A-LevelMathematicsMechanics9709/43 · Oct/Nov 2017 · Q711 Marks

A particle P of mass 0.2 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction between the particle and the plane is 0.3. A force of magnitude T N acts upwards on P at 15° above a line of greatest slope of the plane (see diagram).[Figure 7.1]

(i)Find the least value of T for which the particle remains at rest.[6]
(ii)The force of magnitude T N is now removed. A new force of magnitude 0.25 N acts on P up the plane, parallel to a line of greatest slope of the plane. Starting from rest, P slides down the plane. After moving a distance of 3 m, P passes through the point A. Use an energy method to find the speed of P at A.[5]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

R = 0.2g cos 30 – T sin 15 B1
[F = 0.3 × (0.2g cos 30 - T sin 15)] M1 Use of F = μR
T cos15 + 0.3 × (0.2g cos30 – T sin15) = 0.2gsin30 M1 For resolving along the plane
A1
M1 For solving a 4 term equation for T
T = 0.541 A1
0.3 x 0.2g cos 30 × 3 [= 1.5588 J] B1 WD against F = friction × distance
WD = 0.25 × 3 [= 0.75 J] B1 WD against 0.25 force
0.2g x 3 sin 30 [= 3 J] B1 PE loss = mgh
[½ (0.2) v² = 3 – 1.5588 – 0.75] M1 Work/Energy equation
Speed = 2.63 ms⁻¹ A1

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