| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem combining circular motion with projectile motion. Part (a) requires applying Newton's second law in circular motion and energy conservation—a well-practiced technique. Parts (b) and (c) involve straightforward projectile motion calculations once the initial conditions are established. While it requires multiple steps and careful coordinate work, the methods are entirely standard for Further Maths students and no novel insight is needed. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Energy: 1mu2 1mv2 mga1cos | |
| 2 2 | *M1 | m must be present, dimensionally correct, no |
| Answer | Marks | Guidance |
|---|---|---|
| a | B1 | No reaction when P loses contact. |
| Eliminate v2 | DM1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3ag | A1 | AG |
| Answer | Marks |
|---|---|
| 7(b) | Vertical component of velocity of P when it leaves the sphere: vsin |
| Answer | Marks | Guidance |
|---|---|---|
| | *B1 | Must not come from u. |
| V2 vsin2 2ga1cos | DM1 | Use of 'v2 u2 2as. |
| Answer | Marks | Guidance |
|---|---|---|
| 216 | A1 | AEF |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(c) | 1 847ag 55ag |
| Answer | Marks |
|---|---|
| | M1 |
| Answer | Marks |
|---|---|
| 6 6g g | A1 |
Question 7:
--- 7(a) ---
7(a) | Energy: 1mu2 1mv2 mga1cos
2 2 | *M1 | m must be present, dimensionally correct, no
missing terms.
Allow sin instead of cos.
Allow sign errors.
mv2
N2L: mgcos
a | B1 | No reaction when P loses contact.
Eliminate v2 | DM1
u2 2ag
cos
3ag | A1 | AG
4
--- 7(b) ---
7(b) | Vertical component of velocity of P when it leaves the sphere: vsin
55ag
216
| *B1 | Must not come from u.
V2 vsin2 2ga1cos | DM1 | Use of 'v2 u2 2as.
Allow sin for cos.
Allow sign errors.
847ag
V
216 | A1 | AEF
3
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1 847ag 55ag
t
g 216 216
| M1
1 a a
847 55 1.48
6 6g g | A1
2
A smooth sphere with centre $O$ and of radius $a$ is fixed to a horizontal plane. A particle $P$ of mass $m$ is projected horizontally from the highest point of the sphere with speed $u$, so that it begins to move along the surface of the sphere. The particle $P$ loses contact with the sphere at the point $Q$ on the sphere, where $OQ$ makes an angle $\theta$ with the upward vertical through $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos\theta = \frac{u^2 + 2ag}{3ag}$. [4]
\end{enumerate}
It is given that $\cos\theta = \frac{5}{9}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $a$ and $g$, an expression for the vertical component of the velocity of $P$ just before it hits the horizontal plane to which the sphere is fixed. [3]
\item Find an expression for the time taken by $P$ to fall from $Q$ to the plane. Give your answer in the form $k\sqrt{\frac{a}{g}}$, stating the value of $k$ correct to 3 significant figures. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q7 [9]}}