M4 June 2005 Q4
4. A lorry of mass \(M\) is moving along a straight horizontal road. The engine produces a constant driving force of magnitude \(F\). The total resistance to motion is modelled as having magnitude \(kv^2\), where \(k\) is a constant, and \(v\) is the speed of the lorry.
Given the lorry moves with constant speed \(V\),
(a) show that \(V = \sqrt{\dfrac{F}{k}}\). (2)
Given instead that the lorry starts from rest,
(b) show that the distance travelled by the lorry in attaining a speed of \(\tfrac{1}{2}V\) is \[\frac{M}{2k}\ln\left(\frac{4}{3}\right).\] (9)
| Scheme | Marks |
|---|---|
| For constant speed, \(\ F - kV^2 = 0\) | M1 |
| \(\Rightarrow V = \sqrt{F/k}\ \ *\) | A1 |
| (2) |
Notes
The published mark scheme for this paper is handwritten.
| Scheme | Marks |
|---|---|
| \(F - kv^2 = Ma\) | M1 A1 |
| \(\Rightarrow F - kv^2 = Mv\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | M1 |
| \(\displaystyle\int \mathrm{d}x = M\int \frac{v}{F - kv^2}\,\mathrm{d}v\) | M1 |
| \(x = -\dfrac{M}{2k}\ln\left(F - kv^2\right)\ (+c)\) | A1 |
| \(x = 0,\ v = 0 \Rightarrow c = \dfrac{M}{2k}\ln F\) | M1 A1 |
| \(x = \dfrac{M}{2k}\left\{\ln F - \ln\left(F - kv^2\right)\right\}\) | |
| \(X = \dfrac{M}{2k}\ln\left(\dfrac{F}{F - k.\frac{F}{4k}}\right)\) | M1 |
| \(= \dfrac{M}{2k}\ln\dfrac{4}{3}\ \ *\) | A1 |
| (9) | |
| (11 marks) |