| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Container or open-topped solid |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring systematic application of standard formulas. Part (a) involves calculating CoM of a composite lamina (cylinder surface + base), part (b) requires optimization by differentiation of a CoM expression, and parts (c)-(d) involve CoM calculations with displaced volumes. While lengthy with multiple steps, each component uses routine Further Maths mechanics techniques without requiring novel geometric insight or particularly challenging problem-solving. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Let the centre of mass lie x cm above the base. |
| Total area 4 2 2 4 1 8 ( 1 6 0 ) = + = | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 6 0 x 1 6 0 1 4 4 9 = + | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| x = 8 .1 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | DR | |
| Total mass =400+16h1 (=400+16h) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| ( 4 0 0 1 6 h ) y 4 0 0 8 .1 1 6 h 12 h + = + | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 0 0 1 6 h 5 0 2 h + + | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| d h ( 5 0 2 h 2 ) + | M1* | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| ( 5 0 2 h ) 2 h ( 4 0 5 h 2 ) 2 0 + − + = | M1dep* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| h = 5 .9 0 7 3 6 | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (c) | Let the centre of mass of the water lie d cm above the base. |
| Answer | Marks | Guidance |
|---|---|---|
| 1 6 1 3 .5 6 .7 5 1 6 1 3 .5 43 3 3 d 43 3 3 3 = − + | M1 | 3.1b |
| 1 8 0 d 1 3 5 0 d 7 .5 = = | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (d) | Total mass |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | M1 | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| z = 6 .2 3 3 4 9 1 | A1 | 2.2a |
Question 6:
6 | (a) | Let the centre of mass lie x cm above the base.
Total area 4 2 2 4 1 8 ( 1 6 0 ) = + = | B1 | 1.1 | Expression for correct total area of
container
1 6 0 x 1 6 0 1 4 4 9 = + | M1 | 1.1 | Table of values idea with correct
number of terms
x = 8 .1 | A1 | 1.1 | AG must be convincingly shown.
[3]
(b) | DR
Total mass =400+16h1 (=400+16h) | B1 | 1.1 | Expression for correct total mass of
container and water
( 4 0 0 1 6 h ) y 4 0 0 8 .1 1 6 h 12 h + = + | M1 | 1.1 | Table of values idea with correct
number of terms
( 4 0 0 1 6 h ) y 3 2 4 0 8 h 2 + = +
3 2 4 0 8 h 2 4 0 5 h 2 + +
y = =
4 0 0 1 6 h 5 0 2 h + + | A1 | 2.2a
d y ( 5 0 2 h ) 2 h ( 4 0 5 h 2 ) 2 + − +
=
d h ( 5 0 2 h 2 ) + | M1* | 3.1a | Attempt to differentiate using
quotient rule.
Least value of y occurs when
( 5 0 2 h ) 2 h ( 4 0 5 h 2 ) 2 0 + − + = | M1dep* | 2.1 | Setting the numerator of their
expression to zero.
h 2 5 0 h 4 0 5 0 + − =
h = 5 .9 0 7 3 6 | A1 | 2.2a
[6]
(c) | Let the centre of mass of the water lie d cm above the base.
( )
1 6 1 3 .5 6 .7 5 1 6 1 3 .5 43 3 3 d 43 3 3 3 = − + | M1 | 3.1b
1 8 0 d 1 3 5 0 d 7 .5 = = | A1 | 2.2a | AG must be convincingly shown.
[2]
(d) | Total mass
=400+ ( 1613.5−433 ) +4334 (=400+324)
3 3 | B1 | 1.1
( 400+324)z=4008.1+1350+43343
3 | M1 | 3.1b | Table of values idea with correct
number of terms
z = 6 .2 3 3 4 9 1 | A1 | 2.2a
[3]
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.6 In this question you may use the fact that the volume of a sphere of radius $r$ is $\frac { 4 } { 3 } \pi r ^ { 3 }$.\\
Fig. 6.1 shows a container in the shape of an open-topped cylinder. The cylinder has height 18 cm and radius 4 cm . The curved surface and the base can be modelled as uniform laminae with the same mass per unit area. The container rests on a horizontal surface.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-6_506_342_621_255}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the container lies 8.1 cm above its base.
The mass of the container is 400 grams. Water is poured into the container to reach a height of $h \mathrm {~cm}$ above the base. The centre of mass of the combined container and water lies $y \mathrm {~cm}$ above the base. Water has a density of 1 gram per $\mathrm { cm } ^ { 3 }$.
\item In this question you must show detailed reasoning.
By formulating an expression for $y$ in terms of $h$, determine the value of $h$ for which $y$ is lowest.
More water is now poured into the container. A sphere of radius 3 cm is placed into the container, where it sinks to the bottom. The surface of the water is now 4.5 cm from the top of the container, as shown in Fig. 6.2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-6_432_355_2001_255}
\end{center}
\end{figure}
\item Show that the centre of mass of the water in the container lies 7.5 cm above the base of the container.
The sphere has a density of 4 grams per $\mathrm { cm } ^ { 3 }$.\\
The centre of mass of the combined container, water and sphere lies $z \mathrm {~cm}$ above the base.
\item Determine the value of $z$.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2023 Q6 [14]}}