| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 This is a standard projectiles question requiring trajectory equation manipulation and use of a given trigonometric identity. Part (i) involves algebraic manipulation to form a quadratic in tan θ (routine for M2), part (ii) applies range formula with known angle, and part (iii) is a straightforward sketch. The question is slightly above average difficulty due to the algebraic manipulation required, but follows a well-established M2 template with clear guidance provided. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(3.2T^2 - 16T + 10.2 = 0\) \(T = \frac{3}{4}, \frac{17}{4}\) | B1 M1 A1 A1 [4] | For using \(\cos\theta = 1/\sec\theta\) and the given identity to obtain a quadratic in \(T(\tan\theta)\). AEF AG |
| Answer | Marks | Guidance |
|---|---|---|
| For \(\tan\theta = 0.75\), distance is 38.4 m For \(\tan\theta = 4.25\), distance is 17.8 m | M1 A1 A1 [3] | For solving \(y = 0\) for \(x\) or for using \(R = V^2\sin 2\theta/g\) |
| Answer | Marks |
|---|---|
| For sketching two parabolic arcs which intersect once, both starting at the origin, each with \(y ⩾ 0\) throughout, and each returning to the x-axis, the arc for which the angle of projection is smaller having the greater range. The ranges appear significantly greater than x at the intersection, and slightly greater, respectively. | B1 B1 [2] |
## (i)
$7 = 16\tan\theta - 10 \times 16^2(2-20^2)\cos^2\theta$ $[7 = 16T - 3.2(1 + T^2)]$
$3.2T^2 - 16T + 10.2 = 0$ $T = \frac{3}{4}, \frac{17}{4}$ | B1 M1 A1 A1 [4] | For using $\cos\theta = 1/\sec\theta$ and the given identity to obtain a quadratic in $T(\tan\theta)$. AEF AG
## (ii)
$[x = \tan\theta \cos^2\theta / 0.0125$ or $x = 20^2\sin 2\theta/g]$
For $\tan\theta = 0.75$, distance is 38.4 m For $\tan\theta = 4.25$, distance is 17.8 m | M1 A1 A1 [3] | For solving $y = 0$ for $x$ or for using $R = V^2\sin 2\theta/g$
## (iii)
For sketching two parabolic arcs which intersect once, both starting at the origin, each with $y ⩾ 0$ throughout, and each returning to the x-axis, the arc for which the angle of projection is smaller having the greater range. The ranges appear significantly greater than x at the intersection, and slightly greater, respectively. | B1 B1 [2] |
---A particle is projected from a point $O$ on horizontal ground. The velocity of projection has magnitude $20 \text{ m s}^{-1}$ and direction upwards at an angle $\theta$ to the horizontal. The particle passes through the point which is 7 m above the ground and 16 m horizontally from $O$, and hits the ground at the point $A$.
\begin{enumerate}[label=(\roman*)]
\item Using the equation of the particle's trajectory and the identity $\sec^2 \theta = 1 + \tan^2 \theta$, show that the possible values of $\tan \theta$ are $\frac{4}{3}$ and $\frac{1}{4}$. [4]
\item Find the distance $OA$ for each of the two possible values of $\tan \theta$. [3]
\item Sketch in the same diagram the two possible trajectories. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q5 [9]}}