| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Standard +0.8 This M3 question requires energy methods with elastic strings on an inclined plane, involving both smooth and rough surface cases. Part (a) needs setting up energy conservation with elastic potential energy and gravitational PE, while part (b) requires incorporating work done against friction. The multi-step nature, combination of mechanics concepts, and need to handle the refined model with friction parameter makes this moderately challenging, though it follows standard M3 techniques without requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{3mgx^2}{2l} = 2mgx \sin \alpha\) | M1A1 B1(A1 on e-pen) | |
| \(3x^2 = 4xl \times \frac{3}{5}\) | ||
| \(5x^2 = 4xl\) | ||
| \(x = \frac{4}{5}l\) | DM1A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 2mg \cos \alpha \left(= \frac{8}{5}mg\right)\) | B1 | |
| \(\frac{3mg}{2l} \times \frac{4}{25}l^2 = 2mg \times \frac{2}{5}l \times \frac{3}{5}, \quad \mu = \frac{8}{5}mg \times \frac{2}{5}l\) | M1A1ft. B1ft (A1 on e-pen) | |
| \(6 = 12 - 16\mu\) | ||
| \(16\mu = 6\) | DM1A1 | (6) |
| \(\mu = \frac{3}{8}\) | [11] |
| Answer | Marks |
|---|---|
| M1 | for an energy equation with an EPE term of the form \(\frac{kmgx^2}{l}\) and a GPE term. If a KE term is included it must become 0 later. |
| A1 | for a correct EPE term |
| B1 | for a correct GPE term. This can be in terms of the distance moved down the plane or the vertical distance fallen |
| M1 dep | for solving their equation to obtain the distance moved or using the vertical distance and obtaining the distance moved along the plane. |
| A1 | for \(x = \frac{4}{5}l\) oe eg \(x = \frac{12}{15}l\) |
| Answer | Marks |
|---|---|
| B1 | for resolving perpendicular to the plane to obtain \(R = 2mg \cos \alpha\). May only be seen in an equation. |
| M1 | for an work-energy equation with an EPE term of the form \(\frac{kmgx^2}{l}\), a GPE term and the work done against friction. The work term must include a distance along the plane. |
| A1 | for EPE and GPE terms correct and work subtracted from the GPE |
| B1 ft | for the work term ft their \(R\) |
| M1 dep | for solving to obtain a value for \(\mu\) |
| A1 cso | for \(\mu = \frac{3}{8}\) oe inc 0.375 but not 0.38 |
| Answer | Marks |
|---|---|
| (a) M1A1B0M1A0 | (so 2 penalties for mis-read) |
| (b) | |
| B1 | \(R = mg \cos \alpha\) |
| M1, A1 | Equation, with EPE correct and \(mg \times \frac{2}{5}l \times \frac{3}{5}\) |
| B1 ft | \(\mu = \frac{4mg}{5} \times \frac{2}{5}l\) |
| DM1, A1 | \(\mu = 0\) |
**(a)**
$\frac{3mgx^2}{2l} = 2mgx \sin \alpha$ | M1A1 B1(A1 on e-pen) |
$3x^2 = 4xl \times \frac{3}{5}$ | |
$5x^2 = 4xl$ | |
$x = \frac{4}{5}l$ | DM1A1 | (5)
**(b)**
$R = 2mg \cos \alpha \left(= \frac{8}{5}mg\right)$ | B1 |
$\frac{3mg}{2l} \times \frac{4}{25}l^2 = 2mg \times \frac{2}{5}l \times \frac{3}{5}, \quad \mu = \frac{8}{5}mg \times \frac{2}{5}l$ | M1A1ft. B1ft (A1 on e-pen) |
$6 = 12 - 16\mu$ | |
$16\mu = 6$ | DM1A1 | (6) |
$\mu = \frac{3}{8}$ | [11] |
**Notes for Question 4(a):**
M1 | for an energy equation with an EPE term of the form $\frac{kmgx^2}{l}$ and a GPE term. If a KE term is included it must become 0 later. |
A1 | for a correct EPE term |
B1 | for a correct GPE term. This can be in terms of the distance moved down the plane or the vertical distance fallen |
M1 dep | for solving their equation to obtain the distance moved or using the vertical distance and obtaining the distance moved along the plane. |
A1 | for $x = \frac{4}{5}l$ oe eg $x = \frac{12}{15}l$ |
**(b)**
B1 | for resolving perpendicular to the plane to obtain $R = 2mg \cos \alpha$. May only be seen in an equation. |
M1 | for an work-energy equation with an EPE term of the form $\frac{kmgx^2}{l}$, a GPE term and the work done against friction. The work term must include a distance along the plane. |
A1 | for EPE and GPE terms correct and work subtracted from the GPE |
B1 ft | for the work term ft their $R$ |
M1 dep | for solving to obtain a value for $\mu$ |
A1 cso | for $\mu = \frac{3}{8}$ oe inc 0.375 but not 0.38 |
**If m used instead of 2m, assuming correct otherwise:**
(a) M1A1B0M1A0 | (so 2 penalties for mis-read) |
(b) | |
B1 | $R = mg \cos \alpha$ |
M1, A1 | Equation, with EPE correct and $mg \times \frac{2}{5}l \times \frac{3}{5}$ |
B1 ft | $\mu = \frac{4mg}{5} \times \frac{2}{5}l$ |
DM1, A1 | $\mu = 0$ |
---
4.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[>=latex, scale=1.5]
% Configuration variables
\def\angle{30} % Inclination angle alpha
\def\cpos{4.5} % Distance of point C from origin
\def\apos{7.0} % Distance of point A from origin
\def\totalLen{8.5} % Length of the incline line
% 1. Draw the horizontal dashed baseline
\draw[dashed] (0,0) -- (6,0);
% 2. Draw the main inclined plane line
\draw (0,0) -- (\angle:\totalLen);
% 3. Draw the angle arc and label it alpha
\draw (1.2, 0) arc (0:\angle:1.2);
\node at (15:1.5) {$\alpha$};
% 4. Elements on the incline (using a rotated scope for simplicity)
\begin{scope}[rotate=\angle]
% Coordinate definitions
\coordinate (C) at (\cpos, 0);
\coordinate (A) at (\apos, 0);
% Draw the solid dot at C
\fill ($(C)+(0,.075)$) circle (2pt);
% Draw the perpendicular tick mark at A
\draw (A) ++(0, 0.12) -- ++(0, -0.24);
% Draw the thin rod slightly above the incline line
\draw[thin] (\cpos, 0.08) -- (\apos, 0.08);
% Draw the double-headed dimension arrow for 'l' below the incline
\draw[<->] (\cpos, -0.4) -- (\apos, -0.4) node[midway, fill=white, inner sep=1pt] {$l$};
% Place labels C and A below their respective points
\node[below=14pt] at (C) {$C$};
\node[below=10pt, right=6pt] at (A) {$A$};
\end{scope}
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
One end of a light elastic string, of natural length $l$ and modulus of elasticity $3 m g$, is fixed to a point $A$ on a fixed plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 3 } { 5 }$ A small ball of mass $2 m$ is attached to the free end of the string. The ball is held at a point $C$ on the plane, where $C$ is below $A$ and $A C = l$ as shown in Figure 3. The string is parallel to a line of greatest slope of the plane. The ball is released from rest. In an initial model the plane is assumed to be smooth.
\begin{enumerate}[label=(\alph*)]
\item Find the distance that the ball moves before first coming to instantaneous rest.
In a refined model the plane is assumed to be rough. The coefficient of friction between the ball and the plane is $\mu$. The ball first comes to instantaneous rest after moving a distance $\frac { 2 } { 5 } l$.
\item Find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q4 [11]}}