Oct 2021: AS AS Paper 2 Section B: Mechanics

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

Examiner's comment

(a) Many candidates just found the time to the top, losing both marks.

Common mistakes

• displacement 0 at time T: 0 = 14.7T − 4.9T² gives T = 3
• finding only the time to the top (1.5 s) scores nothing
• use g = 9.8

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

Examiner's comment

(b) There were very few fully correct answers because most candidates did not realise that they had to consider the change in direction and just put t = 4 into a suvat equation. Some found the distance to the top and then used t = 2.5 to find the remaining distance but then did not round the answer to 2 or 3 sf following use of g = 9.8. A few used g = 9.81 or 10 despite the clear instruction on the front of the paper.

Common mistakes

• distance, not displacement: the stone changes direction at t = 1.5, so add 11.025 up and 30.625 down = 41.7 m
• substituting t = 4 into one suvat equation gives the displacement (19.6) only
• 2 or 3 s.f. after g = 9.8

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

Examiner's comment

(c) Candidates had very little idea – considering the mass was the most frequent incorrect response. Some quoted air resistance even though that had been specifically excluded in the question.

Common mistakes

• e.g. take account of the size of the stone (spin), do not model it as a particle, use a more accurate value of g
• air resistance is excluded
• "mass" is wrong
• incorrect extra refinements score B0

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

Examiner's comment

(a) Most candidates knew that they needed to differentiate but integration was also seen once. Some went on to put a = 0, solving to get t = 5 which they went on to use in part (b) but this subsequent working was ignored in part (a).

Common mistakes

• differentiate v to get a = 10 − 2t
• integrating is wrong
• setting a = 0 is not required

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

Examiner's comment

Some went on to put a = 0, solving to get t = 5 which they went on to use in part (b) but this subsequent working was ignored in part (a). (b) The majority used v = 0 when t = 6, found a value of k and obtained a quadratic. Unfortunately, many made sign errors and their expression had +24 instead of -24. If they went on to solve their incorrect quadratic using a correct method, these candidates scored two of the four marks available. There were several completely correct answers.

Common mistakes

• instantaneously at rest means v = 0
• v = 0 at t = 6 gives k = 24
• sign errors (+24 instead of −24)
• solve the quadratic for t = 4

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

Examiner's comment

(c) Some candidates knew that they needed to integrate v but a few tried to use suvat equations, losing all the marks and some did not substitute a numerical value for k. The last two marks were rarely scored as the change of direction was not appreciated.

Common mistakes

• integrate v (suvat is invalid: acceleration not constant)
• v changes sign at t = 4 so split the integral 0 to 4 and 4 to 6 and add magnitudes: 116/3 m
• substitute k = 24 first

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

Examiner's comment

(a) There were many completely correct answers and those who took the acceleration to be positive in a different direction could also score full marks. Some had no idea where g should go so answers like ‘T – 2m = 2mga’ were sometimes seen. The mass m was occasionally left out also. A few quoted suvat equations, connecting time and distance, instead of equations of motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared.

Common mistakes

• T − 2mg = 2ma for P and 5mg − T = 5ma for Q
• g placed wrongly (e.g. T − 2m = 2mga)
• m omitted
• suvat equations are not equations of motion

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

Examiner's comment

motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared. (b)Those who had correct equations of motion generally went on to find the acceleration but some missed g in their answer. Not many progressed successfully after this - some found T because that is what is usually required but in this question, it was not necessary. Some calculated the time taken for Q to reach the ground and used it in an equation for P, not realising that they had to find the velocity that P had acquired. Of those who did make progress, some missed the h in their answer for v, and others only added 2h instead of 3h in the final step.

Common mistakes

• a = 3g/7
• P's speed when Q lands from v² = 2ah (P has risen h)
• then P moves freely under gravity: H = 3h/7
• total height 2h + h + H = 24h/7
• T is not needed
• adding only 2h, or using the time Q takes to fall, are common errors

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

Examiner's comment

(c) A very small number answered correctly. Mass was again a common response.

Common mistakes

• the distance Q falls to the ground would not be exactly h (size of the balls)
• "mass" is a common wrong answer
• incorrect extras score B0

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

Examiner's comment

(d) Again, there was a very small number of correct responses, some of which were spoilt by saying that the reason for the accelerations being different was that the tensions were different because of friction at the pulley.

Common mistakes

• the accelerations of P and Q would not have equal magnitude
• saying the tensions differ because of friction at the pulley is wrong