| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Multiple particles on string |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring systematic application of Newton's second law in horizontal and vertical directions for two connected particles. While it involves multiple bodies and string tensions, the approach is methodical: resolve forces on Q (horizontal centripetal force and vertical equilibrium), then on P (similar resolution with an inclined string). The calculations are straightforward once the force diagrams are drawn, making this slightly easier than average for A-level mechanics. |
| Spec | 3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) TPQ = (0.4g) = 4N | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| TBQ = 0.4 × 5 × 0.3 | M1 | mω2 |
| Answer | Marks |
|---|---|
| TBQ = 3N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) Tcos α = 0.8g + 4 | M1 | Attempts to find either component of T |
| Answer | Marks | Guidance |
|---|---|---|
| Tsin = 0.8x5 x0.3 | A1 | Both components correct |
| Answer | Marks | Guidance |
|---|---|---|
| T = 12 + 6 | M1 | Or any equivalent method to find T |
| Answer | Marks | Guidance |
|---|---|---|
| TAP = 13.4N ( = 6 N) | A1 | |
| α° = tan –1 (6/12) = tan –1 (1/2) = 26.6° | B1ft | |
| OR | α | |
| Tcos = 0.8g + 4 | M1 | Attempts to find either component of T |
| Answer | Marks | Guidance |
|---|---|---|
| Tsin = 0.8x5 x0.3 | A1 | Both components correct |
| Answer | Marks |
|---|---|
| tan = 6/12 | M1 |
| Answer | Marks |
|---|---|
| = 26.6 | A1 |
| TAP = 13.4N | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme: Teachers’ version | Syllabus |
| GCE A LEVEL – October/November 2010 | 9709 | 53 |
Question 3:
3 | (i) TPQ = (0.4g) = 4N | B1
2
TBQ = 0.4 × 5 × 0.3 | M1 | mω2
Uses F = r
TBQ = 3N | A1
[3]
(ii) Tcos α = 0.8g + 4 | M1 | Attempts to find either component of T
α 2
Tsin = 0.8x5 x0.3 | A1 | Both components correct
2 2 2
T = 12 + 6 | M1 | Or any equivalent method to find T
5
TAP = 13.4N ( = 6 N) | A1
α° = tan –1 (6/12) = tan –1 (1/2) = 26.6° | B1ft
OR | α
Tcos = 0.8g + 4 | M1 | Attempts to find either component of T
α 2
Tsin = 0.8x5 x0.3 | A1 | Both components correct
α
tan = 6/12 | M1
α
= 26.6 | A1
TAP = 13.4N | B1ft
[5]
Page 5 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9709 | 53\includegraphics{figure_3}
Particles $P$ and $Q$ have masses $0.8$ kg and $0.4$ kg respectively. $P$ is attached to a fixed point $A$ by a light inextensible string which is inclined at an angle $\alpha°$ to the vertical. $Q$ is attached to a fixed point $B$, which is vertically below $A$, by a light inextensible string of length $0.3$ m. The string $BQ$ is horizontal. $P$ and $Q$ are joined to each other by a light inextensible string which is vertical. The particles rotate in horizontal circles of radius $0.3$ m about the axis through $A$ and $B$ with constant angular speed $5$ rad s$^{-1}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item By considering the motion of $Q$, find the tensions in the strings $PQ$ and $BQ$. [3]
\item Find the tension in the string $AP$ and the value of $\alpha$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q3 [8]}}