| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces and application of F=mrω². Part (i) and (ii) involve straightforward force resolution with given values. Part (iii) requires recognizing that 'point of leaving' means N=0, then solving for v. Slightly above average due to the three-part structure and the conceptual understanding needed in part (iii), but all techniques are routine for M2. |
| Spec | 3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\sin30°\) | B1 | |
| \(T\sin30° = 0.3x0.4x2\) | M1 | resolving horizontally |
| A1 | ||
| \(T = 0.96\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(R + T\cos30° = 0.3x9.8\) | M1 | resolving vertically |
| A1 | ||
| \(R = 2.11\) | A1√ | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_1\sin30° = 0.3 \times \sqrt{v/0.4}\) | M1 | or \(0.3x0.4x\omega^2\) \((T_1 = 1.5v^2)\) |
| \(T_1\cos30° = 0.3 \times 9.8\) | B1 | (\(T_1 = 1.96\sqrt{3} = 3.3948\)) |
| \(R = 0\) | B1 | may be implied or stated |
| \(\tan30°=v^2/(0.4 \times 9.8)\text{for elim of } T_1\) | M1 | and v=0.4ω (ω = 3.76) |
| \(v = 1.50\) | A1 | 6 |
**(i)**
$T\sin30°$ | B1 |
$T\sin30° = 0.3x0.4x2$ | M1 | resolving horizontally
| A1 |
$T = 0.96$ | A1 | 4 |
**(ii)**
$R + T\cos30° = 0.3x9.8$ | M1 | resolving vertically
| A1 |
$R = 2.11$ | A1√ | 3 | $\sqrt{\text{their } T}$ $(2.94-T\cos30°)$ |
**(iii)**
$T_1\sin30° = 0.3 \times \sqrt{v/0.4}$ | M1 | or $0.3x0.4x\omega^2$ $(T_1 = 1.5v^2)$
$T_1\cos30° = 0.3 \times 9.8$ | B1 | ($T_1 = 1.96\sqrt{3} = 3.3948$)
$R = 0$ | B1 | may be implied or stated
$\tan30°=v^2/(0.4 \times 9.8)\text{for elim of } T_1$ | M1 | and v=0.4ω (ω = 3.76)
$v = 1.50$ | A1 | 6 | 13 |7\\
\includegraphics[max width=\textwidth, alt={}, center]{1fbb3693-0beb-47c8-800f-50041f105699-4_782_1006_274_571}
One end of a light inextensible string of length 0.8 m is attached to a fixed point $A$ which lies above a smooth horizontal table. The other end of the string is attached to a particle $P$, of mass 0.3 kg , which moves in a horizontal circle on the table with constant angular speed $2 \mathrm { rad } \mathrm { s } ^ { - 1 } . A P$ makes an angle of $30 ^ { \circ }$ with the vertical (see diagram).\\
(i) Calculate the tension in the string.\\
(ii) Calculate the normal contact force between the particle and the table.
The particle now moves with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is on the point of leaving the surface of the table.\\
(iii) Calculate $v$.
\hfill \mbox{\textit{OCR M2 2007 Q7 [13]}}