A-LevelMathematicsDifferential equations9709/51 · May/June 2019 · Q712 Marks

A particle P of mass 0.5 kg is attached to a fixed point O by a light elastic string of natural length 1 m and modulus of elasticity 16N. The particle P is projected vertically upwards from O with speed 6ms¯¹. A resisting force of magnitude 0.1x² N acts on P when P has displacement x m above O. After projection the upwards velocity of P is v m s¯¹.

(i)Show that, before the string becomes taut, v(dv/dx) = -10 - 0.2x².[2]
(ii)Find the velocity of P at the instant the string becomes taut.[4]
(iii)Find an expression for the acceleration of P while it is moving upwards after the string becomes taut.[2]
(iv)Verify that P comes to instantaneous rest before the extension of the string is 0.5 m.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[M1] Use Newton's Second Law vertically
[A1] v(dv/dx) = -10 - 0.2x²
[M1] Attempt to integrate the expression in part (i)
[A1] v²/2 = -10x - 0.2x³/3 + c
[M1] Either use limits or find c and put x = 1
[A1] v = 3.98 (329...) ms⁻¹
[M1] Use Newton's Second Law vertically when string becomes taut
[A1] v(dv/dx) = -10 - 0.2x² - 32x + 32 = 22 – 32x – 0.2x²
[M1] Attempt to integrate after the string becomes taut
[A1] v²/2 = 22x - 32x²/2 - 0.2x³/3 + k
[M1] Either use limits or find k and put x = 1.5
[A1] As v²/2 cannot be negative, P comes to rest before the extension of the string is 0.5.

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.