M3 June 2011 Q7

7. A particle \(P\) of mass 0.5 kg is attached to the mid-point of a light elastic string of natural length 1.4 m and modulus of elasticity 2 N. The ends of the string are attached to the points \(A\) and \(B\) on a smooth horizontal table, where \(AB = 2\) m. The mid-point of \(AB\) is \(O\) and the point \(C\) is on the table between \(O\) and \(B\) where \(OC = 0.2\) m. At time \(t = 0\) the particle is released from rest at \(C\). At time \(t\) seconds the length of the string \(AP\) is \((1 + x)\) m.

(a) Show that the tension in \(BP\) is \(\dfrac{2}{7}(3 - 10x)\) N. (2)

(b) Find, in terms of \(x\), the tension in \(AP\). (1)

(d) Find the greatest speed of \(P\) during the motion. (2)

The point \(D\) lies between \(O\) and \(A\), where \(OD = 0.1\) m.

(e) Find the time taken by \(P\) to move directly from \(C\) to \(D\). (4)

Scheme	Marks
Total extn. \(= 0.6\) \(T_b = \dfrac{\lambda \times \text{ext}}{l} = \dfrac{2(0.3 - x)}{0.7} = \dfrac{2}{7}(3 - 10x)\) *	M1 A1
	(2)

Scheme	Marks
\(T_a = \dfrac{2(x + 0.3)}{0.7} \qquad \left(= \dfrac{2}{7}(10x + 3)\right)\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(T_b - T_a = 0.5\ddot{x}\) \(\dfrac{2}{7}(3 - 10x) - \dfrac{2}{7}(10x + 3) = 0.5\ddot{x}\)	M1 A1 ft
\(2 \times \left(-\dfrac{20x}{7}\right) = 0.5\ddot{x}\) \(\ddot{x} = -\dfrac{40}{7 \times 0.5}x\) \((\therefore \text{ S.H.M.})\)	M1 A1
Period \(= \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{7 \times 0.5}{40}} = 2\pi\sqrt{\dfrac{7}{80}}\) *	M1 A1
	(6)

Scheme	Marks
\(v_{\max} = a\omega = 0.2\sqrt{\dfrac{80}{7}}\) o.e. or a.w.r.t. 0.68 m s\(^{-1}\)	M1 A1
	(2)

Mark scheme (e)

Scheme	Marks
\(x = a\cos\omega t = 0.2\cos\left(\sqrt{\dfrac{80}{7}}t\right)\)	M1
\(x = -0.1 \qquad -\dfrac{0.1}{0.2} = \cos\left(\sqrt{\dfrac{80}{7}}t\right)\)	A1
\(t = \sqrt{\dfrac{7}{80}}\cos^{-1}(-0.5)\) \(t = \sqrt{\dfrac{7}{80}} \times \dfrac{2\pi}{3} = \dfrac{\pi}{3}\sqrt{\dfrac{7}{20}}\) o.e. (accept a.w.r.t. 0.62) s	M1 A1
	(4)
	(15 marks)