| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard projectiles question requiring routine application of kinematic equations to find x(t) and y(t), elimination of t to get the trajectory, and basic trigonometry for part (ii). The calculations are straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 30\cos 60t\) | B1 | Use horizontal motion |
| \(y = 30\sin 60t - \dfrac{gt^2}{2}\) | B1 | Use \(s = ut + \dfrac{gt^2}{2}\) vertically |
| \(y = \dfrac{30\sin 60x}{30\cos 60} - \dfrac{5x^2}{(30\cos 60)^2}\) | M1 | Attempt to eliminate \(t\) |
| \(y = 1.73x - 0.0222x^2\) or \(y = \sqrt{3}x - \dfrac{x^2}{45}\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = y\) or \(\tan 45 = \dfrac{y}{x}\) | M1 | |
| \(1 = 1.73 - 0.0222x\) or \(1 = \sqrt{3} - \dfrac{x}{45}\) | M1 | \(x\) common to all three terms |
| \(x = 32.9\) | A1 | |
| Total: 3 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 30\cos 60t$ | B1 | Use horizontal motion |
| $y = 30\sin 60t - \dfrac{gt^2}{2}$ | B1 | Use $s = ut + \dfrac{gt^2}{2}$ vertically |
| $y = \dfrac{30\sin 60x}{30\cos 60} - \dfrac{5x^2}{(30\cos 60)^2}$ | M1 | Attempt to eliminate $t$ |
| $y = 1.73x - 0.0222x^2$ or $y = \sqrt{3}x - \dfrac{x^2}{45}$ | A1 | |
| **Total: 4** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = y$ or $\tan 45 = \dfrac{y}{x}$ | M1 | |
| $1 = 1.73 - 0.0222x$ or $1 = \sqrt{3} - \dfrac{x}{45}$ | M1 | $x$ common to all three terms |
| $x = 32.9$ | A1 | |
| **Total: 3** | | |4 A small ball $B$ is projected with speed $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $60 ^ { \circ }$ above the horizontal from a point $O$. At time $t \mathrm {~s}$ after projection the horizontal and vertically upwards displacements of $B$ from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.\\
(i) Express $x$ and $y$ in terms of $t$ and hence find the equation of the trajectory of the ball.\\
(ii) Find the value of $x$ for which $O B$ makes an angle of $45 ^ { \circ }$ above the horizontal.\\
\hfill \mbox{\textit{CAIE M2 2019 Q4 [7]}}