| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Topic | Circular Motion 2 |
| Type | Vertical circle with peg/obstacle |
| Difficulty | Challenging +1.2 This is a standard vertical circle problem with a twist (the peg/bell obstacle). Part (i) requires finding minimum speed for complete revolution using energy conservation and tension condition at the top. Part (ii) involves finding when tension becomes zero during motion, requiring energy methods and solving a transcendental equation. While it requires multiple techniques and careful setup, the methods are well-practiced in Further Maths mechanics courses and the problem follows a familiar template. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
Total: 11 marks
**Part (i):**
At top, $mg (+ T) = mv^2/r$ **M1** N2($\downarrow$) at top
so $v^2 \geqslant 0.6g$ or $6$ **M1A1** Obtain inequality (or equation); correct, a.e.f. [No mention of $T$: M1M0A1]
C of E: $\frac{1}{2}mv^2 + 2mgr = \frac{1}{2}mu^2$ **M1A1** Use conservation of energy; correct equation
$v^2 = u^2 - 4gr;\quad u^2 \geqslant 5gr = 30$ **M1** Solve for $u$
$\Rightarrow u_{\min} = 5.48$ **A1** Answer, awrt 5.48. Condone $\sqrt{30}$. Common Error: $v_{\min} = 0$ at top leading to $u_{\min} = \sqrt{24} = 4.90$. M0M0A0M1A1M1A0 or SC3.
**Part (ii):**
$mg\cos\theta + T = mv^2/r$ **M1** Resolve inwards at general angle, $T$ not needed (NB $-mg\cos\theta$ if with downward vertical). Condone sign errors or $v^2/r$ for M1.
$v^2 = gr\cos\theta$ **A1** Correct condition
$\frac{1}{2}mv^2 = \frac{1}{2}m\cdot5^2 - 0.6mg(1 + \cos\theta)$ **M1A1** C of E for general angle; correct equation. $-\cos\theta$ if with downward vertical. Condone $mgh$ for M1.
$u^2 = gr(2 + 3\cos\theta)\ [25 = 6(2 + 3\cos\theta)]$ **M1** Find value for $\cos\theta\ [= \pm13/18]$
$\theta = \cos^{-1}(0.7222) = 43.8°$ **A1** Answer in range [43.7, 43.8]
Total: **11 marks**13 A fairground game consists of a small ball fixed to one end of a light inextensible string of length 0.6 m . The other end of the string is fixed to a point $O$. A small bell is fixed 0.6 m vertically above $O$. Initially the ball hangs vertically in equilibrium. The object of the game is to project the ball with an initial horizontal velocity $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ so that it moves in a vertical circle and hits the bell.\\
(i) Find the smallest possible value of $u$ for which the ball hits the bell.\\
(ii) Given, instead, that the value of $u$ is 5 , find the angle made by the string with the upward vertical at the moment when the string becomes slack.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2017 Q13 [9]}}