| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.8 This question requires deriving a quadratic in tan θ from the trajectory equation (involving algebraic manipulation with g=10), solving it to find two angles of projection, then calculating range for each angle and finding their difference. It combines multiple mechanics concepts with non-trivial algebra, going beyond routine projectile problems but remains within standard M2 scope. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(15 = 40\tan\theta - \frac{g \cdot 40^2}{2 \times 40^2\cos^2\theta}\) | M1 | Substitutes in projectile equation |
| \(15 = 40\tan\theta - 5\sec^2\theta\) | M1 | Uses \(\sec^2\theta = 1 + \tan^2\theta\) |
| \(\tan^2\theta - 8\tan\theta + 4 = 0\) | A1 AG | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\theta = \tan^{-1}(4 \pm 2\sqrt{3})\) | M1 | Solves quadratic equation for \(\theta\) |
| \(\theta = 28.2°\) or \(82.4°\) | A1 | |
| \(R = \frac{40^2\sin(2 \times 28.2°)}{g}\) or \(R = \frac{40^2\sin(2 \times 82.4°)}{g}\) | M1 | Valid formula for one range; \(0 = R\tan 28.2° - \frac{gR^2}{2 \times 40^2\cos 28.2°}\) or \(0 = r\tan 82.4° - \frac{gr^2}{2 \times 40^2\cos 82.4°}\) |
| \(R = 133\) or \(R = 41.9\) (or 42.0) | A1 | Using exact angles. Allow \(\pm 0.2\) |
| Difference \(= 91.1\) m | A1 | |
| [5] |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $15 = 40\tan\theta - \frac{g \cdot 40^2}{2 \times 40^2\cos^2\theta}$ | M1 | Substitutes in projectile equation |
| $15 = 40\tan\theta - 5\sec^2\theta$ | M1 | Uses $\sec^2\theta = 1 + \tan^2\theta$ |
| $\tan^2\theta - 8\tan\theta + 4 = 0$ | A1 AG | |
| | **[3]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\theta = \tan^{-1}(4 \pm 2\sqrt{3})$ | M1 | Solves quadratic equation for $\theta$ |
| $\theta = 28.2°$ or $82.4°$ | A1 | |
| $R = \frac{40^2\sin(2 \times 28.2°)}{g}$ or $R = \frac{40^2\sin(2 \times 82.4°)}{g}$ | M1 | Valid formula for one range; $0 = R\tan 28.2° - \frac{gR^2}{2 \times 40^2\cos 28.2°}$ or $0 = r\tan 82.4° - \frac{gr^2}{2 \times 40^2\cos 82.4°}$ |
| $R = 133$ or $R = 41.9$ (or 42.0) | A1 | Using exact angles. Allow $\pm 0.2$ |
| Difference $= 91.1$ m | A1 | |
| | **[5]** | |
---3 Two particles $P$ and $Q$ are projected simultaneously with speed $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ on a horizontal plane. Both particles subsequently pass at different times through the point $A$ which has horizontal and vertically upward displacements from $O$ of 40 m and 15 m respectively.\\
(i) By considering the equation of the trajectory of a projectile, show that each angle of projection satisfies the equation $\tan ^ { 2 } \theta - 8 \tan \theta + 4 = 0$.\\
(ii) Calculate the distance between the points at which $P$ and $Q$ strike the plane.
\hfill \mbox{\textit{CAIE M2 2010 Q3 [8]}}