M2 January 2005 Q7
7.
A particle \(P\) is projected from a point \(A\) with speed 32 m s\(^{-1}\) at an angle of elevation \(\alpha\), where \(\sin\alpha = \tfrac{3}{5}\). The point \(O\) is on horizontal ground, with \(O\) vertically below \(A\) and \(OA = 20\) m. The particle \(P\) moves freely under gravity and passes through a point \(B\), which is 16 m above ground, before reaching the ground at the point \(C\), as shown in Figure 4.
Calculate
(a) the time of the flight from \(A\) to \(C\), (5)
(b) the distance \(OC\), (3)
(c) the speed of \(P\) at \(B\), (4)
(d) the angle that the velocity of \(P\) at \(B\) makes with the horizontal. (3)
| Scheme | Marks |
|---|---|
| \(\uparrow\) \(u_y = 32 \times \tfrac{3}{5}\ \ (= 19.2)\) | B1 |
| \(-20 = 19.2t - 4.9t^2\) −1 each error | M1 A2(1, 0) |
| \(t \approx 4.8\) or 4.77 (s) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\rightarrow\) \(u_x = 32 \times \tfrac{4}{5}\ \ (= 25.6)\) | B1 |
| \(d = 25.6 \times 4.77\ldots\) | M1 |
| \(\approx 120\) or 122 (m) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\uparrow\) \(v_y^2 = 19.2^2 + 2 \times 9.8 \times 4\ \ \left[v_y^2 = 447.04,\ v_y \approx 21.14\right]\) | M1 |
| \(V^2 = 447.04 + 25.6^2\) | M1 A1 |
| \(V = 33\) or 33.2 \((\text{m s}^{-1})\) | A1 |
| (4) |
Alternative for (c)
| \(\tfrac{1}{2}m(V^2 - 32^2) = mg \times 4\) | M1 A1 |
| \(V^2 = 1102.4\) | M1 |
| \(V = 33\) or 33.2 \((\text{m s}^{-1})\) | A1 (4) |
| Scheme | Marks |
|---|---|
| \(\tan\theta = \dfrac{21.14}{25.6}\) \(\left(\text{or } \cos\theta = \dfrac{25.6}{33.2}, \ldots\right)\) ft their components or resultant | M1 A1ft |
| \(\theta \approx 40^\circ\) or 39.6\(^\circ\) | A1 |
| (3) | |
| (15 marks) |
Notes
There is a maximum penalty of one mark per question for not rounding to appropriate accuracy.