| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.3 This is a standard circular motion problem with a bead on a string requiring resolution of forces and application of F=mrω². The geometry is given, making it straightforward to find angles and apply Newton's second law. The multi-part structure and calculation requirements place it slightly above average difficulty, but it follows a well-established method taught in M2 with no novel insight required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(T\cos\theta = 0.01 \times 9.8\) | M1 | resolving vertically |
| \(8/10T = 0.01 \times 9.8\) | A1 | with \(\cos\theta = 8/10\) |
| \(T = 0.1225 \text{ N}\) | A1 | 3 |
| (ii) \(T + T\sin\theta = ma\) | M1 | resolving horizontally |
| use of \(mro^2\) | M1 | |
| \(\omega = 5.72 \text{ rads}^{-1}\) | A1 | 3 |
| (iii) \(\text{K.E.} = \frac{1}{2}x0.01x(ro)^2\) | M1 | \(\frac{1}{2}mv^2\) with \(v=rw\) |
| \(\text{K.E.} = 0.0588\) | A1 | 2 |
(i) $T\cos\theta = 0.01 \times 9.8$ | M1 | resolving vertically
$8/10T = 0.01 \times 9.8$ | A1 | with $\cos\theta = 8/10$
$T = 0.1225 \text{ N}$ | A1 | 3 | AG
(ii) $T + T\sin\theta = ma$ | M1 | resolving horizontally
use of $mro^2$ | M1 |
$\omega = 5.72 \text{ rads}^{-1}$ | A1 | 3 |
(iii) $\text{K.E.} = \frac{1}{2}x0.01x(ro)^2$ | M1 | $\frac{1}{2}mv^2$ with $v=rw$
$\text{K.E.} = 0.0588$ | A1 | 2 | $\checkmark 0.0018 \times \text{their } \omega^2$ | 83\\
\includegraphics[max width=\textwidth, alt={}, center]{35477eb8-59e0-4de6-889c-1f5841f65eec-2_451_533_1676_808}
One end of a light inextensible string of length 1.6 m is attached to a point $P$. The other end is attached to the point $Q$, vertically below $P$, where $P Q = 0.8 \mathrm {~m}$. A small smooth bead $B$, of mass 0.01 kg , is threaded on the string and moves in a horizontal circle, with centre $Q$ and radius $0.6 \mathrm {~m} . Q B$ rotates with constant angular speed $\omega$ rad s $^ { - 1 }$ (see diagram).\\
(i) Show that the tension in the string is 0.1225 N .\\
(ii) Find $\omega$.\\
(iii) Calculate the kinetic energy of the bead.
\hfill \mbox{\textit{OCR M2 2005 Q3 [8]}}