| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 4 |
| Topic | Circular Motion 1 |
| Type | String through hole – hanging particle in equilibrium below table |
| Difficulty | Standard +0.3 This is a standard circular motion problem with two particles connected by a string through a ring. It requires resolving forces on both particles (tension equals weight of Q, vertical and horizontal components for P's circular motion) and applying F=mrω². The setup is clearly defined with straightforward force diagrams, making it slightly easier than average despite involving multiple steps and two equations. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
(i) $R(\uparrow_Q)$: $T - 1.5g = 0\ [T = 15]$
$R(\uparrow_P)$: $T\cos\theta - 1.2g = 0$
$[15\cos\theta = 12]$
$\Rightarrow \cos\theta = 0.8$
Distance below $= 0.12/\tan\theta = \mathbf{0.16}$ m
- M1: Resolve vertically for 1.5 kg mass, can be implied e.g. by $T=15$
- M1: Resolve vertically for 1.2 kg mass
- A1: Value of $\cos\theta\ [\theta = 36.9°]$
- A1: Correct value of $h$
**[4]**
(ii) $T = 1.5g$
$R(\rightarrow_P)$: $T\sin\theta = 1.2r\omega^2\ [= 9]$
$\Rightarrow 1.5g\sin\theta = 1.2\times0.12\times\omega^2$
$[a = 7.5]$
$\Rightarrow \omega = \sqrt{62.5} = \mathbf{7.91}$ rad s$^{-1}$
- B1: Value of $T$ used $[= 15]$
- M1: Resolve horiz for $P$ and use $r\omega^2$ or $v^2/r$
- A1: Correct equation, and $v = r\omega$ if $v$ used
- A1: Answer, in range $[7.9, 7.91]$ or $\dfrac{5}{2}\sqrt{10}$
**[4]**9\\
\includegraphics[max width=\textwidth, alt={}, center]{1a89caec-6da8-4b83-9ffa-efc209ecbc8d-4_506_730_625_712}
Particles $P$ and $Q$, of masses 1.2 kg and 1.5 kg respectively, are attached to the ends of a light inextensible string. The string passes through a small smooth ring which is attached to the ceiling but which is free to rotate. $P$ rotates at $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle of radius 0.12 m , and $Q$ hangs vertically in equilibrium (see diagram). Determine\\
(i) the vertical distance below the ring at which $P$ rotates,\\
(ii) the value of $\omega$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q9 [4]}}