A2 October 2021 Q6
6. A light elastic string, of natural length \(l\) and modulus of elasticity \(2mg\), has one end attached to a fixed point \(A\) and the other end attached to a particle \(P\) of mass \(m\). The particle \(P\) hangs in equilibrium at the point \(O\).
The particle \(P\) is pulled down vertically from \(O\) to the point \(B\), where \(OB = l\), and released from rest.
Air resistance is modelled as being negligible.
Using the model,
The particle \(P\) first comes to instantaneous rest at the point \(C\).
Using the model,
| Scheme | Marks | AO |
|---|---|---|
| \(mg = \dfrac{2mge}{l}\) | M1 | 3.1a |
| \(e = \dfrac{1}{2}l\) so \(AO = \dfrac{3l}{2}\) * | A1* | 1.1b |
| (2) |
Notes
M1: Use of Hooke’s Law and \(T = mg\)
A1*: Correct answer fully justified
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion vertically: \(mg - T = m\ddot{x}\). | M1 | 2.1 |
| \(mg - \dfrac{2mg(x + e)}{l} = m\ddot{x}\). | A1 | 1.1b |
| \(-\dfrac{2g}{l}x = \ddot{x}\), so SHM with \(\omega^2 = \dfrac{2g}{l}\) | A1 | 1.1b |
| Use of \(\dfrac{2\pi}{\omega}\) | M1 | 3.1a |
| \(2\pi\sqrt{\dfrac{l}{2g}} = \pi\sqrt{\dfrac{2l}{g}}\) * | A1* | 2.2a |
| (5) |
Notes
M1: All terms needed but allow \(a\) for the acceleration
A1: Correct unsimplified equation including \(\ddot{x}\)
A1: Must mention SHM
M1: Correct method
A1*: Correct answer correctly shown
| Scheme | Marks | AO |
|---|---|---|
| Complete method to find \(h\) | M1 | 3.1a |
| \(mgh = \dfrac{2mg\left(\dfrac{3l}{2}\right)^2}{2l}\) OR \(v^2 = \dfrac{2g}{l}\left(l^2 - \left(-\dfrac{1}{2}l\right)^2\right)\) and \(0 = \dfrac{3gl}{2} - 2gs\) | A1 A1 | 1.1b 1.1b |
| \(h = \dfrac{9l}{4}\) | A1 | 1.1b |
| (4) |
Notes
M1: Correct no. of terms, dimensionally correct in energy equation
OR
use SHM to find \(v^2\) at unstretched position AND then use motion under gravity
Correct no. of terms, dimensionally correct in both equations
A1: Equation with at most one error
A1: Correct equation OR correct equations
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| \(-\dfrac{1}{2}l = l\cos\omega t\) | M1 | 3.1a |
| \(t = \dfrac{2\pi}{3}\sqrt{\dfrac{l}{2g}}\) | A1 | 1.1b |
| \(v = \sqrt{\dfrac{2g}{l}\left(l^2 - \left(-\dfrac{1}{2}l\right)^2\right)}\) OR | M1 | 2.1 |
| \(0 = \sqrt{\dfrac{3gl}{2}} - gt_1 \Rightarrow t_1 = \sqrt{\dfrac{3l}{2g}}\) | M1 | 3.1a |
| Total time \(= \dfrac{2\pi}{3}\sqrt{\dfrac{l}{2g}} + \sqrt{\dfrac{3l}{2g}}\) oe | A1 | 1.1b |
| (5) | ||
| (16 marks) |
Notes
M1: Complete method to find time to reach unstretched position
A1: Correct time
M1: Complete method to find speed at unstretched position
A1: Correct speed
M1: Complete method to find time to rest position
A1: Cao