June 2019: AS AS Paper 2 Section B: Mechanics

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

Examiner's comment

The first part was well done by the majority of candidates who used a suvat equation to find the required speed after 3 seconds.

Common mistakes

• free fall from rest with g = 10: v = 0 + 10 × 3 = 30 m s⁻¹

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

Examiner's comment

Part (b) involved drawing a speed-time graph for the motion of the parachutist. Many correct shapes were seen for the first mark but those who included a solid vertical line at t = T were penalised. Other less common errors included not starting the graph at the origin, not using a straight line for each section and having a sloping line for the final section (when the parachutist is moving with constant speed). The given figures (‘30’, 3, 5, T) were required to be correctly placed for the second mark; occasionally one or two were omitted but many did achieve this mark. In the third part the majority of the candidates

Common mistakes

• start at the origin
• straight line up to 30 at t = 3, straight line down to 6 at t = 5, then horizontal to T
• mark 3, 5, T and 30
• no solid vertical line at t = T
• final section not sloping

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

Examiner's comment

one or two were omitted but many did achieve this mark. In the third part the majority of the candidates realised that the area under the graph represented the distance travelled. However, not all used the correct structure, namely a triangle, rhombus and rectangle (or equivalent) and the relevant correct formulae for their areas. There was often confusion in the use of T (which was defined in the question) so that effectively the final stated answer was 5 less than the actual value. Some used suvat equations for each stage of the motion, often more or less successfully, although a fairly common error was to use a final velocity of zero in the third section. Attempts to use a single suvat equation to cover the entire distance were seen on occasion; these showed a lack of understanding of the situation and achieved no credit. Those with correct working did not always round the final answer to the nearest whole number as required.

Common mistakes

• total area under the speed-time graph = 550: ½ × 3 × 30 + ½(30 + 6) × 2 + 6(T − 5) = 550
• constant speed after t = 5 is 6 m s⁻¹
• use T − 5 (not T) for the last section
• one suvat equation for the whole motion scores nothing
• T = 83 to the nearest whole number

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

Examiner's comment

Part (d) required recognition of the fact that including the effect of air resistance would lead to an increased value of T. Although this was widely understood, some candidates commented that ‘it would take longer’ which did not answer the question as stated.

Common mistakes

• air resistance reduces the speed before the parachute opens, so the new value of T is bigger
• must refer to T ("takes longer" alone is B0)

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

Examiner's comment

Part (e) required a suggestion for a further refinement of the model. There were many possible acceptable answers with using a more accurate value for g, considering wind effects and allowing time for the parachute to open being some of the more popular options. A common suggestion which was not credited was to include the mass of the parachutist. If two or more answers were offered they all had to be correct to achieve the mark; this was to avoid rewarding candidates who were ‘hedging their bets’ by listing everything they could think of.

Common mistakes

• e.g. more accurate g, wind, time for the parachute to open, dimensions of the parachutist, deceleration not constant
• mass of the parachutist, upthrust, air resistance or terminal velocity are B0
• all suggestions offered must be correct

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

Examiner's comment

Although the first part of this question was well done by some candidates, a significant minority failed to write down an equation of motion of the correct structure for each mass. Some included a weight term in the equation for the particle moving horizontally whilst others showed no real understanding of what was required and confused equations of motion with suvat equations or omitted this part completely. Those who managed to produce two valid equations generally had no difficulty in deriving the given answer for the acceleration. Since it was a ‘show that’ question, it was important that the final answer followed from sufficient and entirely correct working; those who went straight from the two simultaneous equations to writing down the answer were not awarded the final mark. Such instances were, however, fairly rare. Some attempted a ‘whole system’ approach which was equivalent to writing down just one equation with the tension eliminated. This was deemed not sufficient for a complete method and scored a maximum of two marks.

Common mistakes

• separate equations of motion: Q: 0.6g − T = 0.6a, P: T = 0.8a (no weight term for P on the horizontal table)
• solve to a = 4.2 showing working
• a whole-system equation alone is incomplete

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

Examiner's comment

Part (b) was also completed with mixed degrees of success. To solve the problem it was necessary to split the motion into two stages, before and after one mass hits the ground. Those who attempted to use a single suvat equation to cover the whole motion could achieve no credit. Some successfully found the time it took for Q to fall to the ground but were then unsure how to proceed. Those who calculated the speed with which Q hit the ground generally used this in considering the subsequent motion of P as it continued to move across the table. However, the most common error at this stage was in not realising that P would move with constant speed since there was no tension and the table surface was smooth; some included an acceleration of 4.2 ms-2 (or even 9.8 ms-2) in the relevant suvat equation or, more 1 commonly, used s = 2 (u + v)t with v = 0. These candidates generally scored three out of a possible six marks. The final part required the identification of a possible limitation of the model. The assumptions of this

Common mistakes

• two stages: 0.4 = ½ × 4.2 × t₁² gives t₁ = 0.436 and v = 1.83
• after Q hits the floor the rope is slack and the table smooth, so P moves at constant speed for the remaining 1.1 m (not with a = 4.2 or 9.8, and not v = 0 at the pulley)
• total time 1.04 s

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

Examiner's comment

The final part required the identification of a possible limitation of the model. The assumptions of this model were listed in the question and so any one of these (such as smooth pulley, light string, inextensible string, modelling the balls as particles) could be used in describing such a limitation. Often other factors were identified which were not part of the specified assumptions for this model and so were not awarded the mark.

Common mistakes

• choose an assumption stated in the question: rope light or inextensible, pulley smooth or small, balls as particles
• factors outside the stated model are not credited

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

Examiner's comment

In the first part, the majority of candidates knew that finding acceleration involved differentiating the given expression for the velocity and, apart from the occasional slip, this was generally done well. However, a fair number then found either t when a = 0 or a when t = 0 without realising that, to find the relevant value of t, it was necessary to solve the quadratic equation obtained by setting v = 0 (when the particle is at instantaneous rest) . Many of those with the correct working lost the final mark for giving a negative answer for the ‘magnitude’ of the acceleration. Most candidates knew they had to integrate for part (b) which they did mostly successfully and many

Common mistakes

• instantaneous rest means v = 0: solve 12 + 4t − t² = 0 to t = 6 (reject −2)
• a = dv/dt = 4 − 2t = −8
• magnitude 8 m s⁻² (positive)
• not a when t = 0 or t when a = 0

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

Examiner's comment

Most candidates knew they had to integrate for part (b) which they did mostly successfully and many went on to substitute t = 3 as required to find the distance travelled. There were a few numerical slips in carrying out the substitution but there were many correct solutions seen to this part of the question.

Common mistakes

• integrate v: s = 12t + 2t² − t³/3
• v > 0 for 0 ≤ t ≤ 3 so distance = s(3) = 45 m