A-LevelMathematicsMechanics9709/52 · Feb/Mar 2018 · Q610 Marks

A small object of mass 0.2 kg rests at a point O on a rough horizontal surface. The coefficient of friction between the object and the surface is 0.5. A force of magnitude P N acting at an angle θ below the horizontal is applied to the object. The velocity of the object is v m s¯¹ away from O at time t s after the force begins to act (see diagram). It is given that tan θ = ¾ and that P = 0.4t for 0 ≤ t ≤ 8.

(i)Find the value of t when the object starts to move.[3]
(ii)Show that, when the force is acting and the object is in motion, dv/dt = t - 5.[2]
(iii)When t = 8 the force of magnitude P N ceases to act. Find the distance travelled by the object after t = 8 before it comes to rest.[5]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.

[M1] Note sinθ = 0.6 and cosθ = 0.8 (θ = 36.87°) Resolve vertically and use F = μR
[M1] Resolve horizontally
[A1] t = 5
[M1] 0.2dv/dt = 0.4t × 0.8 – (1 + 0.12t) Use Newton's Second Law horizontally
[AG] dv/dt = t-5 A1
[M1] ∫dv = ∫(t-5)dt v = t²/2 - 5t + c Attempt to integrate the equation from part(ii)
[A1] v = 0 when t = 5 hence c = 12.5 Finds the constant of integration, c
[A1] v=8²/2-5 × 8 + 12.5 = 4.5 Find v when t = 8
[M1] a = -0.5 × 0.2g / 0.2 = −5 m s⁻¹ and s = 4.52 / (2 × 5) Finds a and uses v² = u² + 2as
[A1] s = 2.025 m

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