M2 June 2014 (R) Q6
6.

A particle \(P\) is projected from a point \(A\) with speed 25 m s\(^{-1}\) at an angle of elevation \(\alpha\), where \(\sin\alpha = \dfrac{4}{5}\). The point \(A\) is 10 m vertically above the point \(O\) which is on horizontal ground, as shown in Figure 4. The particle \(P\) moves freely under gravity and reaches the ground at the point \(B\).
Calculate
The point \(C\) lies on the path of \(P\). The direction of motion of \(P\) at \(C\) is perpendicular to the direction of motion of \(P\) at \(A\).
| Scheme | Marks |
|---|---|
| \(0 = (25\sin\alpha)^2 - 2gs\) | M1 |
| \(s = 400 \div 19.6\ \ (20.4)\) | A1 |
| Height above ground \(= 10 + 400 \div 19.6 = 30\) or 30.4 m | A1 |
| (3) |
Notes
M1 A complete method using suvat to find \(s\)
A1 Correct expression in \(s\) only
A1 30 or 30.4 only
| Scheme | Marks |
|---|---|
| \(10 = -25 \times \dfrac{4}{5}t + \dfrac{1}{2} \times gt^2\) | M1 A1 |
| \(4.9t^2 - 20t - 10 = 0\ \ \ \ \ t = \dfrac{20 \pm \sqrt{400 + 4 \times 4.9 \times 10}}{2 \times 4.9}\) | DM1 |
| \(t = 4.531\ldots\) s | A1 |
| Horiz distance \(= 25\cos\alpha t\ (= 15t\ \text{m})\) | M1 |
| \(= 68\) m | A1 |
| (6) |
Notes
M1 A complete method using suvat to find the total time from \(A\) to \(B\). Condone sign slips.
A1 Correctly substituted equation in \(t\)
DM1 Dependent on the preceding M1. Solve for \(t\)
A1 68 or 68.0 only
| Scheme | Marks |
|---|---|
| At \(C\) horiz speed \(= 15\) m s\(^{-1}\) | |
| Vert speed \(= \dfrac{15}{\tan\alpha}\) | M1 |
| \(= 11.25\) | A1 |
| \(11.25 = -20 + gt\) | DM1 |
| \(t = \dfrac{20 + 11.25}{9.8} = 3.2\) or 3.19 | A1 |
| (4) | |
| (13 marks) |
Notes
M1 Use similar triangles, or equivalent, to find vertical speed at C
DM1 Use suvat to find time from \(A\) to \(C\). Dependent on the preceding M1
A1 3.2 or 3.19 only