| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| 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: horizontal surface motion |
| Difficulty | Standard +0.3 This is a standard M3 elastic energy problem requiring energy conservation with two strings. While it involves multiple strings and finding maximum speed, the setup is straightforward: calculate extensions, apply Hooke's law, use energy conservation (EPE → KE), and recognize symmetry for part (b). The algebra is routine and the conceptual demand is typical for M3. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Elastic PE when \(P\) is at \(X\): \(E = \frac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \frac{4mg\left(\frac{4}{3}l\right)^2}{2l} = \frac{40mgl}{9}\) | M1 A1 | |
| \(\frac{1}{2}mV^2 + 2 \times \frac{4mgl^2}{2l} = \frac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \frac{4mg\left(\frac{4}{3}l\right)^2}{2l}\) | M1 A1=A1ft | |
| \(\frac{1}{2}V^2 + 4gl = \frac{8}{9}gl + \frac{32}{9}gl\) | ||
| \(V^2 = \frac{8gl}{9}\) | M1 | solving for \(V^2\) |
| \(V = \left(\frac{8gl}{9}\right)^{\frac{1}{2}}\) | A1 | or exact equivalents |
| Total | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| The maximum speed occurs when \(a = 0\) | B1 | |
| At \(M\) the particle is in equilibrium (sum of forces is zero) \(\Rightarrow a = 0\) | B1 | |
| Total | (2) [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| HL: \(T_1 = \frac{4mg(l+x)}{l}\), \(T_2 = \frac{4mg(l-x)}{l}\) | ||
| N2L: \(m\ddot{x} = T_2 - T_1 = -\frac{8mg}{l}x\) | M1 A1 | |
| This is SHM, centre \(M\); \(a = \frac{l}{3}\), \(\omega^2 = \frac{8g}{l}\) | A1, A1ft | |
| \(v^2 = \omega^2(a^2 - x^2) \Rightarrow v^2 = \frac{8g}{l}\left(\frac{l^2}{9} - x^2\right)\) | M1 | depends on showing SHM |
| At \(M\), \(x=0\): \(V^2 = \frac{8gl}{9}\), \(V = \left(\frac{8gl}{9}\right)^{\frac{1}{2}}\) | M1, A1 | or exact equivalents |
| Total | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| The particle is performing SHM about the mid-point of \(AB\) | B1 | |
| The maximum speed occurs at the centre of oscillation (when \(x = 0\)) | B1 | |
| Total | (2) [9] |
# Question 3:
## Part (a) — Energy Method:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Elastic PE when $P$ is at $X$: $E = \frac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \frac{4mg\left(\frac{4}{3}l\right)^2}{2l} = \frac{40mgl}{9}$ | M1 A1 | |
| $\frac{1}{2}mV^2 + 2 \times \frac{4mgl^2}{2l} = \frac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \frac{4mg\left(\frac{4}{3}l\right)^2}{2l}$ | M1 A1=A1ft | |
| $\frac{1}{2}V^2 + 4gl = \frac{8}{9}gl + \frac{32}{9}gl$ | | |
| $V^2 = \frac{8gl}{9}$ | M1 | solving for $V^2$ |
| $V = \left(\frac{8gl}{9}\right)^{\frac{1}{2}}$ | A1 | or exact equivalents |
| **Total** | **(7)** | |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| The maximum speed occurs when $a = 0$ | B1 | |
| At $M$ the particle is in equilibrium (sum of forces is zero) $\Rightarrow a = 0$ | B1 | |
| **Total** | **(2) [9]** | |
## Part (a) — Newton's Second Law Alternative:
| Working/Answer | Marks | Guidance |
|---|---|---|
| HL: $T_1 = \frac{4mg(l+x)}{l}$, $T_2 = \frac{4mg(l-x)}{l}$ | | |
| N2L: $m\ddot{x} = T_2 - T_1 = -\frac{8mg}{l}x$ | M1 A1 | |
| This is SHM, centre $M$; $a = \frac{l}{3}$, $\omega^2 = \frac{8g}{l}$ | A1, A1ft | |
| $v^2 = \omega^2(a^2 - x^2) \Rightarrow v^2 = \frac{8g}{l}\left(\frac{l^2}{9} - x^2\right)$ | M1 | depends on showing SHM |
| At $M$, $x=0$: $V^2 = \frac{8gl}{9}$, $V = \left(\frac{8gl}{9}\right)^{\frac{1}{2}}$ | M1, A1 | or exact equivalents |
| **Total** | **(7)** | |
## Part (b) — Newton's Second Law Alternative:
| Working/Answer | Marks | Guidance |
|---|---|---|
| The particle is performing SHM about the mid-point of $AB$ | B1 | |
| The maximum speed occurs at the centre of oscillation (when $x = 0$) | B1 | |
| **Total** | **(2) [9]** | |3. A light elastic string has natural length $2 l$ and modulus of elasticity $4 m g$. One end of the string is attached to a fixed point $A$ and the other end to a fixed point $B$, where $A$ and $B$ lie on a smooth horizontal table, with $A B = 4 l$. A particle $P$ of mass $m$ is attached to the mid-point of the string.
The particle is released from rest at the point of the line $A B$ which is $\frac { 5 l } { 3 }$ from $B$. The speed of $P$ at the mid-point of $A B$ is $V$.
\begin{enumerate}[label=(\alph*)]
\item Find $V$ in terms of $g$ and $L$.
\item Explain why $V$ is the maximum speed of $P$.\\
(Total 9 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2005 Q3 [9]}}