| 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 two-angle projectile problem requiring trajectory equation manipulation and a quadratic in tan θ. While it involves multiple parts and the sec²θ identity, the approach is methodical and well-practiced in M2 syllabi. The algebraic manipulation is straightforward once the trajectory equation is set up, and finding range is routine. Slightly above average due to the algebraic steps, but this is a textbook exercise type. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{1}{4}, \frac{17}{4}\) | \(B1, M1, A1, A1\) [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 \(\tan\theta = 4.25\), 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 |
|---|---|
| The ranges appear significantly greater than x at the intersection, and slightly greater, respectively. | \(B1, B1\) [2] |
**(i)**
$7 = 16\tan\theta - 10 \times 16^2(2 \times 20^2)\cos^2\theta$
$[7 = 16T - 3.2(1 + T^2)]$
$3.2T^2 - 16T + 10.2 = 0$
$T = \frac{1}{4}, \frac{17}{4}$ | $B1, M1, A1, A1$ [4] | For using $\cos\theta = 1/\sec\theta$ and the given identity to obtain a quadratic in $T(\tan\theta)$. AEF. AG
**(ii)**
$[x = \tan\theta \cos^2\theta \times 0.0125$ 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 $y \geq 0$ 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] |
---5 A particle is projected from a point $O$ on horizontal ground. The velocity of projection has magnitude $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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 $A$.\\
(i) 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 $\frac { 3 } { 4 }$ and $\frac { 17 } { 4 }$.\\
(ii) Find the distance $O A$ for each of the two possible values of $\tan \theta$.\\
(iii) Sketch in the same diagram the two possible trajectories.
\hfill \mbox{\textit{CAIE M2 2010 Q5 [9]}}