| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod hinged or with applied force |
| Difficulty | Standard +0.3 This is a standard non-uniform rod equilibrium problem requiring moment equations about a point and resolution of forces. The multi-part structure (finding center of mass location, then new force P, then coefficient of friction) follows a typical progression. While it requires careful setup of moment equations and trigonometry, it's a routine M2 mechanics question with no novel insights needed—slightly easier than average due to its predictable structure. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Notes |
| \(0.3g = 6e/0.8\) | M1 | Uses \(T = \frac{\lambda x}{L}\) |
| \(e = 0.4\) m | A1 | |
| \(EE = 6 \times 0.4^2/(2 \times 0.8)\) | B1 FT | FT for their \(e\) |
| \(0.3v^2/2 - 0.3 \times 2^2/2 = 0.3\,g(0.8 + 0.4) - 6 \times 0.4^2/(2 \times 0.8)\) | M1 | Sets up a 4 term energy equation involving EE, KE and PE |
| \(v = 4.9(0)\) m s\(^{-1}\) or \(2\sqrt{6}\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Notes |
| \(0.3 \times 2^2/2 + 0.3\,gL = 6(L - 0.8)^2/(2 \times 0.8)\) | M1 | Sets up a 3 term energy equation involving EE, KE and PE |
| A1 | ||
| \(L = 2.18\) m | A1 | Ignore answers less than 0.8 |
| Total: 3 |
## Question 5:
**Part (i)**
| Answer | Marks | Notes |
|--------|-------|-------|
| $0.3g = 6e/0.8$ | M1 | Uses $T = \frac{\lambda x}{L}$ |
| $e = 0.4$ m | A1 | |
| $EE = 6 \times 0.4^2/(2 \times 0.8)$ | B1 FT | FT for their $e$ |
| $0.3v^2/2 - 0.3 \times 2^2/2 = 0.3\,g(0.8 + 0.4) - 6 \times 0.4^2/(2 \times 0.8)$ | M1 | Sets up a 4 term energy equation involving EE, KE and PE |
| $v = 4.9(0)$ m s$^{-1}$ or $2\sqrt{6}$ | A1 | |
| **Total: 5** | | |
**Part (ii)**
| Answer | Marks | Notes |
|--------|-------|-------|
| $0.3 \times 2^2/2 + 0.3\,gL = 6(L - 0.8)^2/(2 \times 0.8)$ | M1 | Sets up a 3 term energy equation involving EE, KE and PE |
| | A1 | |
| $L = 2.18$ m | A1 | Ignore answers less than 0.8 |
| **Total: 3** | | |
---5 A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 6 N . The other end of the string is attached to a fixed point $O$. The particle is projected vertically downwards from $O$ with initial speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Calculate the greatest speed of the particle during its descent.\\
(ii) Find the greatest distance of the particle below $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{2b0425b2-2f8f-491a-996c-3d3b589bd7df-12_558_554_260_794}
The end $A$ of a non-uniform rod $A B$ of length 0.6 m and weight 8 N rests on a rough horizontal plane, with $A B$ inclined at $60 ^ { \circ }$ to the horizontal. Equilibrium is maintained by a force of magnitude 3 N applied to the rod at $B$. This force acts at $30 ^ { \circ }$ above the horizontal in the vertical plane containing the rod (see diagram).\\
(i) Find the distance of the centre of mass of the rod from $A$.\\
The 3 N force is removed, and the rod is held in equilibrium by a force of magnitude $P \mathrm {~N}$ applied at $B$, acting in the vertical plane containing the rod, at an angle of $30 ^ { \circ }$ below the horizontal.\\
(ii) Calculate $P$.\\
In one of the two situations described, the $\operatorname { rod } A B$ is in limiting equilibrium.\\
(iii) Find the coefficient of friction at $A$.\\
$7 \quad$ A particle $P$ is projected from a point $O$ with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t \mathrm {~s}$ after projection the horizontal and vertically upwards displacements of $P$ from $O$ are $x \mathrm {~m}$ and $y \mathrm {~m}$ respectively. The equation of the trajectory of $P$ is $y = 2 x - \frac { 25 x ^ { 2 } } { V ^ { 2 } }$.\\
(i) Write down the value of $\tan \theta$, where $\theta$ is the angle of projection of $P$.\\
When $t = 4 , P$ passes through the point $A$ where $x = y = a$.\\
(ii) Calculate $V$ and $a$.\\
(iii) Find the direction of motion of $P$ when it passes through $A$.\\
\hfill \mbox{\textit{CAIE M2 2017 Q5 [8]}}