A2 June 2019 Q2
2. A particle, \(P\), of mass 0.4 kg is moving along the positive \(x\)-axis, in the positive \(x\) direction under the action of a single force. At time \(t\) seconds, \(t \gt 0\), \(P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v\ \text{m s}^{-1}\). The force is acting in the direction of \(x\) increasing and has magnitude \(\dfrac{k}{v}\) newtons, where \(k\) is a constant.
At \(x = 3\), \(v = 2\) and at \(x = 6\), \(v = 2.5\)
The time taken for the speed of \(P\) to increase from \(2\ \text{m s}^{-1}\) to \(2.5\ \text{m s}^{-1}\) is \(T\) seconds.
| Scheme | Marks | AO |
|---|---|---|
| Strategy to find \(v^3\) in terms of \(x\) | M1 | 3.1a |
| Differential equation in \(v\) and \(x\): \(\ 0.4v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{k}{v}\) | M1 | 2.1 |
| \(\Rightarrow 0.4v^2\dfrac{\mathrm{d}v}{\mathrm{d}x} = k,\qquad \dfrac{0.4}{3}v^3 = kx + C\) | A1 | 1.1b |
| \(\begin{aligned}&x = 3,\ v = 2 &&\dfrac{3.2}{3} = 3k + C\\[4pt] &x = 6,\ v = 2.5 &&\dfrac{25}{12} = 6k + C\end{aligned}\) | M1 | 2.1 |
| \(\Rightarrow 3k = \dfrac{25}{12} - \dfrac{3.2}{3},\quad k = \dfrac{61}{180},\quad C = \dfrac{1}{20}\) | A1 | 1.1b |
| \(v^3 = \dfrac{3}{0.4}\left(\dfrac{61x}{180} + \dfrac{1}{20}\right) = \dfrac{61x + 9}{24}\) * | A1* | 2.2a |
| (6) |
Notes
M1: Complete strategy e.g. use of \(F = ma\) with appropriate form for \(a\), and use boundary conditions to confirm given result
M1: Separate variables and integrate. Usual rules for integration. Condone missing \(C\)
A1: Correct integration. Accept equivalent forms. Condone missing \(C\).
M1: Use boundary conditions to form 2 equations in 2 unknowns and solve for \(k\) or \(C\)
A1: Obtain correct values for the constants
A1*: Obtain given answer from correct working
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{5k}{2v} = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \left(\dfrac{61}{72v}\right)\) | M1 | 2.5 |
| \(\displaystyle\int 2v\,\mathrm{d}v = \int 5k\,\mathrm{d}t \quad\Rightarrow\quad v^2 = 5kt + C' \qquad \left(36v^2 = 61t + C'\right)\) | M1 | 2.1 |
| \(\left[v^2\right]_2^{2.5} = \left[5kt\right]_0^T\) \(\left(61T = 36\left(2.5^2 - 2^2\right)\right)\) | M1 | 1.1b |
| \(T = \dfrac{180}{61}\left(\dfrac{9}{20}\right) = \dfrac{81}{61}\) * | A1* | 2.2a |
| (4) | ||
| (10 marks) |
Notes
M1: Select correct form for derivative and form a correct differential equation in \(v\) and \(t\) – follow their \(k\)
M1: Separate and integrate. Condone with no \(+C'\) – follow their \(k\)
M1: Evaluate definite integral of the form \(pv^2 = qt + C'\) or use limits to find value of constant of integration – follow their \(k\)
A1*: Obtain given answer from correct working
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \sqrt[3]{\dfrac{61x + 9}{24}}\) | M1 | 2.5 |
| \(\displaystyle\int (61x + 9)^{-\frac{1}{3}}\,\mathrm{d}x = \int \dfrac{1}{\sqrt[3]{24}}\,\mathrm{d}t,\qquad \dfrac{3}{2\times 61}(61x + 9)^{\frac{2}{3}} = \dfrac{t}{\sqrt[3]{24}} + C''\) | M1 | 2.1 |
| \(T = 2\times\sqrt[3]{3}\times\dfrac{3}{2\times 61}\left(375^{\frac{2}{3}} - 192^{\frac{2}{3}}\right) = \dfrac{3\times\sqrt[3]{3}}{61}\left(\left(5\sqrt[3]{3}\right)^2 - \left(4\sqrt[3]{3}\right)^2\right)\) | M1 | 1.1b |
| \(T = \dfrac{9}{61}(25 - 16) = \dfrac{81}{61}\) * | A1* | 2.2a |
| (4) |
M1: Select correct form for derivative and form a correct differential equation in \(x\) and \(t\)
M1: Separate and integrate. Condone with no \(+C''\)
M1: Evaluate definite integral of the form \(pt = q(61x + 9)^{\frac{2}{3}} + C''\) or use limits to find value of constant of integration
A1*: Obtain given answer from correct working
NB: Both parts have given answers, so check very carefully