June 2023: AS AS Paper 2 Section B: Mechanics

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), common incorrect answers were “took the same time” or explanations about one accelerating faster but running at a slower speed with no reference to distance. However, most seemed to know that the area represented distance and realised that since it was a race, the children had covered the same distance. In the second part, a few used Pythagoras or found the

Common mistakes

• area under a v-t graph is distance: both children cover the same distance (the length of the race)
• "they took the same time" alone scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

children had covered the same distance. In the second part, a few used Pythagoras or found the area but most either used the gradient or a suvat approach. A number went from 5a = 4 to 5 a = 4. Most candidates, in part (c), found the area correctly using Pat’s graph. Some made

Common mistakes

• gradient of the first section = 4/5 = 0.8 m s⁻²
• slips such as 5a = 4 giving a = 5/4

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

5 a = 4. Most candidates, in part (c), found the area correctly using Pat’s graph. Some made 1 errors with lengths of sides or omitted the 2. Very few used suvat for the whole motion and a few left out the triangle thus getting either 90 or 110. There were many poor responses for the final part. Since the shape of the graph used in part (d) was the same shape as the graph used in part (c), it was surprising to see so many candidates using a different approach for finding this area even though they realised that the areas were the same. Many candidates either didn’t even attempt it or gave up after writing one equation in two variables. Of those who realised that the acceleration meant that X = t, or whatever variable they had used, a number were not able to use that to obtain an equation in X only and so scored no marks. There were a few suvat attempts and a common error was to use 22.5 instead of 27.5. Those who were able to get to a correct quadratic equation almost always selected the correct root as their final answer and usually gave it to 3sf although there were a few 3.91’s seen.

Common mistakes

• total area under P's graph = ½ × 5 × 4 + 4 × 22.5 = 100 m (triangle + rectangle or trapezium)
• omitting the ½ or the triangle (90 or 110)
• one suvat formula for the whole motion scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

final part. Since the shape of the graph used in part (d) was the same shape as the graph used

Common mistakes

• acceleration 1 m s⁻² means Sam reaches X after X seconds
• area ½X² + X(27.5 − X) = 100 gives a quadratic in X only
• X = 3.92 (3 s.f., choose the root less than 27.5)
• common errors 22.5 instead of 27.5, one equation in two unknowns

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), there were surprisingly few diagrams seen and this would have helped candidates to understand the situation. The most common error was to set v = 0 leading to a height of 78.4 m. A few used t = 5 s or t = 10 s for both the up part and the down part of the motion. The most successful approach was the “whole motion” one, possibly because it is only the most able students who ever really grasp this. A significant number of candidates gave a negative answer. Many candidates, in the second part, managed to stumble across 2.5 s for half the time m s- or to treat this speed as a height. Some candidates 1 distance from part (a) along with 24.5 didn’t seem to realise that the speed of the stone reduced as it went upwards. In part (c), “air

Common mistakes

• whole-motion s = ut + ½at² with u = 39.2, a = −9.8, t = 10 gives s = −98, height 98 m (must be positive)
• setting v = 0 gives 78.4 (height above O only)
• up-and-down method needs 78.4 and 176.4 combined
• a diagram helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

answer. Many candidates, in the second part, managed to stumble across 2.5 s for half the time but then failed to either find the other half or to double it. Some of the better candidates found t =1.5 s and t = 6.5 s and then subtracted which showed a clear understanding of the situation. Some successfully combined the various elements of times up and times down but the most successful approach was to find half the time and double it. A common error was to use the m s- or to treat this speed as a height. Some candidates 1 distance from part (a) along with 24.5

Common mistakes

• speed ≤ 24.5 either side of the top: 0 = 24.5 − 9.8t gives 2.5 s, double to 5 s (or times 1.5 and 6.5 s and subtract)
• forgetting to double
• treating 24.5 as a height

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

didn’t seem to realise that the speed of the stone reduced as it went upwards. In part (c), “air resistance” was the most common response, followed by “use a more accurate value of g”. A common error was not to reference the model at all, so answers about how accurately the time was measured or throwing the stone indoors to stop the wind having an effect were seen, as well as “consider the mass/weight of the stone”. Some wrongly gave the negative “there is no resistance”.

Common mistakes

• include air resistance, or a more accurate value of g, spin, shape or size of the stone
• must refer to the model
• negatives ("there is no air resistance") and mass/weight of the stone score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), virtually all the candidates achieved the mark either by substitution of t = 3 to show that v = 0 or by putting v = 0 and solving for t. A very small number of candidates failed to reject the second negative solution of their quadratic. Most candidates differentiated correctly

Common mistakes

• substitute t = 3 into v = 15 − t² − 2t to get 0 (or solve v = 0 and reject the negative root)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

- but there were a few who rearranged their v = 0 equation and in part (b) and then obtained 8 - not understanding then differentiated and they got 2t+2 as their derivative. Many stopped at 8 that magnitude had to be positive. A few tried to do some sort of “magnitude” calculation using Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c),

Common mistakes

• a = dv/dt = −2t − 2, at t = 3 a = −8
• magnitude 8 must be positive
• Pythagoras-style "magnitude" calculations are wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Pythagoras and so either sqrt (82 + 82) or even sqrt (82 + 32) was sometimes seen. In part (c), most integrated correctly and scored the first two marks. Very few got the differentiating and integrating the wrong way round this year. But after that, the majority just found the value of the integral at t = 4 (with or without subtracting the value at t = 0). Some realised that they had to split the integral into from 0 to 3 then 3 to 4 but several of these then added the two components instead of subtracting them. Those who drew a v-t graph were generally more successful when seeing the area below the axis seemed to help. There were relatively few correct answers to this part.

Common mistakes

• integrate to 15t − t³/3 − t² (must be seen: calculator-only answers score 0)
• P turns at t = 3, so distance = s(3) + (s(3) − s(4)) = 94/3 m
• most just evaluated at t = 4
• adding the two parts instead of subtracting the negative one
• a v-t sketch helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates made no attempt at part (a). Those who did mostly realised that they needed to use F=ma although many were unable to use the correct combination of forces in their car/trailer/whole system equations. Some never substituted T = 2400 and others omitted the R terms or had extra R’s in their equations of motion. Sign errors in solving their two simultaneous equations were common, especially when solving by substitution. Most who answered the second part were successful although some tied themselves up in knots explaining

Common mistakes

• car: 7400 − 2R − 2400 = 1200a
• trailer: 2400 − R = 400a (or whole system 7400 − 3R = 1600a)
• a = 0.5
• missing or extra R terms, not using T = 2400, sign errors solving simultaneously

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

answered the second part were successful although some tied themselves up in knots explaining their answer even though no reason was asked for. There was a surprisingly large number of candidates who made no attempt at this part. The most common error in the final part was to

Common mistakes

• a₁ is less than a (more mass to accelerate)
• no reason needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

candidates who made no attempt at this part. The most common error in the final part was to make no reference to the modelling of the resistance. Comments about friction and air resistance abounded, along with ones about the rope not being light/inextensible, tyres, road surfaces (which at least had an inkling of the right idea) and the weather. Relatively few scored this final mark.

Common mistakes

• the resistance is unlikely to be constant (e.g. varies with speed)
• must refer to the resistance model
• friction/air resistance, rope, tyres or weather comments score 0