A2 June 2022 Paper 1 Q11
11 In this question use \(g\) as 10 m s−2
A smooth plane is inclined at \(30^\circ\) to the horizontal.
The fixed points \(A\) and \(B\) are 3.6 metres apart on the line of greatest slope of the plane, with \(A\) higher than \(B\)
A particle \(P\) of mass 0.32 kg is attached to one end of each of two light elastic strings.
The other ends of these strings are attached to the points \(A\) and \(B\) respectively.
The particle \(P\) moves on a straight line that passes through \(A\) and \(B\)
The natural length of the string \(AP\) is 1.4 metres.
When the extension of the string \(AP\) is \(e_A\) metres, the tension in the string \(AP\) is \(7e_A\) newtons.
The natural length of the string \(BP\) is 1 metre.
When the extension of the string \(BP\) is \(e_B\) metres, the tension in the string \(BP\) is \(9e_B\) newtons.
The particle \(P\) is held at the point between \(A\) and \(B\) which is 0.2 metres from its equilibrium position and lower than its equilibrium position.
The particle \(P\) is then released from rest.
At time \(t\) seconds after \(P\) is released, its displacement towards \(B\) from its equilibrium position is \(x\) metres.
Fully justify your answer. [5 marks]
During the motion the oil causes a resistive force of \(kv\) newtons to act on the particle, where \(v\) m s−1 is the speed of the particle.
The oil causes critical damping to occur.
| Scheme | Marks | AO |
|---|---|---|
| Forms equilibrium force equation with three correct terms. Condone sign errors | B1 | 3.1b |
| Forms at least one correct expression in \(x\) for the tension ie \(9(e_B - x)\) or \(7(e_A + x)\) Condone their incorrect \(e_A\) or \(e_B\) | B1F | 1.1a |
| Forms general force equation with four terms (with at least two terms correct). Condone “a” for \(\ddot{x}\) Condone sign errors on the terms Condone their incorrect \(e_A\) or \(e_B\) | M1 | 3.1b |
| Forms correct force equation. Can be in terms of \(e_A\) & \(e_B\) Condone “a” for \(\ddot{x}\) Condone their incorrect \(e_A\) or \(e_B\) | A1F | 1.1b |
| Constructs a rigorous argument to show the required result | R1 | 2.1 |
| (5) |
Typical solution
In equilibrium position
\[7e_A = 9e_B + 0.32g\sin 30\]\[1.2 = e_A + e_B\]\[e_A = 0.775,\quad e_B = 0.425\]After release
\[9(e_B - x) + 0.32g\sin 30 - 7(e_A + x) = 0.32\ddot{x}\]\[9(0.425 - x) + 0.32g\sin 30 - 7(0.775 + x) = 0.32\ddot{x}\]\[3.825 - 9x + 1.6 - 5.425 - 7x = 0.32\ddot{x}\]\[-16x = 0.32\ddot{x}\]\[\ddot{x} + 50x = 0\]as required
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains fully correct 2nd order DE or auxiliary equation, any correct form. Condone \(-k\) Allow \(m\) instead of 0.32 | B1 | 2.2a |
| Sets up \(b^2 - 4ac = 0\) from their 2nd order DE or Auxiliary Equation | M1 | 1.2 |
| Obtains correct value of \(k\) from correct working | R1 | 2.1 |
| (3) | ||
| (ii) Obtains correct solution from their three term Auxiliary Equation | M1 | 3.1a |
| Obtains correct solution of their three term differential equation | A1F | 1.1b |
| Uses \(x = 0.2\) when \(t = 0\) to obtain correct \(A\) | B1 | 3.3 |
| Sets their correct \(\dot{x} = 0\) when \(t = 0\) | M1 | 3.3 |
| Obtains their correct \(B\) Ft from \(\lambda^2 + \dfrac{16\sqrt{2}}{5}\lambda + 50 = 0\) | A1F | 1.1b |
| Completes a rigorous argument to obtain the correct result | R1 | 2.1 |
| (6) | ||
| (iii) Forms an equation to find the value of \(t\) at the maximum speed Must start from a damped harmonic model | M1 | 3.1a |
| Obtains a correct equation to find the value of \(t\) at the maximum speed | A1F | 1.1b |
| Obtains their correct \(t\), any form | A1F | 1.1b |
| Uses their value of \(t\) to obtain the velocity Must start from a damped harmonic model | M1 | 3.4 |
| Obtains correct max speed, to at least 1sf or exact Condone missing units (−0.5 m s−1 = A0) | A1 | 3.2a |
| (5) | ||
| (19 marks) |
Typical solution
(i)
\[9(0.425 - x) + 0.32g\sin 30 - 7(0.775 + x) - k\dot{x} = 0.32\ddot{x}\]\[0.32\ddot{x} + k\dot{x} + 16x = 0\]\[0.32\lambda^2 + k\lambda + 16 = 0\]Critical Damping so:
\[b^2 - 4ac = 0\]\[k^2 - 4 \times 0.32 \times 16 = 0\]\[k^2 = \frac{512}{25}\]\[k = \frac{16\sqrt{2}}{5}\](ii)
\[0.32\lambda^2 + \frac{16\sqrt{2}}{5}\lambda + 16 = 0\]\[\lambda = -5\sqrt{2} \text{ twice}\]\[x = A\mathrm{e}^{-5\sqrt{2}\,t} + Bt\mathrm{e}^{-5\sqrt{2}\,t}\]\[x = 0.2,\ t = 0 \Rightarrow A = 0.2\]\[\dot{x} = -5\sqrt{2}A\mathrm{e}^{-5\sqrt{2}\,t} + B\mathrm{e}^{-5\sqrt{2}\,t} - 5\sqrt{2}Bt\mathrm{e}^{-5\sqrt{2}\,t}\]\[\dot{x} = 0,\ t = 0 \Rightarrow B = \sqrt{2}\]\[x = 0.2\mathrm{e}^{-5\sqrt{2}\,t} + \sqrt{2}t\mathrm{e}^{-5\sqrt{2}\,t}\](iii)
\[\begin{aligned}\dot{x} &= -\sqrt{2}\mathrm{e}^{-5\sqrt{2}\,t} + \sqrt{2}\mathrm{e}^{-5\sqrt{2}\,t} - 10t\mathrm{e}^{-5\sqrt{2}\,t} \\ &= -10t\mathrm{e}^{-5\sqrt{2}\,t}\end{aligned}\]\[\ddot{x} = -10\mathrm{e}^{-5\sqrt{2}\,t} + 50\sqrt{2}t\mathrm{e}^{-5\sqrt{2}\,t}\]At Max \(\dot{x}\)
\[\ddot{x} = 0 = -10\mathrm{e}^{-5\sqrt{2}\,t} + 50\sqrt{2}t\mathrm{e}^{-5\sqrt{2}\,t}\]\[5\sqrt{2}t = 1\]\[t = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}\]\[\dot{x} = -\sqrt{2}\mathrm{e}^{-1} = -0.52026\ldots\]Max Speed = 0.5 m s−1