| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Challenging +1.8 This M4 question requires setting up potential energy for a compound framework with elastic string, finding equilibrium via differentiation, and testing stability with second derivative. It involves multiple components (gravitational PE for two rods, elastic PE), coordinate geometry to find string extension, and calculus-based energy methods. The multi-step nature, combination of statics and energy principles, and M4-level content place it well above average difficulty, though the algebraic manipulation follows standard patterns once set up correctly. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Impulse \(= m(a\mathbf{i}+a\mathbf{j}) - m(b\mathbf{i}) = m(a-b)\mathbf{i} + ma\mathbf{j}\) | ||
| Wall is parallel to \(2\mathbf{i}+\mathbf{j}\), so no impulse along wall; impulse perpendicular to wall, i.e. parallel to \((-\mathbf{i}+2\mathbf{j})\) | B1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Normal to wall is in direction \((-\mathbf{i}+2\mathbf{j})\) (perpendicular to \(2\mathbf{i}+\mathbf{j}\)) | B1 | Identifying normal direction |
| Component of velocity before impact normal to wall: \(\frac{(b\mathbf{i})\cdot(-\mathbf{i}+2\mathbf{j})}{ | \mathbf{-i+2j} | } = \frac{-b}{\sqrt{5}}\) |
| Component after impact normal to wall: \(\frac{a(\mathbf{i+j})\cdot(-\mathbf{i}+2\mathbf{j})}{\sqrt{5}} = \frac{a}{\sqrt{5}}\) | M1 A1 | |
| \(e = \frac{a/\sqrt{5}}{b/\sqrt{5}} = \frac{a}{b}\) | M1 A1 | |
| Component along wall before \(=\) component along wall after (smooth wall): \(\frac{2b}{\sqrt{5}} = \frac{3a}{\sqrt{5}}\), so \(b = \frac{3a}{2}\), giving \(e = \frac{2}{3}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| KE before \(= \frac{1}{2}mb^2 = \frac{1}{2}m\cdot\frac{9a^2}{4}\) | M1 | |
| KE after \(= \frac{1}{2}m \cdot 2a^2\) | ||
| Fraction lost \(= \frac{\frac{9a^2}{4} - 2a^2}{\frac{9a^2}{4}} = \frac{\frac{a^2}{4}}{\frac{9a^2}{4}} = \frac{1}{9}\) | M1 A1 |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Impulse $= m(a\mathbf{i}+a\mathbf{j}) - m(b\mathbf{i}) = m(a-b)\mathbf{i} + ma\mathbf{j}$ | | |
| Wall is parallel to $2\mathbf{i}+\mathbf{j}$, so no impulse along wall; impulse perpendicular to wall, i.e. parallel to $(-\mathbf{i}+2\mathbf{j})$ | B1 | Shown |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Normal to wall is in direction $(-\mathbf{i}+2\mathbf{j})$ (perpendicular to $2\mathbf{i}+\mathbf{j}$) | B1 | Identifying normal direction |
| Component of velocity before impact normal to wall: $\frac{(b\mathbf{i})\cdot(-\mathbf{i}+2\mathbf{j})}{|\mathbf{-i+2j}|} = \frac{-b}{\sqrt{5}}$ | M1 A1 | |
| Component after impact normal to wall: $\frac{a(\mathbf{i+j})\cdot(-\mathbf{i}+2\mathbf{j})}{\sqrt{5}} = \frac{a}{\sqrt{5}}$ | M1 A1 | |
| $e = \frac{a/\sqrt{5}}{b/\sqrt{5}} = \frac{a}{b}$ | M1 A1 | |
| Component along wall before $=$ component along wall after (smooth wall): $\frac{2b}{\sqrt{5}} = \frac{3a}{\sqrt{5}}$, so $b = \frac{3a}{2}$, giving $e = \frac{2}{3}$ | M1 A1 | |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| KE before $= \frac{1}{2}mb^2 = \frac{1}{2}m\cdot\frac{9a^2}{4}$ | M1 | |
| KE after $= \frac{1}{2}m \cdot 2a^2$ | | |
| Fraction lost $= \frac{\frac{9a^2}{4} - 2a^2}{\frac{9a^2}{4}} = \frac{\frac{a^2}{4}}{\frac{9a^2}{4}} = \frac{1}{9}$ | M1 A1 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-38_451_1077_315_370}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a framework $A B C$, consisting of two uniform rods rigidly joined together at $B$ so that $\angle A B C = 90 ^ { \circ }$. The $\operatorname { rod } A B$ has length $2 a$ and mass $4 m$, and the $\operatorname { rod } B C$ has length $a$ and mass $2 m$. The framework is smoothly hinged at $A$ to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length $2 a$ and modulus of elasticity $3 m g$, is attached to $A$. The string passes through a small smooth ring $R$ fixed at a distance $2 a$ from $A$, on the same horizontal level as $A$ and in the same vertical plane as the framework. The other end of the string is attached to $B$. The angle $A R B$ is $\theta$, where $0 < \theta < \frac { \pi } { 2 }$.\\
(a) Show that the potential energy $V$ of the system is given by
$$V = 8 a m g \sin 2 \theta + 5 a m g \cos 2 \theta + \text { constant }$$
(b) Find the value of $\theta$ for which the system is in equilibrium.\\
(c) Determine the stability of this position of equilibrium.
\begin{enumerate}
\item A smooth uniform sphere $S$, of mass $m$, is moving on a smooth horizontal plane when it collides obliquely with another smooth uniform sphere $T$, of the same radius as $S$ but of mass $2 m$, which is at rest on the plane. Immediately before the collision the velocity of $S$ makes an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$, with the line joining the centres of the spheres. Immediately after the collision the speed of $T$ is $V$. The coefficient of restitution between the spheres is $\frac { 3 } { 4 }$.\\
(a) Find, in terms of $V$, the speed of $S$\\
(i) immediately before the collision,\\
(ii) immediately after the collision.\\
(b) Find the angle through which the direction of motion of $S$ is deflected as a result of the collision.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q7 [12]}}