| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Maximum speed in SHM |
| Difficulty | Standard +0.8 This is a substantial M3 SHM question requiring multiple techniques: equilibrium analysis with Hooke's law, proving SHM conditions, applying SHM formulas for amplitude/period, and projectile motion after string cutting. While each part follows standard methods, the 15-mark multi-part structure with string tension considerations and the transition from SHM to projectile motion makes it moderately challenging, above average for A-level but within expected M3 scope. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| R (↑): \(T = 2mg\) | B1 | |
| Hooke's law: \(T = \frac{6mge}{3a}\) | M1 | |
| \(2mg = \frac{6mge}{3a}\) | ||
| \(e = a\) | ||
| \(AO = 4a\) | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| 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 | |
| \(-2mg - \frac{2mgx}{a} = 2m\ddot{x}\) | ||
| \(\ddot{x} = -\frac{g}{a}x\) | A1 | |
| period \(2\pi\sqrt{\frac{a}{g}}\) | A1 | (5 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = \omega^2(a^2 - x^2)\) | ||
| \(v_{\max}^2 = \frac{g}{a}\left[\frac{a}{4}^2 - 0\right]\) | M1 A1 | |
| \(v_{\max} = \frac{1}{4}\sqrt{(ga)}\) | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\frac{a}{8}\): \(v^2 = \frac{g}{a}\left[\frac{a^2}{16} - \frac{a^2}{64}\right] = \frac{3ag}{64}\) | M1 | |
| \(v^2 = u^2 + 2as\) | M1 | |
| \(\frac{3ag}{64} = 0 + 2gh\) | ||
| \(h = \frac{3a}{128}\) | A1 | |
| Total height above O \(= \frac{a}{8} + \frac{3a}{128} = \frac{19a}{128}\) | A1 | (4 marks total) |
## Part (a)
R (↑): $T = 2mg$ | B1 |
Hooke's law: $T = \frac{6mge}{3a}$ | M1 |
$2mg = \frac{6mge}{3a}$ | |
$e = a$ | |
$AO = 4a$ | A1 | (3 marks total)
## Part (b)
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 |
$-2mg - \frac{2mgx}{a} = 2m\ddot{x}$ | |
$\ddot{x} = -\frac{g}{a}x$ | A1 |
period $2\pi\sqrt{\frac{a}{g}}$ | A1 | (5 marks total)
## Part (c)
$v^2 = \omega^2(a^2 - x^2)$ | |
$v_{\max}^2 = \frac{g}{a}\left[\frac{a}{4}^2 - 0\right]$ | M1 A1 |
$v_{\max} = \frac{1}{4}\sqrt{(ga)}$ | A1 | (3 marks total)
## Part (d)
$x = -\frac{a}{8}$: $v^2 = \frac{g}{a}\left[\frac{a^2}{16} - \frac{a^2}{64}\right] = \frac{3ag}{64}$ | M1 |
$v^2 = u^2 + 2as$ | M1 |
$\frac{3ag}{64} = 0 + 2gh$ | |
$h = \frac{3a}{128}$ | A1 |
Total height above O $= \frac{a}{8} + \frac{3a}{128} = \frac{19a}{128}$ | A1 | (4 marks total)
**[15 marks total]**A light elastic string, of natural length $3a$ and modulus of elasticity $6mg$, has one end attached to a fixed point $A$. A particle $P$ of mass $2m$ is attached to the other end of the string and hangs in equilibrium at the point $O$, vertically below $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $AO$. [3]
\end{enumerate}
The particle is now raised to point $C$ vertically below $A$, where $AC > 3a$, and is released from rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ moves with simple harmonic motion of period $2\pi\sqrt{\frac{a}{g}}$. [5]
\end{enumerate}
It is given that $OC = \frac{1}{4}a$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the greatest speed of $P$ during the motion. [3]
\end{enumerate}
The point $D$ is vertically above $O$ and $OD = \frac{1}{8}a$. The string is cut as $P$ passes through $D$, moving upwards.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the greatest height of $P$ above $O$ in the subsequent motion. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [15]}}