A2 June 2021 Paper 1 Q15

AQACurrent spec13 marksSecond Order Differentials

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

A particle \(P\) of mass \(m\) is attached to two light elastic strings, \(AP\) and \(BP\).

The other ends of the strings, \(A\) and \(B\), are attached to fixed points which are 4 metres apart on a rough horizontal surface at the bottom of a container.

The coefficient of friction between \(P\) and the surface is 0.68

When the extension of string \(AP\) is \(e_A\) metres, the tension in \(AP\) is \(24me_A\)
When the extension of string \(BP\) is \(e_B\) metres, the tension in \(BP\) is \(10me_B\)
The natural length of string \(AP\) is 1 metre
The natural length of string \(BP\) is 1.3 metres

A horizontal line from A to B with the particle P on it, between A and B

(a) Show that when \(AP = 1.5\) metres, the tension in \(AP\) is equal to the tension in \(BP\). [1 mark]

(b) \(P\) is held at the point between \(A\) and \(B\) where \(AP = 1.9\) metres, and then released from rest.

At time \(t\) seconds after \(P\) is released, \(AP = (1.5 + x)\) metres.

The line from A to B with P on it; the distance from A to P is marked (1.5 + x)

Show that when \(P\) is moving towards \(A\),

\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 34x = 6.664\]

[3 marks]

(c) The container is then filled with oil, and \(P\) is again released from rest at the point between \(A\) and \(B\) where \(AP = 1.9\) metres.

At time \(t\) seconds after \(P\) is released, the oil causes a resistive force of magnitude \(10mv\) newtons to act on the particle, where \(v\) m s⁻¹ is the speed of the particle.

Find \(x\) in terms of \(t\) when \(P\) is moving towards \(A\). [9 marks]

Scheme	Marks	AO
Uses a rigorous argument to obtain the required result	R1	2.1
	(1)

Mark scheme (b)

Scheme	Marks	AO
Obtains one correct tension	B1	1.1a
Uses Newton’s second law to form a four term differential equation with at least two terms correct (allow equivalent notation for derivatives) Condone sign errors on the terms	M1	3.1b
Completes a rigorous argument to give the required differential equation	R1	2.1
	(3)

Typical solution

\[m\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = 10m(1.2 - x) - 24m(0.5 + x) + 6.664m\]\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 34x = 6.664\]

Mark scheme (c)

Scheme	Marks	AO
Obtains correct 2^nd order DE	B1	2.2a
Obtains correct solution to their three term Auxiliary Equation	M1	1.1a
Obtains their correct Complementary Function	A1F	1.1b
Obtains correct Particular Integral ACF	B1	1.1b
Obtains correct general solution (ft their CF, but must have two unknowns)	A1F	2.2a
Uses \(x = 0.4\) when \(t = 0\) to obtain correct \(A\) ACF	B1	3.3
Sets their correct \(\dot{x} = 0\) when \(t = 0\)	M1	1.1a
Obtains correct \(B\) ACF	A1	1.1b
Obtains correct final equation ACF	R1	2.1
	(9)
	(13 marks)

Typical solution

\[m\ddot{x} = 10mv + 6.664m - 34mx\]

But \(v = -\dot{x}\)

\[\ddot{x} + 10\dot{x} + 34x = 6.664\]\[\lambda^2 + 10\lambda + 34 = 0\]\[\lambda = -5 \pm 3\mathrm{i}\]

CF:

\[x = A\mathrm{e}^{-5t}\cos 3t + B\mathrm{e}^{-5t}\sin 3t\]

PI: \(x = 0.196\)

General Solution:

\[x = A\mathrm{e}^{-5t}\cos 3t + B\mathrm{e}^{-5t}\sin 3t + 0.196\]\[t = 0,\ x = 0.4 \Rightarrow A = 0.204\]\[\begin{aligned} \dot{x} = {} & -5A\mathrm{e}^{-5t}\cos 3t - 3A\mathrm{e}^{-5t}\sin 3t \\ & -5B\mathrm{e}^{-5t}\sin 3t + 3B\mathrm{e}^{-5t}\cos 3t \end{aligned}\]\[0 = -5A + 3B\]\[B = 0.34\]\[x = 0.204\mathrm{e}^{-5t}\cos 3t + 0.34\mathrm{e}^{-5t}\sin 3t + 0.196\]