| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Challenging +1.2 This is a standard projectiles problem requiring knowledge of trajectory equations and the relationship between maximum height and range. Students must find the horizontal distance at a specific vertical position (3/4 of max height), which involves substituting into the trajectory equation and solving a quadratic. While it requires multiple steps and careful algebra, the approach is methodical and follows standard Further Maths projectiles techniques without requiring novel insight. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(a) | ( usinθ)2 |
| Answer | Marks |
|---|---|
| 2g | M1A1 |
| At ¾ greatest height, →vcosα=ucosθ | M1 |
| Answer | Marks |
|---|---|
| 4 2g | M1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(b) | 1 4 2 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Answer | Marks |
|---|---|
| 5g | A1 |
| → t =d /ucosθ | M1 |
| Answer | Marks |
|---|---|
| 25g | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/33 Cambridge International AS & A Level – Mark Scheme May/June 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 14
Question | Answer | Marks
--- 6(a) ---
6(a) | ( usinθ)2
Greatest height =
2g | M1A1
At ¾ greatest height, →vcosα=ucosθ | M1
( usinθ)2
At ¾ greatest height, ↑ ( vsinα)2 =( usinθ)2 −2g. 3
4 2g | M1
1
vsinα= usinθ
2 | A1
1
So tanα= tanθ AG
2 | A1
6
Question | Answer | Marks
--- 6(b) ---
6(b) | 1 4 2
tanα= . =
2 3 3
1
↑ usinθ=usinθ−gt
2 | M1
2u
t =
5g | A1
→ t =d /ucosθ | M1
6u2
=
25g | A1
4
Question | Answer | Marks\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that $\tan \theta = \frac{4}{3}$, find the horizontal distance travelled by $P$ when it first reaches three-quarters of its greatest height. Give your answer in terms of $u$ and $g$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q6 [4]}}