OCR M4 2010 June — Question 2 9 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2010
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of lamina by integration
DifficultyChallenging +1.2 This is a standard M4 centre of mass problem requiring integration of exponential functions to find area, first moments, and coordinates. While it involves multiple integrals and exact form answers with logarithms, the technique is routine for Further Maths students and follows a well-practiced algorithm with no conceptual surprises.
Spec6.04d Integration: for centre of mass of laminas/solids
2 The region bounded by the \(x\)-axis, the \(y\)-axis, the line \(x = \ln 3\), and the curve \(y = \mathrm { e } ^ { - x }\) for \(0 \leqslant x \leqslant \ln 3\), is occupied by a uniform lamina. Find, in an exact form, the coordinates of the centre of mass of this lamina.
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Area is \(\int_0^{\ln 3} e^{-x}\,dx\)M1 Limits not required
\(= \left[-e^{-x}\right]_0^{\ln 3} \left(= \frac{2}{3}\right)\)A1 For \(-e^{-x}\)
\(\int xy\,dx = \int_0^{\ln 3} xe^{-x}\,dx\)M1 Limits not required
\(= \left[-xe^{-x} - e^{-x}\right]_0^{\ln 3} \left(= \frac{2}{3} - \frac{1}{3}\ln 3\right)\)M1, A1 Integration by parts; for \(-xe^{-x} - e^{-x}\)
\(\bar{x} = \dfrac{\frac{2}{3} - \frac{1}{3}\ln 3}{\frac{2}{3}} = 1 - \frac{1}{2}\ln 3\)A1
\(\int \frac{1}{2}y^2\,dx = \int_0^{\ln 3} \frac{1}{2}(e^{-x})^2\,dx\)M1 \(\int(e^{-x})^2\,dx\) or \(\int(-\ln y)y\,dy + (\frac{1}{3}\ln 3)\times\frac{1}{6}\)
\(= \left[-\frac{1}{4}e^{-2x}\right]_0^{\ln 3} \left(= \frac{2}{9}\right)\)A1 \(-\frac{1}{4}e^{-2x}\) or \(-\frac{1}{2}y^2\ln y + \frac{1}{4}y^2\) (dep on M1)
\(\bar{y} = \dfrac{\frac{2}{9}}{\frac{2}{3}} = \frac{1}{3}\)A1 [9] Max penalty of 1 mark for correct answers in an unacceptable form (eg decimals) 
# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Area is $\int_0^{\ln 3} e^{-x}\,dx$ | M1 | Limits not required |
| $= \left[-e^{-x}\right]_0^{\ln 3} \left(= \frac{2}{3}\right)$ | A1 | For $-e^{-x}$ |
| $\int xy\,dx = \int_0^{\ln 3} xe^{-x}\,dx$ | M1 | Limits not required |
| $= \left[-xe^{-x} - e^{-x}\right]_0^{\ln 3} \left(= \frac{2}{3} - \frac{1}{3}\ln 3\right)$ | M1, A1 | Integration by parts; for $-xe^{-x} - e^{-x}$ |
| $\bar{x} = \dfrac{\frac{2}{3} - \frac{1}{3}\ln 3}{\frac{2}{3}} = 1 - \frac{1}{2}\ln 3$ | A1 | |
| $\int \frac{1}{2}y^2\,dx = \int_0^{\ln 3} \frac{1}{2}(e^{-x})^2\,dx$ | M1 | $\int(e^{-x})^2\,dx$ or $\int(-\ln y)y\,dy + (\frac{1}{3}\ln 3)\times\frac{1}{6}$ |
| $= \left[-\frac{1}{4}e^{-2x}\right]_0^{\ln 3} \left(= \frac{2}{9}\right)$ | A1 | $-\frac{1}{4}e^{-2x}$ or $-\frac{1}{2}y^2\ln y + \frac{1}{4}y^2$ (dep on M1) |
| $\bar{y} = \dfrac{\frac{2}{9}}{\frac{2}{3}} = \frac{1}{3}$ | A1 [9] | Max penalty of 1 mark for correct answers in an unacceptable form (eg decimals) |

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2 The region bounded by the $x$-axis, the $y$-axis, the line $x = \ln 3$, and the curve $y = \mathrm { e } ^ { - x }$ for $0 \leqslant x \leqslant \ln 3$, is occupied by a uniform lamina. Find, in an exact form, the coordinates of the centre of mass of this lamina.

\hfill \mbox{\textit{OCR M4 2010 Q2 [9]}}
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