| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.2 This is a standard M3 SHM question with elastic strings requiring multiple techniques (equilibrium, SHM period derivation, energy methods, projectile motion after string cut), but follows a predictable structure with clear signposting. The slack phase consideration and multi-part nature elevate it slightly above average difficulty, but all steps use routine M3 methods without requiring novel insight. |
| Spec | 3.02h Motion under gravity: vector form4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(R(\uparrow)\): \(T = 2mg\) | B1 | |
| Hooke's law: \(T = \frac{6mge}{3a}\) | M1 | |
| \(2mg = \frac{6mge}{3a}\), \(e = a\) | A1 | |
| \(AO = 4a\) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| H.L.: \(T = \frac{6mg(a-x)}{3a} = \frac{2mg(a-x)}{a}\) | B1ft | |
| Eqn. of motion: \(-2mg + T = 2m\ddot{x}\) | M1 | |
| \(-2mg + \frac{2mg(a-x)}{a} = 2m\ddot{x}\) | M1 | |
| \(-\frac{2mgx}{a} = 2m\ddot{x}\) | ||
| \(\ddot{x} = -\frac{g}{a}x\) | A1 | |
| Period \(= 2\pi\sqrt{\frac{a}{g}}\) ✱ | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(v^2 = \omega^2(a^2 - x^2)\) | M1 A1 | |
| \(v_{\max}^2 = \frac{g}{a}\left(\left(\frac{a}{4}\right)^2 - 0\right)\) | ||
| \(v_{\max} = \frac{1}{4}\sqrt{ga}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(x = -\frac{a}{8}\): \(v^2 = \frac{g}{a}\left(\frac{a^2}{16} - \frac{a^2}{64}\right) = \frac{3ag}{64}\) | M1 | |
| \(v^2 = u^2 + 2as\): \(0 = \frac{3ag}{64} - 2gh\) | M1 A1 | |
| \(h = \frac{3a}{128}\) | ||
| Total height above \(O = \frac{a}{8} + \frac{3a}{128} = \frac{19a}{128}\) | A1 | (4) [15] |
## Question 7:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $R(\uparrow)$: $T = 2mg$ | B1 | |
| Hooke's law: $T = \frac{6mge}{3a}$ | M1 | |
| $2mg = \frac{6mge}{3a}$, $e = a$ | A1 | |
| $AO = 4a$ | | (3) |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| H.L.: $T = \frac{6mg(a-x)}{3a} = \frac{2mg(a-x)}{a}$ | B1ft | |
| Eqn. of motion: $-2mg + T = 2m\ddot{x}$ | M1 | |
| $-2mg + \frac{2mg(a-x)}{a} = 2m\ddot{x}$ | M1 | |
| $-\frac{2mgx}{a} = 2m\ddot{x}$ | | |
| $\ddot{x} = -\frac{g}{a}x$ | A1 | |
| Period $= 2\pi\sqrt{\frac{a}{g}}$ ✱ | A1 | (5) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $v^2 = \omega^2(a^2 - x^2)$ | M1 A1 | |
| $v_{\max}^2 = \frac{g}{a}\left(\left(\frac{a}{4}\right)^2 - 0\right)$ | | |
| $v_{\max} = \frac{1}{4}\sqrt{ga}$ | A1 | (3) |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| $x = -\frac{a}{8}$: $v^2 = \frac{g}{a}\left(\frac{a^2}{16} - \frac{a^2}{64}\right) = \frac{3ag}{64}$ | M1 | |
| $v^2 = u^2 + 2as$: $0 = \frac{3ag}{64} - 2gh$ | M1 A1 | |
| $h = \frac{3a}{128}$ | | |
| Total height above $O = \frac{a}{8} + \frac{3a}{128} = \frac{19a}{128}$ | A1 | (4) **[15]** |
The image you've shared appears to be a **back/colophon page** of an Edexcel publication — it contains only publisher information (address, contact details, order code, registration info) and **no mark scheme content**.
There are no questions, answers, mark allocations, or mathematical working on this page to extract.
If you have additional pages from the mark scheme, please share those and I'll be happy to extract and format the content for you.\begin{enumerate}
\item A light elastic string, of natural length $3 a$ and modulus of elasticity $6 m g$, has one end attached to a fixed point $A$. A particle $P$ of mass $2 m$ is attached to the other end of the string and hangs in equilibrium at the point $O$, vertically below $A$.\\
(a) Find the distance $A O$.
\end{enumerate}
The particle is now raised to point $C$ vertically below $A$, where $A C > 3 a$, and is released from rest.\\
(b) Show that $P$ moves with simple harmonic motion of period $2 \pi \sqrt { } \left( \frac { a } { g } \right)$.
It is given that $O C = \frac { 1 } { 4 } a$.\\
(c) Find the greatest speed of $P$ during the motion.
The point $D$ is vertically above $O$ and $O D = \frac { 1 } { 8 } a$. The string is cut as $P$ passes through $D$, moving upwards.\\
(d) Find the greatest height of $P$ above $O$ in the subsequent motion.\\
\hfill \mbox{\textit{Edexcel M3 2010 Q7 [15]}}