Question 1
• zero with ‘1’
• negative with ‘2’
• decreasing with ‘3’.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.5.2 — Turning effect of forces (Part (a)(i))
• Topic 1.7.1 — Energy (Part (a)(ii))
• Topic 1.2 — Motion (Part (b)(i))
• Topic 1.2 — Motion (Part (b)(ii))
▶️ Answer/Explanation
(a)(i)
For an object to be in a state of equilibrium, two physical requirements must be satisfied simultaneously. First, there must be no resultant (net) force acting on the object in any direction, meaning all translational forces cancel out. Second, there must be no resultant (net) moment acting on the object, ensuring there is no tendency for it to rotate about any pivot point.
(a)(ii)
For the correct answer:
4.7×10 7 J or 47 MJ
The change in gravitational potential energy is calculated using the formula ΔE p =mgh. By substituting the given values into the equation, we get ΔE p =3200 kg×9.8 m/s 2 ×1500 m. This calculation results in a total energy change of 47,040,000 J, which is expressed as 4.7×10 7 J using appropriate significant figures.
(b)(i)
Point ‘1’ should be marked on any horizontal section of the graph where the gradient is zero, indicating constant speed. Point ‘2’ is located on the section where the speed is rapidly decreasing after the parachute opens at A, representing a negative gradient. Point ‘3’ is placed on the initial curve after t=0 where the slope is becoming less steep as the rate of acceleration drops due to increasing air resistance.
(b)(ii)
Immediately after jumping, the only significant force is weight (W=mg), causing maximum acceleration. As speed increases, the upward air resistance (drag) grows, reducing the resultant downward force according to F net =W−Drag. This causes the acceleration to decrease continuously until the air resistance equals the weight. At this point, the resultant force becomes zero, and the parachutist reaches a constant terminal velocity.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.6$ — Momentum (Parts $\mathrm{(b)}$, $\mathrm{(c)}$)
• Topic $1.7.1$ — Energy (Part $\mathrm{(a)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$25.92 \text{ J}$
Kinetic energy is calculated using the formula $E_k = \frac{1}{2}mv^2$. First, convert the mass to kilograms: $160 \text{ g} = 0.16 \text{ kg}$. Substituting the values, $E_k = \frac{1}{2} \times 0.16 \text{ kg} \times (18 \text{ m/s})^2$, which results in $0.08 \times 324 = 25.92 \text{ J}$.
(b)
For the correct answer:
$24 \text{ N}$
The average force is defined as the rate of change of momentum, given by $F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}$. Here, the change in velocity $\Delta v$ is $18 \text{ m/s} – 0 \text{ m/s} = 18 \text{ m/s}$. Using the mass of $0.16 \text{ kg}$ and time of $0.12 \text{ s}$, the calculation is $F = \frac{0.16 \times 18}{0.12} = \frac{2.88}{0.12} = 24 \text{ N}$.
(c)
For the correct answer:
Reduces the impact force on the hands.
Moving the hands backward increases the time interval $\Delta t$ over which the ball’s momentum is reduced to zero. Since the change in momentum $\Delta p$ remains the same, increasing the time in the equation $F = \frac{\Delta p}{\Delta t}$ results in a smaller resultant force $F$ acting on the hands. This reduction in force minimizes the pressure and potential pain or injury during the catch.
Question 3
Explain why this way of moving across the ice is safer than walking. Use your understanding of pressure in your answer.
The surface area of the pond is $5.0\text{ m}^2$.
The mass of the ice is $690\text{ kg}$.
The density of water is $1000\text{ kg/m}^3$.
Point X is $0.45\text{ m}$ below the ice.
Calculate the pressure at point X due to the ice and the water.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.5.1$ — Effect of forces (Part $\mathrm{(a)}$, $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
Pressure is defined as force per unit area, expressed by the equation $p = \frac{F}{A}$. When a person lies down or crawls, their weight (force) is distributed over a significantly larger surface area compared to standing on two feet. By increasing the area $A$, the pressure $p$ exerted on the ice decreases, which minimizes the stress on the frozen surface and reduces the likelihood of the ice cracking or breaking.
(b)
The total pressure at X is the sum of the pressure from the ice and the liquid pressure. First, find the pressure of the ice: $P_{\text{ice}} = \frac{W}{A} = \frac{mg}{A} = \frac{690 \times 9.8}{5.0} = 1352.4\text{ Pa}$. Next, calculate the water pressure using $P_{\text{water}} = \rho gh = 1000 \times 9.8 \times 0.45 = 4410\text{ Pa}$. Summing these gives $P_{\text{total}} = 1352.4 + 4410 = 5762.4\text{ Pa}$, which rounds to $5.8 \times 10^3\text{ Pa}$.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.1.3 — Gases and the absolute scale of temperature (Part (a))
• Topic 2.1.2 — Particle model (Parts (b)(i), (b)(ii))
• Topic 1.8 — Pressure (Part (c))
▶️ Answer/Explanation
(a)
For the correct answer:
The pressure decreases.
As temperature decreases, the average kinetic energy and speed of the gas particles also decrease. Consequently, particles collide with the container walls less frequently and with less force (F= Δt Δp ). Since pressure is defined as force per unit area (p= A F ), these reduced collisions result in a overall lower pressure exerted by the gas.
(b)(i)
For the correct answer:
−273 ∘ C
Absolute zero is the theoretical point on the temperature scale where all thermal motion ceases. On the Celsius scale, this corresponds to a value of approximately −273 ∘ C. It serves as the starting point (0 K) for the Kelvin scale, which is the absolute scale of temperature used in scientific calculations.
(b)(ii)
For the correct answer:
Lowest possible temperature where particles have least kinetic energy.
In the kinetic particle model, temperature is a measure of the average kinetic energy of the particles. Absolute zero is the lowest possible temperature where the particles of a substance have the minimum possible kinetic energy. At this point, the internal energy of the system is at its minimum, and the particles essentially stop moving.
(c)
For the correct answer:
197 cm 3
According to Boyle’s Law for a fixed mass of gas at constant temperature, p 1 V 1 =p 2 V 2 . Given p 1 =9.0×10 4 Pa, V 1 =350 cm 3 , and p 2 =1.6×10 5 Pa, we rearrange for V 2 : V_2= p_2 p_1V_1 = 1.6×10 5 Pa (9.0×10 4 Pa)×(350 cm 3 ) =196.875 cm 3
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.4 — Density (Part (a)(i))
• Topic 2.2.2 — Specific heat capacity (Part (a)(ii))
• Topic 2.3.4 — Consequences of thermal energy transfer (Parts (a)(iii), (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
79kg
To find the mass, first calculate the room’s volume using V=length×width×height=4.5×6.1×2.4=65.88m 3 . Using the density formula ρ= V m , rearrange it to m=ρ×V. Substituting the values, m=1.2kg/m 3 ×65.88m 3 =79.056kg, which rounds to 79kg as required.
(a)(ii)
For the correct answer:
290s
First, calculate the thermal energy required using ΔE=mcΔθ=79×1000×(20.0−16.0)=316,000J. Since power is the rate of energy transfer, use the equation P= t ΔE rearranged to t= P ΔE . The time taken is t= 1100W 316,000J ≈287.3s, which rounds to 290s to two significant figures.
(a)(iii)
For the correct answer:
thermal energy is lost from the room (to the surroundings)
The calculation assumes all energy from the heater goes into the air, but in reality, thermal energy is lost through walls and windows via conduction and convection. Additionally, some energy is used to heat the objects and furniture within the room rather than just the air itself. Consequently, the actual time taken would be longer than this theoretical minimum.
(b)
For the correct answer:
conduction AND convection
Double glazing uses a trapped layer of air or a vacuum between two glass panes to reduce heat loss. Air is a very poor thermal conductor, which significantly limits heat transfer by conduction. Furthermore, because the gap is narrow and sealed, air cannot circulate freely to form large convection currents, thereby minimizing thermal energy transfer by convection to the outside.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.1 — General properties of waves (Parts (a), (b))
▶️ Answer/Explanation
(a)
For the correct answer:
P-waves: longitudinal; S-waves: transverse
Seismic waves are classified based on the direction of particle vibration relative to the direction of wave travel. P-waves (primary) are longitudinal waves where vibrations are parallel to the direction of propagation, while S-waves (secondary) are transverse waves where vibrations occur at right angles to the direction of travel.
(b)
For the correct answer:
1.6 km
The relationship between wave speed, frequency, and wavelength is given by the wave equation v=fλ. To find the wavelength, rearrange the formula to λ= f v and substitute the given values: λ= 4.5 Hz 7.2 km/s =1.6 km. This calculation determines the physical distance between two successive crests or identical points of the P-wave as it travels through the Earth’s crust.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.2.2 & 3.2.3 — Refraction of light and Speed of light (Parts (a), (b), (c), (d))
▶️ Answer/Explanation
(a)
For the correct answer:
The normal is drawn as a dashed line perpendicular (90 ∘ ) to the oil surface at the point of incidence.
A normal is an imaginary construction line used in optics to represent the perpendicular to the boundary between two media. By definition, it must meet the surface at a right angle at the exact point where the light ray strikes the interface. This line serves as the reference for measuring the angles of incidence and refraction.
(b)
For the correct answer:
34.4 ∘ (or 34.3 ∘ depending on rounding)
The angle of refraction r is found using Snell’s Law: n 1 sin(i)=n 2 sin(r). Given n air =1.0, n oil =1.47, and i=56 ∘ , the equation becomes 1.0×sin(56 ∘ )=1.47×sin(r). Rearranging gives sin(r)= 1.47 sin(56 ∘ ) ≈0.564, resulting in r=arcsin(0.564)≈34.4 ∘ .
(c)
For the correct answer:
3.0×10 8 m/s
Light travels at its maximum possible speed in a vacuum, which is approximately 300,000,000 meters per second. Since air is very thin compared to solids or liquids, the speed of light in air is very nearly the same as in a vacuum. In IGCSE Physics, this value is standardly recalled and used as 3.0×10 8 m/s for all calculations involving air.
(d)
For the correct answer:
2.04×10 8 m/s
The refractive index n is defined as the ratio of the speed of light in a vacuum c to the speed of light in the medium v. Using the formula n= v c , we can rearrange it to find the speed in oil: v= n c . Substituting the values gives v= 1.47 3.0×10 8 m/s ≈2.0408×10 8 m/s, which rounds to 2.04×10 8 m/s to three significant figures.
Question 8
The resistors R 1 , R 2 and R 3 are identical.
The reading on the voltmeter is 6.0 V.
When the diode is conducting, it has zero resistance and zero potential difference (p.d.) across it.
2. Determine the new value of the ratio of the p.d. across R 2 to the p.d. across R 3 when all the cells are reversed.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.1 — Simple phenomena of magnetism (Parts (a)(i), (a)(ii))
• Topic 4.2.1 — Electric charge (Part (b))
• Topic 4.3.3 — Action and use of circuit components (Parts (c)(i), (c)(ii), (c)(iii))
▶️ Answer/Explanation
(a)(i)
A magnetic field is defined as a region in which a magnetic pole (such as the North pole of a compass needle) experiences a magnetic force.
(a)(ii)
The direction of a magnetic field at any point is specifically defined as the direction of the force that would be exerted on the North pole (N pole) of a magnet if it were placed at that point.
(b)
To represent the electric field, draw four straight radial lines that are equally spaced around the sphere and touch its surface. Each line must include an arrowhead pointing towards the center of the negatively charged sphere, as electric field lines represent the path a positive test charge would follow.
(c)(i)
The voltmeter measures the total electromotive force (E total ) of the three identical cells connected in series. Since the total e.m.f. is 6.0 V, the e.m.f. of a single cell is calculated as E cell = 3 6.0 V =2.0 V.
(c)(ii)
With the diode conducting (0Ω), R 2 and R 3 are in series. Since the resistors are identical (R 2 =R 3 ), and potential difference in series is proportional to resistance (V∝R), the ratio of the p.d. across R 2 to the p.d. across R 3 is exactly 1:1 (Note: If R 3 represents two resistors in series, the ratio would be 1:2).
(c)(iii)1.
When the cells are reversed, the diode becomes reverse-biased, acting as an insulator with infinite resistance. This prevents any current from flowing through the branch containing R 1 , so the current in R 1 becomes 0 A.
(c)(iii)2.
Because the reverse-biased diode prevents current from flowing through the entire circuit, there is no flow of charge through the resistors. Since V=IR and I=0, the potential difference across both R 2 and R 3 is 0 V, making the ratio 0:0 or undefined, though often expressed as 1:1 in zero-state logic.
Question 9
| Type of Radiation | Number of Protons | Number of Neutrons | Charge (C) | Stopped By |
|---|---|---|---|---|
| α | 2 | +3.2 × 10-19 | Thin sheet of paper | |
| β | 0 | Thin sheet of aluminium | ||
| γ | 0 |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.2.1 — Detection of Radioactivity (Parts (a), (b), (c), (d))
▶️ Answer/Explanation
(a)
Alpha (α): 2 neutrons; Beta (β): 0 protons, charge −1.6×10 −19 C; Gamma (γ): 0 neutrons, 0 charge, stopped by thick lead/concrete.
An alpha particle is a helium nucleus ( 2 4 He), meaning it has 2 protons and 2 neutrons. A beta particle is an electron ( −1 0 e), which has 0 protons/neutrons and a single negative elementary charge. Gamma radiation consists of electromagnetic waves, which possess neither mass nor charge and require dense shielding like lead to be absorbed.
(b)
A neutron changes into a proton and an electron is emitted.
During beta decay, a neutron within the nucleus decays into a proton and an electron via the weak interaction. The proton remains in the nucleus, increasing the atomic number Z by 1, while the electron (the β-particle) is ejected. The mass number A remains constant throughout this process.
(c)
95 counts/min
Calculate the source’s initial rate: R 0 =550−30=520 counts/min. The number of half-lives is n= 25 75 =3. The remaining source activity is R= 2 3 520 = 8 520 =65 counts/min. Adding back the constant background gives a final detector reading of 65+30=95 counts/min.
(d)
Use long-handled tongs; store in lead-lined containers.
Safety is maintained by increasing the distance from the source using tools like tongs, which reduces the intensity of radiation reaching the body. Storing materials in lead-lined boxes provides effective shielding to absorb ionising radiation, while limiting exposure time further minimizes the total dose received by laboratory personnel.
Question 10
The stage that follows the stable state in the life cycle of the star is the red supergiant stage. It then explodes as a supernova to form a nebula, this leaves behind a neutron star or a black hole.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.1.1 & 6.1.2 — The Earth and the Solar System (Parts (a), (b))
• Topic 6.2.2 & 6.2.3 — Stars and the Universe (Parts (c), (d))
▶️ Answer/Explanation
(a)
For the correct answer:
v= T 2πr where v is average orbital speed, r is average radius, and T is orbital period.
Average orbital speed is defined as the distance travelled in one complete orbit (the circumference of a circle, 2πr) divided by the time taken for that orbit (the orbital period, T). For planets, r represents the average distance from the Sun.
(b)
For the correct answer:
The Earth’s axis is tilted relative to its orbit around the Sun.
As the Earth orbits the Sun, its fixed axial tilt means different hemispheres lean towards or away from the Sun at different times of the year. Regions far from the Equator experience large changes in the angle of incidence of solar radiation and daylight hours, leading to significant seasonal temperature shifts.
(c)
For the correct answer:
red supergiant; nebula; neutron star; black hole
Stars significantly more massive than the Sun expand into red supergiants after their stable hydrogen-burning phase. When their core collapses, they undergo a supernova explosion, dispersing a nebula of gas and dust while leaving behind a highly dense remnant, either a neutron star or a black hole.
(d)
For the correct answer:
1.6×10 9 light-years
Using Hubble’s Law, v=H 0 d, the distance in km is d= H 0 v = 2.2×10 −18 s −1 33000 km/s =1.5×10 22 km. Converting to light-years using 1 ly≈9.46×10 12 km, we calculate d= 9.46×10 12 1.5×10 22 ≈1.58×10 9 light-years, which rounds to 1.6×10 9 light-years.