A2 October 2020 Q3
3. A particle \(P\) of mass 0.5 kg is moving along the positive \(x\)-axis in the direction of \(x\) increasing. At time \(t\) seconds \((t \geqslant 0)\), \(P\) is \(x\) metres from the origin \(O\) and the speed of \(P\) is \(v\ \text{m s}^{-1}\). The resultant force acting on \(P\) is directed towards \(O\) and has magnitude \(kv^2\) N, where \(k\) is a positive constant.
When \(x = 1\), \(v = 4\) and when \(x = 2\), \(v = 2\)
The time taken for the speed of \(P\) to decrease from \(4\ \text{m s}^{-1}\) to \(2\ \text{m s}^{-1}\) is \(T\) seconds.
| Scheme | Marks | AO |
|---|---|---|
| Form differential equation: \(0.5a = 0.5v\dfrac{\mathrm{d}v}{\mathrm{d}x} = -kv^2\) | M1 | 2.5 |
| \(\Rightarrow \displaystyle\int \dfrac{1}{2v}\,\mathrm{d}v = \int -k\,\mathrm{d}x\) | M1 | 2.1 |
| \(\dfrac{1}{2}\ln v = -kx + C\) | A1 | 1.1b |
| \(x = 1,\ v = 4 \qquad \tfrac{1}{2}\ln 4 = -k + C\) \(x = 2,\ v = 2 \qquad \tfrac{1}{2}\ln 2 = -2k + C\) | M1 | 3.1a |
| \(\Rightarrow k = \dfrac{1}{2}(\ln 4 - \ln 2) = \dfrac{1}{2}\ln 2,\quad C = \dfrac{1}{2}\ln 8:\quad \ln v = -x\ln 2 + \ln 8\) | A1 | 1.1b |
| \(\ln v = x\ln\dfrac{1}{2} + \ln 8,\quad v = 8\times\left(\dfrac{1}{2}\right)^x\) | A1 | 2.2a |
| \(\left(a = 8,\ b = \dfrac{1}{2}\right)\) | ||
| (6) |
Notes
M1: Form differential equation in \(v\) and \(x\). Condone sign error
M1: Separate and integrate to form equation in \(v\) and \(x\). Condone missing constant of integration
A1: Any equivalent form. Condone missing constant of integration.
M1: Complete strategy to use the differential equation and boundary conditions to find \(v\)
A1: Correct expression in \(v\) and \(x\) in any form. Accept \(\ln v = -0.693..x + 2.079\ldots\)
A1: Expression in the required form. Do not need to see a separate statement of the values of \(a\) and \(b\).
If mass is omitted from the differential equation can score M0M1A1M1A1A0
| Scheme | Marks | AO |
|---|---|---|
| \(0.5\dfrac{\mathrm{d}v}{\mathrm{d}t} = -kv^2\) (follow their \(k\)) | M1 | 2.5 |
| \(\displaystyle\int \dfrac{1}{v^2}\,\mathrm{d}v = \int -\ln 2\,\mathrm{d}t \qquad \Rightarrow -\dfrac{1}{v} + C' = -t\ln 2\) | M1 | 2.1 |
| \(\Rightarrow \left[-\dfrac{1}{v}\right]_4^2 = \left[-t\ln 2\right]_0^T\) | M1 | 1.1b |
| \(-\dfrac{1}{2} + \dfrac{1}{4} = -T\ln 2,\qquad T = \dfrac{1}{4\ln 2}\) * | A1* | 2.2a |
| (4) | ||
| (10 marks) |
Notes
M1: Differential equation in \(v\) and \(t\) (in \(x\) and \(t\) for alternative solution)
M1: Separate and integrate
M1: Use limits on a definite integral or to find value of \(C'\)
A1*: Obtain given result from correct working
If mass is omitted from the differential equation can score M0M1M1A0
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| \(v = \dfrac{8}{2^x} = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) (follow their \(v\)) | M1 | 2.5 |
| \(\displaystyle\int 2^x\,\mathrm{d}x = \int 8\,\mathrm{d}t \qquad \Rightarrow \dfrac{2^x}{\ln 2} = 8t + C'\) | M1 | 2.1 |
| \(\left[\dfrac{2^x}{\ln 2}\right]_1^2 = \left[8t\right]_0^T\) | M1 | 1.1b |
| \(\dfrac{1}{\ln 2}(4 - 2) = 8T,\qquad T = \dfrac{1}{4\ln 2}\) * | A1* | 2.2a |
| (4) |