A2 June 2019 Q6
6. The points \(A\) and \(B\) lie on a smooth horizontal surface with \(AB = 4.5\) m.
A light elastic string has natural length 1.5 m and modulus of elasticity 15 N. One end of the string is attached to \(A\) and the other end of the string is attached to \(B\). A particle, \(P\), of mass 0.2 kg, is attached to the stretched string so that \(APB\) is a straight line and \(AP = 1.5\) m. The particle rests in equilibrium on the surface.
The particle is now moved directly towards \(A\) and is held on the surface so \(APB\) is a straight line with \(AP = 1\) m.
The particle is released from rest.
| Scheme | Marks | AO |
|---|---|---|
| Use of Hooke’s law seen | B1 | 2.1 |
| Let \(x\) be the displacement from equilibrium, then \(T_B - T_A = -0.2\ddot{x}\). | M1 | 3.1a |
| \(\dfrac{15(2 + x)}{1} - \dfrac{15(1 - x)}{0.5} = -0.2\ddot{x}\) | A1 A1 | 1.1b 1.1b |
| \(\Rightarrow \ddot{x} = -225x = -15^2x\), which is of the form \(\ddot{x} = -\omega^2 x\), so SHM * | A1* | 3.2a |
| (5) |
Notes
B1: Use of Hooke’s law seen at least once (with natural length \(\leqslant 1.5\)).
M1: Form equation of motion involving a difference of 2 tensions. Must be dimensionally correct. Need all terms. Condone incorrect lengths. Condone sign errors.
A1: Unsimplified equation with at most one error
A1: Correct unsimplified equation
A1*: All correct and justification of SHM – comment required
| Scheme | Marks | AO |
|---|---|---|
| \(a = 0.5,\quad \omega = 15\) or their \(\omega\) | B1ft | 1.1b |
| Max speed \(= a\omega = 7.5\ (\text{m s}^{-1})\) | B1ft | 1.2 |
| Max acceleration \(= a\omega^2 = 112.5\ (\text{m s}^{-2})\) | B1ft | 1.2 |
| (3) |
Notes
B1ft: Correct values seen or implied. \(a\) must be correct but follow their \(\omega\) from (a)
B1ft: Follow their \(a\), \(\omega\)
B1ft: Follow their \(a\), \(\omega\)
| Scheme | Marks | AO |
|---|---|---|
| \(x = 0.5\cos 15t \Rightarrow \dot{x} = -7.5\sin 15t\) | B1ft | 2.2a |
| \(|\dot{x}| = 7.5\sin 15t = 5 \Rightarrow \sin 15t = \dfrac{2}{3} \Rightarrow t = \ldots\) | M1 | 1.1b |
| \(t = 0.0486\ldots\) (s) | A1 | 1.1b |
| Complete strategy to find the required time | M1 | 3.1a |
| Speed \(\gt 5\) for \(\dfrac{2\pi}{15} - 4t = 0.22\) (s) | A1 | 1.1b |
| (5) | ||
| (13 marks) |
Notes
B1ft: Use correct equation for speed \((-a\omega\sin\omega t)\)
Follow their \(a\) and \(\omega\). Allow sin or cos form
Can also use correct expression for \(x\) combined with use of \(v^2 = \omega^2\left(a^2 - x^2\right)\)
M1: Find a relevant time e.g. time taken for \(0\ \text{m s}^{-1}\) to \(5\ \text{m s}^{-1}\)
A1: Seen or implied
M1: Complete strategy for the required time – e.g. find value for \(t\) when \(v = 5\) and use symmetry and periodic time
A1: 2 s.f. or better (0.22428…)