| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two projectiles meeting - 2D flight |
| Difficulty | Challenging +1.2 This is a standard two-particle projectile collision problem requiring systematic application of kinematic equations. While it involves multiple particles with different launch times and angles, the solution follows a clear algorithmic approach: resolve velocities using given tan values, equate horizontal displacements, solve the resulting equation, then find collision coordinates. The algebra is moderately involved but straightforward for Further Maths students. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | For Q: xucosT | B1 |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Equate and attempt to rearrange. |
| Answer | Marks | Guidance |
|---|---|---|
| 4uT 21 5T 1 | A1 | AG |
| Answer | Marks |
|---|---|
| 7(b) | Vertical motion to collision: |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 A1 | M1 for both expressions, one correct. |
| Answer | Marks | Guidance |
|---|---|---|
| 2 4 | M1 | Equate and attempt to solve |
| Answer | Marks |
|---|---|
| T 3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 7(c) | x42 | B1 |
| y 24 | M1 | |
| y24 (or 24 m below O) | A1 | Correct sign or in words. |
Question 7:
--- 7(a) ---
7(a) | For Q: xucosT | B1
35
For P: x cosT 1
2 | B1
35
Collision, so cosT 1ucosT
2 | M1 | Equate and attempt to rearrange.
35 3 2
T 1u T
2 5 5
4uT 21 5T 1 | A1 | AG
Shown convincingly.
4
--- 7(b) ---
7(b) | Vertical motion to collision:
1
For Q: yusinT gT2
2
35 1
For P: y sinT 1 g(T 1)2
2 2 | M1 A1 | M1 for both expressions, one correct.
1 1 35 4 1
Equate: u T gT2 T 1 g(T 1)2
5 2 2 5 2
14T 1 1 g T2 2T 1T2 21 T 1
2 4 | M1 | Equate and attempt to solve
16T 3621T 21, 1 55T
T 3 | A1
4
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | x42 | B1
y 24 | M1
y24 (or 24 m below O) | A1 | Correct sign or in words.
3
Particles $P$ and $Q$ are projected in the same vertical plane from a point $O$ at the top of a cliff. The height of the cliff exceeds 50 m. Both particles move freely under gravity. Particle $P$ is projected with speed $\frac{35}{2} \text{ m s}^{-1}$ at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac{4}{3}$. Particle $Q$ is projected with speed $u \text{ m s}^{-1}$ at an angle $\beta$ above the horizontal, where $\tan \beta = \frac{1}{2}$. Particle $Q$ is projected one second after the projection of particle $P$. The particles collide $T$ s after the projection of particle $Q$.
\begin{enumerate}[label=(\alph*)]
\item Write down expressions, in terms of $T$, for the horizontal displacements of $P$ and $Q$ from $O$ when they collide and hence show that $4uT = 21\sqrt{5(T + 1)}$. [4]
\item Find the value of $T$. [4]
\item Find the horizontal and vertical displacements of the particles from $O$ when they collide. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q7 [11]}}