Exampro a-level Physics (7407/7408)

E22. Those who gained one mark usually had the first line correct. Those who gained two marks usually forgot the gravitational PE in the second line.

E23. Most candidates successfully completed part (a) (i).
Part (ii) was also successfully completed by a majority of the candidates.
The use of the equation for uniform decelerated motion was inappropriate in part (iii). Only a minority of the candidates appreciated that the work done had to be equated to the gravitational potential energy that had been transferred to forms other than to kinetic energy.
Only a small minority did not gain some credit in part (iv) and most stated friction and air resistance. A significant proportion of these did not identify where the friction was occurring. Vague answers such as between the skier and the ground were not accepted.
Most candidates completed part (b) (i) successfully.
Part (ii) was also done well by the majority of the candidates.

E24. Part (a) (i) was straightforward and was answered very well.
Again, part (a) (ii) proved to be a particularly accessible question and candidates performed well.
In part (a) (iii), most grasped the concept that energy was wasted but less able candidates did not realise that the cyclist did work, believing it to be a simple transfer of potential energy to kinetic energy. Many candidates perhaps did not realise that marks could be gained by performing the relevant calculation.
In part (b) (i), a surprising number of candidates used time = distance/speed = 160/16 = 10 s. They did not appreciate that the situation involved uniform deceleration and therefore a kinematics equation should be used.
Part (b) (ii) was generally done well, with most calculating the acceleration and then using F = ma. It was possible to gain full marks even with the use of an incorrect answer from the previous question.

E25. Part (a) (i) was a fairly easy two marks for the majority of candidates. Most gave 420000 but about 10% omitted this and lost one mark.
Part (a) (ii) was straight forward for most but a common error was the use of the volume instead of the velocity.
For part (a) (iii), perhaps 10-20% didn’t know ‘M’ was 10⁶. Quite a few divided by 10 × 10^–6. About 14% left this question out, either indicating that they had run out of time or they had forgotten how to calculate efficiency. About 43% gained full marks.
Perhaps candidates pressed for time had spotted part (b) and sensibly decided to skip part (a) (iii) and pick up this little gift instead; still a little surprising that only 56% managed to get both marks even though 95% attempted the question.

E26. It was evident from their attempts at part (a) that during their courses many candidates had considered the application of conservation of momentum to events involving an explosion. It was less clear that they had ever considered an explosion that takes place in a moving object, or considered how conservation of energy applies in an explosion. Consequently, part (a) of the question proved to be difficult, not least because it was unfamiliar territory for so many. Part (b), which was formulaic and involved much less original thinking, brought much more success for the majority.

In part (a) only a very small proportion of the candidates were able to produce answers that were well organised, coherent, detailed and contained correct physics to merit a ‘high level’ mark of five or six. More answers fell into the ‘intermediate level’ (three or four marks) and even more into the ‘low level’ (one or two marks). A major failing in most answers was to overlook the question’s requirement to address the two conservation laws ‘in this instance’. For a high level answer, it was necessary to consider an explosion on a moving space vehicle travelling in a straight line in deep space. All of the italicised section is significant. The system has momentum before exploding (unlike a straightforward recoil example); this momentum has to be conserved because there are no external forces in deep space. Hence the probe speeds up and the capsule must be ejected along the original line of movement (although it may not be possible to tell that this is ‘backwards’ until the calculation has been done). Forces between probe and capsule during the explosion are equal and opposite, but they are internal forces for the system. When considering momentum, it was common for candidates to conclude that ‘momentum must be conserved because momentum is always conserved’.

In the explosion, chemical energy is converted into kinetic energy; this increases the total kinetic energy of the system, which is shared between probe and capsule. Examiners saw many very weak answers that showed total confusion – such as momentum being converted into energy, mass being converted into energy, or energy not being conserved. A serious omission in many answers was that of the word ‘kinetic’ before ‘energy’, whilst many answers referred to the event as an ‘inelastic collision’. There was seldom any reference to conservation of the total energy of the system taken as a whole.

Most candidates recovered from their poor attempts at part (a) to gain all three marks for the calculation in part (b) (i). There were also many awards of full marks in part (b) (ii), where the main mistake was to calculate only the kinetic energy of the system (probe + capsule) after the explosion, and to regard this as the answer to the question. Apparently, the candidates who did this had not realised that the system had an initial kinetic energy.

E27. The majority of candidates were successful in part (a) (i). A few worked backwards by substituting 3.7 and getting 67.15 m. This only received two marks if there was a clear statement that this showed 3.7 to be the correct time. Candidates should be encouraged to write down their answer to more than two significant figures for ‘show that’ questions like this one, although this was not penalised here.
A very large number of candidates expressed their answer to part (a) (ii) to three significant figures (eg 40.5) rather than two significant figures. Many used 67 m instead of 150 m. As a result, few gained both marks.

The majority gained two marks in part (a) (iii). A common incorrect answer was v = 67/3.7 = 18, where candidates were not aware that this was not appropriate for constantly accelerated motion.

Most candidates realised that Pythagoras’ theorem was needed in part (a) (iv). However, a surprising number incorrectly used the values 150 m and 67 m, rather than their answers to parts (a) (ii) and (a) (iii). Many candidates gave bearings from north instead of an angle from the horizontal or vertical. The best way to convey direction here would be to calculate the angle and then show this on a sketch and also write ‘angle from horizontal’. If there is not a clear diagram then the candidate would need to say ‘from the horizontal and downwards to the right’ or words to that effect.

Very few candidates had a problem with part (b) (i).

For part (b) (ii), there was some confusion about energy changes. Most candidates mentioned transfer of PE to KE but did not go beyond ‘energy lost due to air resistance’ and did not gain the second mark. The question asks for ‘energy changes’ so they needed to say kinetic energy changes to internal energy (heat/thermal are also accepted at AS). Some described the motion in terms of forces, velocity and acceleration – these gained no marks. Quite a few described the kinetic energy changing to gravitational as the cannonball emerges from the cannon.

E28. Part (a)(i) was answered correctly by a very high proportion of candidates.
Another straight-forward question followed with part (a)(ii); incorrect or missing units accounted for most of the lost marks. ‘Pa’ and ‘Nm’ were frequently quoted wrong units. Inappropriate use of 9.81 for acceleration was also seen.
There were very few mistakes in part (a)(iii), apart from the occasional use of s = vt = 58 × 3.5 instead of using ‘suvat’.
Candidates found part (a)(iv) tricky, with many using P = Fv with v as the final velocity rather than the average. This would give twice the average power. Many candidates found 20% of the power output and didn’t realise that the power output is 20% of the input power and they should therefore multiply by five rather than divide.
A surprisingly number of candidates made no attempt at part (b) because they did not know where to start. Most of those candidates who knew how to approach the question, gained full marks. The use of ‘suvat’ could only gain two marks out of three. Marks were often lost here due to arithmetic errors in calculations; typically, forgetting to square the speed.

E29.A common error in part (a) (i) was to use t = s/v = 50/6.7. Students are not always aware that, for uniform acceleration, the ‘suvat’ equations must be used. Another common error was to use t = √(2s/a) with the speed used as the acceleration.
Nearly all students were successful with the calculation in part (a) (ii). However, a large number stated an answer to more than two significant figures and lost a mark.