| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.3 This is a straightforward centre of mass question requiring standard moment calculations and basic equilibrium principles. Part (a) is routine moments about a point, part (b) uses simple geometry with the centre of mass below the suspension point, part (c) requires recognizing symmetry of tension forces over a smooth peg, and part (d) applies basic force resolution. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Let AG have length x cm. |
| 7 x = ( 3 .5 0 ) + ( 1 .4 3 0 ) + ( 2 .1 6 0 ) | M1 | 3.4 |
| x=24 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | BD= 402−302 = 700 | M1 |
| Answer | Marks |
|---|---|
| = 3 5 .8 1 3 | M1 |
| A1 | 3.1a |
| 1.1 | Correct attempt to find A D B or |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | The forces that the string exerts at each end of the rod must | |
| be equal (in magnitude), | B1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| since the rod is in equilibrium), the angles must be equal. | B1 | 2.2a |
| Answer | Marks |
|---|---|
| (d) | Let A D = x cm |
| Answer | Marks | Guidance |
|---|---|---|
| are equal, so 1.5x=80−x. | M1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| x = 3 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 3 4 8 | M1 | 3.1a |
| ADC=95.079 so A D G = 4 7 .5 3 9 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2Tcos(∠𝐴𝐷𝐺) = 7𝑔 | M1 | 3.4 |
| T = 5 0 .8 0 9 | A1 | 2.2a |
Question 6:
6 | (a) | Let AG have length x cm.
7 x = ( 3 .5 0 ) + ( 1 .4 3 0 ) + ( 2 .1 6 0 ) | M1 | 3.4
x=24 | A1 | 1.1
[2]
(b) | BD= 402−302 = 700 | M1 | 3.1b | Seen or implied.
A D G = a r c t a n 37 00 − a r c t a n 67
0 0 0
= 3 5 .8 1 3 | M1
A1 | 3.1a
1.1 | Correct attempt to find A D B or
G D B .
[3]
(c) | The forces that the string exerts at each end of the rod must
be equal (in magnitude), | B1 | 2.4
and since only these forces have a horizontal component (and
since the rod is in equilibrium), the angles must be equal. | B1 | 2.2a
[2]
(d) | Let A D = x cm
A r e a ( C D G ) = 1 .5 A r e a ( A D G ) , and angles ADG and CDG
are equal, so 1.5x=80−x. | M1 | 3.1a | Realising that DC must be 32 64 = 1 .5
times AD.
x = 3 2 | A1 | 1.1 | Or for deducing length of CD (48
cm)
c o s ( A D C ) = 3 2 2 + 4 82 2 − 6 0 2
2 3 4 8 | M1 | 3.1a | Using Cosine Rule.
ADC=95.079 so A D G = 4 7 .5 3 9 | A1 | 1.1
Let the tension in the string be T N.
2Tcos(∠𝐴𝐷𝐺) = 7𝑔 | M1 | 3.4
T = 5 0 .8 0 9 | A1 | 2.2a
[6]
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 Fig. 6.1 shows a light rod ABC , of length 60 cm , where B is the midpoint of AC . Particles of masses $3.5 \mathrm {~kg} , 1.4 \mathrm {~kg}$ and 2.1 kg are attached to $\mathrm { A } , \mathrm { B }$ and C respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b624694-edb6-4000-838f-3557e078952d-7_241_1056_367_251}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
The centre of mass is located at a point G along the rod.
\begin{enumerate}[label=(\alph*)]
\item Determine the distance AG .
Two light inextensible strings, each of length 40 cm , are attached to the rod, one at A , the other at C. The other ends of these strings are attached to a fixed point D. The rod is allowed to hang in equilibrium.
\item Determine the angle AD makes with the vertical.
The two strings are now replaced by a single light inextensible string of length 80 cm . One end of the string is attached to A and the other end of the string is attached to C. The string passes over a smooth peg fixed at D. The rod hangs in equilibrium, but is not vertical, as shown in Fig. 6.2.
Fig. 6.2
\item Explain why angle ADG and angle CDG must be equal.
\item Determine the tension in the string.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2022 Q6 [13]}}