Edexcel M3 2013 June — Question 5 13 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2013
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeComposite solid with standard shapes - calculation only
DifficultyStandard +0.8 This is a two-part M3 centre of mass question requiring: (a) volume integration using the disc method with algebraic manipulation of (x+1)^4, and (b) combining two solids with different densities. While the techniques are standard for M3, the algebraic complexity and the density ratio consideration elevate it slightly above average difficulty for A-level.
Spec4.08d Volumes of revolution: about x and y axes6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_622_1186_251_443} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = ( x + 1 ) ^ { 2 }\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 3. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).
  1. Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).
    (8) \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_558_492_1263_703} \captionsetup{labelformat=empty} \caption{Figure 4}
    \end{figure} A uniform solid hemisphere is fixed to \(S\) to form a solid \(T\). The hemisphere has the same radius as the smaller plane face of \(S\) and its plane face coincides with the smaller plane face of \(S\), as shown in Figure 4. The mass per unit volume of the hemisphere is 10 times the mass per unit volume of \(S\). The centre of the circular plane face of \(T\) is \(A\). All lengths are measured in centimetres.
  2. Find the distance of the centre of mass of \(T\) from \(A\).
(a)
M1 for using \(V = \int_0^2 \pi y^2 \, dx = \pi\int_0^2 (x+1)^4 \, dx\). Limits not needed. Attempting the integration by inspection or expansion (algebra must be seen).
A1 for correct integration. Limits not needed.
M1 for substituting the correct limits into their integrated function. No need to simplify.
M1 for attempting to integrate \(\int_0^2 \pi y^2 x \, dx = \pi\int_0^2 x(x+1)^4 \, dx\). Limits not needed. By parts. This mark can be awarded once the integral has been expressed as the difference of an appropriate integrated function and an integral.
A1 for correct, complete integration \(\pi\left[\frac{x(x+1)^5}{5}\right]_0^2 - \pi\left[\frac{(x+1)^6}{30}\right]_0^2\) or \(\frac{2 \times 3^5\pi}{5} - \pi\left[\frac{(x+1)^6}{30}\right]_0^2\). Limits not needed.
M1 for substituting the correct limits into their integrated function. No need to simplify.
Alternative methods for \(\int_0^2 \pi y^2 x \, dx = \pi\int_0^2 x(x+1)^4 \, dx\):
M1 for expanding and integrating or making a suitable substitution and attempting the integration. Limits not needed.
A1 for correct integration. Limits not needed.
M1 for substituting the correct limits into their integrated function. No need to simplify.
M1 for using \(x = \frac{\int \pi y^2 x \, dx}{\int \pi y^2 \, dx}\). Their integrals need not be correct.
A1cao for \(x = 1.5068\ldots\) Accept 1.5, 1.51 or better or \(\frac{547}{363}\).
(b)
B1ft for correct mass ratio, follow through their volume for S. Need \(\pi\) now.
B1ft for correct distances, follow through their distance for S, but remember it must be \(2 -\) answer from (a) if working from A. Distances from the common face are \(-3\), answer from (a), \(x\). Distances from other end are \(5\), \(1 +\) answer from (a), \(x\).
M1 for a dimensionally correct moments equation.
A1ft for a fully correct moments equation, follow through their distances and mass ratio.
A1cao for \(0.7208\ldots\) Accept 0.72 or better. Exact is \(\frac{1191}{1652}\).
(a)

M1 for using $V = \int_0^2 \pi y^2 \, dx = \pi\int_0^2 (x+1)^4 \, dx$. Limits not needed. Attempting the integration by inspection or expansion (algebra must be seen).

A1 for correct integration. Limits not needed.

M1 for substituting the correct limits into their integrated function. No need to simplify.

M1 for attempting to integrate $\int_0^2 \pi y^2 x \, dx = \pi\int_0^2 x(x+1)^4 \, dx$. Limits not needed. By parts. This mark can be awarded once the integral has been expressed as the difference of an appropriate integrated function and an integral.

A1 for correct, complete integration $\pi\left[\frac{x(x+1)^5}{5}\right]_0^2 - \pi\left[\frac{(x+1)^6}{30}\right]_0^2$ or $\frac{2 \times 3^5\pi}{5} - \pi\left[\frac{(x+1)^6}{30}\right]_0^2$. Limits not needed.

M1 for substituting the correct limits into their integrated function. No need to simplify.

**Alternative methods for** $\int_0^2 \pi y^2 x \, dx = \pi\int_0^2 x(x+1)^4 \, dx$:

M1 for expanding and integrating or making a suitable substitution and attempting the integration. Limits not needed.

A1 for correct integration. Limits not needed.

M1 for substituting the correct limits into their integrated function. No need to simplify.

M1 for using $x = \frac{\int \pi y^2 x \, dx}{\int \pi y^2 \, dx}$. Their integrals need not be correct.

A1cao for $x = 1.5068\ldots$ Accept 1.5, 1.51 or better or $\frac{547}{363}$.

(b)

B1ft for correct mass ratio, follow through their volume for S. Need $\pi$ now.

B1ft for correct distances, follow through their distance for S, but remember it must be $2 -$ answer from (a) if working from A. Distances from the common face are $-3$, answer from (a), $x$. Distances from other end are $5$, $1 +$ answer from (a), $x$.

M1 for a dimensionally correct moments equation.

A1ft for a fully correct moments equation, follow through their distances and mass ratio.

A1cao for $0.7208\ldots$ Accept 0.72 or better. Exact is $\frac{1191}{1652}$.

---
5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_622_1186_251_443}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

The shaded region $R$ is bounded by the curve with equation $y = ( x + 1 ) ^ { 2 }$, the $x$-axis, the $y$-axis and the line with equation $x = 2$, as shown in Figure 3. The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid $S$.
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to find the $x$ coordinate of the centre of mass of $S$.\\
(8)

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_558_492_1263_703}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

A uniform solid hemisphere is fixed to $S$ to form a solid $T$. The hemisphere has the same radius as the smaller plane face of $S$ and its plane face coincides with the smaller plane face of $S$, as shown in Figure 4. The mass per unit volume of the hemisphere is 10 times the mass per unit volume of $S$. The centre of the circular plane face of $T$ is $A$. All lengths are measured in centimetres.
\item Find the distance of the centre of mass of $T$ from $A$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2013 Q5 [13]}}
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