A-LevelMathematicsMechanics9709/42 · Feb/Mar 2019 · Q711 Marks

The diagram shows the vertical cross-section PQR of a slide. The part PQ is a straight line of length 8 m inclined at angle α to the horizontal, where sin α = 0.8. The straight part PQ is tangential to the curved part QR, and R is h m above the level of P. The straight part PQ of the slide is rough and the curved part QR is smooth. A particle of mass 0.25 kg is projected with speed 15 m s⁻¹ from P towards Q and comes to rest at R. The coefficient of friction between the particle and PQ is 0.5.

(i)Find the work done by the friction force during the motion of the particle from P to Q.[4]
(ii)Hence find the speed of the particle at Q.[4]
(iii)Find the value of h.[3]

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Mark scheme

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[B1] R = 0.25g × 0.6 [= 1.5]
[M1] Use F = μR [F = 0.5 × 0.25g × 0.6] [F = 0.75]
[M1] Using WD = Force × distance moved in direction of force [WD against friction = F × 8]
[A1] WD = 6 J
[M1] Work-energy equation in the form Initial KE = Final KE + WD against F + PE gain [½ × 0.25×15² = ½ × 0.25 × v² + 6 + 0.25g × 8 × 0.8]
[A1ft] Correct Work-Energy equation for the motion to Q. ft on WD
[M1] Solving the work-energy equation for v
[A1] v = 7 m s⁻¹
[M1] KE lost from Q to R = PE gain from Q to R H is the height of R above Q [½ × 0.25 × 7² = 0.25 × g × H] Or [½ × m × 7² = m × g × H]
[A1] H = 7²/2g = 2.45 m
[A1] Total height h = 6.4 + H = 8.85

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