M3 June 2007 Q7

EdexcelOld spec15 marksHooke's Law Work-Energy Principle

Figure 1: P hanging at the mid-point of a string between A and B, AB = 3l horizontal, P a distance 2l below AB — Figure 1

A light elastic string, of natural length \(3l\) and modulus of elasticity \(\lambda\), has its ends attached to two points \(A\) and \(B\), where \(AB = 3l\) and \(AB\) is horizontal. A particle \(P\) of mass \(m\) is attached to the mid-point of the string. Given that \(P\) rests in equilibrium at a distance \(2l\) below \(AB\), as shown in Figure 1,

(a) show that \(\lambda = \dfrac{15mg}{16}\). (9)

The particle is pulled vertically downwards from its equilibrium position until the total length of the elastic string is \(7.8l\). The particle is released from rest.

(b) Show that \(P\) comes to instantaneous rest on the line \(AB\). (6)

Diagram: P below the mid-point of AB, half-width 1.5l, depth 2l, tensions T along PA and PB at angle alpha to the vertical, weight mg — Mark scheme (a)

Diagram: P a distance h below the mid-point of AB, half-width 1.5l, each half of the string 3.9l — Mark scheme (b)

Scheme	Marks
\(AP = \sqrt{\left((1.5l)^2 + (2l)^2\right)} = 2.5l\)	M1 A1
\(\cos\alpha = \dfrac{4}{5}\)	B1
Hooke’s Law \(T = \dfrac{\lambda(2.5l - 1.5l)}{1.5l}\ \ \left(= \dfrac{2\lambda}{3}\right)\)	M1 A1
\(\uparrow\) \(2T\cos\alpha = mg\ \ \ \ \left(T = \dfrac{5mg}{8}\right)\)	M1 A1
\(2 \times \dfrac{2\lambda}{3} \times \dfrac{4}{5} = mg\ \ \ \ \left(\dfrac{2\lambda}{3} = \dfrac{5mg}{8}\right)\)	M1
\(\lambda = \dfrac{15mg}{16}\) * cso	A1
	(9)

Scheme	Marks
\(h = \sqrt{\left((3.9l)^2 - (1.5l)^2\right)} = 3.6l\)	M1 A1
Energy \(\dfrac{1}{2}mv^2 + mg \times h = 2 \times \dfrac{15mg}{16} \times \dfrac{(2.4l)^2}{2 \times 1.5l}\) ft their \(h\)	M1 A1ft = A1
Leading to \(v = 0\) * cso	A1
	(6)
	(15 marks)