| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Challenging +1.2 This is a standard circular motion problem requiring geometric setup (using Pythagoras to find QR from the constraint that PR + QR = 4a) and then applying Newton's second law with tension resolution. The geometry is straightforward once the constraint is recognized, and the circular motion equation is a direct application of standard techniques. It's moderately above average due to the two-part structure and need to handle two tensions, but follows a well-established problem type in Further Maths mechanics. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
**Question 7(i) and 7(ii)**
**(i)**
$x^2 + 9a^2 = (4a - x)^2 \Rightarrow \ldots \Rightarrow x = \dfrac{7}{8}a$ (AG) — M1A1
**[2]**
**(ii)**
$T\cos\theta = mg$ — B1
$T + T\sin\theta = \dfrac{mv^2}{r}$ — M1*A1 (M1 needs two forces)
$\cos\theta = \dfrac{24}{25}$, $\quad \sin\theta = \dfrac{7}{25}$ or $\tan\theta = \dfrac{7}{24}$ — B1 (One correct, may be implied)
Solve to obtain $v = \sqrt{\dfrac{7ag}{6}}$ — dep*M1, A1
**[6]**7\\
\includegraphics[max width=\textwidth, alt={}, center]{86cc07e7-ea69-4480-96c8-82b818445199-3_599_499_1279_822}
A light inextensible string of length $4 a$ has one end fixed at a point $P$ and the other end fixed at a point $Q$, which is vertically below $P$ and at a distance $3 a$ from $P$. A small smooth ring $R$ of mass $m$ is threaded on the string. $R$ moves in a horizontal circle with centre $Q$ and with the string taut (see diagram).\\
(i) Show that $Q R = \frac { 7 } { 8 } a$.\\
(ii) Find the speed of $R$ in terms of $a$ and $g$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q7 [6]}}