M4 June 2012 Q3
3. Two particles, of masses \(m\) and \(2m\), are connected to the ends of a long light inextensible string. The string passes over a small smooth fixed pulley and hangs vertically on either side. The particles are released from rest with the string taut. Each particle is subject to air resistance of magnitude \(kv^2\), where \(v\) is the speed of each particle after it has moved a distance \(x\) from rest and \(k\) is a positive constant.
| Scheme | Marks |
|---|---|
| \(2mg - T - kv^2 = 2ma\) | M1 A1 |
| \(T - mg - kv^2 = ma\) | M1 A1 |
| Adding, \(mg - 2kv^2 = 3ma\) | |
| \(\dfrac{2g}{3} - \dfrac{4kv^2}{3m} = 2v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | DM1 |
| \(\dfrac{\mathrm{d}(v^2)}{\mathrm{d}x} + \dfrac{4kv^2}{3m} = \dfrac{2g}{3}\ *\) | A1 |
| (6) |
Notes
M1 Equation of motion for particle of mass \(2m\)
A1 aef
M1 Equation of motion for particle of mass m
A1 aef
DM1 Eliminate \(T\), substitute for \(a\) and rearrange. Dependent on both previous M marks.
A1 Reach given answer correctly
| Scheme | Marks |
|---|---|
| \(IF = \mathrm{e}^{\int\frac{4k}{3m}dx} = \mathrm{e}^{\frac{4kx}{3m}}\) | B1 |
| \(v^2\mathrm{e}^{\frac{4kx}{3m}} = \dfrac{2g}{3}\displaystyle\int \mathrm{e}^{\frac{4kx}{3m}}\,\mathrm{d}x = \dfrac{mg}{2k}\mathrm{e}^{\frac{4kx}{3m}}\ (+C)\) | M1 A1 |
| \(v^2 = \dfrac{mg}{2k} + C\mathrm{e}^{\frac{-4kx}{3m}}\) | |
| \(x = 0,\ v = 0 \Rightarrow C = -\dfrac{mg}{2k}\) | M1 |
| \(v^2 = \dfrac{mg}{2k}\left(1 - \mathrm{e}^{\frac{-4kx}{3m}}\right)\) | A1 |
| (5) |
Notes
M1 Use integrating factor to obtain \(\dfrac{d}{dx}\left(v^2e^{\frac{4kx}{3m}}\right) = \dfrac{2g}{3}e^{\frac{4kx}{3m}}\) and integrate
M1 Use initial values to evaluate \(C\) or as limits in a definite integral and find an expression for \(v^2\).
A1 aef.
OR
| Separate variables: \(\displaystyle\int \frac{3m}{2mg - 4kv^2}\,\mathrm{d}v^2 = \int 1\,\mathrm{d}x\) | B1 |
| \(x = -\dfrac{3m}{4k}\ln\left|2mg - 4kv^2\right|\ (+C)\) | M1 A1 |
| \(x = -\dfrac{3m}{4k}\ln\left|\dfrac{2mg}{2mg - 4kv^2}\right|\) | M1 |
| \(v^2 = \dfrac{mg}{2k}\left(1 - \mathrm{e}^{\frac{-4kx}{3m}}\right)\) | A1 |
Printed in the notes column beside the OR method: CF \(v^2 = Ae^{-\frac{4k}{3m}x}\); PI \(v^2 = b \Rightarrow 0 + \dfrac{4k}{3m}b = \dfrac{2g}{3}\); GS \(v^2 = Ae^{-\frac{4k}{3m}x} + \dfrac{mg}{2k}\); \(x = 0,\ v = 0 \Rightarrow A = -\dfrac{mg}{2k}\); \(v^2 = \dfrac{mg}{2k}\left(1 - \mathrm{e}^{\frac{-4kx}{3m}}\right)\)
| Scheme | Marks |
|---|---|
| When \(x = 0,\ T = \dfrac{4mg}{3}\) | M1 A1 |
| As \(x \to \infty,\ T \to \dfrac{9mg}{6} = \dfrac{3mg}{2}\) | M1 A1 |
| Hence, \(\dfrac{4mg}{3} \leqslant T \lt \dfrac{3mg}{2}\). \(*\) | A1 |
| (5) | |
| (16 marks) |
Notes
M1 Substitute \(v = 0\) in the initial equations and solve for \(T\)
M1 For large \(x\), \(v^2 \to \dfrac{mg}{2k}\). Substitute in the initial equations and solve for \(T\)
A1 cwo – answer is given.