June 2023 Paper 2 Q20
20 In this question use \(g = 9.8\) m s−2
Nell and her pet dog Maia are visiting the beach.
The beach surface can be assumed to be level and horizontal.
Nell and Maia are initially standing next to each other.
Nell throws a ball forward, from a height of 1.8 metres above the surface of the beach, at an angle of 60° above the horizontal with a speed of 14 m s−1
Exactly 0.2 seconds after the ball is thrown, Maia sets off from Nell and runs across the surface of the beach, in a straight line with a constant acceleration \(a\) m s−2
Maia catches the ball when it is 0.3 metres above ground level as shown in the diagram below.
Find \(a\) [7 marks]
| Scheme | Marks | AO |
|---|---|---|
| States or uses \(14\cos 60\) for the horizontal component. | B1 | 3.1b |
| States or uses \(14\sin 60\) for the vertical component. | B1 | 3.1b |
| Uses \(s = ut + \dfrac{1}{2}at^2\) with \(u\) = their vertical component of velocity, \(a = -g\) and \(s = \pm 1.5\) OE PI by \(t\) = AWFW [2.54, 2.60] | M1 | 3.3 |
| Obtains \(t\) = 2.592 AWFW [2.54, 2.60] Exact value is \(\dfrac{3\sqrt{10} + 5\sqrt{3}}{7}\) | A1 | 1.1b |
| Multiplies their \(t\) value by their horizontal component provided their \(t\) > 0.2 | M1 | 1.1b |
| Substitutes \(u\) = 0, their \(t\) – 0.2 into \(ut + \dfrac{1}{2}at^2\) to obtain an expression for the horizontal distance travelled by the dog. | M1 | 3.3 |
| Obtains \(6.3\) CAO | A1 | 3.2a |
| (7 marks) |
Typical solution
\[u_H = 14\cos 60 = 7\]\[u_V = 14\sin 60 = 7\sqrt{3}\]\[s = 7\sqrt{3}\,t - \frac{g}{2}t^2\]\[-1.5 = 7\sqrt{3}\,t - 4.9t^2\]\[4.9t^2 - 12.124t - 1.5 = 0\]\(t = 2.592\) seconds
\[7t = 18.144\]\[ut + \frac{1}{2}at^2 = 0.5a(2.392)^2\]\[2.860832a = 18.144\]\[a = 6.3\]