| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a straightforward projectile motion question requiring standard techniques: recognizing that horizontal velocity is constant (giving cos θ directly), then using standard SUVAT equations for maximum height and range. All steps are routine applications of well-practiced formulas with no problem-solving insight required, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For 'speed at greatest height = \(x\)' used | M1 | (\(40 = 50 \cos \theta\)) |
| \(\cos \theta = 0.8\) | A1 | 2 marks |
| (ii) For using \(y = 0\) at maximum height or \(y = 0\) at impact | M1 | |
| [At max height \(t = 50 \times 0.6/10\) or at impact \(t = 50 \times 0.6\) (\(\frac{2}{5} \times 10\))] | ||
| For substituting for \(t\) at max height (\(= 3\)) into | M1 | |
| \(H = 50t \sin \theta - \frac{1}{2}gt^2\) or \(H = \frac{50 \sin \theta + 0}{2} \cdot t\) | ||
| Greatest height is 45m | A1 | |
| For substituting \(t\) at max ht (\(= 3\)) into \(R = 2(50t \cos \theta)\) or \(t\) at impact (\(= 6\)) into \(R = (50t \cos \theta)\) or for using \(R = \frac{V^2\sin 2\theta}{g}\) (directly or by substituting \(y = 0\) in the trajectory equation) | M1 | |
| Distance is 240m | A1 | 5 marks |
| Answer | Marks |
|---|---|
| For using \(y = 0\) in \(y^2 = y_0^2 - 2gH\) or using the principle of conservation of energy (2 non-zero KE terms and a GPE term needed) or using the trajectory equation with \(x = R/2\) (if (b) is answered before (a)). | M1 |
| \(0 = (50 \times 0.6)^2 - 2 \times 10H\) or \(\frac{1}{2}m50^2 = \frac{1}{2}m40^2 + mgH\) or | A1 |
| \(H = 120(0.75) - 10(120)^2/[2(50)^2(0.8)^2]\) | |
| Greatest height is 45m | A1 |
| For using \(R = \frac{V^2\sin 2\theta}{g}\) (directly or by substituting \(y = 0\) in the trajectory equation) | M1 |
| Distance is 240m | A1 |
**(i)** For 'speed at greatest height = $x$' used | M1 | ($40 = 50 \cos \theta$)
$\cos \theta = 0.8$ | A1 | 2 marks
**(ii)** For using $y = 0$ at maximum height or $y = 0$ at impact | M1 |
[At max height $t = 50 \times 0.6/10$ or at impact $t = 50 \times 0.6$ ($\frac{2}{5} \times 10$)] | |
For substituting for $t$ at max height ($= 3$) into | M1 |
$H = 50t \sin \theta - \frac{1}{2}gt^2$ or $H = \frac{50 \sin \theta + 0}{2} \cdot t$ | |
Greatest height is 45m | A1 |
For substituting $t$ at max ht ($= 3$) into $R = 2(50t \cos \theta)$ or $t$ at impact ($= 6$) into $R = (50t \cos \theta)$ or for using $R = \frac{V^2\sin 2\theta}{g}$ (directly or by substituting $y = 0$ in the trajectory equation) | M1 |
Distance is 240m | A1 | 5 marks
**Alternative for methods not involving $t$ at max ht. or $t$ at impact:**
For using $y = 0$ in $y^2 = y_0^2 - 2gH$ or using the principle of conservation of energy (2 non-zero KE terms and a GPE term needed) or using the trajectory equation with $x = R/2$ (if (b) is answered before (a)). | M1 |
$0 = (50 \times 0.6)^2 - 2 \times 10H$ or $\frac{1}{2}m50^2 = \frac{1}{2}m40^2 + mgH$ or | A1 |
$H = 120(0.75) - 10(120)^2/[2(50)^2(0.8)^2]$ | |
Greatest height is 45m | A1 |
For using $R = \frac{V^2\sin 2\theta}{g}$ (directly or by substituting $y = 0$ in the trajectory equation) | M1 |
Distance is 240m | A1 |
---4 A particle is projected from a point $O$ on horizontal ground with speed $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ to the horizontal. Given that the speed of the particle when it is at its highest point is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$,\\
(i) show that $\cos \theta = 0.8$,\\
(ii) find, in either order,
\begin{enumerate}[label=(\alph*)]
\item the greatest height reached by the particle,
\item the distance from $O$ at which the particle hits the ground.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2004 Q4 [7]}}