| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.8 This is a multi-part elastic string problem requiring energy conservation, understanding that maximum speed occurs when acceleration is zero (net force = 0), and solving simultaneous equations. It combines mechanics concepts (elastic potential energy, gravitational PE, kinetic energy) with algebraic manipulation across three connected parts, making it moderately challenging but still within standard M2 scope. |
| Spec | 3.03d Newton's second law: 2D vectors6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Mg = \dfrac{12.5e}{0.8}\) | 1 | M1 |
| \(e = 0.64M\) AG | 1 | A1 |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Mg(0.8 + e) =\) | 1 | M1 |
| \(\dfrac{M \times 44^2}{2} + \dfrac{12.5e^2}{(2 \times 0.8)}\) | 1 | A1 |
| \(10M(0.8 + 0.64M) =\) | 1 | M1 |
| \(9.68M + \dfrac{12.5(0.64M)^2}{1.6}\) | 1 | A1 |
| \(8 + 6.4M = 9.68 + 3.2M\) | 1 | M1 |
| \(M = 0.525\) | 1 | A1 |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.525gd = \dfrac{12.5(d - 0.8)^2}{(2 \times 0.8)}\) | 1 | M1 |
| \(0.672d = d^2 - 1.6d + 0.64\) | 1 | M1 |
| \(d = 1.94\) | 1 | A1 |
| Total | 3 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Mg = \dfrac{12.5e}{0.8}$ | 1 | M1 | Uses $T = \lambda e/l$ |
| $e = 0.64M$ AG | 1 | A1 | |
| **Total** | **2** | | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Mg(0.8 + e) =$ | 1 | M1 | PE/KE/EE conservation |
| $\dfrac{M \times 44^2}{2} + \dfrac{12.5e^2}{(2 \times 0.8)}$ | 1 | A1 | |
| $10M(0.8 + 0.64M) =$ | 1 | M1 | |
| $9.68M + \dfrac{12.5(0.64M)^2}{1.6}$ | 1 | A1 | $8M + 6.4M^2 = 9.68M + 3.2M^2$ |
| $8 + 6.4M = 9.68 + 3.2M$ | 1 | M1 | Attempt to solve equation in $M$ |
| $M = 0.525$ | 1 | A1 | |
| **Total** | **6** | | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.525gd = \dfrac{12.5(d - 0.8)^2}{(2 \times 0.8)}$ | 1 | M1 | PE/EE balance |
| $0.672d = d^2 - 1.6d + 0.64$ | 1 | M1 | |
| $d = 1.94$ | 1 | A1 | |
| **Total** | **3** | | |7 A particle $P$ of mass $M \mathrm {~kg}$ is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 12.5 N . The other end of the string is attached to a fixed point $A$. The particle is released from rest at $A$ and falls vertically until it comes to instantaneous rest at the point $B$. The greatest speed of $P$ during its descent is $4.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the extension of the string is $e \mathrm {~m}$.\\
(i) Show that $e = 0.64 M$.\\
(ii) Find a second equation in $e$ and $M$, and hence find $M$.\\
(iii) Calculate the distance $A B$.\\
\hfill \mbox{\textit{CAIE M2 Q7 [11]}}