| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Compound pendulum oscillations |
| Difficulty | Challenging +1.8 This is a challenging two-part mechanics problem requiring energy conservation, circular motion dynamics, and projectile motion for part (i)-(ii), followed by a complex moment of inertia calculation for a composite rigid body system. The first part requires careful application of energy methods and the condition for string tension becoming zero, while the second part (though cut off) involves calculating moments of inertia about a non-standard axis for multiple bodies. This exceeds typical A-level mechanics in both conceptual depth and computational complexity, placing it firmly in Further Maths territory. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga\cos\theta\) | M1 | Use conservation of energy to slack point \(P_1\) |
| \(mv^2/a - mg\cos\theta = 0\) | M1, A1 | Equate tension at \(P_1\) to 0 by using \(F = ma\); A1 if both eqns correct, with \(m\) included. AG |
| \(v^2 = 2ag - 2ag\cos\theta = ag\cos\theta\) | M1 | Combine to verify \(\cos\theta\) using \(u = \sqrt{2ag}\) |
| \(\cos\theta = 2/3\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_V = v\sin\theta = \sqrt{2ag/3}\,(\sqrt{5}/3)\) or \(v_V^2 = (10/27)\,ag\) | M1 | Find vertical speed \(v_V\) at \(P_1\) |
| \(h = v_V^2/2g = (5/27)\,a\) or \(0.185\,a\) | M1, A1 | Find height risen above \(P_1\) by considering vertical motion |
| \(h + a\cos\theta = (23/27)\,a\) or \(0.852\,a\) | A1 | Find total height risen above level of \(O\) |
| Total: 4 |
## Question 4:
### Part 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mga\cos\theta$ | M1 | Use conservation of energy to slack point $P_1$ |
| $mv^2/a - mg\cos\theta = 0$ | M1, A1 | Equate tension at $P_1$ to 0 by using $F = ma$; A1 if both eqns correct, with $m$ included. AG |
| $v^2 = 2ag - 2ag\cos\theta = ag\cos\theta$ | M1 | Combine to verify $\cos\theta$ using $u = \sqrt{2ag}$ |
| $\cos\theta = 2/3$ | A1 | |
| **Total: 5** | | |
### Part 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_V = v\sin\theta = \sqrt{2ag/3}\,(\sqrt{5}/3)$ or $v_V^2 = (10/27)\,ag$ | M1 | Find vertical speed $v_V$ at $P_1$ |
| $h = v_V^2/2g = (5/27)\,a$ or $0.185\,a$ | M1, A1 | Find height risen above $P_1$ by considering vertical motion |
| $h + a\cos\theta = (23/27)\,a$ or $0.852\,a$ | A1 | Find total height risen above level of $O$ |
| **Total: 4** | | |
---4 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 $O$ and $P$ is held with the string taut and horizontal. The particle $P$ is projected vertically downwards with speed $\sqrt { } ( 2 a g )$ so that it begins to move along a circular path. The string becomes slack when $O P$ makes an angle $\theta$ with the upward vertical through $O$.\\
(i) Show that $\cos \theta = \frac { 2 } { 3 }$.\\
(ii) Find the greatest height, above the horizontal through $O$, reached by $P$ in its subsequent motion.\\
\includegraphics[max width=\textwidth, alt={}, center]{4240c99e-10ba-443e-8021-1872e6e64ccf-10_1051_744_258_696}
A thin uniform $\operatorname { rod } A B$ has mass $\lambda M$ and length $2 a$. The end $A$ of the rod is rigidly attached to the surface of a uniform hollow sphere (spherical shell) with centre $O$, mass $3 M$ and radius $a$. The end $B$ of the rod is rigidly attached to the surface of a uniform solid sphere with centre $C$, mass $5 M$ and radius $a$. The rod lies along the line joining the centres of the spheres, so that $C B A O$ is a straight line. The horizontal axis $L$ is perpendicular to the rod and passes through the point of the rod that is a distance $\frac { 1 } { 2 } a$ from $B$ (see diagram). The object consisting of the rod and the two spheres can rotate freely about $L$.\\
\hfill \mbox{\textit{CAIE FP2 2019 Q4 [9]}}