| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring standard techniques: eliminate t from parametric equations to get trajectory (substitute t = x/4 into y equation), then read off initial velocity components from coefficients. Both parts are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 4t\) and \(y = 6t - 5t^2\) | M1 | Use horizontal and vertical motion and attempt to eliminate \(t\) |
| \(y = \frac{6x}{4} - 5\left(\frac{x}{4}\right)^2 = 1.5x - \frac{5x^2}{16}\) or \(1.5x - 0.3125x^2\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan\theta = 1.5\) | M1 | Use the trajectory equation from the formula sheet |
| \(\theta = 56.3°\) | A1 | |
| \(V^2\cos^2 56.3 = 16\) | M1 | Again use the trajectory equation |
| \(V = 7.21 \text{ m s}^{-1}\) | A1 | |
| OR | ||
| \(V\cos\theta = 4\) and \(V\sin\theta = 6\) | M1 | Initial horizontal and vertical velocities |
| \(V^2\cos^2\theta + V^2\sin^2\theta = 4^2 + 6^2\) OR \(\tan\theta = 6/4\) | M1 | Use Pythagoras's theorem or trigonometry of a right angled triangle |
| \(V = 7.21 \text{ m s}^{-1}\) | A1 | |
| \(\theta = 56.3°\) | A1 | |
| 4 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 4t$ and $y = 6t - 5t^2$ | M1 | Use horizontal and vertical motion and attempt to eliminate $t$ |
| $y = \frac{6x}{4} - 5\left(\frac{x}{4}\right)^2 = 1.5x - \frac{5x^2}{16}$ or $1.5x - 0.3125x^2$ | A1 | |
| | **2** | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = 1.5$ | M1 | Use the trajectory equation from the formula sheet |
| $\theta = 56.3°$ | A1 | |
| $V^2\cos^2 56.3 = 16$ | M1 | Again use the trajectory equation |
| $V = 7.21 \text{ m s}^{-1}$ | A1 | |
| **OR** | | |
| $V\cos\theta = 4$ and $V\sin\theta = 6$ | M1 | Initial horizontal and vertical velocities |
| $V^2\cos^2\theta + V^2\sin^2\theta = 4^2 + 6^2$ OR $\tan\theta = 6/4$ | M1 | Use Pythagoras's theorem or trigonometry of a right angled triangle |
| $V = 7.21 \text{ m s}^{-1}$ | A1 | |
| $\theta = 56.3°$ | A1 | |
| | **4** | |3 A small ball is projected from a point $O$ on horizontal ground. At time $t \mathrm {~s}$ after projection the horizontal and vertically upwards displacements of the ball from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively, where $x = 4 t$ and $y = 6 t - 5 t ^ { 2 }$.\\
(i) Find the equation of the trajectory of the ball.\\
(ii) Hence or otherwise calculate the angle of projection of the ball and its initial speed.\\
\hfill \mbox{\textit{CAIE M2 2019 Q3 [6]}}