| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Radial and transverse acceleration |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics question on circular motion requiring application of no-slip conditions and decomposition of acceleration into radial and transverse components. The calculations are straightforward once the relationships are established, with no novel problem-solving required beyond textbook methods. |
| Spec | 6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5\omega_A = 0.3\omega_B,\; \omega_B = 5/3\; [\text{rad s}^{-1}]\) | M1 A1 | Equate tangential speeds to find \(\omega_B\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5 \times \frac{1}{2} = 0.25\) | M1 A1 | Find tangential acceleration \(r\,d^2\theta/dt^2\) |
| \(0.5 \times 1^2 = 0.5\) | B1 | Find radial acceleration \(r(d\theta/dt)^2\) |
| \(\sqrt{(0.25^2 + 0.5^2)} = \sqrt{5}/4\) or \(0.559\; [\text{m s}^{-2}]\) | M1, A1 | Combine to give magnitude of acceleration |
| \(\tan^{-1}(0.25/0.5) = 0.464\;\text{rad}\) or \(26.6°\) | M1, A1 | Find angle made with \(PA\) (A.E.F.) |
| Part total: 7 marks | Question total: 9 marks |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5\omega_A = 0.3\omega_B,\; \omega_B = 5/3\; [\text{rad s}^{-1}]$ | M1 A1 | Equate tangential speeds to find $\omega_B$ |
**Part total: 2 marks**
## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5 \times \frac{1}{2} = 0.25$ | M1 A1 | Find tangential acceleration $r\,d^2\theta/dt^2$ |
| $0.5 \times 1^2 = 0.5$ | B1 | Find radial acceleration $r(d\theta/dt)^2$ |
| $\sqrt{(0.25^2 + 0.5^2)} = \sqrt{5}/4$ or $0.559\; [\text{m s}^{-2}]$ | M1, A1 | Combine to give magnitude of acceleration |
| $\tan^{-1}(0.25/0.5) = 0.464\;\text{rad}$ or $26.6°$ | M1, A1 | Find angle made with $PA$ (A.E.F.) |
**Part total: 7 marks | Question total: 9 marks**
---4\\
\begin{tikzpicture}[>=Latex, semithick, line cap=round, line join=round]
% Define radii
\def\rA{3} % Radius of circle A (scales to 0.5m)
\def\rB{1.8} % Radius of circle B (scales to 0.3m)
% Define centers
\coordinate (A) at (0,0);
\coordinate (B) at (\rA+\rB, 0);
\coordinate (Contact) at (\rA, 0);
% Draw the two circles
\draw (A) circle (\rA);
\draw (B) circle (\rB);
% Draw horizontal distance arrows
% Arrow for 0.5 m
\draw[<->] (A) -- (Contact) node[midway, above] {$0.5\mathrm{\,m}$};
% Arrow for 0.3 m
\draw[<->] (Contact) -- (B) node[midway, above] {$0.3\mathrm{\,m}$};
% Draw center labels
\node[left=2pt] at (A) {$A$};
\node[right=2pt] at (B) {$B$};
% Draw point P and its label
\coordinate (P) at (60:\rA);
\fill (P) circle (1.2pt);
\node[above right=0pt] at (P) {$P$};
% Draw angular velocity arrow and label
\draw[->] (155:\rA+0.6) arc (155:215:\rA+0.6)
node[pos=0.8, left=2pt] {$\frac{1}{2}t\mathrm{\,rad\,s^{-1}}$};
\end{tikzpicture}
Two coplanar discs, of radii 0.5 m and 0.3 m , rotate about their centres $A$ and $B$ respectively, where $A B = 0.8 \mathrm {~m}$. At time $t$ seconds the angular speed of the larger disc is $\frac { 1 } { 2 } t \mathrm { rad } \mathrm { s } ^ { - 1 }$ (see diagram). There is no slipping at the point of contact. For the instant when $t = 2$, find\\
(i) the angular speed of the smaller disc,\\
(ii) the magnitude of the acceleration of a point $P$ on the circumference of the larger disc, and the angle between the direction of this acceleration and $P A$.
\hfill \mbox{\textit{CAIE FP2 2010 Q4 [9]}}