| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with cone and cylinder |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring the formula for cone COM (h/4 from base) and weighted average calculation. The algebra is straightforward with the given answer to show, making it slightly easier than average for M3 material. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratio: \(\frac{2}{3}\pi r^2 h \quad \frac{1}{3}\pi r^2 kh \quad \frac{2}{3}\pi r^2 h + \frac{1}{3}\pi r^2 kh\) (or \(2:k:(2+k)\)) | B1 | Correct ratio of volumes/masses - any form |
| Dist from \(O\): \(-\frac{1}{2}h \quad \frac{1}{4}kh \quad \bar{x}\) | B1 | Correct distances from \(O\) or vertex. One distance may be negative |
| \(2\left(-\frac{1}{2}h\right) + k\times\frac{k}{4}h = (2+k)\bar{x}\) | M1, A1ft | Forming moments equation; correct equation (all signs correct for chosen point) |
| \(\bar{x} = \frac{(k^2-4)h}{4(2+k)} = \frac{h(k-2)(k+2)}{4(2+k)} = \frac{h}{4}(k-2)\) | A1cso (5) | Correct completion. Factorisation must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta = \frac{\bar{x}}{r}\) | M1 | Form expression for \(\tan\theta\) using given \(\bar{x}\). \(\bar{x}=h \Rightarrow\) correct \(\bar{x}\) used |
| \(\tan\theta = \frac{32h}{4\times 8\times 3h}\) | A1 | Substitute for \(r\) and \(k\) to obtain correct numerical value for \(\tan\theta\) |
| \(\theta = 18.43\ldots°\) or \(0.321\ldots\) rad; accept \(18°\) or \(0.32\) rad or better | A1 (3) | Correct angle, may be degrees or radians |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio: $\frac{2}{3}\pi r^2 h \quad \frac{1}{3}\pi r^2 kh \quad \frac{2}{3}\pi r^2 h + \frac{1}{3}\pi r^2 kh$ (or $2:k:(2+k)$) | B1 | Correct ratio of volumes/masses - any form |
| Dist from $O$: $-\frac{1}{2}h \quad \frac{1}{4}kh \quad \bar{x}$ | B1 | Correct distances from $O$ or vertex. One distance may be negative |
| $2\left(-\frac{1}{2}h\right) + k\times\frac{k}{4}h = (2+k)\bar{x}$ | M1, A1ft | Forming moments equation; correct equation (all signs correct for chosen point) |
| $\bar{x} = \frac{(k^2-4)h}{4(2+k)} = \frac{h(k-2)(k+2)}{4(2+k)} = \frac{h}{4}(k-2)$ | A1cso (5) | Correct completion. Factorisation **must** be shown |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = \frac{\bar{x}}{r}$ | M1 | Form expression for $\tan\theta$ using given $\bar{x}$. $\bar{x}=h \Rightarrow$ correct $\bar{x}$ used |
| $\tan\theta = \frac{32h}{4\times 8\times 3h}$ | A1 | Substitute for $r$ and $k$ to obtain correct numerical value for $\tan\theta$ |
| $\theta = 18.43\ldots°$ or $0.321\ldots$ rad; accept $18°$ or $0.32$ rad or better | A1 (3) | Correct angle, may be degrees or radians |
---4.
A uniform solid $S$ consists of two right circular cones of base rate\\
has height $2 h$ and the centre of the plane face of this cone is $O$. T\\
$k h$ where $k > 2$. The two cones are joined so that their plane fac\\
Figure 2 .\\
(a) Show that the distance of the centre of mass of $S$ from $O$ is\\
$\frac { h } { 4 } ( k - 2 )$\\
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\hfill \mbox{\textit{Edexcel M3 2016 Q4 [8]}}