| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.8 This is a standard M4 centre of mass problem requiring integration to find coordinates of a lamina under a parabola. Part (i) uses the standard formula for x̄ with straightforward polynomial integration. Part (ii) adds a constraint requiring ȳ calculation and solving for k. While methodical and multi-step, it follows textbook procedures without requiring novel insight, placing it moderately above average difficulty. |
| Spec | 1.08a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \int_1^2 kx^2\,dx = \left[\frac{kx^3}{3}\right]_1^2 = \frac{7k}{3}\) | *M1 A1 | M1 attempt to integrate to find area. Limits not required for M mark |
| \(A\bar{x} = \int_1^2 x(kx^2)\,dx = \left[\frac{1}{4}kx^4\right]_1^2 = \frac{15k}{4}\) | *M1 A1 | M1 for \(\int x^3\,dx\) and attempt to integrate. Limits not required for M mark |
| \(\bar{x} = \left(\frac{15k}{4}\right)\left(\frac{3}{7k}\right)\) | M1 dep* | M1 for \(\bar{x} = \frac{A\bar{x}}{A}\) |
| \(\bar{x} = \frac{45}{28}\) | A1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A\bar{y} = \frac{1}{2}\int_1^2 (kx^2)^2\,dx = \frac{1}{2}k^2\left[\frac{x^5}{5}\right]_1^2 = \frac{31k^2}{10}\) | *M1 A1 | M1 for \(\frac{1}{2}\int y^2\,dx\) and attempt to integrate |
| \(\left(\frac{31k^2}{10}\right)\left(\frac{3}{7k}\right) = \frac{45}{28}\) | M1 dep* | \(\text{cv}(\bar{x}) = \text{cv}(\bar{y})\) - this mark is dependent on scoring all M marks in (i) and (ii) |
| \(k = \frac{75}{62}\) | A1 [4] | Cao |
| Or \(A\bar{y} = \frac{1}{4\sqrt{k}}\int_k^{4k} y^{\frac{3}{2}}\,dy = \frac{1}{4\sqrt{k}}\left[\frac{2}{5}y^{\frac{5}{2}}\right]_k^{4k} = \frac{31k^2}{10}\) | M1 for \(\frac{1}{4\sqrt{k}}\int y^{\frac{3}{2}}\,dy\) and attempt to integrate |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \int_1^2 kx^2\,dx = \left[\frac{kx^3}{3}\right]_1^2 = \frac{7k}{3}$ | *M1 A1 | M1 attempt to integrate to find area. Limits not required for M mark |
| $A\bar{x} = \int_1^2 x(kx^2)\,dx = \left[\frac{1}{4}kx^4\right]_1^2 = \frac{15k}{4}$ | *M1 A1 | M1 for $\int x^3\,dx$ and attempt to integrate. Limits not required for M mark |
| $\bar{x} = \left(\frac{15k}{4}\right)\left(\frac{3}{7k}\right)$ | M1 dep* | M1 for $\bar{x} = \frac{A\bar{x}}{A}$ |
| $\bar{x} = \frac{45}{28}$ | A1 **[6]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A\bar{y} = \frac{1}{2}\int_1^2 (kx^2)^2\,dx = \frac{1}{2}k^2\left[\frac{x^5}{5}\right]_1^2 = \frac{31k^2}{10}$ | *M1 A1 | M1 for $\frac{1}{2}\int y^2\,dx$ and attempt to integrate |
| $\left(\frac{31k^2}{10}\right)\left(\frac{3}{7k}\right) = \frac{45}{28}$ | M1 dep* | $\text{cv}(\bar{x}) = \text{cv}(\bar{y})$ - this mark is dependent on scoring all M marks in (i) and (ii) |
| $k = \frac{75}{62}$ | A1 **[4]** | Cao |
| Or $A\bar{y} = \frac{1}{4\sqrt{k}}\int_k^{4k} y^{\frac{3}{2}}\,dy = \frac{1}{4\sqrt{k}}\left[\frac{2}{5}y^{\frac{5}{2}}\right]_k^{4k} = \frac{31k^2}{10}$ | | M1 for $\frac{1}{4\sqrt{k}}\int y^{\frac{3}{2}}\,dy$ and attempt to integrate |
---2 The region bounded by the $x$-axis, the lines $x = 1$ and $x = 2$, and the curve $y = k x ^ { 2 }$, where $k$ is a positive constant, is occupied by a uniform lamina.\\
(i) Find the exact $x$-coordinate of the centre of mass of the lamina.\\
(ii) Given that the $x$ - and $y$-coordinates of the centre of mass of the lamina are equal, find the exact value of $k$.
\hfill \mbox{\textit{OCR M4 2015 Q2 [10]}}