| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Ratio of tensions/forces |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics question requiring energy conservation, circular motion equations at two positions, and algebraic manipulation to find cos θ from a tension ratio. Part (b) requires identifying the lowest point as maximum speed. The multi-step reasoning and Further Maths context place it well above average difficulty. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(A\): \(T_A - mg\cos\theta = m\times\frac{5ag}{a}\) | B1 | N2L |
| At \(B\): \(T_B + mg\cos\theta = m\times\frac{v^2}{a}\) | B1 | N2L |
| \(\frac{1}{2}m\times 5ag - \frac{1}{2}mv^2 = mga\times 2\cos\theta\) | M1 | Energy equation with correct number of terms |
| \(v^2 = 5ag - 4ga\cos\theta\) | A1 | Accept multiplied by \(m\) and/or divided by \(a\) |
| Use ratio of tensions \(= 9:5\) | M1 | Use ratio and simplify to an expression in \(\cos\theta\) |
| \(\cos\theta = \frac{2}{5}\) | A1 | CAO |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Greatest speed at lowest point: \(-\frac{1}{2}m\times 5ag + \frac{1}{2}mV^2 = mga\times(1-\cos\theta)\) | M1 | Energy equation including lowest point, correct number of terms |
| \(V = \sqrt{\frac{31ag}{5}}\) | A1 FT | Ft their \(\cos\theta\) from part (a) |
| Total | 2 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| At $A$: $T_A - mg\cos\theta = m\times\frac{5ag}{a}$ | B1 | N2L |
| At $B$: $T_B + mg\cos\theta = m\times\frac{v^2}{a}$ | B1 | N2L |
| $\frac{1}{2}m\times 5ag - \frac{1}{2}mv^2 = mga\times 2\cos\theta$ | M1 | Energy equation with correct number of terms |
| $v^2 = 5ag - 4ga\cos\theta$ | A1 | Accept multiplied by $m$ and/or divided by $a$ |
| Use ratio of tensions $= 9:5$ | M1 | Use ratio and simplify to an expression in $\cos\theta$ |
| $\cos\theta = \frac{2}{5}$ | A1 | CAO |
| **Total** | **6** | |
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## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Greatest speed at lowest point: $-\frac{1}{2}m\times 5ag + \frac{1}{2}mV^2 = mga\times(1-\cos\theta)$ | M1 | Energy equation including lowest point, correct number of terms |
| $V = \sqrt{\frac{31ag}{5}}$ | A1 FT | Ft their $\cos\theta$ from part (a) |
| **Total** | **2** | |6 A particle $P$, 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 particle $P$ moves in complete vertical circles about $O$ with the string taut. The points $A$ and $B$ are on the path of $P$ with $A B$ a diameter of the circle. $O A$ makes an angle $\theta$ with the downward vertical through $O$ and $O B$ makes an angle $\theta$ with the upward vertical through $O$. The speed of $P$ when it is at $A$ is $\sqrt { 5 a g }$.
The ratio of the tension in the string when $P$ is at $A$ to the tension in the string when $P$ is at $B$ is $9 : 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\cos \theta$.
\item Find, in terms of $a$ and $g$, the greatest speed of $P$ during its motion.\\
\includegraphics[max width=\textwidth, alt={}, center]{b10c65ef-dafd-4746-be5b-789130b7d030-12_613_718_251_676}
The smooth vertical walls $A B$ and $C B$ are at right angles to each other. A particle $P$ is moving with speed $u$ on a smooth horizontal floor and strikes the wall $C B$ at an angle $\alpha$. It rebounds at an angle $\beta$ to the wall $C B$. The particle then strikes the wall $A B$ and rebounds at an angle $\gamma$ to that wall (see diagram). The coefficient of restitution between each wall and $P$ is $e$.\\
(a) Show that $\tan \beta = e \tan \alpha$.\\
(b) Express $\gamma$ in terms of $\alpha$ and explain what this result means about the final direction of motion of $P$.\\
As a result of the two impacts the particle loses $\frac { 8 } { 9 }$ of its initial kinetic energy.
\item Given that $\alpha + \beta = 90 ^ { \circ }$, find the value of $e$ and the value of $\tan \alpha$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q6 [8]}}