| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Folded lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes (rectangles and a triangle) with given formulae, followed by a suspension problem and basic framework analysis. The decomposition into standard shapes is straightforward, and the calculations are routine applications of taught methods with no novel problem-solving required. |
| Spec | 6.04a Centre of mass: gravitational effect6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(300\binom{\bar{x}}{\bar{y}} = 72\binom{-6}{3} + 192\binom{4}{-6} + 36\binom{10}{-4}\) | B1 | Correctly identifying position of c.m. of triangle EFH \((10, -4)\) |
| M1 | A systematic method for at least 1 cpt | |
| B1 | Either all \(x\) or all \(y\) values correct or 2 vector terms correct or allow one common error in both components | |
| \(\binom{\bar{x}}{\bar{y}} = \binom{696}{-1080}\) | ||
| so \(\bar{x} = 2.32\) | A1 | |
| \(\bar{y} = -3.6\) | A1 | Allow FT for either if only error is common to both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| centre of mass is at G; \(\tan\alpha = \frac{9.6}{14.32}\) | M1* | Identifying correct angle. May be implied |
| B1 | At least 1 relevant distance found. FT (i) | |
| M1dep* | Use of \(\arctan\frac{9.6}{14.32}\) or \(\arctan\frac{14.32}{9.6}\) o.e. | |
| so \(\alpha = 33.8376\ldots\) so \(33.8°\) (3 s.f.) | A1 | cao or \(180° - 33.8°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Marking given tension and thrust | B1 | Each labelled with magnitude and correct direction |
| Marking all other forces internal to rods acting on A, B and C (as T or C) | B1 | Need ALL forces at \(A\), \(B\) and \(C\). Need pairs of arrows on \(AB\), \(AC\) and \(BC\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equilibrium at A \(\uparrow\): \(T_{AB}\cos 30 - 18 = 0\) | M1 | Equilibrium at one pin-joint |
| \(T_{AB} = 12\sqrt{3}\). Force in AB: \(12\sqrt{3}\) N (T) | A1 | 20.8. Sign consistent with tension on their diagram |
| \(A \leftarrow\): \(T_{AC} + T_{AB}\cos 60 + 5 = 0\) | M1 | |
| \(T_{AC} = -(5 + 6\sqrt{3})\). Force in AC: \((5 + 6\sqrt{3})\) N (C) | F1 | FT their \(T_{AB}\) |
| At B: \(T_{BR}\cos 60 - T_{AB} = 0\), so \(T_{BR} = 24\sqrt{3}\) | M1 | Allow FT. Other methods possible but award M1 only for complete method leading to \(T_{BC}\) |
| At B in direction BC: \(T_{BC} - T_{BR}\cos 30 = 0\) | ||
| \(T_{BC} = 36\). Force in BC: 36 N (T) | F1 | |
| A1 | cao WWW T/C all correct |
# Question 3:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $300\binom{\bar{x}}{\bar{y}} = 72\binom{-6}{3} + 192\binom{4}{-6} + 36\binom{10}{-4}$ | B1 | Correctly identifying position of c.m. of triangle EFH $(10, -4)$ |
| | M1 | A systematic method for at least 1 cpt |
| | B1 | Either all $x$ or all $y$ values correct or 2 vector terms correct or allow one common error in both components |
| $\binom{\bar{x}}{\bar{y}} = \binom{696}{-1080}$ | | |
| so $\bar{x} = 2.32$ | A1 | |
| $\bar{y} = -3.6$ | A1 | Allow FT for either if only error is common to both |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| centre of mass is at G; $\tan\alpha = \frac{9.6}{14.32}$ | M1* | Identifying correct angle. May be implied |
| | B1 | At least 1 relevant distance found. FT (i) |
| | M1dep* | Use of $\arctan\frac{9.6}{14.32}$ or $\arctan\frac{14.32}{9.6}$ o.e. |
| so $\alpha = 33.8376\ldots$ so $33.8°$ (3 s.f.) | A1 | cao or $180° - 33.8°$ |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Marking given tension and thrust | B1 | Each labelled with magnitude and correct direction |
| Marking all other forces internal to rods acting on A, B and C (as T or C) | B1 | Need ALL forces at $A$, $B$ and $C$. Need pairs of arrows on $AB$, $AC$ and $BC$ |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equilibrium at A $\uparrow$: $T_{AB}\cos 30 - 18 = 0$ | M1 | Equilibrium at one pin-joint |
| $T_{AB} = 12\sqrt{3}$. Force in AB: $12\sqrt{3}$ N (T) | A1 | 20.8. Sign consistent with tension on their diagram |
| $A \leftarrow$: $T_{AC} + T_{AB}\cos 60 + 5 = 0$ | M1 | |
| $T_{AC} = -(5 + 6\sqrt{3})$. Force in AC: $(5 + 6\sqrt{3})$ N (C) | F1 | FT their $T_{AB}$ |
| At B: $T_{BR}\cos 60 - T_{AB} = 0$, so $T_{BR} = 24\sqrt{3}$ | M1 | Allow FT. Other methods possible but award M1 only for complete method leading to $T_{BC}$ |
| At B in direction BC: $T_{BC} - T_{BR}\cos 30 = 0$ | | |
| $T_{BC} = 36$. Force in BC: 36 N (T) | F1 | |
| | A1 | cao WWW T/C all correct |
---
3
\begin{enumerate}[label=(\alph*)]
\item You are given that the position of the centre of mass, G , of a right-angled triangle cut from thin uniform material in the position shown in Fig. 3.1 is at the point $\left( \frac { 1 } { 3 } a , \frac { 1 } { 3 } b \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-4_328_382_360_845}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
A plane thin uniform sheet of metal is in the shape OABCDEFHIJO shown in Fig. 3.2. BDEA and CDIJ are rectangles and FEH is a right angle. The lengths of the sides are shown with each unit representing 1 cm .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-4_862_906_1032_584}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the coordinates of the centre of mass of the metal sheet, referred to the axes shown in Fig. 3.2.
The metal sheet is freely suspended from corner B and hangs in equilibrium.
\item Calculate the angle between BD and the vertical.
\end{enumerate}\item Part of a framework of light rigid rods freely pin-jointed at their ends is shown in Fig. 3.3. The framework is in equilibrium.
All the rods meeting at the pin-joints at $\mathrm { A } , \mathrm { B }$ and C are shown. The rods connected to $\mathrm { A } , \mathrm { B }$ and C are connected to the rest of the framework at $\mathrm { P } , \mathrm { Q } , \mathrm { R } , \mathrm { S }$ and T .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-5_499_734_493_662}
\captionsetup{labelformat=empty}
\caption{Fig. 3.3}
\end{center}
\end{figure}
There is a tension of 18 N in rod AP and a thrust (compression) of 5 N in rod AQ.
\begin{enumerate}[label=(\roman*)]
\item Show the forces internal to the rods acting on the pin-joints at $\mathrm { A } , \mathrm { B }$ and C .
\item Calculate the forces internal to the rods $\mathrm { AB } , \mathrm { BC }$ and CA , stating whether each rod is in tension or compression. [You may leave your answers in surd form. Your working in this part should be consistent with your diagram in part (i).]\\
$4 P$ and $Q$ are circular discs of mass 3 kg and 10 kg respectively which slide on a smooth horizontal surface. The discs have the same diameter and move in the line joining their centres with no resistive forces acting on them. The surface has vertical walls which are perpendicular to the line of centres of the discs. This information is shown in Fig. 4 together with the direction you should take as being positive.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ea3c0177-bf3b-4475-9ab1-ae628aeb0bf0-6_430_1404_443_328}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item For what time must a force of 26 N act on P to accelerate it from rest to $13 \mathrm {~ms} ^ { - 1 }$ ?
P is travelling at $13 \mathrm {~ms} ^ { - 1 }$ when it collides with Q , which is at rest. The coefficient of restitution in this collision is $e$.
\item Show that, after the collision, the velocity of P is $( 3 - 10 e ) \mathrm { ms } ^ { - 1 }$ and find an expression in terms of $e$ for the velocity of Q.
\item For what set of values of $e$ does the collision cause P to reverse its direction of motion?
\item Determine the set of values of $e$ for which P has a greater speed than Q immediately after the collision.
You are now given that $e = \frac { 1 } { 2 }$. After P and Q collide with one another, each has a perfectly elastic collision with a wall. P and Q then collide with one another again and in this second collision they stick together (coalesce).
\item Determine the common velocity of P and Q .
\item Determine the impulse of Q on P in this collision.
\end{enumerate}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M2 2012 Q3 [18]}}