A2 June 2023 Q8
8.
The fixed points \(A\) and \(B\) lie on a smooth horizontal surface with \(AB = 6\) m.
A particle \(P\) has mass 0.3 kg.
One end of a light elastic string, of natural length 2 m and modulus of elasticity 20 N, is attached to \(P\), and the other end is attached to \(A\).
One end of another light elastic string, of natural length 2 m and modulus of elasticity 40 N, is attached to \(P\) and the other end is attached to \(B\).
The particle \(P\) is at rest in equilibrium at the point \(E\) on the surface, as shown in Figure 7.
The particle \(P\) is now held at the midpoint of \(AB\) and released from rest.
The time between the instant when \(P\) is released and the instant when it first returns to the point \(E\) is \(S\) seconds.
| Scheme | Marks | AO |
|---|---|---|
 | ||
| In equilibrium: \(T_A = T_B\) | M1 | 3.3 |
| \(\dfrac{40e}{2} = \dfrac{20(2 - e)}{2}\) | A1 | 1.1b |
| \(2e = 2 - e \quad \Rightarrow \quad e = \dfrac{2}{3} \quad \Rightarrow \quad EB = \dfrac{8}{3}\ \text{(m)}\) * | A1* | 2.2a |
| (3) |
Notes
M1: Form equation for equilibrium of forces acting on \(P\). Dimensionally correct.
A1: Correct unsimplified equation in one unknown.
A1*: Obtain given answer from correct working. Condone missing units.
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion for \(P\) when displaced \(x\) from equilibrium in direction of \(A\) | M1 | 2.1 |
| \(\dfrac{40(e + x)}{2} - \dfrac{20(2 - e - x)}{2} = -0.3\ddot{x}\) | A1 A1 | 1.1b 1.1b |
| \(\ddot{x} = -100x\) Of the form \(\ddot{x} = -\omega^2x\) hence SHM* | A1* | 3.2a |
| (4) |
Notes
M1: Equation of motion about the equilibrium. Need all terms and dimensionally correct.
Condone sign errors.
A1: Unsimplified equation with \(e\) or given \(e\) with at most one error.
A1: Correct unsimplified equation with \(e\) or given \(e\)
A1*: Reach given conclusion in correct form from correct working.
| Scheme | Marks | AO |
|---|---|---|
| Periodic time \(= \dfrac{2\pi}{\text{their}\ \omega}\left(= \dfrac{\pi}{5}\ \text{(s)}\right)\) | M1 | 3.4 |
| \(\Rightarrow S = \dfrac{1}{4} \times \dfrac{2\pi}{\text{their}\ \omega}\) | M1 | 3.1a |
| \(S = \dfrac{\pi}{20}\) | A1 | 1.1b |
| (3) |
Notes
M1: Use the model to find the periodic time for \(\omega\) obtained correctly.
M1: Correct method to find the required time for \(\omega\) obtained correctly.
A1: Correct only.
Alternative (c)
| Scheme | Marks | AO |
|---|---|---|
| Use of \(x = a\cos(\text{their}\,\omega)t\) or \(x = a\sin(\text{their}\,\omega)t\) | M1 | |
| \(0 = \tfrac{1}{3}\cos(\text{their}\,\omega)S\) or \(\tfrac{1}{3} = \tfrac{1}{3}\sin(\text{their}\,\omega)S\) | M1 | |
| \(S = \dfrac{\pi}{20}\) | A1 | |
| (3) |
| Scheme | Marks | AO |
|---|---|---|
| Use of \(v = a\omega\cos\omega t\) or \(v = a\omega\sin\omega t\) | M1 | 3.4 |
| \(2 = \dfrac{\omega}{3}\cos\omega t \quad (t = 0.0927...)\) or \(2 = \dfrac{\omega}{3}\sin\omega t \quad (t = 0.06435...)\) | A1ft | 1.1b |
| time \(= 4 \times 0.0927..\) or time \(= \dfrac{\pi}{5} - 4 \times 0.06435..\) | DM1 | 3.1a |
| time \(= 0.37\) (s) \((0.371\ \text{(s)})\) | A1 | 1.1b |
| (4) | ||
| (14 marks) |
Notes
M1: Use a correct model for the speed or displacement of \(P\) for \(\omega\) obtained correctly
A1ft: Follow their \(\omega\). NB \(v = 2\) when \(x = \dfrac{4}{15}\)
DM1: Complete method to find the required time.
A1: 0.37 or better (0.370918….)