| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| 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 standard vertical elastic string energy conservation problem requiring setup of energy equation with gravitational PE and elastic PE, then finding maximum speed by differentiating or recognizing it occurs at equilibrium position. The two-part structure and need to handle both elastic PE and gravitational PE simultaneously elevates it slightly above average difficulty, but the method is well-practiced in M2 courses. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4g(0.5 + x) = \frac{6x^2}{(2\times 0.5)}\) | M1 | Set up an energy equation |
| \(6x^2 - 4x - 2 = 0\) or \(3x^2 - 2x - 1 = 0\) | M1 | Attempt to solve a 3 term quadratic equation |
| \(x = 1\) (ignore \(-\frac{1}{3}\) if seen) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4g = \frac{6e}{0.5}\) | M1 | Use \(T = \frac{\lambda x}{l}\) to find the extension at the equilibrium position |
| \(e = \frac{1}{3}\) | A1 | |
| \(PE\) change \(= 0.4g\left(0.5 + \frac{1}{3}\right)\) | B1ft | Ft for candidate's \(e\) |
| \(\frac{0.4V^2}{2} = 0.4g\left(0.5 + \frac{1}{3}\right) - \frac{6\left(\frac{1}{3}\right)^2}{(2\times 0.5)}\) | M1 | Set up a three term energy equation |
| \(V = 3.65 \text{ ms}^{-1}\) | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4g(0.5 + x) = \frac{6x^2}{(2\times 0.5)}$ | M1 | Set up an energy equation |
| $6x^2 - 4x - 2 = 0$ or $3x^2 - 2x - 1 = 0$ | M1 | Attempt to solve a 3 term quadratic equation |
| $x = 1$ (ignore $-\frac{1}{3}$ if seen) | A1 | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4g = \frac{6e}{0.5}$ | M1 | Use $T = \frac{\lambda x}{l}$ to find the extension at the equilibrium position |
| $e = \frac{1}{3}$ | A1 | |
| $PE$ change $= 0.4g\left(0.5 + \frac{1}{3}\right)$ | B1ft | Ft for candidate's $e$ |
| $\frac{0.4V^2}{2} = 0.4g\left(0.5 + \frac{1}{3}\right) - \frac{6\left(\frac{1}{3}\right)^2}{(2\times 0.5)}$ | M1 | Set up a three term energy equation |
| $V = 3.65 \text{ ms}^{-1}$ | A1 | |
---5 A particle $P$ of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 6 N . The other end of the string is attached to a fixed point $O$. The particle $P$ is released from rest at the point $( 0.5 + x ) \mathrm { m }$ vertically below $O$. The particle $P$ comes to instantaneous rest at $O$.\\
(i) Find $x$.\\
(ii) Find the greatest speed of $P$.\\
\hfill \mbox{\textit{CAIE M2 2019 Q5 [8]}}