M3 January 2013 Q3
3. A particle \(P\) of mass 0.6 kg is moving along the \(x\)-axis in the positive direction. At time \(t = 0\), \(P\) passes through the origin \(O\) with speed 15 m s\(^{-1}\). At time \(t\) seconds the distance \(OP\) is \(x\) metres, the speed of \(P\) is \(v\) m s\(^{-1}\) and the resultant force acting on \(P\) has magnitude \(\dfrac{12}{(t + 2)^2}\) newtons. The resultant force is directed towards \(O\).
(a) Show that \(v = 5\left(\dfrac{4}{t + 2} + 1\right)\). (5)
(b) Find the value of \(x\) when \(t = 5\) (5)
| Scheme | Marks |
|---|---|
| \(0.6a = -\dfrac{12}{(t + 2)^2}\) | M1 |
| \(0.6\displaystyle\int \mathrm{d}v = -\int \frac{12}{(t + 2)^2}\,\mathrm{d}t\) | |
| \(0.6v = \dfrac{12}{(t + 2)}\ \ \ (+c)\) | M1depA1 |
| \(t = 0\ \ v = 15\ \ \ \ 0.6 \times 15 = 6 + c \Rightarrow c = 3\) | M1dep |
| \(0.6v = \dfrac{12}{(t + 2)} + 3\ \ \ \ \ v = \dfrac{20}{(t + 2)} + 5 = 5\left(\dfrac{4}{t + 2} + 1\right)\) * | A1 |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 5\left(\dfrac{4}{t + 2} + 1\right)\) | M1 |
| \(x = \displaystyle\int 5\left(\frac{4}{t + 2} + 1\right)\mathrm{d}t\) | |
| \(x = 5\left(4\ln(t + 2) + t\right)\ \ (+c')\) | M1depA1 |
| \(t = 0,\ x = 0\ \ c' = -20\ln 2\) | |
| \(t = 5\ \ \ \ x = 5\left(4\ln 7 + 5\right) - 20\ln 2 = 50.05\ldots = 50.1\) or better | M1dep |
| or \(20\ln\left(\dfrac{7}{2}\right) + 25\) | A1 |