| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Standard +0.3 This is a standard projectile motion question requiring derivation of the trajectory equation (routine manipulation of kinematic equations), finding the maximum point using calculus or symmetry, and solving a quadratic equation for angle. All techniques are well-practiced in Further Maths mechanics with no novel insight required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | → x=ucosα t |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| ucosα 2 ucosα | M1 | Eliminate |
| Answer | Marks | Guidance |
|---|---|---|
| 2u2 | A1 | AG |
| Answer | Marks |
|---|---|
| 5(b) | ( usinα)2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2g 4g | M1 A1 | Accept alternative methods, for example differentiate expression in |
| Answer | Marks | Guidance |
|---|---|---|
| 2g | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(c) | Use greatest height displacements in trajectory equation |
| Answer | Marks | Guidance |
|---|---|---|
| 4g 2g 2u24g2 | M1 | Use equation of trajectory (substitute coordinates of Q |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| tan2α−4tanα+3=0 | M1 | Obtain a three-term quadratic in tanα |
| tanα=1, 3 so α=71.6° | A1 | Both solutions needed |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | → x=ucosα t
1
↑ y=usinα t− gt2
2 | B1 | Both
2
x 1 x
Eliminate t: y=usinα. − g
ucosα 2 ucosα | M1 | Eliminate
gx2
y=xtanα− sec2α
2u2 | A1 | AG
3
--- 5(b) ---
5(b) | ( usinα)2
u2
Greatest height = =
2g 4g | M1 A1 | Accept alternative methods, for example differentiate expression in
(a) and equate to 0.
u2
t =usin45/g so d =ucos45.usin45/g =
2g | A1 | AG
3
Question | Answer | Marks | Guidance
--- 5(c) ---
5(c) | Use greatest height displacements in trajectory equation
u2 u2 gu4
= tanα− sec2α
4g 2g 2u24g2 | M1 | Use equation of trajectory (substitute coordinates of Q
u2
u2 =2u2tanα− (1+tan2α)
2 | M1 | ( )
Use of sec2α= 1+tan2α
tan2α−4tanα+3=0 | M1 | Obtain a three-term quadratic in tanα
tanα=1, 3 so α=71.6° | A1 | Both solutions needed
4
Question | Answer | Marks | GuidanceA particle $P$ is projected with speed $u$ at an angle $\alpha$ 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 $t$ are denoted by $x$ and $y$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Derive the equation of the trajectory of $P$ in the form
$$y = x\tan\alpha - \frac{gx^2}{2u^2}\sec^2\alpha.$$ [3]
\end{enumerate}
The point $Q$ is the highest point on the trajectory of $P$ in the case where $\alpha = 45°$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the $x$-coordinate of $Q$ is $\frac{u^2}{2g}$. [3]
\item Find the other value of $\alpha$ for which $P$ would pass through the point $Q$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q5 [10]}}