| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a sophisticated 3D circular motion problem requiring resolution of forces in two directions (horizontal for centripetal force, vertical for equilibrium), geometric reasoning to find the radius at different angles, and working with two tension forces simultaneously. Part (ii) requires showing a specific result algebraically. More complex than standard conical pendulum questions due to the two-string constraint geometry. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(B\cos30 - A\sin30 = 0.2g\) | B1 | Resolving vertically for P |
| \(B\cos60 - A\cos30 = 0.2 \times 1.2^2/0.4\) | M1 A1 | 2 components of tension, N2L with \(\text{accn} = v^2/r\) |
| M1 | Attempts to eliminate one unknown | |
| \(A = 0.753\) N | A1 | |
| \(B = 2.74\) N | A1 (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = 0.8\sin60\) | B1 | |
| M1 | Resolves vertically or uses N2L horizontally | |
| \(B\cos60 - A\cos30 = 0.2g\) | A1 | |
| \(B\cos30 - A\cos60 = 0.2 \times 5^2 \times 0.8\sin60\) | A1 | |
| M1 | For solving to find \(A\) | |
| \(A = 0\) | A1 (6) |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $B\cos30 - A\sin30 = 0.2g$ | B1 | Resolving vertically for P |
| $B\cos60 - A\cos30 = 0.2 \times 1.2^2/0.4$ | M1 A1 | 2 components of tension, N2L with $\text{accn} = v^2/r$ |
| | M1 | Attempts to eliminate one unknown |
| $A = 0.753$ N | A1 | |
| $B = 2.74$ N | A1 (6) | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 0.8\sin60$ | B1 | |
| | M1 | Resolves vertically or uses N2L horizontally |
| $B\cos60 - A\cos30 = 0.2g$ | A1 | |
| $B\cos30 - A\cos60 = 0.2 \times 5^2 \times 0.8\sin60$ | A1 | |
| | M1 | For solving to find $A$ |
| $A = 0$ | A1 (6) | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{f8633b64-b20c-4471-9641-ccc3e6854f2c-4_479_499_255_824}\\
$O A$ is a rod which rotates in a horizontal circle about a vertical axis through $O$. A particle $P$ of mass 0.2 kg is attached to the mid-point of a light inextensible string. One end of the string is attached to the $\operatorname { rod }$ at $A$ and the other end of the string is attached to a point $B$ on the axis. It is given that $O A = O B$, angle $O A P =$ angle $O B P = 30 ^ { \circ }$, and $P$ is 0.4 m from the axis. The rod and the particle rotate together about the axis with $P$ in the plane $O A B$ (see diagram).\\
(i) Calculate the tensions in the two parts of the string when the speed of $P$ is $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The angular speed of the rod is increased to $5 \mathrm { rad } \mathrm { s } ^ { - 1 }$, and it is given that the system now rotates with angle $O A P =$ angle $O B P = 60 ^ { \circ }$.\\
(ii) Show that the tension in the part $A P$ of the string is zero.
{www.cie.org.uk} after the live examination series.
}
\hfill \mbox{\textit{CAIE M2 2016 Q6 [12]}}