| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard M2 projectile question requiring application of trajectory equation and kinematic formulas with given angle and point coordinates. While it involves multiple parts and some algebraic manipulation, the methods are routine and well-practiced for M2 students with no novel problem-solving required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Substitutes \(\theta = 30°\), \(x = 10\) and \(y = 2\) into the correct general equation for the trajectory \([2 = 10 \tan 30° - \frac{100g}{2V^2 \cos^2 30°}]\) or eliminates \(T\) from \(10 = \frac{VT\sqrt{3}}{2}\), \(2 = \frac{VT}{2} - 5T^2\) (direct & correct equation) | B1 | |
| Transposes to obtain a numerical expression for \(V^2\) (or \(\frac{1000}{2(0.75)(\frac{10}{\sqrt{3}} - 2)}\)) \(( = 176.6705)\) | M1 | |
| Obtains \(V = 13.3\) (13.29175) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Substitutes for \(V\) in \(10 = \frac{VT\sqrt{3}}{2}\) or \(2 = \frac{VT}{2} - 5T^2\) and solves for \(T\) | M1 | |
| Obtains \(T' = 0.869\) (0.868735) (If vertical motion is considered, the A mark is awarded only if verification that the value of \(T\) found corresponds to (10, 20) rather than (5.3, 2) takes place) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(\tan \alpha = \pm \frac{v}{x}\) or \(\tan \alpha = \pm \frac{dy}{dx}\) (Allow \(F\) scale \(\theta\), \(\frac{v_y}{v_x}\)) | M1 | |
| Obtains \(\dot{x} = 13.3\sqrt{3}/2\) [13.3 \(\cos 3°S\) or \(\frac{dy}{dx} = \tan 30° - \frac{gx}{(176.67)(0.75)}\) \([0.57735 - 0.07547x]\) | B1 ft | |
| or \(\frac{dy}{dx} = \tan 30° - \frac{10g}{(176.67)(0.75)}\) \([0.57735 - 0.7547]\) | B1 ft | |
| Obtains angle as \(\approx 59°\) (169.9432) or \((\underline{\pm} 10.1°\) (10.0568) | A1 | 4 marks |
**(i)**
| Substitutes $\theta = 30°$, $x = 10$ and $y = 2$ into the correct general equation for the trajectory $[2 = 10 \tan 30° - \frac{100g}{2V^2 \cos^2 30°}]$ or eliminates $T$ from $10 = \frac{VT\sqrt{3}}{2}$, $2 = \frac{VT}{2} - 5T^2$ (direct & correct equation) | B1 |
| Transposes to obtain a numerical expression for $V^2$ (or $\frac{1000}{2(0.75)(\frac{10}{\sqrt{3}} - 2)}$) $( = 176.6705)$ | M1 |
| Obtains $V = 13.3$ (13.29175) | A1 | 3 marks |
**(ii)**
| Substitutes for $V$ in $10 = \frac{VT\sqrt{3}}{2}$ or $2 = \frac{VT}{2} - 5T^2$ and solves for $T$ | M1 |
| Obtains $T' = 0.869$ (0.868735) (If vertical motion is considered, the A mark is awarded only if verification that the value of $T$ found corresponds to (10, 20) rather than (5.3, 2) takes place) | A1 | 2 marks |
**(iii)**
| Uses $\tan \alpha = \pm \frac{v}{x}$ or $\tan \alpha = \pm \frac{dy}{dx}$ (Allow $F$ scale $\theta$, $\frac{v_y}{v_x}$) | M1 |
| Obtains $\dot{x} = 13.3\sqrt{3}/2$ [13.3 $\cos 3°S$ or $\frac{dy}{dx} = \tan 30° - \frac{gx}{(176.67)(0.75)}$ $[0.57735 - 0.07547x]$ | B1 ft |
| or $\frac{dy}{dx} = \tan 30° - \frac{10g}{(176.67)(0.75)}$ $[0.57735 - 0.7547]$ | B1 ft |
| Obtains angle as $\approx 59°$ (169.9432) or $(\underline{\pm} 10.1°$ (10.0568) | A1 | 4 marks |7 A ball is projected from a point $O$ with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at an angle of $30 ^ { \circ }$ above the horizontal. At time $T$ s after projection, the ball passes through the point $A$, whose horizontal and vertically upward displacements from $O$ are 10 m and 2 m respectively.\\
(i) By using the equation of the trajectory, or otherwise, find the value of $V$.\\
(ii) Find the value of $T$.\\
(iii) Find the angle that the direction of motion of the ball at $A$ makes with the horizontal.
\hfill \mbox{\textit{CAIE M2 2002 Q7 [9]}}