| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Deriving standard centre of mass formulae by integration |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question with multiple parts requiring systematic application of learned techniques: proving the cone COM formula (bookwork), combining two bodies using volume ratios, and applying equilibrium conditions. While it requires careful calculation across 4 parts (17 marks total), each step follows established methods without requiring novel insight or particularly challenging problem-solving. |
| Spec | 6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{rx}{h}\) | ||
| \((\pi)\int_0^h xy^2dx = (\pi)\int_0^h \frac{r^2x^3}{h^2}dx\) | M1A1 | |
| \((\pi)\left[\frac{r^2x^4}{4h^2}\right]_0^h = (\pi)\frac{r^2h^2}{4}\) | A1 | |
| \(\bar{x} = \frac{\int xy^2dx}{\int y^2dx} = \frac{\frac{r^2h^2}{4} \div 3}{= \frac{3}{4}h\) * | M1A1cso | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass | ||
| \(\frac{1}{3}\pi \times 5r^3\) | B1 | Correct mass ratio Correct distances from \(O\) or any other point. Both known distances may be positive here, but one must have a minus both here and in the equation or just in the equation. |
| \(\frac{2}{3}\pi r^3\) | ||
| Dist from O | \(\frac{1}{4} \times 5r\) | B1 |
| \(-\frac{3}{8}r\) | ||
| \(\frac{25r}{4} - \frac{3r}{4} = 7\bar{x}\) | M1A1ft | Attempt a moments equation using their mass ratio and distances (signs may be incorrect) Correct equation, follow through their mass ratio and distances but signs to be correct |
| \(\bar{x} = \frac{11r}{14}\) | A1 cao | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \frac{\frac{11r}{14}}{r} = \frac{11}{14}\) | M1A1ft | Using tan to find the required angle. Ratio can be either way up. Tan ratio to be correct way up and have correct numbers, follow through their ans to (b) |
| \(\theta = 38.15...°\) (38 or better) (0.6659... rad; 0.67 or better) | A1 cao | (3) Correct size of angle in degrees or radians, 2 sf or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass of cone \(= \frac{5}{2}M\) or Mass of \(S = \frac{7}{2}M\) or mass of the particle | B1 | Correct mass for cone or for \(S\) |
| \(= \frac{2}{3}\pi r^3\rho\) and work with masses in volume form | ||
| \(kMr(g) = \frac{7}{2}M \times \frac{11r}{14}(g)\) | M1A1ft | |
| \(k = \frac{11}{4}\) or 2.75 | A1 cao | (4) [17] |
| Answer | Marks |
|---|---|
| ALT 1 | Use \(M = \frac{2}{3}\pi r^3\) so mass of particle is \(\frac{2}{3}\pi r^3 k\) and use volumes in the work. Mass of the particle scores B1 provided all the work is using volumes. |
| ALT 2 | Can work with the cone, hemisphere and particle so the equation will have 3 terms (or 4 if \(\bar{x}\) for the c of m of \(S +\) particle is not assumed to be zero. |
| Answer | Marks |
|---|---|
| Aspect | Guidance |
| Integration | Attempting \((\pi)\int_0^h xy^2dx\) \(x\) not needed. No integration needed for this mark but must move to a function of \(x\) only. limits not needed. |
| Correct integral | \(\int_0^h \frac{r^2x^3}{h^2}dx\) with or without \(\pi\) limits needed |
| Integration and limits | Integrate correctly and substitute correct limits |
| Centre of mass formula | Using \(\bar{x} = \frac{\int xy^2dx}{\int y^2dx}\) \(\pi\) in both or neither integral. Denominator need not be in integral form. If mass/unit volume included, must appear in both or neither. |
| Final answer | Correct GIVEN answer cso so \(y = \frac{rx}{h}\) must have been used. If \(h = r\) is assumed: M marks only available. If \(h = 5r\) is assumed max available is 4/5 (Deduct final A mark) |
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{rx}{h}$ | | |
| $(\pi)\int_0^h xy^2dx = (\pi)\int_0^h \frac{r^2x^3}{h^2}dx$ | M1A1 | |
| $(\pi)\left[\frac{r^2x^4}{4h^2}\right]_0^h = (\pi)\frac{r^2h^2}{4}$ | A1 | |
| $\bar{x} = \frac{\int xy^2dx}{\int y^2dx} = \frac{\frac{r^2h^2}{4} \div 3}{= \frac{3}{4}h$ * | M1A1cso | (5) |
**Key Notes:**
- (a)M1: Attempting $(\pi)\int_0^h xy^2dx$ $x$ not needed. No integration needed for this mark but must move to a function of $x$ only. limits not needed.
- (a)A1: Correct integral $\int_0^h \frac{r^2x^3}{h^2}dx$ with or without $\pi$ limits needed
- (a)A1: Integrate correctly and substitute correct limits
- (a)M1: Using $\bar{x} = \frac{\int xy^2dx}{\int y^2dx}$ $\pi$ in both or neither integral. Denominator need not be in integral form. If mass/unit volume included, must appear in both or neither.
- (a)A1cso: Correct GIVEN answer cso so $y = \frac{rx}{h}$ must have been used. If $h = r$ is assumed: M marks only available. If $h = 5r$ is assumed max available is 4/5 (Deduct final A mark)
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Mass** | | |
| $\frac{1}{3}\pi \times 5r^3$ | B1 | Correct mass ratio Correct distances from $O$ or any other point. Both known distances may be positive here, but one must have a minus both here and in the equation or just in the equation. |
| $\frac{2}{3}\pi r^3$ | | |
| **Dist from O** | $\frac{1}{4} \times 5r$ | B1 | |
| $-\frac{3}{8}r$ | | |
| $\frac{25r}{4} - \frac{3r}{4} = 7\bar{x}$ | M1A1ft | Attempt a moments equation using their mass ratio and distances (signs may be incorrect) Correct equation, follow through their mass ratio and distances but signs to be correct |
| $\bar{x} = \frac{11r}{14}$ | A1 cao | (5) |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{\frac{11r}{14}}{r} = \frac{11}{14}$ | M1A1ft | Using tan to find the required angle. Ratio can be either way up. Tan ratio to be correct way up and have correct numbers, follow through their ans to (b) |
| $\theta = 38.15...°$ (38 or better) (0.6659... rad; 0.67 or better) | A1 cao | (3) Correct size of angle in degrees or radians, 2 sf or better |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass of cone $= \frac{5}{2}M$ or Mass of $S = \frac{7}{2}M$ or mass of the particle | B1 | Correct mass for cone or for $S$ |
| $= \frac{2}{3}\pi r^3\rho$ and work with masses in volume form | | |
| $kMr(g) = \frac{7}{2}M \times \frac{11r}{14}(g)$ | M1A1ft | |
| $k = \frac{11}{4}$ or 2.75 | A1 cao | (4) [17] |
**Key Notes:**
- (b)B1: Correct mass ratio
- (b)B1: Correct distances from $O$ or any other point. Both known distances may be positive here, but one must have a minus both here and in the equation or just in the equation.
- (b)M1: Attempt a moments equation using their mass ratio and distances (signs may be incorrect)
- (b)A1ft: Correct equation, follow through their mass ratio and distances but signs to be correct
- (b)A1cao: Correct distance for the c of m
- (c)M1: Using tan to find the required angle. Ratio can be either way up.
- (c)A1ft: Tan ratio to be correct way up and have correct numbers, follow through their ans to (b)
- (c)A1cao: Correct size of angle in degrees or radians, 2 sf or better
- (d)B1: Correct mass for cone or for $S$
- (d)M1: Moments equation about $O$ or another point. This may include an $\bar{x}$ for the c of m of $S +$ particle
- (d)A1ft: Correct nos in the equation, follow through their ans to (b) and mass of $S \neq M$ i to be 0 now
- (d)A1cao: Correct value, 11/4 or 2.75
**ALTs For (d):**
| | |
|---|---|
| **ALT 1** | Use $M = \frac{2}{3}\pi r^3$ so mass of particle is $\frac{2}{3}\pi r^3 k$ and use volumes in the work. Mass of the particle scores B1 provided all the work is using volumes. |
| **ALT 2** | Can work with the cone, hemisphere and particle so the equation will have 3 terms (or 4 if $\bar{x}$ for the c of m of $S +$ particle is not assumed to be zero. |
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**Page Detail Guidance for Question 7(a):**
| Aspect | Guidance |
|---|---|
| Integration | Attempting $(\pi)\int_0^h xy^2dx$ $x$ not needed. No integration needed for this mark but must move to a function of $x$ only. limits not needed. |
| Correct integral | $\int_0^h \frac{r^2x^3}{h^2}dx$ with or without $\pi$ limits needed |
| Integration and limits | Integrate correctly and substitute correct limits |
| Centre of mass formula | Using $\bar{x} = \frac{\int xy^2dx}{\int y^2dx}$ $\pi$ in both or neither integral. Denominator need not be in integral form. If mass/unit volume included, must appear in both or neither. |
| Final answer | Correct GIVEN answer cso so $y = \frac{rx}{h}$ must have been used. If $h = r$ is assumed: M marks only available. If $h = 5r$ is assumed max available is 4/5 (Deduct final A mark) |\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to show that the centre of mass of a uniform solid right circular cone of height $h$ is at a distance $\frac{3}{4}h$ from the vertex of the cone.
[You may assume that the volume of a cone of height $h$ and base radius $r$ is $\frac{1}{3}\pi r^2 h$] [5]
\end{enumerate}
\includegraphics{figure_2}
A uniform solid $S$ consists of a right circular cone, of radius $r$ and height $5r$, fixed to a hemisphere of radius $r$. The centre of the plane face of the hemisphere is $O$ and this plane face coincides with the base of the cone, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the distance of the centre of mass of $S$ from $O$. [5]
\end{enumerate}
The point $A$ lies on the circumference of the base of the cone. The solid is suspended by a string attached at $A$ and hangs freely in equilibrium.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the size of the angle between $OA$ and the vertical. [3]
\end{enumerate}
The mass of the hemisphere is $M$. A particle of mass $kM$ is fixed to the surface of the hemisphere on the axis of symmetry of $S$. The solid is again suspended by the string attached at $A$ and hangs freely in equilibrium. The axis of symmetry of $S$ is now horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the value of $k$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q7 [17]}}