| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.8 This M4 question requires finding centres of mass for non-standard laminas using integration. Part (i) involves standard formulas with exponential functions (6 marks), while part (ii) requires handling a piecewise function and computing moment integrals (7 marks). The calculations are lengthy and require careful algebraic manipulation of exponentials, but the techniques are standard for M4. The 13 total marks and multi-step integration place it moderately above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| \(A_1 = \int_0^2 2e^{2x}dx\) | M1* | Attempt at integration to find area; Limits not required for M mark |
| \(= \left[4e^{2x}\right]_0^{\frac{1}{2}} = 4(e - 1)\) | A1 | |
| \(A_1\bar{y} = \frac{1}{2}\int_0^{\frac{1}{2}}\left(2e^{2x}\right)^2 dx = \frac{1}{2}\int_0^2 4e^x dx\) | M1* | Attempt at integration; Limits not required for M and A marks |
| \(= \frac{1}{2}\left[4e^x\right]_0^2 = (2e^2 - 2)\) | A1 | |
| \(\bar{y} = \frac{2e^2 - 2}{4(e - 1)} = \frac{e + 1}{2}\) | M1dep* A1 | M1 for \(\bar{y} = \frac{A_1\bar{y}}{A_1}\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| \(A_\lambda = \frac{1}{2}(3)(2e)\) | B1 | Or by integration |
| \(A_\lambda\bar{x} = \int_0^2 x\left(2e^{-\frac{1}{x}}\right)dx\) | M1* | Integration by parts; Clear indication of integrating exponential terms and differentiating \(x\) term |
| \(= \left[4xe^{-\frac{1}{x}}\right]_0^2 - \int_0^2 4e^{-\frac{1}{x}}dx\) | ||
| \(= 4xe^{-\frac{1}{x}} - 8e^{-\frac{1}{x}}\) (= 8) | A1 A1 | All terms integrated correctly (A1 for one error) |
| \((A_1 + A_\lambda)\bar{x} = cv(8) + 3A_\lambda\) | M1dep* | M1 for table of values idea |
| \(\bar{x}(4e - 4 + 3e) = 8 + 3(3e)\) | A1 | Two terms correct |
| \(x = \frac{9e + 8}{7e - 4}\) | A1 | oe |
| [7] |
**(i)**
$A_1 = \int_0^2 2e^{2x}dx$ | M1* | Attempt at integration to find area; Limits not required for M mark
$= \left[4e^{2x}\right]_0^{\frac{1}{2}} = 4(e - 1)$ | A1 |
$A_1\bar{y} = \frac{1}{2}\int_0^{\frac{1}{2}}\left(2e^{2x}\right)^2 dx = \frac{1}{2}\int_0^2 4e^x dx$ | M1* | Attempt at integration; Limits not required for M and A marks
$= \frac{1}{2}\left[4e^x\right]_0^2 = (2e^2 - 2)$ | A1 |
$\bar{y} = \frac{2e^2 - 2}{4(e - 1)} = \frac{e + 1}{2}$ | M1dep* A1 | M1 for $\bar{y} = \frac{A_1\bar{y}}{A_1}$
| [6] |
**(ii)**
$A_\lambda = \frac{1}{2}(3)(2e)$ | B1 | Or by integration
$A_\lambda\bar{x} = \int_0^2 x\left(2e^{-\frac{1}{x}}\right)dx$ | M1* | Integration by parts; Clear indication of integrating exponential terms and differentiating $x$ term
$= \left[4xe^{-\frac{1}{x}}\right]_0^2 - \int_0^2 4e^{-\frac{1}{x}}dx$ |
$= 4xe^{-\frac{1}{x}} - 8e^{-\frac{1}{x}}$ (= 8) | A1 A1 | All terms integrated correctly (A1 for one error)
$(A_1 + A_\lambda)\bar{x} = cv(8) + 3A_\lambda$ | M1dep* | M1 for table of values idea
$\bar{x}(4e - 4 + 3e) = 8 + 3(3e)$ | A1 | Two terms correct
$x = \frac{9e + 8}{7e - 4}$ | A1 | oe
| [7] |The region bounded by the curve $y = 2e^{\frac{1}{2}x}$ for $0 \leq x \leq 2$, the $x$-axis, the $y$-axis and the line $x = 2$, is occupied by a uniform lamina.
\begin{enumerate}[label=(\roman*)]
\item Find the exact value of the $y$-coordinate of the centre of mass of the lamina. [6]
\end{enumerate}
As shown in the diagram below, a uniform lamina occupies the closed region bounded by the $x$-axis, the $y$-axis and the curve $y = f(x)$ where
$$f(x) = \begin{cases}
2e^{\frac{1}{2}x} & 0 \leq x \leq 2, \\
\frac{2}{3}(5-x)e & 2 \leq x \leq 5.
\end{cases}$$
\includegraphics{figure_4}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the exact value of the $x$-coordinate of the centre of mass of the lamina. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR M4 2016 Q4 [13]}}