| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring moments about a pivot point, elastic string tension formula (T = λx/l), and energy conservation. While it involves several concepts (toppling condition, Hooke's law, kinematics), each step follows standard A-level mechanics procedures without requiring novel insight. The 'show that' part guides students to the answer, and the energy equation in part (ii) is a routine application once the extension is known. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| C of M is 0.1 m from AB | B1 | |
| \(0.05T = 20(0.25 - 0.1)\) | M1 | Moments about D |
| \(T = 60\) N | A1 | [3 marks] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(60 = 48e/0.6\) | M1 | Toppling \(T = \lambda\text{ext/nat length}\) |
| \(e = 0.75\) m | A1 | |
| \(2.5v^2/2 + 48(0.75)^2/(2 \times 0.6) = 2.5g(0.75 + 0.6)\) | M1, A1 | KE/EE/PE balance |
| \(v = 3\ \text{ms}^{-1}\) | A1 | [5 marks] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| C of M is 0.1 m from AB | B1 | |
| $0.05T = 20(0.25 - 0.1)$ | M1 | Moments about D |
| $T = 60$ N | A1 | **[3 marks]** AG |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $60 = 48e/0.6$ | M1 | Toppling $T = \lambda\text{ext/nat length}$ |
| $e = 0.75$ m | A1 | |
| $2.5v^2/2 + 48(0.75)^2/(2 \times 0.6) = 2.5g(0.75 + 0.6)$ | M1, A1 | KE/EE/PE balance |
| $v = 3\ \text{ms}^{-1}$ | A1 | **[5 marks]** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{8f8492a7-8a83-4eb2-81ee-99b4a385b704-3_876_483_260_840}
A uniform triangular prism of weight 20 N rests on a horizontal table. $A B C$ is the cross-section through the centre of mass of the prism, where $B C = 0.5 \mathrm {~m} , A B = 0.4 \mathrm {~m} , A C = 0.3 \mathrm {~m}$ and angle $B A C = 90 ^ { \circ }$. The vertical plane $A B C$ is perpendicular to the edge of the table. The point $D$ on $A C$ is at the edge of the table, and $A D = 0.25 \mathrm {~m}$. One end of a light elastic string of natural length 0.6 m and modulus of elasticity 48 N is attached to $C$ and a particle of mass 2.5 kg is attached to the other end of the string. The particle is released from rest at $C$ and falls vertically (see diagram).\\
(i) Show that the tension in the string is 60 N at the instant when the prism topples.\\
(ii) Calculate the speed of the particle at the instant when the prism topples.
\hfill \mbox{\textit{CAIE M2 2015 Q5 [8]}}