| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Challenging +1.8 This is a challenging mechanics problem requiring understanding of projectile motion, oblique impacts, and coefficient of restitution. The key insight is recognizing that the velocity component perpendicular to the inclined plane before impact relates to the vertical rebound through the restitution equation. While it involves multiple concepts (projectiles, vector resolution, impacts), the mathematical steps are relatively standard once the setup is understood. The 4-mark allocation and 'show that' format indicate moderate difficulty within Further Maths, though the geometric reasoning about velocity components at an oblique impact elevates it above routine exercises. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03k Newton's experimental law: direct impact |
| Answer | Marks |
|---|---|
| 7(a) | When P strikes plane, velocity is →ucos, |
| Answer | Marks | Guidance |
|---|---|---|
| 5 5 | M1 | 3 |
| Answer | Marks |
|---|---|
| 5 5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 5 | M1 | |
| tan=1/etan, e tan2 =1 | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | (usin)2 8u2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2g 25g | M1A1 | Note: alternative methods. |
| Answer | Marks |
|---|---|
| 5 | M1 |
| Answer | Marks |
|---|---|
| 25 16 | M1 |
| Answer | Marks |
|---|---|
| Substitute to find e | M1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
Question 7:
--- 7(a) ---
7(a) | When P strikes plane, velocity is →ucos,
3 3
Before impact: parallel to inclined plane ucos, perpendicular to plane usin
5 5 | M1 | 3
u
5
3 3
After impact: components ucos (parallel) and eusin (perpendicular)
5 5 | A1
3 3
Since velocity is vertical after impact, tan= ucos/ eusin
5 5 | M1
tan=1/etan, e tan2 =1 | A1 | AG
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | (usin)2 8u2
Greatest height of P before impact: H = =
2g 25g | M1A1 | Note: alternative methods.
3
After impact, vertical speed of P is u (cos)2 +e2(sin)2
5 | M1
3
Use V2 =U2 +2as to greatest height, equal to H
16
9 ( ) 3
u2 (cos)2 +e2(sin)2 =2g H
25 16 | M1
1 e 1
Use part (a): tan= , cos= , sin=
e 1+e 1+e
Substitute to find e | M1
1
3e2 +2e−1=0, e=
3 | A1
6A particle $P$ is projected with speed $u$ at an angle $\tan^{-1}\left(\frac{4}{3}\right)$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. When $P$ is moving horizontally, it strikes a smooth inclined plane at the point $A$. This plane is inclined to the horizontal at an angle $\alpha$, and the line of greatest slope through $A$ lies in the vertical plane through $O$ and $A$.
As a result of the impact, $P$ moves vertically upwards. The coefficient of restitution between $P$ and the inclined plane is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that $e \tan^2 \alpha = 1$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q7 [4]}}