| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a standard Further Maths M4 centre of mass problem requiring integration of exponential functions. While it involves multiple steps (finding area, first moments, and coordinates), the integrals are straightforward (∫e^x and ∫e^{2x}), and part (i) is scaffolded as a 'show that' question. The algebraic manipulation is routine for FM students, making this slightly above average difficulty but well within expected M4 competency. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Mass calculation | M1 | \(m = \rho\int_0^{\ln 5} e^x dx = 4\rho\) |
| First moment (x-coordinate) | M1 | \(4\rho\bar{x} = \int_0^{\ln 5} \rho xe^x dx = \rho[xe^x]_0^{\ln 5} - \int_0^{\ln 5} e^x dx = \rho \cdot 5\ln 5 - 4\rho\) |
| x-coordinate of centroid | A1 | \(\bar{x} = \frac{5}{4}\ln 5 - 4\) (shown) |
| First moment (y-coordinate) | M1 | \(4\rho\bar{y} = \int_0^{\ln 5} (\frac{1}{2}y) \cdot \rho y dx = \int_0^{\ln 5} \frac{1}{2}\rho e^{2x} dx = \rho[\frac{1}{4}e^{2x}]_0^{\ln 5} = \frac{1}{4}\rho(25-1) = 6\rho\) |
| y-coordinate of centroid | A1 | \(\bar{y} = 1 \cdot 5\) |
**Mass calculation** | M1 | $m = \rho\int_0^{\ln 5} e^x dx = 4\rho$
**First moment (x-coordinate)** | M1 | $4\rho\bar{x} = \int_0^{\ln 5} \rho xe^x dx = \rho[xe^x]_0^{\ln 5} - \int_0^{\ln 5} e^x dx = \rho \cdot 5\ln 5 - 4\rho$
**x-coordinate of centroid** | A1 | $\bar{x} = \frac{5}{4}\ln 5 - 4$ (shown)
**First moment (y-coordinate)** | M1 | $4\rho\bar{y} = \int_0^{\ln 5} (\frac{1}{2}y) \cdot \rho y dx = \int_0^{\ln 5} \frac{1}{2}\rho e^{2x} dx = \rho[\frac{1}{4}e^{2x}]_0^{\ln 5} = \frac{1}{4}\rho(25-1) = 6\rho$
**y-coordinate of centroid** | A1 | $\bar{y} = 1 \cdot 5$
---5 The region bounded by the $x$-axis, the $y$-axis, the line $x = \ln 5$ and the curve $y = \mathrm { e } ^ { x }$ for $0 \leqslant x \leqslant \ln 5$, is occupied by a uniform lamina.\\
(i) Show that the centre of mass of this lamina has $x$-coordinate
$$\frac { 5 } { 4 } \ln 5 - 1$$
(ii) Find the $y$-coordinate of the centre of mass.
\hfill \mbox{\textit{OCR M4 2002 Q5 [8]}}