M3 January 2006 Q6
6. One end of a light inextensible string of length \(l\) is attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\) which is hanging freely at rest at point \(B\). The particle \(P\) is projected horizontally from \(B\) with speed \(\sqrt{(3gl)}\). When \(AP\) makes an angle \(\theta\) with the downward vertical and the string remains taut, the tension in the string is \(T\).
(a) Show that \(T = mg(1 + 3\cos\theta)\). (6)
(b) Find the speed of \(P\) at the instant when the string becomes slack. (3)
(c) Find the maximum height above the level of \(B\) reached by \(P\). (5)
| Scheme | Marks |
|---|---|
| Energy \(\dfrac{1}{2}m\left(u^2 - v^2\right) = mgl(1 - \cos\theta)\) | M1 A1 |
| \(\left[v^2 = gl + 2gl\cos\theta\right]\) | |
| N2L \(T - mg\cos\theta = \dfrac{mv^2}{l}\) | M1 A1 |
| \(= \dfrac{mgl(1 + 2\cos\theta)}{l}\) | M1 |
| \(T = mg(1 + 3\cos\theta)\) * cso | A1 |
| (6) |
Notes
The scheme links the first two M1 marks to the third M1 with arrows: the third M1 depends on both.
| Scheme | Marks |
|---|---|
| \(T = 0 \Rightarrow \cos\theta = -\dfrac{1}{3}\) | B1 |
| \(v^2 = gl - \dfrac{2}{3}gl \Rightarrow v = \left(\dfrac{gl}{3}\right)^{\frac{1}{2}}\) | M1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\uparrow\ \ v_y = \left(\dfrac{gl}{3}\right)^{\frac{1}{2}}\sin\theta\ \ \left[= \left(\dfrac{gl}{3}\right)^{\frac{1}{2}} \cdot \dfrac{2\sqrt{2}}{3}\right]\) | M1 |
| \(v^2 = u^2 - 2gh \Rightarrow 2gh = \dfrac{gl}{3} \cdot \dfrac{8}{9} \Rightarrow h = \dfrac{4l}{27}\) | M1 A1 |
| \(H = l(1 - \cos\theta) + \dfrac{4l}{27} = \dfrac{40l}{27}\) | M1 A1 |
| (5) | |
| (14 marks) |
Notes
The scheme links the M1 marks with arrows: each later M1 depends on the first M1.