OCR M4 2011 June — Question 2 7 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2011
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeVariable density rod or lamina
DifficultyStandard +0.8 This is a standard Further Maths M4 centre of mass problem with variable density requiring integration by parts twice (for both mass and moment integrals). While the exponential density function and integration by parts elevate it above typical A-level, it follows a well-established template that M4 students practice extensively, making it moderately challenging but not exceptional.
Spec6.04d Integration: for centre of mass of laminas/solids
2 A straight \(\operatorname { rod } A B\) has length \(a\). The rod has variable density, and at a distance \(x\) from \(A\) its mass per unit length is \(k \mathrm { e } ^ { - \frac { x } { a } }\), where \(k\) is a constant. Find, in an exact form, the distance of the centre of mass of the rod from \(A\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(m = \int_0^a k e^{-\frac{x}{a}} \mathrm{d}x\)M1 For \(\int e^{-\frac{x}{a}} \mathrm{d}x\)
\(= k\left[-ae^{-\frac{x}{a}}\right]_0^a \left(= ka(1-e^{-1})\right)\)A1 For \(-ae^{-\frac{x}{a}}\)
\(m\bar{x} = \int_0^a xk e^{-\frac{x}{a}} \mathrm{d}x\)M1 For \(\int xe^{-\frac{x}{a}} \mathrm{d}x\)
\(= k\left[-axe^{-\frac{x}{a}} - a^2e^{-\frac{x}{a}}\right]_0^a\)M1, A1 Integration by parts; For \(-axe^{-\frac{x}{a}} - a^2e^{-\frac{x}{a}}\)
\(= ka^2(1-2e^{-1})\)A1 For \(a^2(1-2e^{-1})\) or exact equivalent
\(\bar{x} = \frac{ka^2(1-2e^{-1})}{ka(1-e^{-1})} = \frac{a(1-2e^{-1})}{1-e^{-1}} = \frac{a(e-2)}{e-1}\)A1
[7] 
# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m = \int_0^a k e^{-\frac{x}{a}} \mathrm{d}x$ | M1 | For $\int e^{-\frac{x}{a}} \mathrm{d}x$ |
| $= k\left[-ae^{-\frac{x}{a}}\right]_0^a \left(= ka(1-e^{-1})\right)$ | A1 | For $-ae^{-\frac{x}{a}}$ |
| $m\bar{x} = \int_0^a xk e^{-\frac{x}{a}} \mathrm{d}x$ | M1 | For $\int xe^{-\frac{x}{a}} \mathrm{d}x$ |
| $= k\left[-axe^{-\frac{x}{a}} - a^2e^{-\frac{x}{a}}\right]_0^a$ | M1, A1 | Integration by parts; For $-axe^{-\frac{x}{a}} - a^2e^{-\frac{x}{a}}$ |
| $= ka^2(1-2e^{-1})$ | A1 | For $a^2(1-2e^{-1})$ or exact equivalent |
| $\bar{x} = \frac{ka^2(1-2e^{-1})}{ka(1-e^{-1})} = \frac{a(1-2e^{-1})}{1-e^{-1}} = \frac{a(e-2)}{e-1}$ | A1 | |
| | [7] | |

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2 A straight $\operatorname { rod } A B$ has length $a$. The rod has variable density, and at a distance $x$ from $A$ its mass per unit length is $k \mathrm { e } ^ { - \frac { x } { a } }$, where $k$ is a constant. Find, in an exact form, the distance of the centre of mass of the rod from $A$.

\hfill \mbox{\textit{OCR M4 2011 Q2 [7]}}
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