| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.8 This M3 question requires setting up equilibrium with elastic strings (resolving forces with Hooke's law), then applying energy conservation with elastic potential energy. The geometry is non-trivial (finding extended lengths via Pythagoras), and part (b) requires careful energy accounting across multiple forms. More demanding than standard M1/M2 mechanics but typical for M3 level. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Length of string/half string \(= 10\) m \(/ 5\) m (or extn \(= 5\) m) | B1 | Correct length of complete or half string or correct extension (need not be shown explicitly) |
| \(T = \frac{\lambda x}{l} = \frac{20 \times 5}{5} = 20\) | M1, A1 | M1 apply Hooke's law \(x \neq l\); A1 correct tension obtained |
| \(2T\cos\alpha = mg\) | M1 | Resolving vertically, both tensions resolved |
| \(2 \times 20 \times \frac{4}{5} = mg\) | A1ft | Substitute their tension and \(\cos\alpha = \frac{4}{5}\) |
| Weight \(= 32\) N (Accept \(mg = 32\)) | A1 (6) | Correct weight obtained (no ft) |
| (b) PE lost \(= {"}mg{"} \times 4\) | ||
| EPE gained \(= \frac{20 \times 5^2}{2 \times 5} - \frac{20 \times 1^2}{2 \times 5}\) | ||
| \(\frac{1}{2}mv^2 = {"}mg{"} \times 4 - \left(\frac{20 \times 5^2}{2\times 5} - \frac{20 \times 1^2}{2\times 5}\right)\) | M1A1A1 | M1 energy equation with KE, PE and two EPE terms - all calculated with correct formulae; A1A1 deduct one A mark per error (if \(m\) is substituted early, ft their \(m\)) |
| \(\frac{16}{g}v^2 = 32 \times 4 - \left(\frac{20\times 5^2}{2\times 5} - \frac{20\times 1^2}{2\times 5}\right)\) | DM1 | M1 substitute their mass (not weight) |
| \(v^2 = 5g\) | ||
| \(v = 7\), \(7.0\) or \(7.00\) | A1 (5) [11] | 7, 7.0 or 7.00 only acceptable |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Length of string/half string $= 10$ m $/ 5$ m (or extn $= 5$ m) | B1 | Correct length of complete or half string or correct extension (need not be shown explicitly) |
| $T = \frac{\lambda x}{l} = \frac{20 \times 5}{5} = 20$ | M1, A1 | M1 apply Hooke's law $x \neq l$; A1 correct tension obtained |
| $2T\cos\alpha = mg$ | M1 | Resolving vertically, both tensions resolved |
| $2 \times 20 \times \frac{4}{5} = mg$ | A1ft | Substitute their tension and $\cos\alpha = \frac{4}{5}$ |
| Weight $= 32$ N (Accept $mg = 32$) | A1 (6) | Correct weight obtained (no ft) |
| **(b)** PE lost $= {"}mg{"} \times 4$ | | |
| EPE gained $= \frac{20 \times 5^2}{2 \times 5} - \frac{20 \times 1^2}{2 \times 5}$ | | |
| $\frac{1}{2}mv^2 = {"}mg{"} \times 4 - \left(\frac{20 \times 5^2}{2\times 5} - \frac{20 \times 1^2}{2\times 5}\right)$ | M1A1A1 | M1 energy equation with KE, PE and two EPE terms - all calculated with correct formulae; A1A1 deduct one A mark per error (if $m$ is substituted early, ft their $m$) |
| $\frac{16}{g}v^2 = 32 \times 4 - \left(\frac{20\times 5^2}{2\times 5} - \frac{20\times 1^2}{2\times 5}\right)$ | DM1 | M1 substitute their mass (not weight) |
| $v^2 = 5g$ | | |
| $v = 7$, $7.0$ or $7.00$ | A1 (5) [11] | 7, 7.0 or 7.00 only acceptable |
---\begin{enumerate}
\item A light elastic string has natural length 5 m and modulus of elasticity 20 N . The ends of the string are attached to two fixed points $A$ and $B$, which are 6 m apart on a horizontal ceiling. A particle $P$ is attached to the midpoint of the string and hangs in equilibrium at a point which is 4 m below $A B$.\\
(a) Calculate the weight of $P$.
\end{enumerate}
The particle is now raised to the midpoint of $A B$ and released from rest.\\
(b) Calculate the speed of $P$ when it has fallen 4 m .\\
\hfill \mbox{\textit{Edexcel M3 2015 Q4 [11]}}