M3 June 2006 Q5

EdexcelOld spec12 marksHooke's Law Work-Energy Principle

5. Two light elastic strings each have natural length 0.75 m and modulus of elasticity 49 N. A particle \(P\) of mass 2 kg is attached to one end of each string. The other ends of the strings are attached to fixed points \(A\) and \(B\), where \(AB\) is horizontal and \(AB = 1.5\) m.

Figure 2: P at the mid-point of AB, AB = 1.5 m — Figure 2

The particle is held at the mid-point of \(AB\). The particle is released from rest, as shown in Figure 2.

(a) Find the speed of \(P\) when it has fallen a distance of 1 m. (6)

Given instead that \(P\) hangs in equilibrium vertically below the mid-point of \(AB\), with \(\angle APB = 2\alpha\),

(b) show that \(\tan\alpha + 5\sin\alpha = 5\). (6)

Diagram: P a distance 1 m below the mid-point of AB, half-width 0.75 m — Mark scheme (a)

Force diagram: tensions T along PA and PB at angle alpha to the vertical, string length y, weight 2g — Mark scheme (b)

Scheme	Marks
\(AP = \sqrt{\left(0.75^2 + 1^2\right)} = 1.25\)	M1 A1
Conservation of energy
\(\dfrac{1}{2} \times 2 \times v^2 + 2 \times \dfrac{49 \times 0.5^2}{2 \times 0.75} = 2g \times 1\) −1 for each incorrect term	M1 A2 (1, 0)
Leading to \(v \approx 1.8\) (m s\(^{-1}\)) accept 1.81	A1
	(6)

Scheme	Marks
\(R(\uparrow)\) \(2T\cos\alpha = 2g\)	M1 A1
\(y = \dfrac{0.75}{\sin\alpha}\)
Hooke’s Law \(T = \dfrac{49}{0.75}\left(\dfrac{0.75}{\sin\alpha} - 0.75\right) = 49\left(\dfrac{1}{\sin\alpha} - 1\right)\)	M1 A1
Eliminating \(T\) \(\dfrac{9.8}{\cos\alpha} = 49\left(\dfrac{1}{\sin\alpha} - 1\right)\)	M1
\(\tan\alpha = 5(1 - \sin\alpha)\)
\(5 = \tan\alpha + 5\sin\alpha\) * cso	A1
	(6)
	(12 marks)