A2 June 2024 Paper 1 Q18

AQACurrent spec12 marksSecond Order Differentials

18 In this question use \(g = 9.8\) m s⁻²

Two light elastic strings each have one end attached to a small ball \(B\) of mass 0.5 kg

The other ends of the strings are attached to the fixed points \(A\) and \(C\), which are 8 metres apart with \(A\) vertically above \(C\)

The whole system is in a thin tube of oil, as shown in the diagram below.

A narrow vertical tube between fixed point A at the top and fixed point C at the bottom, with the ball B on the string between them

The string connecting \(A\) and \(B\) has natural length 2 metres, and the tension in this string is \(7e\) newtons when the extension is \(e\) metres.

The string connecting \(B\) and \(C\) has natural length 3 metres, and the tension in this string is \(3e\) newtons when the extension is \(e\) metres.

(a) Find the extension of each string when the system is in equilibrium. [3 marks]

(b) It is known that in a large bath of oil, the oil causes a resistive force of magnitude \(4.5v\) newtons to act on the ball, where \(v\) m s⁻¹ is the speed of the ball.

Use this model to answer part (b)(i) and part (b)(ii).

(i) The ball is pulled a distance of 0.6 metres downwards from its equilibrium position towards \(C\), and released from rest.

Show that during the subsequent motion the particle satisfies the differential equation

\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 9\frac{\mathrm{d}x}{\mathrm{d}t} + 20x = 0\]

where \(x\) metres is the displacement of the particle below the equilibrium position at time \(t\) seconds after the particle is released. [3 marks]

(ii) Find \(x\) in terms of \(t\) [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms equilibrium force equation with three correct terms. Condone sign errors.	M1	3.1b
Uses the equation \(3 = e_A + e_C\) and their force equation to obtain the extensions.	M1	3.1b
Obtains 1.39 and 1.61	A1	1.1b
	(3)

Typical solution

\[7e_A = 3e_C + 0.5g\]\[3 = e_A + e_C\]\[e_A = 1.39,\quad e_C = 1.61\]

Mark scheme (b)

Scheme	Marks	AO
(i) Forms at least one correct expression in \(x\) for the tension ie \(7(e_A + x)\) or \(3(e_C - x)\) FT their \(e_A\) or \(e_C\)	B1F	3.1b
Forms an equation of motion with five terms (with at least two terms correct). Accept “\(a\)” for \(\ddot{x}\) And “\(v\)” for \(\dot{x}\) Condone sign errors on the terms.	M1	3.1b
Completes a reasoned argument to obtain \(\ddot{x} + 9\dot{x} + 20x = 0\) OE AG	R1	2.1
	(3)
(ii) Obtains solution from their three term Auxiliary Equation.	M1	3.1a
Obtains \(A\mathrm{e}^{-4t} + B\mathrm{e}^{-5t}\)	A1	1.1b
Uses \(x = 0.6\) when \(t = 0\) to find an equation in two variables	M1	3.4
Sets their correct \(\dot{x} = 0\) when \(t = 0\) to find an equation in two variables	M1	3.3
Obtains \(x = 3\mathrm{e}^{-4t} - 2.4\mathrm{e}^{-5t}\) OE	A1	1.1b
	(5)

Typical solution

(i)

\[0.5\ddot{x} = 0.5g + 3(e_C - x) - 7(e_A + x) - 4.5\dot{x}\]\[0.5\ddot{x} = 0.5g + 3(1.61 - x) - 7(1.39 + x) - 4.5\dot{x}\]\[0.5\ddot{x} = 4.9 + 4.83 - 3x - 9.73 - 7x - 4.5\dot{x}\]\[\ddot{x} + 9\dot{x} + 20x = 0\]

(ii)

\[0 = \lambda^2 + 9\lambda + 20\]\[\lambda = -4 \text{ or } \lambda = -5\]\[x = A\mathrm{e}^{-4t} + B\mathrm{e}^{-5t}\]\[\dot{x} = -4A\mathrm{e}^{-4t} - 5B\mathrm{e}^{-5t}\]\[x = 0.6,\ \dot{x} = 0,\ t = 0\]\[0.6 = A + B\]\[0 = -4A - 5B\]\[\Rightarrow A = 3 \text{ and } B = -2.4\]\[x = 3\mathrm{e}^{-4t} - 2.4\mathrm{e}^{-5t}\]

Mark scheme (c)

Scheme	Marks	AO
Gives a valid limitation with reference to the size of the tube.	E1	3.5b
	(1)
	(12 marks)

Typical solution

The resistance force in a thin tube might not be the same as in a large bath.