| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on circular wire/arc |
| Difficulty | Challenging +1.2 This is a standard circular motion problem requiring energy conservation and Newton's second law in the radial direction. While it involves multiple parts and careful geometry (finding angles and heights), the techniques are routine for Further Maths students: applying conservation of energy, using the centripetal force equation, and finding the point where the normal reaction becomes zero. The geometry with sin θ = 1/4 requires some care but is straightforward. This is slightly above average difficulty due to the multi-step nature and geometric setup, but uses entirely standard methods without requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga\) | B1 | Use conservation of energy at lowest point |
| \(R - mg = mv^2/a\) | B1 | Use \(F = ma\) radially at lowest point |
| \(R = mu^2/a + 3mg = 3.3mg\) | B1 | Eliminate \(v^2\) to find \(R\) \([v^2 = 2.3ga]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 + mga\sin\theta\) | M1 A1 | Use conservation of energy at \(B\) to find \(V_B\) |
| \(V_B^2 = (0.3 + 0.5)ga\), \(V_B = \sqrt{(0.8ga)}\) | ||
| or \(2\sqrt{(ga/5)}\) or \(0.894\sqrt{(ga)}\) | A1 | (A.E.F.) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_B = V_B\cos\theta\ [= \frac{1}{4}\sqrt{15}V_B = \sqrt{(\frac{3}{4}ga)}]\) | M1 | Use vertical component \(v_B\) of speed \(V_B\) at \(B\) |
| \(h = v_B^2/2g = 3a/8\) | M1 A1 | Find height \(h\) reached above \(B\) |
| \(h - a\sin\theta = 3a/8 - \frac{1}{4}a = a/8\) A.G. | A1 | Find height \(h\) reached above level of \(O\) |
# Question 4:
**Part (i):**
| $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mga$ | B1 | Use conservation of energy at lowest point |
| $R - mg = mv^2/a$ | B1 | Use $F = ma$ radially at lowest point |
| $R = mu^2/a + 3mg = 3.3mg$ | B1 | Eliminate $v^2$ to find $R$ $[v^2 = 2.3ga]$ |
**Part (ii):**
| $\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 + mga\sin\theta$ | M1 A1 | Use conservation of energy at $B$ to find $V_B$ |
| $V_B^2 = (0.3 + 0.5)ga$, $V_B = \sqrt{(0.8ga)}$ | | |
| or $2\sqrt{(ga/5)}$ or $0.894\sqrt{(ga)}$ | A1 | (A.E.F.) |
**Part (iii):**
| $v_B = V_B\cos\theta\ [= \frac{1}{4}\sqrt{15}V_B = \sqrt{(\frac{3}{4}ga)}]$ | M1 | Use vertical component $v_B$ of speed $V_B$ at $B$ |
| $h = v_B^2/2g = 3a/8$ | M1 A1 | Find height $h$ reached above $B$ |
| $h - a\sin\theta = 3a/8 - \frac{1}{4}a = a/8$ **A.G.** | A1 | Find height $h$ reached above level of $O$ |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{ab5f2781-e5ce-4fce-bc95-9d7f55ea66d9-2_515_583_1388_781}
A smooth wire is in the form of an $\operatorname { arc } A B$ of a circle, of radius $a$, that subtends an obtuse angle $\pi - \theta$ at the centre $O$ of the circle. It is given that $\sin \theta = \frac { 1 } { 4 }$. The wire is fixed in a vertical plane, with $A O$ horizontal and $B$ below the level of $O$ (see diagram). A small bead of mass $m$ is threaded on the wire and projected vertically downwards from $A$ with speed $\sqrt { } \left( \frac { 3 } { 10 } g a \right)$.\\
(i) Find the reaction between the bead and the wire when the bead is vertically below $O$.\\
(ii) Find the speed of the bead as it leaves the wire at $B$.\\
(iii) Show that the greatest height reached by the bead is $\frac { 1 } { 8 } a$ above the level of $O$.
\hfill \mbox{\textit{CAIE FP2 2014 Q4 [10]}}