M4 June 2014 Q6

EdexcelOld spec13 marksFurther Dynamics Variable Force & Kinematics

6. A particle of mass \(m\) kg is attached to one end of a light elastic string of natural length \(a\) metres and modulus of elasticity \(5ma\) newtons. The other end of the string is attached to a fixed point \(O\) on a smooth horizontal plane. The particle is held at rest on the plane with the string stretched to a length \(2a\) metres and then released at time \(t = 0\). During the subsequent motion, when the particle is moving with speed \(v\) m s\(^{-1}\), the particle experiences a resistance of magnitude \(4mv\) newtons. At time \(t\) seconds after the particle is released, the length of the string is \((a + x)\) metres, where \(0 \leqslant x \leqslant a\).

(a) Show that, from \(t = 0\) until the string becomes slack, \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 5x = 0\] (3)

(b) Hence express \(x\) in terms of \(a\) and \(t\). (6)

(c) Find the speed of the particle at the instant when the string first becomes slack, giving your answer in the form \(ka\), where \(k\) is a constant to be found correct to 2 significant figures. (4)

Mark scheme (a)

Scheme	Marks
\(m\ddot{x} = 4mv - \dfrac{5ma \times x}{a} \qquad v = -\dot{x}\)	M1 M1
\(\ddot{x} + 4\dot{x} + 5x = 0\) **	A1
	(3)

Notes

M1 Equation of motion as far as \(m\ddot{x} = \pm 4mv - T\)

M1 Use of \(v = -\dot{x}\)

A1 Reach given answer correctly.

Mark scheme (b)

Scheme	Marks
AE \(m^2 + 4m + 5 = 0,\ \ m = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 5}}{2} = -2 \pm \mathrm{i}\)	M1
\(x = \mathrm{e}^{-2t}(A\cos t + B\sin t)\)	A1
\(t = 0,\ x = a = A\)	M1 A1
\(\dot{x} = -2\mathrm{e}^{-2t}(a\cos t + B\sin t) + \mathrm{e}^{-2t}(-a\sin t + B\cos t)\)	M1
\(t = 0,\ \ \dot{x} = 0 = -2a + B\ \ \ x = \mathrm{e}^{-2t}(a\cos t + 2a\sin t)\)	A1
	(6)

Notes

M1 Solve AE to find GS

M1 Use \(t = 0,\ x = a\) to find A

M1 Differentiate and use boundary conditions to find B

Mark scheme (c)

Scheme	Marks
String goes slack when \(x = \mathrm{e}^{-2t}(a\cos t + 2a\sin t) = 0\)
\(\cos t = -2\sin t,\ \ \tan t = -\dfrac{1}{2}\)	M1 A1
\(\dot{x} = -2\mathrm{e}^{-2t}(a\cos t + 2a\sin t) + \mathrm{e}^{-2t}(-a\sin t + 2a\cos t)\)	M1
\(= \mathrm{e}^{-2t}(-5a\sin t) = -0.01\ldots a \qquad\) Speed \(= 0.011a\) (m s\(^{-1}\))	A1
	(4)
	(13 marks)

Notes

M1 Set \(x = 0\) and solve for \(t\) or \(\tan t\)

M1 Substitute a positive value of t to find the speed. An answer of 0.88... indicates a negative \(t\).

A1 The question specifies 2 sf