parts $\text{(a)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
suitable scale recorded (e.g. $2\text{ cm} : 25\text{ N}$)
two vectors correctly drawn by eye AND correct resultant
$130\text{ N}$
(vertically) upwards
To find the resultant, I’d draw two vectors to scale, each $75\text{ N}$, tail-to-tail at the correct angle from the diagram. I’d then complete the parallelogram and draw the diagonal from the tail point. Measuring this diagonal with my chosen scale gives the magnitude. Since the boat is hanging symmetrically, the horizontal components cancel out, and the resultant is the vector sum of the vertical components, acting straight upwards to support the boat’s weight.
(a)(ii)
For the correct answer:
$13\text{ kg}$
The boat is in equilibrium, so the resultant upward force from the ropes must equal the downward weight of the boat. From (a)(i), the resultant force is $130\text{ N}$, so the weight $W = 130\text{ N}$. Weight is related to mass by $W = mg$, giving mass $m = \frac{W}{g} = \frac{130}{10} = 13\text{ kg}$.
(b)
For the correct answer:
acceleration
momentum
Vector quantities have both magnitude and direction. Acceleration is the rate of change of velocity (which is also a vector), so it has a direction. Momentum is the product of mass and velocity ($p=mv$), and since velocity is a vector, momentum is too. Density, energy, mass, power, and refractive index only have magnitude, so they are scalars.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.1.2$ — Particle model (Parts $\text{(a)(i)}$, $\text{(a)(ii)}$, $\text{(b)(i)}$, $\text{(b)(ii)}$)
• Topic $2.2.2$ — Specific heat capacity (Parts $\text{(b)(i)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
zig zag motion / random changes of direction
random length of path in each direction
When looking through the microscope, a single bright speck doesn’t move smoothly. Its path looks like a chaotic, random zigzag. Each segment of the path has a different random direction and a different random length, showing no predictable pattern at all.
(a)(ii)
For the correct answer (any four from):
• air molecules bombard smoke particles
• air molecules are small (compared to smoke particles) / have small(er) mass
• air molecules are very fast moving
• air molecules move in random directions
• (collisions exert unbalanced) forces on smoke particles
The smoke particles are massive compared to individual air molecules. The surrounding air molecules are constantly moving at high speeds in completely random directions. At any instant, more air molecules might collide with one side of a smoke particle than the other, creating a tiny, unbalanced net force. This force pushes the smoke particle in a random direction, and this process repeats continuously, causing the observed jerky motion.
(b)(i)
For the correct answer:
kinetic energy (and potential energy) of molecules increase (hence internal energy increases)
Internal energy is the sum of the kinetic and potential energies of all the molecules. When we increase the temperature, the air molecules move faster on average. This directly increases their kinetic energy. For many substances, the increased molecular jostling also slightly increases the average separation of molecules, increasing their potential energy. The total internal energy therefore rises.
(b)(ii)
For the correct answer:
bigger change in momentum of molecules OR molecules hit (the walls) harder
(molecules hit) more often / more frequently
Gas pressure is caused by molecules repeatedly colliding with the container walls. A higher temperature means molecules are moving faster. When they hit a wall, they rebound with a larger change in momentum, meaning they exert a greater force per impact (they hit harder). Since they are also moving faster, they travel between walls more quickly, leading to more frequent collisions. Both these effects combine to increase the overall force on the wall area, and thus the pressure goes up.
Question 5
Calculate the energy given out as the water cools from $5.0^{\circ}\text{C}$ to ice at $-18.0^{\circ}\text{C}$.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.2.2$ — Specific heat capacity (Parts $\text{(a)}$, $\text{(b)(ii)}$)
• Topic $1.4$ — Density (Parts $\text{(b)(i)}$)
• Topic $2.2.3$ — Melting, boiling and evaporation (Parts $\text{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
energy required to raise the temperature of $1\text{ kg}$ / $1\text{ g}$ / unit mass of a substance by $1^{\circ}\text{C}$ / unit temperature
Specific heat capacity is a property of a material that tells you how much energy is needed to heat it up. The formal definition must always include the idea of a unit mass (like $1\text{ kg}$) and a unit temperature change (like $1^{\circ}\text{C}$). It essentially measures a substance’s resistance to changing temperature when energy is added.
(b)(i)
For the correct answer:
$0.50\text{ kg}$
The mass can be found from the volume and density using the formula $\rho = \frac{m}{V}$, which rearranges to $m = \rho V$. The volume is $500\text{ cm}^3$. To use standard units, we convert this to $5.0 \times 10^{-4}\text{ m}^3$. The density of water is $1000\text{ kg/m}^3$. So, $m = 1000 \times 5.0 \times 10^{-4} = 0.50\text{ kg}$.
(b)(ii)
For the correct answer:
$190\,000\text{ J}$ / $1.9 \times 10^5\text{ J}$ / $190\text{ kJ}$
This is a three-stage cooling process. First, the water cools from $5.0^{\circ}\text{C}$ to $0^{\circ}\text{C}$: energy $E_1 = mc\Delta\theta = 0.5 \times 4200 \times 5 = 10\,500\text{ J}$. Second, the water freezes at $0^{\circ}\text{C}$: energy $E_2 = mL = 0.5 \times 3.3 \times 10^5 = 165\,000\text{ J}$. Third, the ice cools from $0^{\circ}\text{C}$ to $-18.0^{\circ}\text{C}$: energy $E_3 = mc\Delta\theta = 0.5 \times 2100 \times 18 = 18\,900\text{ J}$. The total energy released is the sum: $10\,500 + 165\,000 + 18\,900 = 194\,400\text{ J}$, which rounds to $1.9 \times 10^5\text{ J}$.
Question 6
- three crests of the water wave in the shallow water
- the direction of travel of the wave in the shallow water.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.1$ — General properties of waves (Parts $\text{(a)(i)}$, $\text{(a)(ii)}$)
• Topic $3.4$ — Sound (Parts $\text{(c)(i)}$, $\text{(c)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
wavefronts semicircles or part semicircles centred on gap
wavelength of waves to right of barrier same as wavelength of incident wave
When a plane wave passes through a narrow gap, it undergoes diffraction. This means it spreads out from the gap as if the gap were a new point source. The wavefronts to the right of the barrier should therefore be drawn as concentric semicircles centred on the gap. It’s important to note that the speed and frequency of the wave don’t change, so the wavelength (distance between crests) must remain the same as the incident waves.
(a)(ii)1 & 2
For the correct answer:
1. wavelength shorter
correct refraction
2. direction of travel perpendicular to wavefronts
As the waves enter the shallow water, they slow down. Because the frequency of the wave must stay the same, the wave equation $v=f\lambda$ tells us the wavelength must decrease (get shorter). The change in speed also causes a change in direction unless the boundary is approached head-on; this bending is called refraction. The direction of travel is always perpendicular to the wavefronts, so an arrow drawn at $90^{\circ}$ to the new wavefronts in the shallow region shows the new path.
(b)
any two from:
- particles (in transverse waves) vibrate perpendicular to the direction of travel (of the wave) OR particles in longitudinal waves vibrate parallel to the direction of travel of the wave
- longitudinal waves have compressions and rarefactions
- transverse waves have troughs and crests
Transverse and longitudinal waves can be distinguished by the direction of particle vibration. In transverse waves, particles vibrate perpendicular to the direction the wave travels, whereas in longitudinal waves, they vibrate parallel. Additionally, their structures differ—longitudinal waves consist of compressions and rarefactions, while transverse waves are characterized by troughs and crests.
(c)(i)
For the correct answer:
$1000\text{ m/s}$ < value < $2000\text{ m/s}$
Sound travels much faster in water than in air. A typical textbook value for the speed of sound in water is around $1500\text{ m/s}$. Any value in the range from $1000\text{ m/s}$ to $2000\text{ m/s}$ is a good estimate.
(c)(ii)
For the correct answer:
molecules closer together / water has greater density
Sound is transmitted by particles vibrating and passing energy to their neighbours. In liquids like water, the molecules are packed much closer together compared to the widely spaced molecules in a gas like air. This tighter coupling means the vibrational energy is transmitted from one molecule to the next far more quickly, resulting in a higher speed of sound.
Question 7
State the name of one region of the electromagnetic spectrum in which the waves have:
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.2.1$ — Reflection of light (Parts $\text{(a)}$, $\text{(b)}$)
• Topic $3.3$ — Electromagnetic spectrum (Parts $\text{(c)(i)}$, $\text{(c)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
ray from left hand corner of the mirror to the eye
angle of incidence = angle of reflection
To find the limiting point P, I’d need to draw a ray from the extreme edge of the mirror (the left corner) to the person’s eye at point X. That’s the reflected ray. Then, at the point where this ray hits the mirror surface, I’d use the law of reflection. The angle of incidence must equal the angle of reflection. The incident ray corresponding to this reflected ray will point to the point P on wall AB, showing the closest point to A that can be seen.
(b)
For the correct answer (any two):
virtual
upright
same size as object
laterally inverted
An image in a plane mirror has a specific set of characteristics. It’s always virtual, meaning you cannot project it onto a screen; the light rays only appear to come from behind the mirror. It’s also upright (the same way up as the object), it appears the same size as the object, and it’s laterally inverted (your left becomes your right).
(c)(i)
For the correct answer:
ultraviolet / X-rays / gamma rays
The electromagnetic spectrum is ordered by wavelength. Visible light sits in a narrow band. Regions with higher frequency and higher energy than visible light also have shorter wavelengths. Ultraviolet is immediately next to visible light, with X-rays and gamma rays having even progressively shorter wavelengths.
(c)(ii)
For the correct answer:
infrared / microwaves / radio (waves)
Looking the other way on the spectrum, the regions with lower frequency and lower energy than visible light have longer wavelengths. Infrared radiation lies just beyond red light in the spectrum. Past that, microwaves and radio waves have the longest wavelengths of all.
Question 8
On Fig. 8.1, draw the symbol for the voltmeter and its connections to the circuit.
Component Y has a resistance of 400 Ω.
In a brightly lit room, the resistance of component X is 350 Ω.
2. Calculate the p.d. across component Y.
State the effect this will have on the p.d. across component Y.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.3.1$ — Circuit diagrams and circuit components (Parts $\text{(a)(i)}$, $\text{(a)(ii)}$, $\text{(b)}$)
• Topic $4.2.4$ — Resistance (Parts $\text{(a)(iii)1}$, $\text{(a)(iii)2}$, $\text{(a)(iv)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
light-dependent resistor / LDR
Component X has its resistance labeled as varying with light. A component whose resistance decreases dramatically when light shines on it is a light-dependent resistor (LDR).
(a)(ii)
For the correct answer:
voltmeter connected in parallel with component Y
To measure the potential difference (p.d.) across any component, a voltmeter must be connected in parallel with it. For component Y, this means connecting the voltmeter leads to the two terminals of Y so it measures the difference in electrical potential from one side to the other.
(a)(iii)1
For the correct answer:
$0.016\text{ A}$
The circuit is a series circuit, so the total resistance is the sum of the two resistors: $R_{total} = 350 + 400 = 750\Omega$. The total voltage is $12\text{ V}$. Using Ohm’s law, the current $I = \frac{V}{R} = \frac{12}{750} = 0.016\text{ A}$.
(a)(iii)2
For the correct answer:
$6.4\text{ V}$
Component Y is the $400\Omega$ resistor. Since the current is the same everywhere in a series circuit, the current through Y is $0.016\text{ A}$. Applying Ohm’s law again, the potential difference across Y is $V = IR = 0.016 \times 400 = 6.4\text{ V}$.
(a)(iv)
For the correct answer:
(in a dark room the p.d. across component Y) decreases
Component X is an LDR. In the dark, its resistance becomes very high. This dramatically increases the total circuit resistance, causing the total current to become very small. Since the p.d. across Y is given by $V_Y = I R_Y$, a much smaller current through Y leads to a much smaller p.d. across it. The fixed $12\text{ V}$ supply is now mostly dropped across the high-resistance component X.
(b)
For the correct answer:
one named practical application of LDR e.g. switch on street lights (at night) / turn on security light (at night)
LDRs are used in circuits that need to respond to light levels. A classic example is in a street light. The LDR’s high resistance in darkness can be used to trigger an electronic switch to turn the lamp on, and its low resistance in daylight can switch it off automatically.
Question 9
Describe and explain what happens to the needle of the sensitive galvanometer.
State the effect of using a stronger magnet on what happens to the needle of the galvanometer.
Calculate the number of turns on the primary coil.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.5.1$ — Electromagnetic induction (Parts $\text{(a)(i)}$, $\text{(a)(ii)}$)
• Topic $4.5.6$ — The transformer (Parts $\text{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer (any four from):
• needle oscillates (as magnet moves up and down)
• coil cuts magnetic field / magnetic field changes (as magnet moves)
• changing (magnetic) field induces voltage/current
• induced voltage/current opposes the motion/change causing it
• force, magnetic field and induced current are mutually perpendicular
When the magnet is pulled down and let go, it oscillates up and down on the spring. As the magnet moves, the magnetic field passing through the coil is constantly changing. Faraday’s law says that a changing magnetic field induces an e.m.f., which drives a current around the coil. The galvanometer needle will therefore swing back and forth, oscillating around the centre-zero mark. The current’s direction changes because the induced e.m.f. always opposes the change causing it (Lenz’s law).
(a)(ii)
For the correct answer:
larger (maximum) deflection
The size of the induced e.m.f. depends on the strength of the magnetic field. A stronger magnet creates a denser magnetic field. When this stronger magnet oscillates, the rate of change of the magnetic flux through the coil is greater, which induces a larger peak voltage. This larger voltage pushes more current through the galvanometer, making the needle swing out to a larger maximum deflection.
(b)
For the correct answer:
$2300$
For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. We need to find the primary turns $N_p$. Rearranging gives $N_p = \frac{V_p}{V_s} \times N_s$. The primary voltage $V_p = 25\,000\text{ V}$, the secondary voltage $V_s = 400\,000\text{ V}$, and the secondary turns $N_s = 36\,000$. So, $N_p = \frac{25\,000}{400\,000} \times 36\,000 = \frac{1}{16} \times 36\,000 = 2\,250$. The calculated answer is $2\,250$, but the mark scheme allows for $2\,300$ depending on rounding or initial values used.
Question 10
The readings on the detector, taken every 20 s, are recorded in Table 10.1.
Protactinium-234 decays to an isotope of uranium (U) by $\beta$-emission.
Write down the nuclide equation for this decay of protactinium-234.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $5.2.1$ — Detection of radioactivity (Parts $\text{(a)}$)
• Topic $5.2.4$ — Half-life (Parts $\text{(b)}$)
• Topic $5.2.3$ — Radioactive decay (Parts $\text{(c)}$)
• Topic $5.1.2$ — The nucleus (Parts $\text{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
background radiation (present in values in Table 10.1)
(background radiation) is removed (before plotting) OR (background radiation) not present in the graph values
The detector reading in the table is a raw count. It records radiation from the protactinium sample plus a certain amount of naturally occurring background radiation from the environment. The graph, however, plots the corrected count rate, which is the data from the sample only. To get the corrected rate, the constant background radiation count is subtracted from each of the table’s readings.
(b)
For the correct answer:
$70 \le \text{half-life} \le 76\text{ (s)}$
The half-life is the time it takes for the count rate to drop by half. Looking at the graph, we can pick a clear starting count rate, e.g., $100\text{ counts/min}$, and see when it falls to $50\text{ counts/min}$. The time difference between these two points on the horizontal axis gives the half-life. Using this method on the plotted curve gives a half-life in the range of $70$ to $76$ seconds.
(c)
For the correct answer:
$\frac{234}{91}\mathrm{Pa} \rightarrow \frac{234}{92}\mathrm{U} + {}^{0}_{-1}\beta$
This is a beta-minus decay equation. A neutron in the protactinium nucleus turns into a proton, increasing the atomic number by 1 (from $91$ to $92$, which is uranium), but the mass number stays the same at $234$. A beta particle (an electron) is emitted from the nucleus, represented as ${}^{0}_{-1}\beta$. On the right-hand side, the total mass number is $234 + 0 = 234$, and the total atomic number is $92 + (-1) = 91$, matching the parent nucleus on the left.