A2 October 2021 Q5
5. A light inextensible string of length \(a\) has one end attached to a fixed point \(O\). The other end of the string is attached to a small stone of mass \(m\). The stone is held with the string taut and horizontal. The stone is then projected vertically upwards with speed \(U\).
The stone is modelled as a particle and air resistance is modelled as being negligible.
Assuming that the string does not break, use the model to
The string will break if the tension in it is equal to \(\dfrac{11mg}{2}\)
Given that \(U = 2\sqrt{ag}\), use the model to
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion along the string at the top of the circle | M1 | 3.1b |
| \(T + mg = \dfrac{mv^2}{a}\) | A1 | 1.1b |
| Conservation of energy | M1 | 3.1b |
| \(\dfrac{1}{2}mU^2 - \dfrac{1}{2}mv^2 = mga\) | A1 | 1.1b |
| Overall strategy to solve these equations for \(T\) and use \(T = 0\) | M1 | 3.1b |
| \(U = \sqrt{3ga}\) | A1 | 1.1b |
| (6) |
Notes
M1: Correct number of terms
A1: Correct equation
M1: All terms needed and dimensionally correct
A1: Correct equation
M1: Solve for \(T\) and use \(T = 0\) (allow \(T \geqslant 0\))
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion along the string at instant string breaks | M1 | 3.1b |
| \(\dfrac{11mg}{2} - mg\cos\alpha = \dfrac{mv^2}{a}\) | A1 | 1.1b |
| Conservation of energy | M1 | 3.1b |
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m.4ag = mga\cos\alpha\) | A1 | 1.1b |
| Solve these equations for \(\cos\alpha \quad \left(= \dfrac{1}{2}\right)\) | M1 | 1.1b |
| Angle turned through is \(210^\circ\) | A1 | 1.1b |
| (6) |
Notes
M1: Correct no. of terms with \(mg\) resolved and correct acceleration component
A1: Correct equation
M1: All terms needed and dimensionally correct
A1: Correct equation
M1: Solve for \(\cos\alpha\)
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Find radial component of acceleration: \(\dfrac{v^2}{a} \quad (= 5g)\) | M1 | 2.1 |
| Find tangential component of acceleration: \(g\sin\alpha \quad \left(= \dfrac{\sqrt{3}}{2}g\right)\) | M1 | 2.1 |
| Square, add and square root | M1 | 3.1b |
| \(\dfrac{\sqrt{103}}{2}g\) or 49.7 (\(\text{m s}^{-2}\)) or 50 (\(\text{m s}^{-2}\)) | A1 | 1.1b |
| (4) | ||
| (16 marks) |
Notes
M1: Uses their value of \(v\) from part (b)
M1: Equation of motion along the tangent oe
M1: Find the magnitude of the resultant acceleration
A1: cao