M3 January 2010 Q5
5.
One end \(A\) of a light inextensible string of length \(3a\) is attached to a fixed point. A particle of mass \(m\) is attached to the other end \(B\) of the string. The particle is held in equilibrium at a distance \(2a\) below the horizontal through \(A\), with the string taut. The particle is then projected with speed \(\sqrt{(2ag)}\), in the direction perpendicular to \(AB\), in the vertical plane containing \(A\) and \(B\), as shown in Figure 4. In the subsequent motion the string remains taut. When \(AB\) is at an angle \(\theta\) below the horizontal, the speed of the particle is \(v\) and the tension in the string is \(T\).
(a) Show that \(v^2 = 2ag(3\sin\theta - 1)\). (5)
(b) Find the range of values of \(T\). (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}m \times 2ag - \dfrac{1}{2}mv^2 = mg(2a - 3a\sin\theta)\) | M1 A1=A1 |
| leading to \(\quad v^2 = 2ga(3\sin\theta - 1)\) * cso | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| minimum value of \(T\) is when \(v = 0 \quad\Rightarrow\quad \sin\theta = \dfrac{1}{3}\) | B1 |
| \(T = mg\sin\theta = \dfrac{mg}{3}\) | M1 A1 |
| maximum value of \(T\) is when \(\theta = \dfrac{\pi}{2} \qquad \left(v^2 = 4ag\right)\) | |
| \(\uparrow \quad T = \dfrac{mv^2}{3a} + mg\) | M1 A1 |
| \(= \dfrac{7mg}{3}\) | A1 |
| \(\left(\dfrac{mg}{3} \leqslant T \leqslant \dfrac{7mg}{3}\right)\) | |
| (6) | |
| (11 marks) |