M3 January 2009 Q2

EdexcelOld spec9 marksHooke's Law

Figure 1: string OP at an angle to the vertical through O, horizontal force four thirds mg at P — Figure 1

A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(3mg\). The other end of the string is attached to a fixed point \(O\). The particle \(P\) is held in equilibrium by a horizontal force of magnitude \(\tfrac{4}{3}mg\) applied to \(P\). This force acts in the vertical plane containing the string, as shown in Figure 1. Find

(a) the tension in the string, (5)

(b) the elastic energy stored in the string. (4)

Force diagram: tension T along the string at angle theta to the vertical, four thirds mg horizontal, mg down — Mark scheme (a)

Scheme	Marks
\((\leftarrow)\) \(T\sin\theta = \dfrac{4}{3}mg\)	M1 A1
\((\uparrow)\) \(T\cos\theta = mg\)	A1
\(T^2 = \left(\dfrac{4}{3}mg\right)^2 + (mg)^2\)	M1
Leading to \(T = \dfrac{5}{3}mg\)	A1
	(5)

Mark scheme (b)

Scheme	Marks
HL \(T = \dfrac{\lambda x}{a} \Rightarrow \dfrac{5}{3}mg = \dfrac{3mge}{a}\) ft their \(T\)	M1 A1ft
\(e = \dfrac{5}{9}a\)
\(E = \dfrac{\lambda x^2}{2a} = \dfrac{3mg}{2a} \times \left(\dfrac{5}{9}a\right)^2 = \dfrac{25}{54}mga\)	M1 A1
	(4)
	(9 marks)