| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with circular arc or semicircular arc components |
| Difficulty | Standard +0.8 This is a non-trivial centre of mass problem requiring students to apply the formula for the centroid of a semicircular arc (2r/π) to two different arcs, use the composite body formula with appropriate coordinate systems, then apply equilibrium conditions. The geometry is moderately complex with the smaller arc's endpoint at the larger arc's centre, requiring careful setup and calculation. More challenging than standard uniform lamina questions but still follows established methods. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (ii) | CoM(large) = 0.6/(π/2) or |
| Answer | Marks |
|---|---|
| θ = 15.3° | B1 |
| Answer | Marks |
|---|---|
| A1 | 3 |
| 4 | OR (2+1)D = 2(1.2/π) – 1(0.6/π) |
| Answer | Marks |
|---|---|
| (iii) | 0.25vdv/dx = 2 + 0.3x2 |
| Answer | Marks |
|---|---|
| Force is 0.5 + 0.75x N towards O | M1 |
| Answer | Marks |
|---|---|
| A1 | 2 |
| Answer | Marks |
|---|---|
| 2 | Allow c = 0 without working |
| Answer | Marks |
|---|---|
| (iii) | (0.9a + 0.9a/2)Y = |
| Answer | Marks |
|---|---|
| a = 0.75 | M1 |
| Answer | Marks |
|---|---|
| A1 | 2 |
| Answer | Marks |
|---|---|
| 3 | 1.5Y = 1 x 0.45 + 0.5 x 0.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 5 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2016 | 9709 | 53 |
Question 2:
--- 2 (i)
(ii) ---
2 (i)
(ii) | CoM(large) = 0.6/(π/2) or
CoM(small) = 0.3/(π/2)
(π x 0.6 + π x 0.3)D =
π x 0.6(1.2/π) – π x 0.3(0.6/π)
D = 0.191m AG
(π x 0.6 + π x 0.3)H =
π x 0.6 x 0.6 + π x 0.3 x 0.9
H = 0.7
tanθ = 0.191/0.7
θ = 15.3° | B1
M1
A1
M1
A1
M1
A1 | 3
4 | OR (2+1)D = 2(1.2/π) – 1(0.6/π)
Moments about ACB
OR 3H = 2 x 0.6 + 1 x 0.9
Moments about A
3 (i)
(ii)
(iii) | 0.25vdv/dx = 2 + 0.3x2
vdv/dx = 1.2 x2 + 8 AG
∫v dv =∫(1.2x2 +8) dx
v2/2 = 0.4x3 + 8x ( + c)
v = 5.17
0.25vdv/dx = 0.3x2 + 1.5 – 0.75x
Force is 0.5 + 0.75x N towards O | M1
A1
M1
A1
A1
M1
A1 | 2
3
2 | Allow c = 0 without working
4 (i)
(ii)
(iii) | (0.9a + 0.9a/2)Y =
0.9a x 0.45 + 0.45a x 0.9 x 2/3
Y = 0.5 m
(0.9a + 0.9a/2)X =
0.9a x a/2 + 0.45a x (a + a/3)
X = 7a/9
0.5 x 6 = (a – 7a/9) x 18
a = 0.75 | M1
A1
M1
A1
M1
A1
A1 | 2
2
3 | 1.5Y = 1 x 0.45 + 0.5 x 0.6
Moments about AD
1.5X = 1 x a/2 + 0.5 x 4a/3
Ft [Yi and (a–Xii)]
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2016 | 9709 | 53\includegraphics{figure_2}
A uniform wire is bent to form an object which has a semicircular arc with diameter $AB$ of length 1.2 m, with a smaller semicircular arc with diameter $BC$ of length 0.6 m. The end $C$ of the smaller arc is at the centre of the larger arc (see diagram). The two semicircular arcs of the wire are in the same plane.
\begin{enumerate}[label=(\roman*)]
\item Show that the distance of the centre of mass of the object from the line $ACB$ is 0.191 m, correct to 3 significant figures. [3]
\end{enumerate}
The object is freely suspended at $A$ and hangs in equilibrium.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the angle between $ACB$ and the vertical. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q2 [7]}}