| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle with peg/obstacle |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation across two different circular motion centers, tension equations at multiple points, and solving a complex trigonometric equation. The multi-stage setup (initial circle, peg contact, new circle, matching tensions) demands careful bookkeeping and extended algebraic manipulation, placing it well above average difficulty but within reach of strong FM students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods |
| Answer | Marks |
|---|---|
| 5(i) | ½mv 2 = ½mu2 + mga cos α |
| Answer | Marks | Guidance |
|---|---|---|
| 1 1 | M1 A1 | Verify v for string horizontal by consvn of energy (A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Total: | 2 | |
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(ii) | T + mg cos α = m (√ag)2 /a, T = mg (1 – cos α) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 A1 | Find tension T at A from F = ma radially |
| Answer | Marks |
|---|---|
| 1 | M1 A1 |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| C | M1 A1 | Find tension T at C from F = ma radially |
| Answer | Marks | Guidance |
|---|---|---|
| mg (1 – cos α) = 3mg (⅓ + 2 cos α)/2 – ½mg | M1 A1 | Find cos α from T = T and substituting for v 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 – cos α = 3 cos α, cos α = ¼ | A1 | |
| Total: | 10 | |
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | ½mv 2 = ½mu2 + mga cos α
1
v 2 = ag + 2 ag cos α, v = √ (ag (1 + 2 cos α)) AG
1 1 | M1 A1 | Verify v for string horizontal by consvn of energy (A0
1
if no m)
Total: | 2
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | T + mg cos α = m (√ag)2 /a, T = mg (1 – cos α)
A A
½mv 2 = ½mu2 + mga cos α – mg ⅔ a cos 60°
2 | M1 A1 | Find tension T at A from F = ma radially
A
Find v 2 at C by consvn. of energy (A0 if no m)
2
or ½mv 2 – mg ⅔ a cos 60°
1 | M1 A1
v 2 = ag + 2ag cos α – ⅔ ag = ag (⅓ + 2 cos α)
2 | A1
T + mg cos 60° = m v 2 / ⅔ a [= 3m v 2 / 2 a]
C 2 2
[T = 3mg cos α]
C | M1 A1 | Find tension T at C from F = ma radially
C
mg (1 – cos α) = 3mg (⅓ + 2 cos α)/2 – ½mg | M1 A1 | Find cos α from T = T and substituting for v 2
A C 2
1 – cos α = 3 cos α, cos α = ¼ | A1
Total: | 10
Question | Answer | Marks | Guidance\includegraphics{figure_5}
A particle of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The point $A$ is such that $OA = a$ and $OA$ makes an angle $\alpha$ with the upward vertical through $O$. The particle is held at $A$ and then projected downwards with speed $\sqrt{(ag)}$ so that it begins to move in a vertical circle with centre $O$. There is a small smooth peg at the point $B$ which is at the same horizontal level as $O$ and at a distance $\frac{5}{4}a$ from $O$ on the opposite side of $O$ to $A$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that, when the string first makes contact with the peg, the speed of the particle is $\sqrt{(ag(1 + 2\cos\alpha))}$. [2]
\end{enumerate}
The particle now begins to move in a vertical circle with centre $B$. When the particle is at the point $C$ where angle $CBO = 150°$, the tension in the string is the same as it was when the particle was at the point $A$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the value of $\cos\alpha$. [10]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q5 [12]}}