M3 January 2013 Q7
7. A particle \(P\) of mass 1.5 kg is attached to the mid-point of a light elastic string of natural length 0.30 m and modulus of elasticity \(\lambda\) newtons. The ends of the string are attached to two fixed points \(A\) and \(B\), where \(AB\) is horizontal and \(AB = 0.48\) m. Initially \(P\) is held at rest at the mid-point, \(M\), of the line \(AB\) and the tension in the string is 240 N.
(a) Show that \(\lambda = 400\) (3)
The particle is now held at rest at the point \(C\), where \(C\) is 0.07 m vertically below \(M\). The particle is released from rest at \(C\).
(b) Find the magnitude of the initial acceleration of \(P\). (6)
(c) Find the speed of \(P\) as it passes through \(M\). (6)
| Scheme | Marks |
|---|---|
| \(T = \dfrac{\lambda x}{l} \Rightarrow 240 = \dfrac{\lambda \times 18}{30}\) | M1A1 |
| \(\lambda = 400\) | A1 |
| Scheme | Marks |
|---|---|
| Extension = 10 cm or 20 cm (used in (b) or (c)) | B1 |
| \(T = \dfrac{400 \times 10}{15} = \left(\dfrac{800}{3}\right)\) | M1A1ft |
| R\((\uparrow)\) \(2T\cos\theta - 1.5g = (\pm)1.5a\) | M1A1 |
| \(\dfrac{1600}{3} \times \dfrac{7}{25} - 1.5 \times 9.8 = (\pm)1.5a\) | |
| \(a = 89.75\ldots\ \ \ a = 90\) m s\(^{-2}\) or 89.8 (positive) | A1 |
| Scheme | Marks |
|---|---|
| E.P.E. \(= \dfrac{1}{2} \times 400 \times \dfrac{0.2^2}{0.3}\) | B1ft (any correct EPE) |
| \(1.5g \times 0.07 + \dfrac{1}{2} \times 1.5v^2 = 200 \times \dfrac{0.2^2}{0.3} - \dfrac{200 \times 0.18^2}{0.3}\) | M1A1A1 |
| \(v^2 = \dfrac{1}{0.75}\left(200 \times \dfrac{0.2^2}{0.3} - \dfrac{200 \times 0.18^2}{0.3} - 1.5g \times 0.07\right)\) | M1dep |
| \(v = 2.32\ldots = 2.3\) m s\(^{-1}\) | A1 |