A2 October 2021 Paper 2 Q8
8 A particle \(P\) of mass 2 kg can only move along the straight line segment \(OA\), where \(OA\) is on a rough horizontal surface. The particle is initially at rest at \(O\) and the distance \(OA\) is 0.9 m.
When the time is \(t\) seconds the displacement of \(P\) from \(O\) is \(x\) m and the velocity of \(P\) is \(v\) m s−1.
\(P\) is subject to a force of magnitude \(4\mathrm{e}^{-2t}\) N in the direction of \(A\) for any \(t \geqslant 0\). The resistance to the motion of \(P\) is modelled as being proportional to \(v\).
At the instant when \(t = \ln 2\), \(v = 0.5\) and the resultant force on \(P\) is 0 N.
| Scheme | Marks | AO |
|---|---|---|
| \(F = ma = 2\dfrac{\mathrm{d}v}{\mathrm{d}t} = 4\mathrm{e}^{-2t} - kv\) | M1 | 3.3 |
| \(t = \ln 2,\ v = 0.5,\ F = 0 \Rightarrow 0 = 1 - 0.5k\) | M1 | 2.2a |
| \(k = 2 \Rightarrow 2\dfrac{\mathrm{d}v}{\mathrm{d}t} = 4\mathrm{e}^{-2t} - 2v \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} + v = 2\mathrm{e}^{-2t}\) | A1 | 1.1 |
| [3] |
Notes
M1: (first) Use of NII with \(m\) and \(a\) replaced and with 2 forces, the given force and \(kv\)
\(F = ma\) can be implicit here
M1: (second) Use of given conditions to derive an equation in \(k\)
Can be done first
A1: AG. Complete argument including \(F = ma\)
| Scheme | Marks | AO |
|---|---|---|
| IF \(= \mathrm{e}^{\int 1\,\mathrm{d}t} = \mathrm{e}^t\) | *B1 | 1.1 |
| \(\mathrm{e}^t\dfrac{\mathrm{d}v}{\mathrm{d}t} + \mathrm{e}^t v = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^t v\right) = \mathrm{e}^t \times 2\mathrm{e}^{-2t}\) | *M1 | 1.1 |
| \(\displaystyle\mathrm{e}^t v = \int 2\mathrm{e}^{-t}\,\mathrm{d}t = -2\mathrm{e}^{-t} + c\) | A1 | 1.1 |
| \(t = 0,\ v = 0 \Rightarrow c = 2\) | dep*M1 | 3.4 |
| \(v = 2\mathrm{e}^{-t} - 2\mathrm{e}^{-2t}\) | A1 | 3.4 |
| [5] |
Notes
*B1: Or CF
*M1: Multiplying by IF and writing LHS as an exact derivative
Or subst correct PI into DE
A1: (first) “\(+\,c\)” required
Or GS \(v = A\mathrm{e}^{-t} - 2\mathrm{e}^{-2t}\)
dep*M1: Use of initial conditions to derive a value for \(c\)
Or using alternative boundary condition
| Scheme | Marks | AO |
|---|---|---|
| As \(t \to \infty\), \(v \to 0\) | M1 | 3.4 |
| So speed starts at 0 and ends at 0 (and is continuous and positive between) so must reach a maximum somewhere in \(t \gt 0\) | A1 | 2.4 |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(v\) is max when \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = 0\) so \(t = \ln 2\) | M1 | 2.2a |
| So \(v_{\max} = 0.5\) (given) (or \(v_{\max} = 2\mathrm{e}^{-\ln 2} - 2\mathrm{e}^{-2\ln 2} = 1 - \dfrac{2}{4} = \dfrac{1}{2}\)) | A1 | 3.4 |
| [2] |
Notes
M1: Deducing time when \(v\) is maximum
Or by finding expression for \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\) and solving \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(v = \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\mathrm{e}^{-t} - 2\mathrm{e}^{-2t} \Rightarrow x = -2\mathrm{e}^{-t} + \mathrm{e}^{-2t} + d\) | M1 | 3.3 |
| \(t = 0,\ x = 0 \Rightarrow 0 = -2 + 1 + d \Rightarrow d = 1\) | M1 | 3.3 |
| \(0.9 = -2\mathrm{e}^{-t} + \mathrm{e}^{-2t} + 1\) | M1 | 3.5a |
| \(\left(\mathrm{e}^{-t}\right)^2 - 2\mathrm{e}^{-t} + 0.1 = 0 \Rightarrow \mathrm{e}^{-t} = \dfrac{10 \pm 3\sqrt{10}}{10}\) \(\Rightarrow t = \ln\left(\dfrac{10}{10 - 3\sqrt{10}}\right) = 2.97\) (3 sf) | A1 | 2.3 |
| [4] |
Notes
M1: (first) Integrating to find expression for \(x\)
M1: (second) Using initial conditions to find value of (new) constant
Or definite integral with correct lower limit…
M1: (third) Recognising that the model is only valid when \(x\) lies between 0 and 0.9
…and upper limit
A1: Rejecting \(t = \ln\left(\dfrac{10}{10 + 3\sqrt{10}}\right) \lt 0\) (can be implicit)