| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | November |
| 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 | Challenging +1.2 This is a multi-part elastic string problem requiring energy conservation and understanding that maximum speed occurs when acceleration is zero (tension equals weight). While it involves several steps and the modulus/extension relationship, the techniques are standard for M2 level with clear signposting through the parts. The 'show that' in part (i) guides students to the key insight, making it moderately above average difficulty but not requiring novel problem-solving approaches. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(Mg = \frac{12.5e}{0.8}\) | M1 | Uses \(T = \lambda e/l\) |
| \(e = 0.64M\) | A1 AG | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(Mg(0.8 + e) = \frac{M \times 44^2}{2} + \frac{12.5e^2}{(2 \times 0.8)}\) | M1 A1 | PE/KE/EE conservation |
| \(10M(0.8 + 0.64M) = 9.68M + \frac{12.5(0.64M)^2}{1.6}\) | M1 A1 | \(8M + 6.4M^2 = 9.68M + 3.2M^2\) |
| \(8 + 6.4M = 9.68 + 3.2M\) | M1 | Attempt to solve equation in \(M\) |
| \(M = 0.525\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(0.525gd = \frac{12.5(d-0.8)^2}{(2 \times 0.8)}\) | M1 | PE/EE balance |
| \(0.672d = d^2 - 1.6d + 0.64\) | M1 | |
| \(d = 1.94\) | A1 | [3] |
## Question 7(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $Mg = \frac{12.5e}{0.8}$ | M1 | Uses $T = \lambda e/l$ |
| $e = 0.64M$ | A1 AG | **[2]** |
## Question 7(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $Mg(0.8 + e) = \frac{M \times 44^2}{2} + \frac{12.5e^2}{(2 \times 0.8)}$ | M1 A1 | PE/KE/EE conservation |
| $10M(0.8 + 0.64M) = 9.68M + \frac{12.5(0.64M)^2}{1.6}$ | M1 A1 | $8M + 6.4M^2 = 9.68M + 3.2M^2$ |
| $8 + 6.4M = 9.68 + 3.2M$ | M1 | Attempt to solve equation in $M$ |
| $M = 0.525$ | A1 | **[6]** |
## Question 7(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $0.525gd = \frac{12.5(d-0.8)^2}{(2 \times 0.8)}$ | M1 | PE/EE balance |
| $0.672d = d^2 - 1.6d + 0.64$ | M1 | |
| $d = 1.94$ | A1 | **[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 2015 Q7 [11]}}