M2 January 2011 Q6
6. [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in a vertical plane, \(\mathbf{i}\) being horizontal and \(\mathbf{j}\) being vertically upwards.]

At time \(t = 0\), a particle \(P\) is projected from the point \(A\) which has position vector \(10\mathbf{j}\) metres with respect to a fixed origin \(O\) at ground level. The ground is horizontal. The velocity of projection of \(P\) is \((3\mathbf{i} + 5\mathbf{j})\) m s\(^{-1}\), as shown in Figure 3. The particle moves freely under gravity and reaches the ground after \(T\) seconds.
(a) For \(0 \leqslant t \leqslant T\), show that, with respect to \(O\), the position vector, \(\mathbf{r}\) metres, of \(P\) at time \(t\) seconds is given by \[\mathbf{r} = 3t\mathbf{i} + (10 + 5t - 4.9t^2)\mathbf{j}\] (3)
(b) Find the value of \(T\). (3)
(c) Find the velocity of \(P\) at time \(t\) seconds \((0 \leqslant t \leqslant T)\). (2)
When \(P\) is at the point \(B\), the direction of motion of \(P\) is 45\(^\circ\) below the horizontal.
(d) Find the time taken for \(P\) to move from \(A\) to \(B\). (2)
(e) Find the speed of \(P\) as it passes through \(B\). (2)
| Scheme | Marks |
|---|---|
| Using \(\ s = ut + \dfrac{1}{2}at^2\) Method must be clear | M1 |
| \(\mathbf{r} = (3t)\mathbf{i} + \left(10 + 5t - 4.9t^2\right)\mathbf{j}\) Answer given | A1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathbf{j}\) component \(= 0:\ 10 + 5t - 4.9t^2\) | M1 |
| quadratic formula: \(\ t = \dfrac{5 \pm \sqrt{25 + 196}}{9.8} = \dfrac{5 \pm \sqrt{221}}{9.8}\) | DM1 |
| \(T = 2.03\ (\text{s}),\ 2.0\ (\text{s})\) positive solution only. | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Differentiating the position vector (or working from first principles) | M1 |
| \(\mathbf{v} = 3\mathbf{i} + (5 - 9.8t)\mathbf{j}\ \ \left(\text{ms}^{-1}\right)\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| At \(B\) the \(\mathbf{j}\) component of the velocity is the negative of the \(\mathbf{i}\) component: \(\ 5 - 9.8t = -3,\ \ 8 = 9.8t\), | M1 |
| \(t = 0.82\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{v} = 3\mathbf{i} - 3\mathbf{j}\), speed \(= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24\ \left(\text{m s}^{-1}\right)\) | M1A1 |
| (2) | |
| (12 marks) |