M8.5 Resultant forces; motion in a plane

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

Examiner's comment

F = mu R and tried to use it in some way. Part (d) involved finding the magnitude of the resultant force acting at A. It was not always well answered and sometimes omitted entirely. A significant minority of responses were, however, fully correct and there were many who, having made an error in part (c), were still able to apply Pythagoras’ theorem

Common mistakes

• resultant at A combines F and R by Pythagoras: (Mg/3)√13 ≈ 1.2Mg
• adding F and R, or giving R = Mg alone, is wrong

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

Examiner's comment

In part (a), the majority of candidates were able to achieve the first mark for adding the two forces and collecting the i and j components. Many just equated this resultant force to the vector (3i + j) rather than using (3i + j) to define the direction. Some set up equations λ + 4 = 3 and mu - 1 = 1 and proceeded to solve them simultaneously gaining no further credit. Those who did use ratio usually went on to derive the required equation correctly; however, it should be remembered that, if required to derive a given answer, this must be written as a final conclusion in the form exactly as stated in the question. Another successful approach was to equate the i and j coefficients to 3k and k and solve as simultaneous equations. A small minority of candidates applied F = 4a or used integration to find v at this stage and then used ratio appropriately to obtain the required answer. In the second part, many candidates were able to find the resultant force and

Common mistakes

• resultant (4 + λ)i + (−1 + μ)j is parallel to 3i + j so use ratios
• equating the resultant to 3i + j (λ + 4 = 3, μ − 1 = 1) scores no more
• final line must be λ − 3μ + 7 = 0 exactly

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

Examiner's comment

answer. In the second part, many candidates were able to find the resultant force and acceleration vectors. Some candidates found the magnitude of the force or acceleration at this stage and proceeded correctly with suvat formulae from there, while others continued with vectors. A number of candidates did not identify the initial velocity as zero, reading back to the information given at the start of the question being important here, so failed to use suvat formulae correctly. Use of u = 3i + j was not uncommon. Although most used suvat formulae, some successfully integrated the constant acceleration to find the velocity and displacement vectors. The majority of candidates knew how to find the magnitude of a vector using Pythagoras. The most significant errors seen included finding the resultant force but then using it as a displacement or velocity, showing a lack of understanding of the situation. A small number of candidates found an incorrect value for mu when substituting λ and there was occasional use of only one of the forces.

Common mistakes

• μ = 3 so resultant 6i + 2j and a = (6i + 2j)/4
• u = 0 as P starts at rest (u = 3i + j is wrong)
• using the force as a displacement
• using only one force
• AB = √160 = 4√10 ≈ 12.6 m

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