| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.8 This is a substantial multi-part centre of mass question requiring: (i) composite lamina calculation with multiple rectangles, (ii) suspended equilibrium angle calculation, (iii) finding mass and position to achieve specified centre of mass, and (iv) 3D stability analysis after folding. The 3D folding aspect and stability region identification elevate this beyond routine M2 questions, though the individual techniques are standard. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(15\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 2\begin{pmatrix}-1\\2.5\end{pmatrix} + 9\begin{pmatrix}1.5\\1.5\end{pmatrix} + 2\begin{pmatrix}4\\0.5\end{pmatrix} + 2\begin{pmatrix}4.5\\-1\end{pmatrix}\) | M1 | A systematic method for at least 1 cpt |
| \(= \begin{pmatrix}28.5\\17.5\end{pmatrix}\) | A1 | Either all \(x\) or all \(y\) values correct, or 2 vector terms correct on RHS |
| All components completely correct | A1 | Need not be explicit |
| \(\bar{x} = 1.9\) | A1 | Accept any form |
| \(\bar{y} = \frac{7}{6}\) | A1 | Accept any form (1.17, 1.2) but not 1.16 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| G is \(2 + 1.9\) to right of K and \(3 - \frac{7}{6} = \frac{11}{6}\) below K | B1 | FT from (i). May be implied |
| When hanging, G is vertically below K | B1 | May be implied |
| Angle is \(\arctan\left(\frac{\frac{11}{6}}{3.9}\right)\) | M1 | o.e. FT their values but must be attempting to find the appropriate angle |
| \(= 25.2°\) (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New c.m. is at (1.5, 1.5) and mass of object is 0.3 kg | B1 | Do not penalise below if mass of lamina taken as 15 |
| For \(\bar{x}\): \((0.3 + m) \times 1.5 = 0.3 \times 1.9 + m \times 0\) | M1 | Recognising need to produce equation in terms of \(m\) for \(x\)-component |
| Must be 0 not \(x\) | M1 | |
| \(m = 0.08\) | A1 | FT their 1.9 from (i). If 15 used, accept \(m = 4\) |
| For \(\bar{y}\): \((0.3 + 0.08) \times 1.5 = 0.3 \times \frac{7}{6} + 0.08y\) | M1 | cao. Condone no reference to \(x\) component. Allow obtained using 15 |
| Particle should be at \((0, 2.75)\) | A1 | Allow 2.74, 2.7375, 2.775, 2.625 from various roundings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The c.m. must lie inside KFDL as seen in the plan in Fig. 3.2 | E1 | Some indication of this is what is required. Accept a closed region with KF correct and sides parallel to KL and FD. |
| E1 | Correct. Accept freehand. | |
| The c.m. shown to be in this region | M1 | Recognition that com is at \((1.7, \bar{y})\) and is related to their critical region even if region is incorrect, or calculation with at least 1 correct equation \((3y + 2x = 9\) and \(3y + 4x = 6)\). Do NOT award simply for a recalculation of com as \((1.7, 7/6)\) |
| E1 | Properly established including a statement. (i.e. correct region, correct com marked and statement of stability) | |
| [4] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $15\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 2\begin{pmatrix}-1\\2.5\end{pmatrix} + 9\begin{pmatrix}1.5\\1.5\end{pmatrix} + 2\begin{pmatrix}4\\0.5\end{pmatrix} + 2\begin{pmatrix}4.5\\-1\end{pmatrix}$ | M1 | A systematic method for at least 1 cpt |
| $= \begin{pmatrix}28.5\\17.5\end{pmatrix}$ | A1 | Either all $x$ or all $y$ values correct, or 2 vector terms correct on RHS |
| All components completely correct | A1 | Need not be explicit |
| $\bar{x} = 1.9$ | A1 | Accept any form |
| $\bar{y} = \frac{7}{6}$ | A1 | Accept any form (1.17, 1.2) but not 1.16 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| G is $2 + 1.9$ to right of K and $3 - \frac{7}{6} = \frac{11}{6}$ below K | B1 | FT from (i). May be implied |
| When hanging, G is vertically below K | B1 | May be implied |
| Angle is $\arctan\left(\frac{\frac{11}{6}}{3.9}\right)$ | M1 | o.e. FT their values but must be attempting to find the appropriate angle |
| $= 25.2°$ (3 s.f.) | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New c.m. is at (1.5, 1.5) and mass of object is 0.3 kg | B1 | Do not penalise below if mass of lamina taken as 15 |
| For $\bar{x}$: $(0.3 + m) \times 1.5 = 0.3 \times 1.9 + m \times 0$ | M1 | Recognising need to produce equation in terms of $m$ for $x$-component |
| Must be 0 not $x$ | M1 | |
| $m = 0.08$ | A1 | FT their 1.9 from (i). If 15 used, accept $m = 4$ |
| For $\bar{y}$: $(0.3 + 0.08) \times 1.5 = 0.3 \times \frac{7}{6} + 0.08y$ | M1 | cao. Condone no reference to $x$ component. Allow obtained using 15 |
| Particle should be at $(0, 2.75)$ | A1 | Allow 2.74, 2.7375, 2.775, 2.625 from various roundings |
# Question 3(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The c.m. must lie inside KFDL as seen in the plan in Fig. 3.2 | E1 | Some indication of this is what is required. Accept a closed region with KF correct and sides parallel to KL and FD. |
| | E1 | Correct. Accept freehand. |
| The c.m. shown to be in this region | M1 | Recognition that com is at $(1.7, \bar{y})$ and is related to their critical region even if region is incorrect, or calculation with at least 1 correct equation $(3y + 2x = 9$ and $3y + 4x = 6)$. Do NOT award simply for a recalculation of com as $(1.7, 7/6)$ |
| | E1 | Properly established including a statement. (i.e. correct region, correct com marked and statement of stability) |
| **[4]** | | |
---3 The object shown shaded in Fig. 3.1 is cut from a flat sheet of thin rigid uniform material; LMJK, OAIJ, AEFH and CDEB are rectangles. The grid-lines in Fig. 3.1 are 1 cm apart.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-4_825_1077_210_822}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Calculate the coordinates of the centre of mass of the object referred to the axes shown in Fig. 3.1. [5]
The object is freely suspended from the point K and hangs in equilibrium.\\
(ii) Calculate the angle that KI makes with the vertical.
The mass of the object is 0.3 kg .\\
A particle of mass $m \mathrm {~kg}$ is attached to the object at a point on the line OJ so that the new centre of mass is at the centre of the square OAIJ.\\
(iii) Calculate the value of $m$ and the position of the particle referred to the axes shown in Fig. 3.1.
The extra particle is now removed and the object shown in Fig. 3.1 is folded: LMJK is folded along JM so that it is perpendicular to OAIJ; ABCDEFH is folded along AH so that it is perpendicular to OAIJ and on the same side of OAIJ as LMJK. The folded object is placed on a horizontal table with the edges KL and FED in contact with the table. A plan view and a 3D representation are shown in Fig. 3.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-4_609_648_1836_246}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42b6ee17-f0ae-4687-8392-281ba724a607-4_332_695_2001_1144}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
(iv) On the plan, indicate the region corresponding to positions of the centre of mass for which the folded object is stable.
You are given that the $x$-coordinate of the centre of mass of the folded object is 1.7 . Determine whether the object is stable.
\hfill \mbox{\textit{OCR MEI M2 2013 Q3 [19]}}