M2 January 2005 Q4
4. A particle \(P\) of mass 0.4 kg is moving under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds, the velocity of \(P\), \(\mathbf{v}\) m s\(^{-1}\), is given by
\[\mathbf{v} = (6t + 4)\mathbf{i} + (t^2 + 3t)\mathbf{j}.\]When \(t = 0\), \(P\) is at the point with position vector \((-3\mathbf{i} + 4\mathbf{j})\) m. When \(t = 4\), \(P\) is at the point \(S\).
(a) Calculate the magnitude of \(\mathbf{F}\) when \(t = 4\). (4)
(b) Calculate the distance \(OS\). (5)
| Scheme | Marks |
|---|---|
| \(\ddot{\mathbf{r}} = 6\mathbf{i} + (2t + 3)\mathbf{j}\) | B1 |
| \(\mathbf{F} = 0.4(6\mathbf{i} + 11\mathbf{j})\) 0.4×something obtained by differentiation, with \(t = 4\) | M1 |
| \(|\mathbf{F}| = \sqrt{(2.4^2 + 4.4^2)}\) modulus of a vector | M1 |
| \(\approx 5.0\) accept more accurate answers | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = (3t^2 + 4t)\mathbf{i} + \left(\tfrac{1}{3}t^3 + \tfrac{3}{2}t^2\right)\mathbf{j}\ (+\mathbf{C})\) | M1 |
| Using boundary values, \(\mathbf{r} = (3t^2 + 4t - 3)\mathbf{i} + \left(\tfrac{1}{3}t^3 + \tfrac{3}{2}t^2 + 4\right)\mathbf{j}\) | A1 |
| \(t = 4\), \(\mathbf{r} = 61\mathbf{i} + 49\tfrac{1}{3}\mathbf{j}\) | A1 |
| \(OS = \sqrt{\left(61^2 + 49\tfrac{1}{3}^2\right)} \approx 78\) (m) accept more accurate answers | M1 A1 |
| (5) | |
| (9 marks) |