| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.8 Part (a) is routine algebraic manipulation using a given formula. Part (b) requires setting up a quadratic in tan θ and using the discriminant to prove impossibility—a standard technique but requiring careful algebra and insight that negative discriminant means no solution. This is above-average difficulty for A-level, typical of Further Maths mechanics questions requiring multi-step reasoning and proof. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Quote trajectory equation from MF19 and use \(\cos\theta = 1/\sec\theta\): \(y = x\tan\theta - \frac{gx^2}{2u^2}(1 + \tan^2\theta)\) | B1 | Must include step with \(\sec^2\theta\); Allow derived from first principles; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(16 = 20 - \frac{10\times100}{2u^2}(1+4)\) | M1 | Substitute into result (a) |
| \(u^2 = 625,\quad (u=25)\) | A1 | |
| Use equation again: \(30 = 18\tan\theta - \frac{10\times324}{2\times625}\left(1+(\tan\theta)^2\right)\) | M1 | |
| \(2.592(\tan\theta)^2 - 18\tan\theta + 32.592 = 0\) | A1 | 3 term quadratic. Alternatives include: \(54t^2 - 375t + 679 = 0\), \(324t^2 - 2250t + 4074 = 0\) |
| Discriminant \(= 324 - 4\times2.592\times32.592 = -13.91\) | M1 | Discriminant for alternatives: \(-6039\) and \(-217404\) |
| As this is less than 0, no real solutions for \(\theta\) | A1 | CWO |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Quote trajectory equation from MF19 and use $\cos\theta = 1/\sec\theta$: $y = x\tan\theta - \frac{gx^2}{2u^2}(1 + \tan^2\theta)$ | B1 | Must include step with $\sec^2\theta$; Allow derived from first principles; AG |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $16 = 20 - \frac{10\times100}{2u^2}(1+4)$ | M1 | Substitute into result (a) |
| $u^2 = 625,\quad (u=25)$ | A1 | |
| Use equation again: $30 = 18\tan\theta - \frac{10\times324}{2\times625}\left(1+(\tan\theta)^2\right)$ | M1 | |
| $2.592(\tan\theta)^2 - 18\tan\theta + 32.592 = 0$ | A1 | 3 term quadratic. Alternatives include: $54t^2 - 375t + 679 = 0$, $324t^2 - 2250t + 4074 = 0$ |
| Discriminant $= 324 - 4\times2.592\times32.592 = -13.91$ | M1 | Discriminant for alternatives: $-6039$ and $-217404$ |
| As this is less than 0, no real solutions for $\theta$ | A1 | CWO |
---5 A particle $P$ is projected with speed $u \mathrm {~ms} ^ { - 1 }$ at an angle of $\theta$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of $P$ from $O$ at a subsequent time $t \mathrm {~s}$ are denoted by $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Starting from the equation of the trajectory given in the List of formulae (MF19), show that
$$\mathrm { y } = \mathrm { x } \tan \theta - \frac { \mathrm { gx } ^ { 2 } } { 2 \mathrm { u } ^ { 2 } } \left( 1 + \tan ^ { 2 } \theta \right)$$
When $\theta = \tan ^ { - 1 } 2 , P$ passes through the point with coordinates $( 10,16 )$.
\item Show that there is no value of $\theta$ for which $P$ can pass through the point with coordinates $( 18,30 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q5 [7]}}