A-LevelMathematicsMechanics9709/51 · May/June 2018 · Q69 Marks

A particle P of mass 0.2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a particle Q of mass 0.3 kg. The string passes through a small hole H in a smooth horizontal surface. A light elastic string of natural length 0.3 m and modulus of elasticity 15 N joins Q to a fixed point A which is 0.4 m vertically below H. The particle P moves on the surface in a horizontal circle with centre H (see diagram).

(i)Calculate the greatest possible speed of P for which the elastic string is not extended.[4]
(ii)Find the distance HP given that the angular speed of P is 8 rad s⁻¹.[5]

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Mark scheme

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

[B1] r [= 0.6 - (0.4-0.3)]= 0.5
[B1] T = 0.3g; Resolve vertically for Q
[M1] 0.2v² / 0.5 = 0.3g; Use Newton's Second Law horizontally for P
[A1] v = 2.74 ms¯¹
[B1] r = 0.5 + e; e = extension of the string
[B1] T = 15e / 0.3 = 50e; Use T = λx/l
[M1] 0.2×8²(0.5 + e) = 50e + 0.3g; Use Newton's Second Law horizontally with a = rω²
[A1] e = (6.4-3) / (50-12.8) (= 0.0914)
[A1] HP = 0.591 m

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