| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring: (a) symmetry argument for centre of mass location, and (b) multi-step equilibrium analysis involving finding the combined centre of mass position (using 3h/4 for cone), applying moment equilibrium about a point, and using Hooke's law with extension calculation. The horizontal axis orientation and geometric constraints add complexity beyond standard moments questions, but the solution path is systematic once set up correctly. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | (Because the cone has) symmetry (about this |
| axis) | B1 | |
| [1] | 2.4 | |
| (b) | Moments taken (centre of base) |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| M1* | 1.1 | Attempts to balance moments |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3.1b | Correct moments equation |
| Answer | Marks | Guidance |
|---|---|---|
| V | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| l | M1dep | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1dep | |
| * | M1dep | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| l = 1.8 (m) | A1 | 2.2a |
Question 8:
8 | (a) | (Because the cone has) symmetry (about this
axis) | B1
[1] | 2.4
(b) | Moments taken (centre of base)
1
8𝑚𝑔( ℎ)+2𝑚𝑔ℎ = ℎ𝑇
𝑉
4 | M1 | 3.3 | Moments about vertex
3
M1 8𝑚𝑔( ℎ) = ℎ𝑇
𝐴
4
At least two correct moment
terms seen
M1* | 1.1 | Attempts to balance moments | Attempts to balance moments | A1 𝑇 = 6𝑚𝑔
𝐴
A1 | 3.1b | Correct moments equation | M1 For finding T + T = 10mg
V A
and combining with simultaneous
equations. May also find moments
about a second point.
T = 4mg
V | A1 | 1.1 | T = 4mg
V
8mgx
4mg =
l | M1dep | 3.4 | 3.4 | Use of Hooke’s law | Use of Hooke’s law
*
1
x = l
2 | M1dep
* | M1dep | 1.1 | Using x=0.5l and/or using x =
2.7-l to find an expression in
terms of l or x only.
*
3
𝑥 +𝑙 = 𝑙 = 1.6+1.1 = 2.7
2
l = 1.8 (m) | A1 | 2.2a
[7]
Moments taken (centre of base)
1
8𝑚𝑔( ℎ)+2𝑚𝑔ℎ = ℎ𝑇
𝑉
4
M1
Moments about vertex
3
M1 8𝑚𝑔( ℎ) = ℎ𝑇
𝐴
4
Using x=0.5l and/or using x =
2.7-l to find an expression in
terms of l or x only.
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.8 A shape, $S$, is formed by attaching a particle of mass $2 m \mathrm {~kg}$ to the vertex of a uniform solid cone of mass $8 m \mathrm {~kg}$. The height of the cone is $h \mathrm {~m}$ and the radius of the base of the cone is 1.1 m .
\begin{enumerate}[label=(\alph*)]
\item Explain why the centre of mass of $S$ must lie on the central axis of the cone.
Two strings are attached to $S$, one at the vertex of the cone and one at $A$ which is a point on the edge of the base of $S$. The other ends of the strings are attached to a horizontal ceiling in such a way that the strings are both vertical. The string attached to $S$ at $A$ is inextensible and has length 1.6 m . The string attached to $S$ at the vertex is elastic with modulus of elasticity 8 mgN .
Shape $S$ is in equilibrium with its axis horizontal (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{05b479a4-4087-4332-924b-43b1aedbb4f2-6_654_1541_879_244}
\item Determine the natural length of the elastic string.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q8 [8]}}