| 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 when P moves horizontally at A, its velocity is purely horizontal, and after impact it moves vertically upward. Students must resolve velocities normal and tangential to the inclined plane, apply the restitution formula correctly, and use the constraint that the tangential component is zero after impact (smooth plane). While the algebra is manageable once set up correctly, the conceptual understanding of oblique impacts and the geometric reasoning required make this significantly harder than average A-level mechanics questions. |
| 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]}}