A-LevelMathematicsMechanics9709/43 · Oct/Nov 2024 · Q58 Marks

A particle of mass 12kg is going to be pulled across a rough horizontal plane by a light inextensible string. The string is at an angle of 30° above the plane and has tension TN (see diagram). The coefficient of friction between the particle and the plane is 0.5 . [Figure 5]

(a)Given that the particle is on the point of moving, find the value of T.[5]
(b)Given instead that the particle is accelerating at 0.2ms⁻², find the value of T.[3]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

[*M1] Correct number of relevant terms with T resolved; allow sign errors; allow sin/cos mix. Can score M1 for any F=Tcos30. Do not allow g missing in the equation for R. Must have 12, not just m. Could see R as part of an equation for F. E.g. F = 0.5(12g - T sin 30).
[A1] Both correct.
[*DM1] Allow sign errors in R; allow consistent sin/cos mix in R but no other errors. Must be two term R as a linear combination of weight and a component of T, and F must be a single term which is a component of T. Do not allow g missing. If correct T cos30 = 0.5(120-Tsin30) or (√3/2)T=60-0.25T. If no working shown to eliminate T or R, then DM2 for getting T value correct for their equations and A1 if fully correct. Could use 0.5R=Tcos30 and solve simultaneously.
[DM1] Allow consistent sin/cos mix and allow sign errors. Must get to 'T='.
[A1] [53.7622...] Note: For sign errors: R-Tsin30=12g answer should be 97.3985... R-Tsin30=-12g answer should be -97.3985... R+Tsin30=-12g answer should be -53.7622...... Each of the above would usually get M1A0M1M1A0.
[*M1] Attempt at N2L; correct number of relevant terms with T resolved; allow sign errors; allow sin/cos mix, but can be F or any reasonable attempt at friction.
[DM1] Must be a two term R as a linear combination of weight and a component of T. Allow sign errors and consistent sin/cos mix. Must get to 'T='. The equations if correct should be T cos30-0.5(120 – Tsin 30) = 12×0.2 or (√3/2)T-60+0.25T = 2.4 or T((√3/2)+0.25) = 62.4 and these must be solved. Any use of T or R from part (a) scores DM0 here.
[A1] T=55.9127... Note: For sign errors: R-Tsin30=12g answer should be 101.294... R-Tsin30=-12g answer should be –93.5026... R+Tsin30=-12g answer should be -51.6117... Each of the above would usually get M1M1A0.