IGCSE Physics October November 2025 Paper 4 Variant 1

Added: 2026-05-14

[00:00:00]Hello everyone, this is John Hashmut and welcome to Physics Simply. In this video, I will be solving the paper 4 exam for October November 2025 variant one. So, let's get started. Question one says, a train travels with a constant velocity of 56 m/s on horizontal track.

[00:00:15]The mass of the train is 440,000 kg.

[00:00:19]State the difference between the velocity of the train and its speed. We say that velocity has a direction and speed does not have direction. Pab says, "Calculate the kinetic energy stored in the moving train. Kinetic energy is calculated as half the mass multiplied by the speed squared." So half 440,000 multiplied by the speed 56^ 2. That gives an answer of 6.9 * 10 ^ 8.

[00:00:48]Part C says a train has a uniform deceleration of 1.2 m/s squared.

[00:00:53]Calculate the constant braking force which brings the train to the stop. We can use F= to m a. The mass is 440,000 and the acceleration against the motion is 1.2. That gives an answer of 5.3 * 10 ^ 5 Newton. Double I says calculate the distance traveled by the train as it comes to rest. We can either use work is equal to force time distance or we use chyntics. Since we have constant acceleration, the initial velocity is 56, the final velocity is zero and the acceleration is -1.2 2 because it's a deceleration and we know want to calculate the distance. So we can use v ^2 = u ^2 + 2 a s. So 0 is equal to 56 ^ 2 + 2 * -1.2 * the unknown s. We can use 56^ squar on the other side with a negative sign. So negative will cancel out with negative. And 2 * 1.2 is 2.4. S is equal to 56^2. Then we divide 56^2 by 2.4. 4 to get an answer of 1,300 m. The answer is 1,36.6 recurring, but we approximate it to two significant figures since the numbers in the question were to two significant figures.

[00:02:07]Question two says, table 2.1 contains information about the planet Mars.

[00:02:11]Define gravitational field strength. We have the mass, gravitational field strength, and the average dynasty.

[00:02:16]Gravitational field of strength is the force per unit mass or the force of gravity per unit mass. But BI says an object has weight of 42 newtons at the surface of the earth. Calculate the weight of the object at the surface of Mars. So the mass is the same by the way and mass is equal to the weight divided by the gravity field strength. So the weight on the earth is 42 and the gravity field strength on the earth is 9.8. That would be the same mass for Mars except we do not know the mass on Mars. And we have the gravity field of strength from the table above is 3.7 Newtons per kilogram. So this is 3.7.

[00:02:54]Now we can use cross multiplication. We multiply 42 by 3.7 and then we divide by 9.8. That gives an answer of 15.857 which is approximately 16 newtons.

[00:03:07]Double I says calculate the volume of Mars. So if denasty is equal to mass divided by volume then volume is equal to mass divided by denasty. And we have these values in the table above. We have the mass 6.4 * 10^ of 23 kg and the average dynasty 3,900 kg per me cubed.

[00:03:26]So we divide 6.4 * of 23 by the 3,900.

[00:03:31]We get an answer of 1.641025 * of 20 which is approximately 1.6. And since the dynasty was kg per me cubed, then this unit is in meters cubed.

[00:03:44]Question C says, "Figure 2.1 shows a space buggy that is tested on Earth. The buggy is traveling at constant speed in a straight line. The driving force on the buggy is 30 Ntons. Draw and label one arrow on figure 2.1 to show the size and direction of the resistive forces on the buggy. So the resistive force is opposite to the motion. And since we are moving with constant speed, then the backward force or the resistive force must be equal to the driving force. So this is also 30 Ntons. Double I says air resistance on Mars is less than air resistance on Earth. The same driving force of 30 Ntons is exerted on the buggy on Mars. State the effect this has on the resultant force on the buggy on Mars. So now there is a resultant force forward since the backward force decreased. Part two says state the relationship between resistive forces, driving force and resultant force. So we say that the resultant force is equal to the driving force minus the resistive forces.

[00:04:45]Question three says figure 3.1 shows a simplified diagram of a solar cell.

[00:04:49]Describe the energy transfer in the solar cell. It is converted from light to electric energy. Part B says suggest how the black coating allows a solar cell to transfer more energy. We say that black surfaces are good absorbers of energy. Proc says 0.72 kilowatt of light is incident on the solar cell. In figure 3.1 the cell has an efficiency of 75%. Calculate the output power of the cell. So we say that the efficiency is equal to the output power divided by the input power multiplied by 100. So we multiply the efficiency by the input power and then we divide by 100. The power output is equal to 75 * 0.72 but time 10 ^ 3 for kilo and then we divide that by 100. That gives an answer of 540 watt. Double I says state the meaning of the term kilowatt hour. We say that it is a unit of energy and energy is equal to power multiplied by time and power p is in kilo watt and time t is in hours.

[00:05:54]So we multiply power by time to get an energy. Triple I says energy is produced by each solar cell for an average of 6 hours per day. A household uses approximately 7,400 kilowatt hour of electric energy per year. Calculate the number of solar cells needed to produce energy for one household. Give your answer as a whole number of solar cells.

[00:06:15]So we can calculate the total energy. We have 7,400 ultiplied by 1,000 for kilowatt and then multiplied by 60 by 60 to convert from hour to seconds. That is the total power during the whole year. And then we divide by the time for 365 days. We get 72.9863 * 10 ^ of 6. That is the total power for 1 day. And the power for one cell was 540. That's per second. We multiply that by 6 hours and then 60 by 60 to get the total energy. Since this was the total energy per day, we need also to get the total energy per day for the cell. That is approximately 1.1664 * 10^ of 7. So this is one cell. And then we divide the total energy by the energy produced by one cell. So we divide 72.986 * 10^ 6 by 1.1664 * 7. That gives an answer of 6.25. 257. So we need more than six cells. So we need seven cells.

[00:07:22]Question 4 A says, a 12vt 50 W immersion heater is used to heat 0.15 kg of water in a beaker. The water is initially at room temperature of 20° C. The specific heat capacity of the water is 4,200 JW per kilog°C. Calculate the energy supply to raise the temperature of the water from 20° to 58°. We can use energy is equal to mc delta theta. The mass and the specific heat capacity has the same unit kilogram. So we do not need to convert 0.15 * 4,200. And then we multiply by the change in temperature 58us 20. That gives an answer of 23,940.

[00:08:03]So we can approximate that to 24,000 Jew or 2.4 * 4 Jew. Pab says, "The immersion heater is removed from the beaker. One metal rod and one plastic rod are placed in the beaker of hot water as shown in figure 4.1. The rods are at room temperature 20°C before they are placed into the beaker. Describe how the temperature of point X on each rod changes after the rods are placed in the beaker and explain your answer. So we have one metal and one non-metal. We say that they both will get hotter. So both temperatures will increase but much faster in the metal or the metal rod since it has free electrons and is a better conductor than plastic. So I increase the information of the free electrons because we have a lot of marks here. Question five says a dolphin communicates with other dolphins underwater by emitting sounds in the range of 7 to 15 kHz. So 7 kHz to 15 kHz. State the value of the speed of sound in air and state how the speed of sound in water differs from the speed of sound in air. So it would be about 340 m/s. It is anything between 300 and 360.

[00:09:17]And the speed of sound in water is faster. It is approximately 1,500 m/s.

[00:09:24]Babis says state and explain if humans with normal hearing can hear all sounds emitted by the dolphin. So humans can hear from 20 hertz to 20 kilohz. So 7,000 to 15,000 is within this range. So yes, we can hear everything emitted from a dolphin. Part C says complete table 5.1 to describe differences in loudness and pitch of two different dolphin sounds. So we have frequencies 14 and 8 and large amplitude and small amplitude.

[00:09:55]For the 14 kHz, the loudness is large due to large amplitude and the pitch is high since 14 is greater than 8. As for the loudness for the 8 kHz, small amplitude means small loudness or low loudness and the pitch is also low since frequency 8 is less than 14. Part D says complete the sentences to describe how sound is transmitted through water.

[00:10:20]Sound waves are made of vibrating particles which produce compressions and rare factions. A compression is a region of high pressure and rare factions is a region of low pressure. The sound waves travel in what direction to the direction of vibrations that would be parallel to the direction of vibrations.

[00:10:40]Question six says figure 6.1 shows part of an optical fiber used in high-speed broadband communication. State two, advantages of using optic fibers in high-speed data transmission compared to electrical signal sent in a copper wire.

[00:10:54]We can say that light travels much faster than electricity and fiber cables are cheaper. We can also say that it is more secure or there are less energy losses. Partbi says the optic fiber is made of glass with a refractive index of 1.4. Calculate the critical angle. We can say that the refractive index is equal to 1 / sin the critical angle. So sin the critical angle is equal to 1 / the refractive index 1.4. And then we calculate that and use shift sign answer to calculate the angle because this is sign the angle not the angle itself. So 1 / 1.4 and then shift sign answer gives an angle of 45.58469 which is approximately 46°.

[00:11:40]Double I is state the meaning of critical angle. We say that is an angle of incidence in a denser medium that gives angle of refraction 90° in a less dense medium. Triple I says on figure 6.1 label the angle of incidence of the ray of light as it hits the wall of the glass and then draw the continuation of the ray until it leaves the glass fiber.

[00:12:05]First we need to draw a normal which is 90° on the surface of the glass at this point and this is the angle of incidence I and the angle of incidence is always greater than the critical angle in the optic fiber. So there will be total internal reflection. So we can draw a reflected ray with an angle of reflection almost equal to the angle of incidence and keep doing that bouncing of the light until it goes out of the fiber. So here I drew two bounces.

[00:12:34]Another one may draw other rays or it can bounce from this point going up then down and then out. You do not have to actually measure the angles of incidence and reflection. So this angle of reflection is equal to this angle of incidence and this angle is equal to this angle and so on.

[00:12:53]Question seven says figure 7.1 shows a circuit containing a 6volt battery of cells and three identical resistors.

[00:13:00]Part A says S_sub_1 is closed and S_UB_2 is open. The current in the emter is 0.080 amp. So now only the current is flowing in this part of the circuit. The lower part is not working. Calculate the resistance R1. So we can calculate the total resistance is equal to the total voltage divided by the total current 0.080. And then we divide that by two since the resistors are identical. So this is twice the resistance of R1. So we divided by two to get R1. So 6 / 0.080.

[00:13:33]And then we divide that by two. We get an answer of 37.5 which is approximately 38 ohms. Part B says S1 and S2 are both closed. Determine the reading on the meter. Show your working. We can say that the full voltage is still 6 volts.

[00:13:51]But the total resistance now is the resistance of the two resistors above in series which was 37.5 * 2. So 75 in parallel with the resistor on the bottom which is also 37.5. So we can use the equation product over sum to calculate the total resistance. 75 * 37.5 over 75 + 37.5. That gives an answer of 25 ohms.

[00:14:19]So we have the total voltage and the total resistance. The total current is equal to the total voltage divided by total resistance 6 over 25. That gives an answer of 0.24 ampere. Double I says explain in terms of work done and the potential difference why there is a larger heating effect in R3 than R1. R3 was the resistor which was connected alone and R1 was connected in series. So it takes half the voltage of R3. So we can say that R1 has half the potential difference and half the current. So R1 has less work done on it.

[00:14:57]Question 8 says, figure 8.1 shows a solenoid. We have the solenoid connected to a cell and a switch. Part AI says, draw on figure 8.1 four complete magnetic field lines that show the pattern and direction of the magnetic field inside and outside the solenoid.

[00:15:12]First we need to know the direction of the current. So the current comes out of the positive side of the battery and then goes to the left and then turns upwards from the front and downwards from the back. So using the right hand grip rule, the fingers are pointing in this direction. The thumb would be pointing to the left. So we would draw field lines which are pointing to the left from outside then inside then outside again and an identical field line like this also going from right to left and then maybe one complete loop for a magnetic field line like this and an identical field line on the upper side of the solenoid also coming out from the left side and going into the right side. Double I says, "Mark a point inside the box on figured 8.1 where the magnetic field is strong. Label this point B. Explain how the diagram shows that the magnetic field is strong at B."

[00:16:07]So the strongest part would be in the middle here. And the reason that this part of the field is the strongest is because the field lines are close to each other. The closer the lines, the stronger the magnetic field. Part B says, "Figure 8.1 shows a solenoid in an electric circuit for a bell. We have a soft iron arm, a spring metal and contacts to keep the circuit closed. We have a striker connected to the metal arm and the bell and the solenoid next to the metal arm. Part I says complete the circuit in figure 8.2 with the symbol for a direct current power supply. So we can draw a battery like this with unknown number of cells or we can just draw two circles but mention that one of them is positive and the other one is negative. Double I says, explain why the soft iron arm pivots, making the striker hit the bell when the switch is closed. We say that when current flows in the solenoid, it creates a magnetic field and attracts the iron arm because iron is a magnetic material. Triple I says, explain why the arm pivots back to its original position after the striker hits the bell. In this diagram, if the arm moves downwards, these contacts will no longer be touching each other. So we now have an open circuit. So we say the contacts move apart and open the circuit. So no more magnetic field and the iron arm is not attracted to the solenoid. So the springy metal pulls it back in place.

[00:17:39]Question 9 a says describe the structure of an atom of helium 4. And we have the proton number and the nucleon number two and four. So first from the proton number we have two protons. And if we subtract 4 minus 2, we get also two neutrons in the nucleus. But we want to describe an atom. So we say also two electrons outside the nucleus or orbiting the nucleus. Part B says the sun is a medium-sized star powered by nuclear fusion reactions which releases energy. State what happens during nuclear fusion reactions which form helium. We say that two hydrogen nuclei join together to form a helium nucleus.

[00:18:22]Double I say state two regions of electromagnetic spectrum by which the sun radiates most of its energy. We can say visible light and infrared there is also ultraviolet. Part says describe what happens to a star when most of the fuel in its center has been converted to helium. We say that it expands to form a red joint or a red super joint if the mass was much greater than the mass of the sun.

[00:18:50]Question 10 says figure 10.1 shows the orbit of the earth and the orbit of a comet around the sun. State which orbit A or B is the orbit of the comet and explain your answer. So for these orbits B is the orbit of the comet. And the explanation is that comets have much more elliptical orbits than planets like the Earth. Part B says, "Describe and explain how the motion of the comet changes as it orbits the sun. We say that its speed increases moving closer as it loses gravitational potential energy and gains kinetic energy. And the opposite when it moves away from the sun, it gains gravitational potential energy and loses kinetic energy. Part says at one position in its orbit the comet is 6.6 * 10^ of -6 light years away from the earth. State the meaning of a lightyear. It is the distance moved or traveled by light in one year or during one year. Double I says determine the distance in meters between the comet and the earth. We have the distance in light years 6.6 * 10^ of -6. And the distance for one light year is multiplied as speed multiplied by time.

[00:20:01]Speed of light is 3 * 10^ of 8 multiplied by the time of 1 year 365 days by 24 hours by 60 minutes by 60 seconds. That gives 6.244 * 10^ of 10. We can approximate that to 6.2 * 10 ^ of 10 and the unit is already given in meters.

[00:20:22]So this was the end of the exam. I hope you enjoyed this video and found it useful. Keep practicing and I will see you in another video.