A2 June 2023 Q2
2. A particle of mass 2 kg is moving in a straight line on a smooth horizontal surface under the action of a horizontal force of magnitude \(F\) newtons.
At time \(t\) seconds \((t \gt 0)\),
- the particle is moving with speed \(v\ \text{m s}^{-1}\)
- \(F = 2 + v\)
The time taken for the speed of the particle to increase from \(5\ \text{m s}^{-1}\) to \(10\ \text{m s}^{-1}\) is \(T\) seconds.
The distance moved by the particle as its speed increases from \(5\ \text{m s}^{-1}\) to \(10\ \text{m s}^{-1}\) is \(D\) metres.
| Scheme | Marks | AO |
|---|---|---|
| \(2\dfrac{\mathrm{d}v}{\mathrm{d}t} = 2 + v\) | M1 | 2.1 |
| \(\Rightarrow \displaystyle\int \dfrac{2}{2 + v}\,\mathrm{d}v = \int 1\,\mathrm{d}t\) | DM1 | 1.1b |
| \(2\ln|2 + v| = t\ (+c)\) | A1 | 1.1b |
| \(T = 2\ln 12 - 2\ln 7 = 2\ln\dfrac{12}{7}\) * | A1* | 2.2a |
| (4) |
Notes
M1: Obtain a differential equation in \(v\) and \(t\)
Allow \(\pm\) for the acceleration
M1: Separate variables and integrate.
A1: Or equivalent unsimplified form. Allow without modulus signs.
A1*: Use limits to eliminate constant of integration, or in a definite integral, to obtain given answer from full and correct working.
Alternative 1 (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} - \dfrac{1}{2}v = 1\) | M1 | |
| Use integrating factor \(\mathrm{e}^{-\frac{1}{2}t}\) and integrate | DM1 | |
| Obtain \(v\mathrm{e}^{-\frac{1}{2}t} = -2\mathrm{e}^{-\frac{1}{2}t}\ (+C)\) | A1 | |
| \(\Rightarrow v = -2 + 7\mathrm{e}^{\frac{1}{2}t},\ \ T = 2\ln\dfrac{12}{7}\) * | A1* | |
| (4) |
Alternative 2 (a)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} - \dfrac{1}{2}v = 1\) | M1 | |
| Auxiliary equation is \(\lambda - \dfrac{1}{2} = 0\) so CF is \(v = A\mathrm{e}^{\frac{1}{2}t}\) and PI is \(v = -2\) | DM1 | |
| \(v = A\mathrm{e}^{\frac{1}{2}t} - 2\) | A1 | |
| \(t = 0, v = 5 \rightarrow A = 7 \Rightarrow 10 = 7\mathrm{e}^{\frac{1}{2}T} - 2 \Rightarrow T = 2\ln\dfrac{12}{7}\) | A1 | |
| (4) |
| Scheme | Marks | AO |
|---|---|---|
| \(v\dfrac{\mathrm{d}v}{\mathrm{d}x} = 1 + \dfrac{1}{2}v\) | M1 | 3.4 |
| \(\Rightarrow \displaystyle\int \dfrac{2v}{2 + v}\,\mathrm{d}v = \int 1\,\mathrm{d}x = \int 2 - \dfrac{4}{2 + v}\,\mathrm{d}v\) | DM1 | 1.1b |
| \(x\ (+A) = 2v - 4\ln|2 + v|\) | A1 | 1.1b |
| \(D = 2(10 - 5) - 4\ln 12 + 4\ln 7 = 10 - 4\ln\left(\dfrac{12}{7}\right)\) | A1 | 1.1b |
| (4) | ||
| (8 marks) |
Notes
M1: Obtain a differential equation in \(v\) and \(x\)
Allow \(\pm\) for the acceleration
M1: Separate variables and integrate.
If using integration by parts they need to complete the integration to score M1.
A1: Or equivalent unsimplified form. Allow without modulus signs.
A1: Use limits to eliminate constant of integration, or in a definite integral, to obtain exact distance from exact working. Any equivalent simplified form.
Alternative 1 (b)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2 + 7\mathrm{e}^{\frac{1}{2}t}\) | M1 | |
| Integrate | DM1 | |
| Obtain \(x = -2t + 14\mathrm{e}^{\frac{1}{2}t}\ (+A)\) | A1 | |
| \(D = -4\ln\left(\dfrac{12}{7}\right) + 14 \times \dfrac{12}{7} - 14 = 10 - 4\ln\left(\dfrac{12}{7}\right)\) | A1 | |
| (4) |
Alternative 2 (b)
| Scheme | Marks | AO |
|---|---|---|
| \(2\ddot{x} - \dot{x} = 2\) | M1 | |
| AE: \(2m^2 - m = 0\), CF: \(x = A\mathrm{e}^{\frac{1}{2}t} + B\), PI: \(x = -2t\) | M1 | |
| \(x = A\mathrm{e}^{\frac{1}{2}t} + B - 2t\) | A1 | |
| \(t = 0,\ x = 0,\ \dot{x} = 5 \Rightarrow x = 14\mathrm{e}^{\frac{1}{2}t} - 14 - 2t \Rightarrow D = 10 - 4\ln\left(\dfrac{12}{7}\right)\) | A1 | |
| (4) |