| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.8 This is a substantial multi-part centre of mass question requiring 3D coordinate geometry, composite body calculations with different densities (faces vs edges), and applications to equilibrium/tipping conditions. The progression from basic COM finding through to tipping analysis and friction coefficients involves multiple sophisticated techniques, though each individual step follows standard M2 methods without requiring novel insight. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let the mass of each face be \(m\); \(5m\bar{z} = 4m \times \frac{a}{2} + 0 \times m\) | M1 | Any complete method. Accept no mention of \(m\) oe |
| A1 | ||
| \(\bar{z} = \frac{2a}{5}\) | A1 | CLOSED/bottomless box is NOT a MR. Mark as per scheme giving method mark if appropriate: max 1/3 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| By symmetry \(\bar{x} = 0\) or by calculation | M1 A1 | Can be awarded if closed/bottomless box used |
| \(8m\bar{y} = 5m\frac{a}{2} + 2m\frac{a}{2} + ma\) | M1 | Any complete method. Accept no mention of \(m\) oe |
| \(= \frac{9a}{2}\) so \(\bar{y} = \frac{9a}{16}\) | E1 | Shown (answer given) |
| \(8m\bar{z} = 5m\frac{2a}{5} + 3ma\) | M1 | Any complete method. Accept no mention of \(m\) oe |
| \(= 5a\) so \(\bar{z} = \frac{5a}{8}\) | A1 | cao. CLOSED/bottomless box: max M1A1M1A0M1A0 |
| [6] | ||
| Alternative: \(8m\begin{pmatrix}\bar{x}\\\bar{y}\\\bar{z}\end{pmatrix} = m\begin{pmatrix}0\\0\\a/2\end{pmatrix} + m\begin{pmatrix}0\\a/2\\0\end{pmatrix} + m\begin{pmatrix}a/2\\a/2\\a/2\end{pmatrix} + m\begin{pmatrix}0\\a\\a/2\end{pmatrix} + m\begin{pmatrix}-a/2\\a/2\\a/2\end{pmatrix} + m\begin{pmatrix}-a/2\\a/2\\a\end{pmatrix} + m\begin{pmatrix}0\\a\\a\end{pmatrix} + m\begin{pmatrix}a/2\\a/2\\a\end{pmatrix}\) | M1 A1 | Each coordinate; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram showing G vertically above bottom edge | B1 | G vertically above bottom edge |
| Use of their \(\bar{z}\) and \(a - \bar{y}\) oe | B1 | Use of their \(\bar{z}\) and \(a - \bar{y}\) oe |
| Use of tan (or equivalent) with either \(\bar{z}\) or \(a - \bar{z}\) and \(\bar{y}\) or \(a - \bar{y}\) | M1 | |
| \(\tan\theta = \dfrac{\frac{7a}{16}}{\frac{5a}{8}} = 0.7\) | M1 | (or equivalent) |
| \(\theta = 34.992...\) so \(35°\) (3 s.f.) | A1 | cao. 55 as answer can get B1B1M1M0A0: 3/5 |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Friction \(F\) N, normal reaction \(R\) N | Allow \(8a^2\) as \(M\) throughout | |
| \(R = Mg\cos 30\) | B1 | |
| N2L down plane: \(Mg\sin 30 - F = 2M\) | M1 | Attempt to use N2L with all terms (allow a missing \(g\)). Allow sign errors |
| A1 | ||
| \(F = \mu R\) | M1 | Used correctly |
| \(\mu = \dfrac{g\sin 30 - 2}{g\cos 30} = 0.34169...\) | ||
| so \(0.342\) (3 s.f.) | A1 | |
| [5] |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let the mass of each face be $m$; $5m\bar{z} = 4m \times \frac{a}{2} + 0 \times m$ | M1 | Any complete method. Accept no mention of $m$ oe |
| | A1 | |
| $\bar{z} = \frac{2a}{5}$ | A1 | CLOSED/bottomless box is NOT a MR. Mark as per scheme giving method mark if appropriate: max 1/3 |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| By symmetry $\bar{x} = 0$ or by calculation | M1 A1 | Can be awarded if closed/bottomless box used |
| $8m\bar{y} = 5m\frac{a}{2} + 2m\frac{a}{2} + ma$ | M1 | Any complete method. Accept no mention of $m$ oe |
| $= \frac{9a}{2}$ so $\bar{y} = \frac{9a}{16}$ | E1 | Shown (answer given) |
| $8m\bar{z} = 5m\frac{2a}{5} + 3ma$ | M1 | Any complete method. Accept no mention of $m$ oe |
| $= 5a$ so $\bar{z} = \frac{5a}{8}$ | A1 | cao. CLOSED/bottomless box: max M1A1M1A0M1A0 |
| **[6]** | | |
| **Alternative:** $8m\begin{pmatrix}\bar{x}\\\bar{y}\\\bar{z}\end{pmatrix} = m\begin{pmatrix}0\\0\\a/2\end{pmatrix} + m\begin{pmatrix}0\\a/2\\0\end{pmatrix} + m\begin{pmatrix}a/2\\a/2\\a/2\end{pmatrix} + m\begin{pmatrix}0\\a\\a/2\end{pmatrix} + m\begin{pmatrix}-a/2\\a/2\\a/2\end{pmatrix} + m\begin{pmatrix}-a/2\\a/2\\a\end{pmatrix} + m\begin{pmatrix}0\\a\\a\end{pmatrix} + m\begin{pmatrix}a/2\\a/2\\a\end{pmatrix}$ | M1 A1 | Each coordinate; cao |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram showing G vertically above bottom edge | B1 | G vertically above bottom edge |
| Use of their $\bar{z}$ and $a - \bar{y}$ oe | B1 | Use of their $\bar{z}$ and $a - \bar{y}$ oe |
| Use of tan (or equivalent) with either $\bar{z}$ or $a - \bar{z}$ and $\bar{y}$ or $a - \bar{y}$ | M1 | |
| $\tan\theta = \dfrac{\frac{7a}{16}}{\frac{5a}{8}} = 0.7$ | M1 | (or equivalent) |
| $\theta = 34.992...$ so $35°$ (3 s.f.) | A1 | cao. 55 as answer can get B1B1M1M0A0: 3/5 |
| **[5]** | | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Friction $F$ N, normal reaction $R$ N | | Allow $8a^2$ as $M$ throughout |
| $R = Mg\cos 30$ | B1 | |
| N2L down plane: $Mg\sin 30 - F = 2M$ | M1 | Attempt to use N2L with all terms (allow a missing $g$). Allow sign errors |
| | A1 | |
| $F = \mu R$ | M1 | Used correctly |
| $\mu = \dfrac{g\sin 30 - 2}{g\cos 30} = 0.34169...$ | | |
| so $0.342$ (3 s.f.) | A1 | |
| **[5]** | | |
---2 Fig. 2.1 shows the positions of the points $\mathrm { P } , \mathrm { Q } , \mathrm { R } , \mathrm { S } , \mathrm { T } , \mathrm { U } , \mathrm { V }$ and W which are at the vertices of a cube of side $a$; Fig. 2.1 also shows coordinate axes, where O is the mid-point of PQ .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-3_510_494_365_788}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
An open box, A, is made from thin uniform material in the form of the faces of the cube with just the face TUVW missing.\\
(i) Find the $z$-coordinate of the centre of mass of A .
Strips made of a thin heavy material are now fixed to the edges TW, WV and VU of box A, as shown in Fig. 2.2. Each of these three strips has the same mass as one face of the box. This new object is B.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-3_488_476_1388_797}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
(ii) Find the $x$-and $z$-coordinates of the centre of mass of B and show that the $y$-coordinate is $\frac { 9 a } { 16 }$.
Object B is now placed on a plane which is inclined at $\theta$ to the horizontal. B is positioned so that face PQRS is on the plane with SR at right angles to a line of greatest slope of the plane and with PQ higher than SR , as shown in Fig. 2.3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{334b2170-3708-46db-bff7-bcad7d5fab00-3_237_284_2087_1555}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}
(iii) Assuming that B does not slip, find $\theta$ if B is on the point of tipping.
B is now placed on a different plane which is inclined at $30 ^ { \circ }$ to the horizontal. When B is released it accelerates down the plane at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iv) Calculate the coefficient of friction between B and the inclined plane.
\hfill \mbox{\textit{OCR MEI M2 2014 Q2 [19]}}