| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring students to match the given trajectory equation y = ax - bx² with the standard form to extract V and θ, then apply routine formulas for range and maximum height. While it involves multiple parts and algebraic manipulation, the techniques are well-practiced and follow directly from standard M2 content with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) | M1 | For using \(\tan\theta = 0.75\) |
| \(\theta = 36.9\) | A1 | |
| \([10/(2 \times 0.8^2 V^2) = 0.02]\) | M1 | For using \(\dfrac{g}{2V^2\cos^2\theta} = 0.02\) |
| \(V = 19.8\) | A1 [4] | |
| (ii) \([x(0.75 - 0.02x) = 0]\) | M1 | For solving \(y = 0\) or using \(R = V^2\sin 2\theta/g\) |
| Distance is \(37.5\ \text{m}\) | A1 [2] | |
| (iii) \([y_{\max} = 0.75 \times 18.75 - 0.02 \times 18.75^2]\) | M1 | For using \(y\) is greatest when \(x = R/2\) or \(H = V^2\sin^2\theta/2g\) |
| Greatest height is \(7.03\ \text{m}\) | A1 [2] | Ft \(0.375R - 0.005R^2\) |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** | M1 | For using $\tan\theta = 0.75$ |
| $\theta = 36.9$ | A1 | |
| $[10/(2 \times 0.8^2 V^2) = 0.02]$ | M1 | For using $\dfrac{g}{2V^2\cos^2\theta} = 0.02$ |
| $V = 19.8$ | A1 [4] | |
| **(ii)** $[x(0.75 - 0.02x) = 0]$ | M1 | For solving $y = 0$ or using $R = V^2\sin 2\theta/g$ |
| Distance is $37.5\ \text{m}$ | A1 [2] | |
| **(iii)** $[y_{\max} = 0.75 \times 18.75 - 0.02 \times 18.75^2]$ | M1 | For using $y$ is greatest when $x = R/2$ or $H = V^2\sin^2\theta/2g$ |
| Greatest height is $7.03\ \text{m}$ | A1 [2] | Ft $0.375R - 0.005R^2$ |
---5 A small stone is projected from a point $O$ on horizontal ground with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta ^ { \circ }$ above the horizontal. Referred to horizontal and vertically upwards axes through $O$, the equation of the stone's trajectory is $y = 0.75 x - 0.02 x ^ { 2 }$, where $x$ and $y$ are in metres. Find\\
(i) the values of $\theta$ and $V$,\\
(ii) the distance from $O$ of the point where the stone hits the ground,\\
(iii) the greatest height reached by the stone.
\hfill \mbox{\textit{CAIE M2 2009 Q5 [8]}}