M3 January 2007 Q4
4.
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a point \(O\). The point \(A\) is vertically below \(O\), and \(OA = a\). The particle is projected horizontally from \(A\) with speed \(\sqrt{(3ag)}\). When \(OP\) makes an angle \(\theta\) with the upward vertical through \(O\) and the string is still taut, the tension in the string is \(T\) and the speed of \(P\) is \(v\), as shown in Figure 2.
(a) Find, in terms of \(a\), \(g\) and \(\theta\), an expression for \(v^2\). (3)
(b) Show that \(T = (1 - 3\cos\theta)mg\). (3)
The string becomes slack when \(P\) is at the point \(B\).
(c) Find, in terms of \(a\), the vertical height of \(B\) above \(A\). (2)
After the string becomes slack, the highest point reached by \(P\) is \(C\).
(d) Find, in terms of \(a\), the vertical height of \(C\) above \(B\). (5)
| Scheme | Marks |
|---|---|
| Energy: \(\tfrac{1}{2}m.3ag - \tfrac{1}{2}mv^2 = mga(1 + \cos\theta)\) | M1 A1 |
| \(v^2 = ag(1 - 2\cos\theta)\) (o.e.) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(T + mg\cos\theta = m\dfrac{v^2}{a}\) | M1 A1 |
| Hence \(T = (1 - 3\cos\theta)mg\) (*) | A1 cso |
| (3) |
| Scheme | Marks |
|---|---|
| Using \(T = 0\) to find \(\cos\theta\) | M1 |
| Hence height above \(A = \tfrac{4}{3}a\) Accept \(1.33a\) (but must have 3+ s.f.) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(v^2 = \tfrac{1}{3}ag\) (o.e.) f.t. using \(\cos\theta = \tfrac{1}{3}\) in \(v^2\) | B1ft |
| consider vert motion: \((v\sin\theta)^2 = 2gh\) (with \(v\) resolved) | M1 A1 |
| \(\sin^2\theta = \tfrac{8}{9}\) (or \(\theta = 70.53\), \(\sin\theta = 0.943\)) and solve for \(h\) (as \(ka\)) | ↓ M1 |
| \(h = \tfrac{4}{27}a\) or \(0.148a\) (awrt) | A1 |
| (5) | |
| (13 marks) |
Notes
↓ marks a mark that depends on the M mark above it (an arrow in the scheme).
OR
| consider energy: \(\tfrac{1}{2}m(v\cos\theta)^2 + mgh = \tfrac{1}{2}mv^2\) (3 non-zero terms) | M1 A1 |
| Sub for \(v\), \(\theta\) and solve for \(h\) | ↓ M1 |
| \(h = \tfrac{4}{27}a\) or \(0.148a\) (awrt) | A1 |