| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | November |
| 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 requiring resolution of forces and application of F=mrω². Part (i) involves setting up two tension equations from geometry and the centripetal force equation—straightforward but multi-step. Part (ii) is a direct application of circular motion formulas once the geometry is established. Slightly above average due to the geometric setup with two angles, but follows standard M2 procedures without requiring novel insight. |
| Spec | 6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| (ii) | Tcos30 – Tcos45 = 0.3g |
| Answer | Marks |
|---|---|
| T = 5.53 N | M1 |
| Answer | Marks |
|---|---|
| A1 | 5 |
| 3 | Resolves vertically |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2015 | 9709 | 52 |
| Answer | Marks |
|---|---|
| (iii) | 0.05 |
| Answer | Marks |
|---|---|
| v = 2.93 ms | M1 |
| Answer | Marks |
|---|---|
| A1 | 2 |
| Answer | Marks |
|---|---|
| 3 | θ = angle of string with vertical |
| Answer | Marks |
|---|---|
| (iii) | Mass of disc = π (1.2 2 – 0.4 2 – 0.3 2 ) |
| Answer | Marks |
|---|---|
| θ = 68.1° | B1 |
| Answer | Marks |
|---|---|
| A1 | 4 |
| Answer | Marks |
|---|---|
| 2 | 1.19π (or in (ii)) |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2015 | 9709 | 52 |
Question 4:
--- 4 (i)
(ii) ---
4 (i)
(ii) | Tcos30 – Tcos45 = 0.3g
T = 18.9
0.3v2
18.9sin30 + 18.9sin45 =
0.6
–1
v = 6.75 ms
0.6 0.6
L = +
sin30 sin45
2
0.3×3 (2.05sinθ) = Tsinθ
T = 5.53 N | M1
A1
M1
A1
A1
B1
M1
A1 | 5
3 | Resolves vertically
T =6 3+6 2
Resolves horizontally,
2
Acceleration = v /r
2.0485…
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9709 | 52
5 (i)
(ii)
(iii) | 0.05
0.2g = R + 21×
0.75
R = 0.6 N
0.8
21 /(0.75cosθ)=0.2g
cosθ −0.75
e = 0.0735
OR
21e 0.8
× =0.2g
0.75 (e+0.75)
e = 0.073529…
0.2(3)2 21(0.05)2
0.2v2 21×0.07352
+ = +
2 (2×0.75) 2 1.5
–1
v = 2.93 ms | M1
A1
M1
A1
A1
M1
A1
A1
M1
A1
A1 | 2
3
3 | θ = angle of string with vertical
Comp of tension = weight
θ = 13.7(291…)
e = 0.8/cosθ – 0.75 = 0.073529…
e = extension
Comp of tension = weight
Uses EE/KE balance
6 (i)
(ii)
(iii) | Mass of disc = π (1.2 2 – 0.4 2 – 0.3 2 )
0 = π (1.2 2 – 0.4 2 – 0.3 2 )y –
(0.4 2 ) × 0.7
y = 0.0941 m
0 = π (1.2 2 – 0.4 2 – 0.3 2 )x–π(0.3 2 ).5
x = 0.0378 m
0.0941176
tanθ =
0.0378151
θ = 68.1° | B1
M1
A1
A1
M1
A1
A1
M1
A1 | 4
3
2 | 1.19π (or in (ii))
LHS = π (1.2 2 – 0.3 2 )×0
LHS = π (1.2 2 – 0.4 2 )×0
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9709 | 52\includegraphics{figure_4}
One end of a light inextensible string is attached to a fixed point $A$. The string passes through a smooth bead $B$ of mass $0.3\,\text{kg}$ and the other end of the string is attached to a fixed point $C$ vertically below $A$. The bead $B$ moves with constant speed in a horizontal circle of radius $0.6\,\text{m}$ which has its centre between $A$ and $C$. The string makes an angle of $30°$ with the vertical at $A$ and an angle of $45°$ with the vertical at $C$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Calculate the speed of $B$. [5]
\end{enumerate}
The lower end of the string is detached from $C$, and $B$ is now attached to this end of the string. The other end of the string remains attached to $A$. The bead is set in motion so that it moves with angular speed $3\,\text{rad s}^{-1}$ in a horizontal circle which has its centre vertically below $A$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the tension in the string. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q4 [8]}}