| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string vertical motion |
| Difficulty | Standard +0.8 This is a multi-stage elastic string problem requiring energy conservation across three distinct scenarios: finding maximum extension, calculating speed at a specific point, and analyzing motion after an inelastic collision with energy loss. While the individual techniques (EPE, GPE, KE) are standard M2 content, the three-part structure with the 96% energy loss and the constraint that the string is slack requires careful bookkeeping and problem-solving beyond routine exercises. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.4gd = 32(d-0.8)^2/(2\times0.8)\) | M1, A1 | PE/EE balance |
| \(20d^2 - 36d + 12.8 = 0\) | M1 | Solves 3 term quadratic |
| \(d = 1.31\) m only | A1 [4] | Other value \(0.4876...\) |
| OR \(0.4g(0.8+e) = 32e^2/(2\times0.8)\) | M1 | PE/EE balance |
| \(20e^2 - 4e + 3.2 = 0\) | A1 | |
| \(e = 0.5(1)\) (also \(-3.12\)) | M1 | Solves 3 term quadratic |
| \(d = 1.31\) m only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.4v^2/2 = 0.4g\times1 - 32(1-0.8)^2/(2\times0.8)\) | M1, A1 | EE/KE/PE balance |
| \(v = 4\ \text{ms}^{-1}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rebound \(v = 0.8\) | B1ft | \(ftcv(v(ii)) \times \sqrt{(1-0.96)} = 0.2v(ii)\) |
| \(0 = 0.4\times0.8^2/2 + 32\times0.2^2/1.6 - 0.4gh\) | M1 | EE/PE/KE balance, \(h = 0.232\) |
| \(OP\ (=1-h) = 0.768\) m | A1 [3] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.4gd = 32(d-0.8)^2/(2\times0.8)$ | M1, A1 | PE/EE balance |
| $20d^2 - 36d + 12.8 = 0$ | M1 | Solves 3 term quadratic |
| $d = 1.31$ m only | A1 [4] | Other value $0.4876...$ |
| **OR** $0.4g(0.8+e) = 32e^2/(2\times0.8)$ | M1 | PE/EE balance |
| $20e^2 - 4e + 3.2 = 0$ | A1 | |
| $e = 0.5(1)$ (also $-3.12$) | M1 | Solves 3 term quadratic |
| $d = 1.31$ m only | A1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.4v^2/2 = 0.4g\times1 - 32(1-0.8)^2/(2\times0.8)$ | M1, A1 | EE/KE/PE balance |
| $v = 4\ \text{ms}^{-1}$ | A1 [3] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rebound $v = 0.8$ | B1ft | $ftcv(v(ii)) \times \sqrt{(1-0.96)} = 0.2v(ii)$ |
| $0 = 0.4\times0.8^2/2 + 32\times0.2^2/1.6 - 0.4gh$ | M1 | EE/PE/KE balance, $h = 0.232$ |
| $OP\ (=1-h) = 0.768$ m | A1 [3] | |7 A particle $P$ of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 32 N . The other end of the string is attached to a fixed point $O$. The particle is released from rest at $O$.\\
(i) Calculate the distance $O P$ at the instant when $P$ first comes to instantaneous rest.
A horizontal plane is fixed at a distance 1 m below $O$. The particle $P$ is again released from rest at $O$.\\
(ii) Calculate the speed of $P$ immediately before it collides with the plane.\\
(iii) In the collision with the plane, $P$ loses $96 \%$ of its kinetic energy. Calculate the distance $O P$ at the instant when $P$ first comes to instantaneous rest above the plane, given that this occurs when the string is slack.
\hfill \mbox{\textit{CAIE M2 2013 Q7 [10]}}