| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Standard +0.3 This is a standard M2 moments question with a composite rigid body (straight rod + circular arc). Part (i) requires finding the center of mass of a circular arc using standard formulas, part (ii) involves taking moments about the hinge (routine equilibrium), and part (iii) requires resolving forces. The projectile motion question is basic kinematics with standard SUVAT equations. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(OG = 2\sin(\pi/8) \div (\pi/8)\) | B1 | \((= 1.94899)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[2\sin(\pi/8) \div (\pi/8)\right] \times \sin(\pi/8)\) | B1 ft | \((= 0.74585)\) ie. horiz cpt of candidates \(OG\) |
| \((\sqrt{8} - 16\sin^2(\pi/8)) \div \pi = 2.82843 - 0.74585\) | M1 | For attempting to find \(OA -\) distance from \(G\) to \(OC\) (subtract two horizontal distances) |
| Distance is \(2.083\) m | A1 | 4 (from figures which give required accuracy) |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For taking moments about \(A\) (3 terms required) | |
| \(4 \times 1\sin 45° + 3 \times 2.083 = T \times 2\) | A1 | |
| \(T = 4.54\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal component is \(4.54\) N and vertical component is \(7\) N | B1 ft | 1 |
## Question 6:
**(i)**
$OG = 2\sin(\pi/8) \div (\pi/8)$ | B1 | $(= 1.94899)$
Distance from $G$ go $OC =$
$\left[2\sin(\pi/8) \div (\pi/8)\right] \times \sin(\pi/8)$ | B1 ft | $(= 0.74585)$ ie. horiz cpt of candidates $OG$
$(\sqrt{8} - 16\sin^2(\pi/8)) \div \pi = 2.82843 - 0.74585$ | M1 | For attempting to find $OA -$ distance from $G$ to $OC$ (subtract two horizontal distances)
Distance is $2.083$ m | A1 | 4 (from figures which give required accuracy)
**(ii)**
| M1 | For taking moments about $A$ (3 terms required)
$4 \times 1\sin 45° + 3 \times 2.083 = T \times 2$ | A1 |
$T = 4.54$ | A1 | 3
**(iii)**
Horizontal component is $4.54$ N and vertical component is $7$ N | B1 ft | 1
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{6fe2c5e0-0496-4fb4-95d2-354b90607b5b-4_620_899_644_623}
A rigid rod consists of two parts. The part $B C$ is in the form of an arc of a circle of radius 2 m and centre $O$, with angle $B O C = \frac { 1 } { 4 } \pi$ radians. $B C$ is uniform and has weight 3 N . The part $A B$ is straight and of length 2 m ; it is uniform and has weight 4 N . The part $A B$ of the rod is a tangent to the arc $B C$ at $B$. The end $A$ of the rod is freely hinged to a fixed point of a vertical wall. The rod is held in equilibrium, with the straight part $A B$ making an angle of $\frac { 1 } { 4 } \pi$ radians with the wall, by means of a horizontal string attached to $C$. The string is in the same vertical plane as the rod, and the tension in the string is $T \mathrm {~N}$ (see diagram).\\
(i) Show that the centre of mass $G$ of the part $B C$ of the rod is at a distance of 2.083 m from the wall, correct to 4 significant figures.\\
(ii) Find the value of $T$.\\
(iii) State the magnitude of the horizontal component and the magnitude of the vertical component of the force exerted on the rod by the hinge.\\
\includegraphics[max width=\textwidth, alt={}, center]{6fe2c5e0-0496-4fb4-95d2-354b90607b5b-5_579_1118_264_516}
A particle $A$ is released from rest at time $t = 0$, at a point $P$ which is 7 m above horizontal ground. At the same instant as $A$ is released, a particle $B$ is projected from a point $O$ on the ground. The horizontal distance of $O$ from $P$ is 24 m . Particle $B$ moves in the vertical plane containing $O$ and $P$, with initial speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and initial direction making an angle of $\theta$ above the horizontal (see diagram). Write down\\
(i) an expression for the height of $A$ above the ground at time $t \mathrm {~s}$,\\
(ii) an expression in terms of $V , \theta$ and $t$ for
\begin{enumerate}[label=(\alph*)]
\item the horizontal distance of $B$ from $O$,
\item the height of $B$ above the ground.
At time $t = T$ the particles $A$ and $B$ collide at a point above the ground.\\
(iii) Show that $\tan \theta = \frac { 7 } { 24 }$ and that $V T = 25$.\\
(iv) Deduce that $7 V ^ { 2 } > 3125$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2005 Q6 [8]}}