M3 January 2006 Q5

5. A light elastic string of natural length \(l\) has one end attached to a fixed point \(A\). A particle \(P\) of mass \(m\) is attached tot he other end of the string and hangs in equilibrium at the point \(O\), where \(AO = \tfrac{5}{4}l\).

(a) Find the modulus of the elasticity of the string. (2)

The particle \(P\) is then pulled down and released from rest. At time \(t\) the length of the string is \(\dfrac{5l}{4} + x\).

(b) Prove that, while the string is taut, \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\frac{4gx}{l}.\] (5)

When \(P\) is released, \(AP = \tfrac{7}{4}l\). The point \(B\) is a distance \(l\) vertically below \(A\).

(d) Describe briefly the motion of \(P\) after it has passed through \(B\) for the first time until it next passes through \(O\). (2)

Scheme	Marks
HL \(T = mg = \dfrac{\lambda \times \frac{1}{4}l}{l} \Rightarrow \lambda = 4mg\)	M1 A1
	(2)

Mark scheme (b)

Scheme	Marks
N2L \(mg - T = m\ddot{x}\)	M1
\(mg - \dfrac{4mg\left(\frac{1}{4}l + x\right)}{l} = m\ddot{x}\)	M1 A1
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\dfrac{4g}{l}x\) * cso	M1 A1
	(5)

Notes

The scheme links the first M1 to the last M1 with an arrow: the last M1 depends on the first.

Mark scheme (c)

Scheme	Marks
\(v^2 = \omega^2\left(a^2 - x^2\right) = \dfrac{4g}{l}\left(\dfrac{l^2}{4} - \dfrac{l^2}{16}\right)\)	M1 A1
Leading to \(v = \tfrac{1}{2}\sqrt{(3gl)}\)	M1 A1
	(4)

Notes

The scheme links the two M1 marks with an arrow: the second M1 depends on the first.

or energy, \(\dfrac{1}{2}\dfrac{4mg \cdot \frac{9l^2}{16}}{l} = \dfrac{1}{2}mv^2 + mg \cdot \dfrac{3l}{4}\) for the first M1 A1 in (c)

(Corrected from the printed mark scheme: \(\frac{9l^2}{16}\) is printed as \(\frac{gl^2}{16}\).)

Scheme	Marks
\(P\) first moves freely under gravity,	B1
then (part) SHM.	B1
	(2)
	(13 marks)