OCR M4 2013 June — Question 2 7 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2013
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeVariable density rod or lamina
DifficultyStandard +0.8 This is a standard variable density centre of mass problem requiring integration with a substitution (due to the square root term). While it involves multiple steps (finding total mass, then first moment, then dividing), the setup is straightforward and the integration technique is routine for M4 students. It's moderately harder than average due to the algebraic manipulation required, but follows a well-practiced template.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums6.04b Find centre of mass: using symmetry
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 given by \(k \left( 4 - \sqrt { \frac { x } { a } } \right)\), where \(k\) is a constant. Find the distance from \(A\) of the centre of mass of the rod.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(M = \int_0^a k\!\left(4 - \sqrt{\dfrac{x}{a}}\right)dx\)M1 For \(\int\!\left(4 - \sqrt{\dfrac{x}{a}}\right)dx\)
\(= k\!\left[4x - \dfrac{2}{3}a^{-\frac{1}{2}}x^{\frac{3}{2}}\right]_0^a \left(= \dfrac{10}{3}ka\right)\)A1 For \(4x - \dfrac{2}{3}a^{-\frac{1}{2}}x^{\frac{3}{2}}\)
\(M\bar{x} = \int_0^a k\!\left(4 - \sqrt{\dfrac{x}{a}}\right)x\,dx\)M1 For \(\int\!\left(4 - \sqrt{\dfrac{x}{a}}\right)x\,dx\)
\(= k\!\left[2x^2 - \dfrac{2}{5}a^{-\frac{1}{2}}x^{\frac{5}{2}}\right]_0^a \left(= \dfrac{8}{5}ka^2\right)\)A2 For \(2x^2 - \dfrac{2}{5}a^{-\frac{1}{2}}x^{\frac{5}{2}}\) — Give A1 for one correct term
\(\bar{x} = \dfrac{\frac{8}{5}ka^2}{\frac{10}{3}ka}\)M1 Dependent on previous M1M1
\(= \dfrac{12}{25}a = 0.48a\)A1
[7] 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M = \int_0^a k\!\left(4 - \sqrt{\dfrac{x}{a}}\right)dx$ | M1 | For $\int\!\left(4 - \sqrt{\dfrac{x}{a}}\right)dx$ |
| $= k\!\left[4x - \dfrac{2}{3}a^{-\frac{1}{2}}x^{\frac{3}{2}}\right]_0^a \left(= \dfrac{10}{3}ka\right)$ | A1 | For $4x - \dfrac{2}{3}a^{-\frac{1}{2}}x^{\frac{3}{2}}$ |
| $M\bar{x} = \int_0^a k\!\left(4 - \sqrt{\dfrac{x}{a}}\right)x\,dx$ | M1 | For $\int\!\left(4 - \sqrt{\dfrac{x}{a}}\right)x\,dx$ |
| $= k\!\left[2x^2 - \dfrac{2}{5}a^{-\frac{1}{2}}x^{\frac{5}{2}}\right]_0^a \left(= \dfrac{8}{5}ka^2\right)$ | A2 | For $2x^2 - \dfrac{2}{5}a^{-\frac{1}{2}}x^{\frac{5}{2}}$ — Give A1 for one correct term |
| $\bar{x} = \dfrac{\frac{8}{5}ka^2}{\frac{10}{3}ka}$ | M1 | Dependent on previous M1M1 |
| $= \dfrac{12}{25}a = 0.48a$ | 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 given by $k \left( 4 - \sqrt { \frac { x } { a } } \right)$, where $k$ is a constant. Find the distance from $A$ of the centre of mass of the rod.

\hfill \mbox{\textit{OCR M4 2013 Q2 [7]}}
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