M3 June 2014 (R) Q6

6. A light elastic string, of natural length \(l\) and modulus of elasticity \(4mg\), has one end attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\). The particle hangs freely at rest in equilibrium at the point \(E\). The distance of \(E\) below \(A\) is \((l + e)\).

(a) Find \(e\) in terms of \(l\). (2)

At time \(t = 0\), the particle is projected vertically downwards from \(E\) with speed \(\sqrt{gl}\).

(b) Prove that, while the string is taut, \(P\) moves with simple harmonic motion. (5)

(d) Find the time at which the string first goes slack. (4)

Scheme	Marks
\(\dfrac{4mge}{l} = mg\)	M1
\(e = \dfrac{1}{4}l\)	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(mg - T = m\ddot{x}\)	M1 A1
\(mg - \dfrac{4mg}{l}\left(x + \dfrac{1}{4}l\right) = m\ddot{x}\)	M1
\(-\dfrac{4g}{l}x = \ddot{x}\)	A1
SHM, \(\left(\text{with } \omega = \sqrt{\dfrac{4g}{l}}\right)\)	A1
	(5)

Notes

M1 using NL2 vertically

A1 correct equation

M1 using Hooke’s law to replace \(T\) with an expression for \(x\). These 3 marks can be gained with \(a\) instead of \(\ddot{x}\)

A1 fully correct, simplified equation

A1 conclusion with all work correct

Scheme	Marks
\(\sqrt{gl} = a\sqrt{\dfrac{4g}{l}}\)	M1 A1
\(a = \dfrac{1}{2}l\)	A1
	(3)

Mark scheme (d)

Scheme	Marks
\(-\dfrac{1}{4}l = \dfrac{1}{2}l\sin\sqrt{\dfrac{4g}{l}}\,t\)	M1 A1
\(t = \dfrac{7\pi}{12}\sqrt{\dfrac{l}{g}}\)	M1 A1
	(4)
	(14 marks)

Notes

M1 for an equation to find required time

A1 correct equation

M1 solving their equation must be in radians and must give a positive value

A1 correct time decimal equivalent acceptable.