M2 January 2009 Q6
6.
A cricket ball is hit from a point \(A\) with velocity of \((p\mathbf{i} + q\mathbf{j})\) m s\(^{-1}\), at an angle \(\alpha\) above the horizontal. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are respectively horizontal and vertically upwards. The point \(A\) is 0.9 m vertically above the point \(O\), which is on horizontal ground.
The ball takes 3 seconds to travel from \(A\) to \(B\), where \(B\) is on the ground and \(OB = 57.6\) m, as shown in Figure 3. By modelling the motion of the cricket ball as that of a particle moving freely under gravity,
(a) find the value of \(p\), (2)
(b) show that \(q = 14.4\), (3)
(c) find the initial speed of the cricket ball, (2)
(d) find the exact value of \(\tan\alpha\). (1)
(e) Find the length of time for which the cricket ball is at least 4 m above the ground. (6)
(f) State an additional physical factor which may be taken into account in a refinement of the above model to make it more realistic. (1)
| Scheme | Marks |
|---|---|
| Horizontal distance: \(57.6 = p \times 3\) | M1 |
| \(p = 19.2\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Use \(s = ut + \dfrac{1}{2}at^2\) for vertical displacement. | M1 |
| \(-0.9 = q \times 3 - \dfrac{1}{2}g \times 3^2\) | A1 |
| \(-0.9 = 3q - \dfrac{9g}{2} = 3q - 44.1\) | |
| \(q = \dfrac{43.2}{3} = 14.4\) *AG* | A1 cso |
| (3) |
| Scheme | Marks |
|---|---|
| initial speed \(\sqrt{p^2 + 14.4^2}\) (with their \(p\)) | M1 |
| \(= \sqrt{576} = 24\) (m s\(^{-1}\)) | A1 cao |
| (2) |
| Scheme | Marks |
|---|---|
| \(\tan\alpha = \dfrac{14.4}{p}\ \left(= \dfrac{3}{4}\right)\) (with their \(p\)) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| When the ball is 4 m above ground: | |
| \(3.1 = ut + \dfrac{1}{2}at^2\) used | M1 |
| \(3.1 = 14.4t - \dfrac{1}{2}gt^2\) o.e. \((4.9t^2 - 14.4t + 3.1 = 0)\) | A1 |
| \(\Rightarrow t = \dfrac{14.4 \pm \sqrt{(14.4)^2 - 4(4.9)(3.1)}}{2(4.9)}\) seen or implied | M1 |
| \(t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8} = 0.23389\ldots\) or \(2.70488\ldots\) awrt 0.23 and 2.7 | A1 |
| duration \(= 2.70488\ldots - 0.23389\ldots\) | M1 |
| \(= 2.47\) or 2.5 (seconds) | A1 |
| (6) |
Notes
(Corrected from the printed mark scheme: the smaller root is printed as 0.023389…)
or 6 (e)
| M1A1M1 as above | |
| \(t = \dfrac{14.4 \pm \sqrt{146.6}}{9.8}\) | A1 |
| Duration \(\ 2 \times \dfrac{\sqrt{146.6}}{9.8}\) o.e. | M1 |
| \(= 2.47\) or 2.5 (seconds) | A1 |
(6)
| Scheme | Marks |
|---|---|
| Eg. : Variable ‘\(g\)’, Air resistance, Speed of wind, Swing of ball, The ball is not a particle. | B1 |
| (1) | |
| (15 marks) |