A2 June 2024 Paper 1 Q10
10 A particle \(B\), of mass 3 kg, moves in a straight line and has velocity \(v\,\mathrm{m\,s^{-1}}\).
At time \(t\) seconds, where \(0 \leqslant t \lt \frac{1}{4}\pi\), a variable force of \(-(15\sin 4t + 6v\tan 2t)\) Newtons is applied to \(B\). There are no other forces acting on \(B\). Initially, when \(t = 0\), \(B\) has velocity \(4.5\,\mathrm{m\,s^{-1}}\).
The motion of \(B\) can be modelled by the differential equation \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + P(t)v = Q(t)\) where \(P(t)\) and \(Q(t)\) are functions of \(t\).
| Scheme | Marks | AO |
|---|---|---|
| \(F = ma\) so So \(-(15\sin 4t + 6v\tan 2t) = 3\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 | 3.3 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + 2v\tan 2t = -5\sin 4t \quad\) (so \(P(t) = 2\tan 2t\) and \(Q(t) = -5\sin 4t\)) | A1 | 1.1 |
| [2] |
Notes
M1: Using \(F = 3a\) and \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\) to form a differential equation - allow minor slips or sign errors but intention must be clear. Their \(F\) must be two terms only.
A1: The correct differential equation in the form \(\frac{\mathrm{d}v}{\mathrm{d}t} + P(t)v = Q(t)\) can imply this mark. \(P(t)\) and \(Q(t)\) do not need to be explicitly stated. ISW once correct form seen. If not written in the form, \(\frac{\mathrm{d}v}{\mathrm{d}t} + P(t)v = Q(t)\) then \(P(t)\) and \(Q(t)\) must be explicitly stated.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{I}(t) = \mathrm{e}^{\int 2\tan 2t\,\mathrm{d}t}\) | M1* | 1.1 |
| \(= \mathrm{e}^{-\ln(\cos 2t)} \quad \left(= \sec 2t\right)\) | M1 | 1.1 |
| \(\left(\frac{\mathrm{d}v}{\mathrm{d}t} + 2v\tan 2t\right) \times \mathrm{I}(t) = -5\sin 4t \times \mathrm{I}(t) \Rightarrow \frac{\mathrm{d}}{\mathrm{d}t}(v \times \mathrm{I}(t)) = -5\sin 4t \times \mathrm{I}(t)\) | ||
| \(v \times \sec 2t = -5\displaystyle\int (\sin 4t \times \sec 2t)\,\mathrm{d}t\) | M1dep* | 1.1 |
| \((v\sec 2t =) -10\displaystyle\int \frac{\sin 2t\cos 2t}{\cos 2t}\,\mathrm{d}t = -10\int \sin 2t\,\mathrm{d}t\) | M1 | 1.1 |
| \(v\sec 2t = 5\cos 2t \; (+c)\) | A1 | 1.1 |
| \(v(0) = 4.5 \Rightarrow c = -0.5\) | M1 | 3.3 |
| \(0 = 5\cos 2t - 0.5 \Rightarrow \cos 2t = 0.1\) | M1dep* | 3.4 |
| So, \(B\) stationary after 0.735 seconds. | A1 | 2.2a |
| [8] |
Notes
M1*: For \(\mathrm{I}(t) = \mathrm{e}^{\int P(t)\,\mathrm{d}t}\) for their \(P(t)\)
M1: For \(\mathrm{I}(t) = \mathrm{e}^{\pm k\ln(\cos 2t)}\) or \(\mathrm{e}^{\pm k\ln(\sec 2t)}\) or \(\pm k\cos 2t\) or \(\pm k\sec 2t\) or \(\pm a\cos^k 2t\) or \(\pm a\sec^k 2t\) for \(a, k \neq 0\).
M1dep*: For \(v \times \mathrm{I}(t) = k_1\displaystyle\int \sin 4t \times \mathrm{I}(t)\,\mathrm{d}t\) with their \(\mathrm{I}(t)\) (in any form) and \(k_1 \neq 0\).
M1: For simplifying \(RHS\) to \(k_2\displaystyle\int \sin 2t\,\mathrm{d}t\) for any \(k_2 \neq 0\) - dependent on all previous M marks.
A1: For correct general solution (any equivalent form). Condone lack of \(+c\).
M1: Using \(v(0) = 4.5\) to find constant term – dependent on first three M marks and an attempt at integration.
M1dep*: For a two-term equation of the form \(\cos 2t = k_3\) where \(|k_3| \lt 1\) and \(k_3 \neq 0\) - dependent on all previous M marks.
A1: Ignore if any other solution(s) found. Allow awrt 0.735 – for reference 0.7353144…
42.1 seconds (calculated in degrees) scores A0.