| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces, Pythagoras to find geometry, and application of F=mrω². While it involves multiple steps (geometry, vertical equilibrium, horizontal circular motion equation), the approach is methodical and follows a well-established template for rotating bead problems. Slightly easier than average due to straightforward geometry and clear physical setup. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
**Question 9(i)**
- $\cos/\sin\theta = 0.8/0.6$ **B1** 3:4:5 triangle implied
- $T\cos\theta = T\sin\theta + 0.2g$ **M1** resolving and balancing forces vertically (trig values not nec)
- $0.8T = 0.6T + 2$ **A1** correct equation with values
- $T = g = 10$ N **A1** $T = g$ or 10
**Question 9(ii)**
- $r = 0.48$ **B1** radius 0.48, can be implied or seen in (i) (or correct cancellation)
- $T\sin\theta + T\cos\theta = mr\omega^2$ or $mv^2/r$ **M1** NII horizontally $= mr\omega^2$
- $0.6T + 0.8T = 0.2\times0.48\omega^2$ or $0.2v^2/0.48$ **A1** FT for correct equation
- $\omega = 12.1$ or $v = 5.80$ (or 5.81 or $\sqrt{33.6}$) **A1** correct $v$ or $\omega$
- Time $2\pi \div \omega = 0.520$ seconds **A1** 0.52 or awrt 0.5209\\
\includegraphics[max width=\textwidth, alt={}, center]{09939c3a-7829-4784-8e6d-ee5356c22cd7-4_433_428_1219_863}
A light inextensible string of length 1.4 m has its ends attached to two points $A$ and $C$, where $A$ is 1 m vertically above $C$. A smooth bead $B$ of mass 0.2 kg is threaded on the string and rotates in a horizontal circle with the string taut. The distance $B A$ is 0.8 m (see diagram). Find\\
(i) the tension in the string,\\
(ii) the time taken for the bead to perform one complete circle.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2018 Q9 [9]}}