M3 June 2014 Q6
6. A particle \(P\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point. The particle is hanging freely at rest, with the string vertical, when it is projected horizontally with speed \(U\). The particle moves in a complete vertical circle.
As \(P\) moves in the circle the least tension in the string is \(T\) and the greatest tension is \(kT\). Given that \(U = 3\sqrt{ag}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}mU^2 - \dfrac{1}{2}mv^2 = 2mga\) | M1A1 |
| \(T + mg = m\dfrac{v^2}{a}\) | M1A1 |
| \(T = \dfrac{\left(mU^2 - 4mga\right)}{a} - mg\) | DM1 |
| \(T = \dfrac{mU^2 - 5mga}{a}\) | A1 |
| \(T \geqslant 0 \Rightarrow U^2 \geqslant 5ga\) | DM1 |
| \(U \geqslant \sqrt{5ag}\) * | A1 |
| (8) |
Notes
M1 for an energy equation, from the bottom to the top. A difference of KE terms and a PE term needed. From bottom to a general point gets M0 until a value for \(\theta\) at the top is used. \(v^2 = u^2 + 2as\) scores M0
A1 for all terms correct (inc signs)
M1 for NL2 along the radius at the top. Two forces and mass x acceleration needed. Accel can be in either form here. But see NB at end of (a)
A1 for a fully correct equation. Acceleration should be \(\dfrac{v^2}{a}\) now.
M1 dep for eliminating \(v\) (vel at top) between the two equations. Dependent on both previous M marks. If \(v\) is set = 0, award M0
A1 for a correct expression for \(T\)
M1 dep for using \(T \geqslant 0\) to obtain an inequality for \(U^2\) or \(U\). Allow with \(>\) Dependent on all previous M marks.
A1 cso for \(U \geqslant \sqrt{5ag}\) * Watch square root! Give A0 if \(>\) seen on previous line.
NB: The second and fourth M marks (and their As if earned) can be given together if \(mg \leqslant m\dfrac{v^2}{a}\) is seen
| Scheme | Marks |
|---|---|
| At top: \(T = \dfrac{9mga - 5mga}{a} = 4mg\) | M1(either tension)A1 |
| At bottom: \(T' - mg = \dfrac{mU^2}{a}\) | A1 |
| \(kT = mg + \dfrac{9mag}{a} = 10mg\) | DM1 |
| \(k = \dfrac{10mg}{4mg} = \dfrac{5}{2}\) | A1 |
| (5) | |
| (13 marks) |
Notes
M1 for obtaining an expression for the tension at the top or at the bottom, no need to substitute for \(U\) yet.
A1 Substitute for \(U\) and obtain one correct tension (\(4mg\) at top or \(10mg\) at bottom)
A1 for the other tension correct
M1 dep for using tension at bottom = \(k\) x tension at the top and solving for \(k\)
A1 cso for \(k = \dfrac{5}{2}\) oe