M2 January 2012 Q7
7. [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively.]
The point \(O\) is a fixed point on a horizontal plane. A ball is projected from \(O\) with velocity \((6\mathbf{i} + 12\mathbf{j})\) m s\(^{-1}\), and passes through the point \(A\) at time \(t\) seconds after projection. The point \(B\) is on the horizontal plane vertically below \(A\), as shown in Figure 3. It is given that \(OB = 2AB\).
Find
(a) the value of \(t\), (7)
(b) the speed, \(V\) m s\(^{-1}\), of the ball at the instant when it passes through \(A\). (5)
At another point \(C\) on the path the speed of the ball is also \(V\) m s\(^{-1}\).
(c) Find the time taken for the ball to travel from \(O\) to \(C\). (3)
| Scheme | Marks |
|---|---|
| \(\mathbf{i} \rightarrow\) distance \(= 6t\) | B1 |
| \(\mathbf{j} \uparrow\ \) distance \(= 12t - \dfrac{1}{2}gt^2\) | M1 A1 |
| At \(B\), \(\ 2\left(12t - \dfrac{1}{2}gt^2\right) = 6t\) | M1 A1 |
| \((24 - 6)t = gt^2\) | DM1 |
| \(18 = gt,\ t = \dfrac{18}{g}\ (= 1.84\text{s})\) | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(\mathbf{i} \rightarrow\) speed \(= 6\) | B1 |
| \(\mathbf{j} \uparrow\ \) velocity \(= 12 - gt = -6\) | M1 A1 |
| \(\therefore\) speed at \(A\) \(= \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}\ (= 8.49)\left(\text{ms}^{-1}\right)\) | M1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\uparrow\ \) speed \(= 12 - gt = +6\) | M1 A1 ft |
| \(t = \dfrac{6}{g}\ (= 0.61\text{s})\) | A1 |
| (3) | |
| (15 marks) |