| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 15 |
| 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 multiple techniques: resolving forces in equilibrium to find λ, calculating acceleration from forces in a non-standard geometry, and applying energy conservation with elastic potential energy. The geometry (particle on string between two fixed points) requires careful consideration of extensions and angles, making it more challenging than routine mechanics problems, though the individual steps are standard for M3 level. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = \frac{\lambda x}{l} \Rightarrow 30 = \frac{\lambda \times 0.3}{0.5}\) | M1A1 | Use Hooke's Law with \(T=30\) |
| \(\lambda = 50\) * | A1* | Obtain given value from fully correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Extension \(= 0.5\) m | B1 | Correct extension, seen explicitly or used in (b) or (c); 0.25 if half string used |
| \(T = \frac{50\times 0.5}{0.5} = (50)\) | M1A1ft | Use HL with \(\lambda=50\) and their extension; correct equation ft |
| \(R(\uparrow)\): \(2T\cos\theta - 1.2g = 1.2a\) | M1 | Equation of motion with \(T\) resolved; \(a\) could be negative; must have 3 terms |
| \(100\times\frac{3}{5} - 1.2\times 9.8 = 1.2a\) | A1ft | Correct equation with their value of \(T\) |
| \(a = 40.2\), \(a = 40\) or \(40.2\) ms\(^{-2}\) (positive) | A1 | Correct value of acceleration (positive); must be 2 or 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| E.P.E. \(= \frac{1}{2}\times 50\times\frac{0.5^2}{0.5}\) | B1ft | Either EPE correct, follow through their extension |
| \(1.2g\times 0.3 + \frac{1}{2}\times 1.2v^2 = \frac{1}{2}\times 50\times\frac{0.5^2}{0.5} - \frac{1}{2}\times 50\times\frac{0.3^2}{0.5}\) | M1A1A1 | Energy equation from \(C\) to ceiling; PE, KE and 2 EPE terms required; EPE of form \(k\frac{x^2}{l}\); both EPE terms correct; completely correct equation |
| \(v^2 = \frac{1}{0.6}\left(25\times\frac{0.5^2}{0.5} - 25\times\frac{0.3^2}{0.5} - 1.2g\times 0.3\right) (= 7.452)\) | DM1 | Solve for \(v^2\) or \(v\) |
| \(v = 2.730... = 2.7\) or \(2.73\) ms\(^{-1}\) | A1 | Correct value of \(v\); must be 2 or 3 sf |
# Question 6:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = \frac{\lambda x}{l} \Rightarrow 30 = \frac{\lambda \times 0.3}{0.5}$ | M1A1 | Use Hooke's Law with $T=30$ |
| $\lambda = 50$ * | A1* | Obtain **given** value from fully correct working |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Extension $= 0.5$ m | B1 | Correct extension, seen explicitly or used in (b) or (c); 0.25 if half string used |
| $T = \frac{50\times 0.5}{0.5} = (50)$ | M1A1ft | Use HL with $\lambda=50$ and their extension; correct equation ft |
| $R(\uparrow)$: $2T\cos\theta - 1.2g = 1.2a$ | M1 | Equation of motion with $T$ resolved; $a$ could be negative; must have 3 terms |
| $100\times\frac{3}{5} - 1.2\times 9.8 = 1.2a$ | A1ft | Correct equation with their value of $T$ |
| $a = 40.2$, $a = 40$ or $40.2$ ms$^{-2}$ (positive) | A1 | Correct value of acceleration (positive); must be 2 or 3 sf |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| E.P.E. $= \frac{1}{2}\times 50\times\frac{0.5^2}{0.5}$ | B1ft | Either EPE correct, follow through their extension |
| $1.2g\times 0.3 + \frac{1}{2}\times 1.2v^2 = \frac{1}{2}\times 50\times\frac{0.5^2}{0.5} - \frac{1}{2}\times 50\times\frac{0.3^2}{0.5}$ | M1A1A1 | Energy equation from $C$ to ceiling; PE, KE and 2 EPE terms required; EPE of form $k\frac{x^2}{l}$; both EPE terms correct; completely correct equation |
| $v^2 = \frac{1}{0.6}\left(25\times\frac{0.5^2}{0.5} - 25\times\frac{0.3^2}{0.5} - 1.2g\times 0.3\right) (= 7.452)$ | DM1 | Solve for $v^2$ or $v$ |
| $v = 2.730... = 2.7$ or $2.73$ ms$^{-1}$ | A1 | Correct value of $v$; must be 2 or 3 sf |
---\begin{enumerate}
\item A particle $P$ of mass 1.2 kg is attached to the midpoint of a light elastic string of natural length 0.5 m and modulus of elasticity $\lambda$ newtons.
\end{enumerate}
The fixed points $A$ and $B$ are 0.8 m apart on a horizontal ceiling. One end of the string is attached to $A$ and the other end of the string is attached to $B$.
Initially $P$ is held at rest at the midpoint $M$ of the line $A B$ and the tension in the string is 30 N .\\
(a) Show that $\lambda = 50$
The particle is now held at rest at the point $C$, where $C$ is 0.3 m vertically below $M$. The particle is released from rest.\\
(b) Find the magnitude of the initial acceleration of $P$\\
(c) Find the speed of $P$ at the instant immediately before it hits the ceiling.
\hfill \mbox{\textit{Edexcel M3 2022 Q6 [15]}}