| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 11 |
| 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=mv²/r. Part (i) and (ii) involve straightforward force resolution with given values. Part (iii) requires recognizing that the particle leaves the table when the normal force becomes zero, which is a common textbook scenario. The problem is slightly above average difficulty due to the three-dimensional geometry and the conceptual understanding needed in part (iii), but all techniques are standard M2 material with no novel insight required. |
| Spec | 3.01b Derived quantities and units3.03n Equilibrium in 2D: particle under forces6.04e Rigid body equilibrium: coplanar forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(a = 1.5^2/(0.7\sin40°)\) | B1 | |
| \([T\sin40° = 0.25a]\) | M1 | For using Newton's second law horizontally |
| Tension is \(1.94\) N | A1 | [3] |
| (ii) | M1 | For resolving forces vertically |
| \(T\cos40° + R = 0.25\,g\) | A1 | |
| Force exerted is \(1.01\) N | A1\(\sqrt{}\) | ft \(2.5 - T\cos40°\) [3] |
| (iii) | M1 | For using Newton's second law horizontally and \(a = r\omega^2\) |
| \(T\sin40° = 0.25(0.7\sin40°)\,\omega^2\) | A1 | |
| \(T\cos40° = 0.25\,g\) \((T = 3.2635\ldots)\) | B1 | |
| \([\tan40° = 0.7\sin40°\,\omega^2/g\) or \(3.2635\ldots\sin40° = 0.25(0.7\sin40°)\,\omega^2]\) | M1 | For eliminating \(T\) or substituting for \(T\) |
| Maximum value of \(\omega\) is \(4.32\) | A1 | [5] |
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $a = 1.5^2/(0.7\sin40°)$ | B1 | |
| $[T\sin40° = 0.25a]$ | M1 | For using Newton's second law horizontally |
| Tension is $1.94$ N | A1 | **[3]** |
| **(ii)** | M1 | For resolving forces vertically |
| $T\cos40° + R = 0.25\,g$ | A1 | |
| Force exerted is $1.01$ N | A1$\sqrt{}$ | ft $2.5 - T\cos40°$ **[3]** |
| **(iii)** | M1 | For using Newton's second law horizontally and $a = r\omega^2$ |
| $T\sin40° = 0.25(0.7\sin40°)\,\omega^2$ | A1 | |
| $T\cos40° = 0.25\,g$ $(T = 3.2635\ldots)$ | B1 | |
| $[\tan40° = 0.7\sin40°\,\omega^2/g$ or $3.2635\ldots\sin40° = 0.25(0.7\sin40°)\,\omega^2]$ | M1 | For eliminating $T$ or substituting for $T$ |
| Maximum value of $\omega$ is $4.32$ | A1 | **[5]** |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{68acf474-5da2-4949-b3b2-fc42cd73bd4a-3_504_878_1557_632}
One end of a light inextensible string of length 0.7 m is attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of mass 0.25 kg . The particle $P$ moves in a circle on a smooth horizontal table with constant speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The string is taut and makes an angle of $40 ^ { \circ }$ with the vertical (see diagram). Find\\
(i) the tension in the string,\\
(ii) the force exerted on $P$ by the table.\\
$P$ now moves in the same horizontal circle with constant angular speed $\omega \operatorname { rad~s } ^ { - 1 }$.\\
(iii) Find the maximum value of $\omega$ for which $P$ remains on the table.
\hfill \mbox{\textit{CAIE M2 2009 Q6 [11]}}