| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Velocity direction at specific time/point |
| Difficulty | Challenging +1.2 This is a standard projectile motion problem requiring systematic application of kinematic equations to relate angle of projection to angle at a specific height. The 6 marks indicate multiple steps, but the approach is methodical: use v² = u² + 2as to find vertical velocity at 3h/4, then apply tan α = vy/vx with constant horizontal velocity. While requiring careful algebra and understanding of projectile motion principles, it follows a well-established problem-solving template without requiring novel geometric insight or proof techniques beyond A-level mechanics. |
| 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 | MarksA particle $P$ is projected with speed $u$ at an angle $\theta$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The direction of motion of $P$ makes an angle $\alpha$ above the horizontal when $P$ first reaches three-quarters of its greatest height.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \alpha = \frac{1}{2}\tan \theta$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q6 [6]}}