A-LevelMathematicsMechanics9709/51 · Oct/Nov 2018 · Q48 Marks

A particle P of mass 0.5 kg is projected along a smooth horizontal surface towards a fixed point A. Initially P is at a point O on the surface, and after projection, P has a displacement from O of x m and velocity vms¯¹. The particle P is connected to A by a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The distance OA is 1.6 m (see diagram). The motion of P is resisted by a force of magnitude 24x² N.

(i)Show that v dv/dx = 32 − 40x − 48x² while P is in motion and the string is stretched.[3]
(ii)The maximum value of v is 4.5. Find the initial value of v.[5]

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Mark scheme

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

[B1] T = 16(1.6-0.8 – x)/0.8 ( = 16-20x) (Use T = λx/L)
[M1] 0.5vdv/dx = 16(1.6 – 0.8 – x)/0.8 – 48x² (Use Newton's Second Law horizontally)
[A1] vdv/dx = 32 - 40x – 48x² (AG)
[M1] 48x² + 40x − 32 = 0 (Put acceleration = 0 for maximum velocity)
[A1] x = 0.5
[M1] ∫vdv = ∫ (32 – 40x – 48x²)dx (Attempt to integrate the equation from part (i))
[M1] 4.5²/2 = 32×0.5 – 20 ×0.5² – 16×0.5³ + c, c = 1.125 (Substitute x = 0.5, v = 4.5 to find c)
[A1] v = 1.5 (Use x = 0)

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