June 2019: Paper 3 Section B: Mechanics

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

Examiner's comment

In the first part, the majority of candidates attempted to differentiate the given vector expression for velocity to find the acceleration as required. This was completed mostly successfully with just the occasional slip. Occasionally the expression was integrated or the fact that it was a vector just ignored with the coefficients of i and j being added; however, such instances were relatively rare. Those who completed the differentiation generally substituted t = 4 to obtain the acceleration and many correct answers were seen. Some went on to calculate the magnitude which was not actually required here but generally the marks had already been achieved for a correct vector. Some candidates worked in column vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration in part (a).

Common mistakes

• a = dv/dt: differentiate both components (powers decrease by 1)
• do not integrate or add the i and j coefficients
• answer 6i − 15j as a vector (finding the magnitude is not needed)

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

Examiner's comment

vectors throughout and some in i and j components; both are equally acceptable. In part (b) integration was required in order to find the position vector. Mostly the indices were dealt with correctly although errors in the coefficients were more common. Some stopped there without either using the initial conditions or substituting t = 4. Many failed to achieve the final mark as a result of numerical/sign errors. A minority attempted to use suvat formulae to solve the problem despite having used calculus

Common mistakes

• r = ∫v dt then add r₀ = −20i + 20j
• coefficient slips (3t² and 2t^(5/2))
• suvat is invalid as acceleration is not constant
• answer 28i − 44j

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

Examiner's comment

In part (a), although many candidates realised that finding an expression for the velocity was relevant to determining the direction of motion, only a minority could use the fact that it was specified as being in the direction of 3i – 4j. The most common error was in equating the actual vectors rather than either equating the ratios of components or equating to a multiple of 3i – 4j. A fair number attempted to use a displacement vector showing a lack of understanding of the situation. The second part was generally

Common mistakes

• v = (−1 + 2T)i + (4 − 3T)j
• parallel to 3i − 4j means ratio of components (4 − 3T)/(−1 + 2T) = −4/3, not equating the vectors themselves
• using displacement is wrong
• T = 8

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

Examiner's comment

displacement vector showing a lack of understanding of the situation. The second part was generally handled more successfully with many correct answers seen for the displacement vector (using s = ut 1 2 at2) although not all proceeded to find the distance by calculating its magnitude. Although most used + suvat formulae throughout the question, some successfully integrated the given constant acceleration to find the velocity and displacement. Use of magnitudes of vectors throughout and some invalid multiplication/division of vectors were seen on occasion.

Common mistakes

• s = ut + ½at² with u = −i + 4j and t = 4 gives 12i − 8j
• distance is its magnitude 4√13 ≈ 14.4 m
• 4(2i − 3j) is not the displacement

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

Examiner's comment

In part (a), many candidates earned the first mark for correctly finding R, but there were a number who resolved incorrectly, using sin instead of cos . Almost all candidates obtained the second mark, 2 F R very often from simply stating somewhere on their paper. Most candidates who found the 3 equations of motion correctly did proceed to find T. The most successful methods were either adding the two initial equations to eliminate T, finding a and substituting back into one of the equations or setting both equations to 6ma and finding T directly. Candidates who rearranged their equations to give a and then equated the quotients more frequently made errors when solving for T. Common errors were inconsistent use or loss of m, including an extra g in the acceleration term or forgetting to include g in their weight and, on the RHS of the equation, putting 3a and 2a rather than 3ma and 2ma. A significant number of candidates set their equation of motion to zero, finding the tension which would keep A at rest on the slope and received little credit. It was surprising how many candidates simply wrote 12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.

Common mistakes

• R = 2mg cos α (not sin α)
• F = ⅔R
• separate equations for A and B (or whole system) with 2ma and 3ma, not 2a, 3a or = 0
• exact trig ratios from tan α = 5/12 (no decimal angle)
• do not just write the given answer

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

Examiner's comment

12 mg T at the end of their incorrect working, assuming it would not be checked. In the second part, 5 many candidates tried to describe what would happen without any supporting calculations. Many talked about the continued motion of A, ignoring the actual question. Others simply said that it would continue moving as B had more mass than A, ignoring the effects of the plane. Many who showed calculations did successfully compare the component of the weight acting down the slope with the maximum friction, although others continued to include tension or actually compared friction with tension, ignoring the weight altogether. A few candidates showed a deeper understanding of the problem and simply Some candidates did not notice that the terms required were available from the compared and tan . equation of motion for A. Some candidates did not read the question carefully and considered what happened immediately after B reached the ground, rather than when block A came to rest. Some used an equation of motion rather than considering forces separately. A minority showed their lack of understanding by using tension in their explanation. The final part was generally well done, although who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.

Common mistakes

• compare Fmax = 16mg/13 with weight component down the slope 10mg/13 (or μ with tan α) numerically
• no tension once B has landed
• consider A at rest, not just after B lands

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

Examiner's comment

understanding by using tension in their explanation. The final part was generally well done, although candidates still persist in giving more answers than are required, which very often means that they include a wrong one and so lose marks. A few just restated the model – “light string” etc. Candidates who didn’t score well in parts (a) and (b) were often able to gain a mark or two in (c) with correct refinements of the model.

Common mistakes

• e.g. extensible or heavy string, rough pulley, blocks not particles, air resistance
• restating the model (light string) scores nothing
• extra wrong answers are penalised

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

Examiner's comment

In part (a), the majority of candidates either produced a geometric reason for the reaction acting perpendicular to the ramp (such as the ramp being tangential to the circular drum) or gave an argument about action being equal and opposite to reaction. Many seemed to think that a reaction is, by definition, at right angles to the surface. Very few recognised the significance of the drum being smooth with the consequence that there is no frictional force and hence only a normal component of the reaction. The second part involved finding the components of the resultant force on the ramp at the point of contact

Common mistakes

• the drum is smooth so there is no friction at C
• "tangent to the circle" or "action and reaction" alone is not the reason

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

Examiner's comment

second part involved finding the components of the resultant force on the ramp at the point of contact with the ground and hence its magnitude. This proved challenging for many and some made little valid progress. A clearly labelled diagram would have helped to identify forces and distances. Most used a normal reaction and a horizontal frictional force at A; however, the latter was sometimes omitted which significantly reduced the number of marks available for subsequent equations. Some employed components parallel and perpendicular to the ramp; such attempts were rare but generally successful. A variety of resolution and moments equations were attempted but sometimes components were omitted, forces not resolved or perpendicular distances not used. Working was often difficult to decipher. Those who attempted vertical and horizontal resolutions and a ‘moments about A’ equation tended to be the more successful although a significant number stopped when they had found the normal reaction only. Some introduced a coefficient of friction which was not necessary to answer the question (even giving it the value 2 from the previous question). Occasionally those who completed a correct solution by 3 squaring and adding components to find the magnitude of the resultant failed to achieve the final accuracy mark for not rounding to 2 or 3 significant figures following the use of g = 9.8 m s-2. Some who had achieved very few marks for the rest of the question managed to secure the mark in part (c)

Common mistakes

• friction at A is horizontal and must be included
• at least one moments equation (e.g. M(A): 20g × 4cos θ = 5N) with perpendicular distances
• three equations then magnitude √(F² + R²) ≈ 67 N (2 or 3 s.f. after g = 9.8)
• μ is not needed
• stopping at R

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

Examiner's comment

who had achieved very few marks for the rest of the question managed to secure the mark in part (c) for stating that the reaction at C decreases if the centre of mass is assumed to be closer to A than to B. No reason was required although some chose to give one. A minority claimed it increased or remained unchanged (although the explanation sometimes implied that candidates were considering the reaction at A rather than at C).

Common mistakes

• reaction at C decreases (moments about A)
• confusing the reaction at C with the reaction at A

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

Examiner's comment

In part (a), the majority of candidates obtained the first mark for the horizontal component and most candidates also made a good attempt to find the vertical component, although sometimes it was seen as part of their working for part (b). Of those who found the horizontal and vertical components of the

Common mistakes

• horizontal 20cos 30° and vertical 20sin 30° − 2g = −9.6
• need both speed (19.8 m/s) and direction (29° below horizontal)
• do not give the answer in i and j (not defined)

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

Examiner's comment

part of their working for part (b). Of those who found the horizontal and vertical components of the velocity, only a proportion proceeded to calculate the speed and of those, very few found the direction. This was despite the clear lead from the question which had given each velocity as a speed with a direction. A sizeable minority broke the universal rule that answers must be given in terms of quantities which are given in the question, giving their velocity in terms of i and j, which had not been defined in this question. The second part proved to be quite discriminating although a significant number of candidates did successfully find the correct values of both and u. The most common error was for candidates to simply equate both the horizontal and vertical displacements at the time of collision and ignore the 50 m horizontal distance altogether. Others assumed that the particles met halfway between A and B. However, only a minority of candidates made errors when resolving their velocities. In the final part, a substantial number of candidates correctly identified a limitation of the model with the size

Common mistakes

• horizontal distances sum to 50 m at t = 2 (not equal, not meeting halfway)
• vertical displacements equal gives u sin θ = 20 sin 30°
• θ = 52°, u = 12.6

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

Examiner's comment

final part, a substantial number of candidates correctly identified a limitation of the model with the size of the balls being the most common response. Wind, spin and an inaccurate value for g were also relatively common. A significant number lost the mark by referring to mass or weight and some referred to air resistance even though the question specified “other than air resistance’’.

Common mistakes

• e.g. size of the balls, spin, wind, g not exactly 9.8
• not air resistance (excluded)
• mass or weight of the balls scores B0