| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.3 This is a standard vertical circle problem requiring energy conservation to find speed, then resolving forces for tension, and finally calculating resultant acceleration. While it involves multiple steps and careful angle work (including the cos^(-1)(4/5) = 37° setup), the techniques are routine for Further Maths students: conservation of energy, T - mg cos θ = mv²/r for tension, and combining radial/tangential components for acceleration. The numerical values are designed to work out neatly (3-4-5 triangle). This is slightly easier than average due to its straightforward application of standard methods. |
| Spec | 3.02g Two-dimensional variable acceleration6.02i Conservation of energy: mechanical energy principle |
(i) Conservation of energy:
$50 \cdot 10\left(\frac{4}{5} - \frac{1}{2}\right) = \frac{1}{2} \times 5v^2$ **M1A1**
$v^2 = 60 \Rightarrow v = \sqrt{60}$ (= 7.75) **A1**
Speed is 7.75 ms$^{-1}$. **A1**
Resolving along string:
$T - 50 \times \frac{4}{5} = \frac{5}{10} \times 60$ **M1A1**
$\Rightarrow T = 70$
Tension is 70 N **A1** [7]
(ii) Acceleration towards centre:
$\frac{60}{10} = 6$ **B1**
Newton II along tangent:
$50 \times \frac{3}{5} = 5a \Rightarrow a = \frac{30}{5} = 6$ **M1A1**
Magnitude of acceleration
$6\sec 45° = 6\sqrt{2}$ ms$^{-2}$ (AG) (CWO) **A1** [4]11\\
\includegraphics[max width=\textwidth, alt={}, center]{742ef62b-bd72-45b4-88e3-70399632e9d6-4_384_524_587_808}
One end of a light inextensible string of length 10 m is attached to a fixed point $O$. A particle of mass 5 kg is attached to the other end of the string. The particle rests in equilibrium below $O$. The particle is pulled aside until the string makes an angle of $60 ^ { \circ }$ with the downward vertical and released from rest (see diagram). At the instant when the string makes an angle $\cos ^ { - 1 } \left( \frac { 4 } { 5 } \right)$ with the downward vertical,\\
(i) find the speed of the particle and the tension in the string,\\
(ii) show that the magnitude of the acceleration of the particle is $6 \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q11 [11]}}