June 2025: 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), the majority of candidates made an attempt to equate the area under the graph to the distance covered. Some used the formula for the area of a trapezium to form an equation in T but the parallel sides were not always identified correctly as T and (T -20). However, the most common approach was to use the area of a triangle, rectangle and triangle respectively for each of the three stages of the motion. Many who used this approach found the time for the constant speed was 70 seconds and then added 5 and 15 to find T= 90 as required. Those who just left their final answer as 70 gained no credit as there was no complete method to find T. A few added the distances rather than the times to 70 showing confusion in their understanding. A small number of candidates used suvat equations for each of the three stages but those using this approach seemed more likely to make an error with one or more of the values. Occasionally attempts were seen using a single suvat equation with s = 400.

Common mistakes

• total area under the speed-time graph = 400: ½(T + (T − 20)) × 5 = 400 or triangle + rectangle + triangle gives T = 90
• the parallel sides of the trapezium are T and T − 20
• leaving 70 (the constant-speed time) scores 0
• adding distances to 70 instead of the times 5 and 15
• the ½ is needed for the M mark

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

Examiner's comment

In part (b), the reasons given for the actual time being different from T were many and varied but usually gained the mark by stating, for example, either a physical factor that could affect the runner or a limitation of the model such as the speed was unlikely to be exactly constant for a given period of time. The road was specified as being straight and horizontal in the question so comments relating to those factors were not relevant here.

Common mistakes

• a limitation of the model: changes of acceleration not instantaneous, speed or acceleration not constant, wind resistance, runner may not run 400 m
• the road is given as straight and horizontal so comments about it (not straight, not horizontal, friction) score 0

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

Examiner's comment

Part (a) was mostly correctly answered, but some candidates were confused as to whether to add or subtract the forces, with 9i + 14j being the most common incorrect answer. This was usually carried through to part (b) but then often replaced in part (c) with the correct answer of 3i – 4j. Many candidiates worked with column vectors but most gave the answer in the required format. Very occasionally the answer was seen in (xi, yj) format or the column vector form of this.

Common mistakes

• add the force vectors: (−3i + 4j) N
• 9i + 14j (subtracting) was the most common error
• answer must be in i, j form

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

Examiner's comment

In part (b), most candidates used Pythagoras correctly but failed to use Newton’s 2nd Law. A significant number, of those that did, multiplied the resultant force by 0.25 instead of dividing it by 0.25.

Common mistakes

• Newton's second law F = ma with m = 0.25 then Pythagoras: magnitude of a = 4 × 5 = 20 m s⁻²
• most used Pythagoras but forgot F = ma, or multiplied by 0.25 instead of dividing
• 20 then "√20" loses the A mark

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

Examiner's comment

Part (c) not always correct even if the rest of the question was and often left blank. There was the occasional correct follow through from part (a).

Common mistakes

• constant velocity means the resultant force is zero (Newton's first law), so F3 = −(their (a)) = (3i − 4j) N
• often left blank

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

Examiner's comment

In part (a), there were many correct answers seen for the equation of motion of Q. However, a significant number of candidates did not seem to know what constituted an equation of motion, sometimes making an attempt at a suvat equation, despite then using it correctly in part (b).

Common mistakes

• Q: 3g − T = 3a (two terms on the left, weight in g)
• a suvat equation is not an equation of motion
• including m (3mg − T = 3ma) loses the A mark

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

Examiner's comment

The most common error in part (b) was to include a weight term in the horizontal equation of motion for P i.e. writing T-2g = 2a rather than T = 2a. Sometimes the direction of the acceleration was not treated consistently in the two equations resulting in a sign error in one of them. A relatively small number of candidates used a whole system approach but the requirement to use a combined mass in the ‘ma’ term ruled out some attempts.

Common mistakes

• P on the horizontal table: T = 2a (no weight term: T − 2g = 2a was the most common error)
• or whole system 3g = 5a with combined mass 5
• a = 3g/5 = 5.9 m s⁻² (2 or 3 s.f.)
• inconsistent directions in the two equations

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

Examiner's comment

Part (c) was generally well attempted with the majority of candidates using v2 = u2+2as to find the required speed. Although most used their acceleration found in part (b) as required, a small minority substituted a = g despite it not following from previous work. A few confused u and v by assuming the final velocity when Q hit the ground was zero, but such instances were rare. The answer was required to 2 or 3 significant figures following the use of g = 9.8 ms-2.

Common mistakes

• v² = 2 × their a × 0.4 gives 2.2 m s⁻¹ (2 or 3 s.f.)
• use their a from (b), not g
• v is not 0 at the floor

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

Examiner's comment

Part (d) required finding the time taken by P to move across the table to the pulley once Q had hit the ground. This proved to be the most challenging part on the paper and few valid approaches were seen. A significant number used the acceleration from the first stage of the motion, or even g, in a suvat equation without realising that P would now travel with constant velocity since the table was modelled as smooth and there was no longer a tension acting. Other common errors seen were to use the distance as 1.2 m, 0.4 m or even 1.6 m rather than 0.8 m. This again showed a lack of appreciation of the mechanics of the situation.

Common mistakes

• after Q hits the floor the rope is slack and the table smooth, so P moves at constant speed 2.17 m/s for 1.2 − 0.4 = 0.8 m: K = 0.8/2.17 = 0.37
• common errors: using the earlier acceleration or g, distance 1.2, 0.4 or 1.6 m
• hardest part on the paper

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

Examiner's comment

In part (e), almost all candidates stated, with a reason, that the acceleration would be less as a result of including a resistance. Those who lost the mark tended to not give a reason or omit this part.

Common mistakes

• acceleration is less, with a reason (resistance opposes the motion, reducing the resultant force)
• no reason scores 0

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

Examiner's comment

Again, part (f) was generally answered well. The assumptions of the model were clearly stated as bullet points in the question. To gain credit, any suggested refinement had to relate specifically to one of those, and common valid responses included considering the mass of the ropes or modelling the particles with dimensions. Using a more accurate value for g was not credited as it was not a given modelling assumption. A further refinement was required so including resistance/friction or air resistance also did not gain the mark.

Common mistakes

• refine one of the listed assumptions: packages not particles (dimensions), mass/weight or friction of the pulley, mass or elasticity of the rope
• a more accurate g, "make it 3D", or air resistance/table friction are not credited

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

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (a), the majority of candidates were able to differentiate s correctly and equate it to zero to produce a quadratic.However, a significant proportion of students did not show any method to justify their solutions from their quadratic, ignoring the instruction at the top of the question and lost two marks. A significant number of candidates did not find a derivative, choosing instead to equate s to zero. This led to t values of 0 and 3 which then had a knock on effect on the marks that could be scored in part (b).

Common mistakes

• v = ds/dt = 6t² − 24t + 18 = 0, t = 1 or t = 3
• working (e.g. factorising) must be shown, not just calculator roots
• setting s = 0 instead of v = 0 (giving t = 0 and 3)

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

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. There was the occasional candidate who incorrectly differentiated, missing out the “+18” term from their answers. This also impacted the marks that could be scored for part (b). Part (b) was challenging and candidates who sketched a graph of the function and considered areas were usually the most successful. The most common error was to just use 0s and 4s while a number integrated s. Some who had made progress used 8 - for the middle section, forgetting that the distance was required. A significant number made no attempt at this part.

Common mistakes

• P changes direction at t = 1 and t = 3: s₁ = 8, s₃ = 0, s₄ = 8, so total distance = 8 + 8 + 8 = 24 m
• most used only s₀ and s₄ (displacement)
• integrating s
• using −8 for the middle section
• a sketch helps

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

Examiner's comment

At the top of this question is a calculator warning which some candidates chose to ignore and lost marks as a result. In part (c), many found the acceleration but a number then failed to equate it to zero and so did not achieve any marks. Some did obtain t = 2 successfully but then did nothing with it. A few used the symmetry or completed the square and got to the required answer very quickly. The odd negative answer for speed was seen.

Common mistakes

• maximum speed where a = dv/dt = 12t − 24 = 0 (or by symmetry or completing the square v = 6(t − 2)² − 6): t = 2, speed 6 m s⁻¹ (must be positive)
• many found a but did not set it to 0, or stopped at t = 2