| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Toppling on inclined plane |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a trapezium cross-section with routine calculations. Part (i) requires standard centroid formula for a trapezium (show that result), part (ii) involves basic toppling condition (centre of mass above base), and part (iii) applies the same principle on an inclined plane. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{3}a\) | B1 | Centre of mass of triangle |
| \((25 + 2.5a)x_G = 25 \times 2.5 + 2.5a \times (5 + \frac{1}{3}a)\) | M1 | Table of values idea, using any fixed axis |
| A1 | ||
| A1 | Relative to the axis they are using | |
| \(x_G = \dfrac{a^2 + 15a + 75}{3(a+10)}\) AG | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{a^2 + 15a + 75}{3(a+10)} = 5\) | *M1 | Substitute \(x_G\) as \(5\) |
| Solving for \(a\) | dep*M1 | |
| \(a = 8.66\) or \(5\sqrt{3}\) | A1 | \(a \leq 8.66\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((25 + 2.5a)y_G = 25 \times 2.5 + 2.5a \times (\frac{2}{3} \times 5)\) | *M1 | Method to find centre of mass from \(AB\) (or \(CD\)) with or without \(a\) substituted |
| \(y_G = \dfrac{10a + 75}{3(a+10)}\) or \(2.89\) | A1ft | ft their \(a\) from (ii), from \(CD\): \(y_G = 2.11\) |
| \(\tan\theta = x_G / y_G = 5/y_G\) | dep*M1 | Using trig to find an appropriate angle, e.g. complement of \(\theta\) |
| A1ft | ft their \(a\) from (ii), but not an incorrect \(y_G\) | |
| \(\theta = 60°\) | A1 | \(\theta \leq 60°\) (anything that rounds to \(60\)) |
| [6] |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{3}a$ | B1 | Centre of mass of triangle |
| $(25 + 2.5a)x_G = 25 \times 2.5 + 2.5a \times (5 + \frac{1}{3}a)$ | M1 | Table of values idea, using any fixed axis |
| | A1 | |
| | A1 | Relative to the axis they are using |
| $x_G = \dfrac{a^2 + 15a + 75}{3(a+10)}$ **AG** | A1 | |
| **[5]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{a^2 + 15a + 75}{3(a+10)} = 5$ | *M1 | Substitute $x_G$ as $5$ |
| Solving for $a$ | dep*M1 | |
| $a = 8.66$ or $5\sqrt{3}$ | A1 | $a \leq 8.66$ |
| **[3]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(25 + 2.5a)y_G = 25 \times 2.5 + 2.5a \times (\frac{2}{3} \times 5)$ | *M1 | Method to find centre of mass from $AB$ (or $CD$) with or without $a$ substituted |
| $y_G = \dfrac{10a + 75}{3(a+10)}$ or $2.89$ | A1ft | ft their $a$ from (ii), from $CD$: $y_G = 2.11$ |
| $\tan\theta = x_G / y_G = 5/y_G$ | dep*M1 | Using trig to find an appropriate angle, e.g. complement of $\theta$ |
| | A1ft | ft their $a$ from (ii), but not an incorrect $y_G$ |
| $\theta = 60°$ | A1 | $\theta \leq 60°$ (anything that rounds to $60$) |
| **[6]** | | |
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\includegraphics[max width=\textwidth, alt={}, center]{d1eb99a1-04e5-43bc-87b4-d0f7c962135c-4_353_579_248_744}
The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a trapezium $A B C D$ with $A B$ and $C D$ perpendicular to $A D$. The lengths of $A B$ and $A D$ are each 5 cm and the length of $C D$ is $( a + 5 ) \mathrm { cm }$.\\
(i) Show the distance of the centre of mass of the prism from $A D$ is
$$\frac { a ^ { 2 } + 15 a + 75 } { 3 ( a + 10 ) } \mathrm { cm } .$$
The prism is placed with the face containing $A B$ in contact with a horizontal surface.\\
(ii) Find the greatest value of $a$ for which the prism does not topple.
The prism is now placed on an inclined plane which makes an angle $\theta ^ { \circ }$ with the horizontal. $A B$ lies along a line of greatest slope with $B$ higher than $A$.\\
(iii) Using the value for $a$ found in part (ii), and assuming the prism does not slip down the plane, find the greatest value of $\theta$ for which the prism does not topple.
\hfill \mbox{\textit{OCR M2 2012 Q7 [14]}}