| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 13 |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring resolution of forces with elastic strings in 2D, Hooke's law applications, elastic potential energy calculations, and a reconfiguration problem. While the geometry is straightforward (3-4-5 triangle), it demands systematic application of multiple concepts across equilibrium analysis and energy, typical of Further Maths but not requiring exceptional insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_A = \frac{\lambda_A \times 2}{2}\) and \(T_B = \frac{\lambda_B \times 1}{2}\) | B1 | Hooke's law used for both strings |
| \(\cos PAB = \frac{4}{5}\) and \(\cos PBA = \frac{3}{5}\) oe | B1 | Correct trig for both angles used |
| \(T_A \cos PAB = T_B \cos PBA\) | M1 | Equating horizontal components |
| \(\frac{2\lambda_A}{2} \times \frac{4}{5} = \frac{\lambda_B}{2} \times \frac{3}{5} \Rightarrow \lambda_p = \frac{8}{3}\lambda_A\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_A \sin PAB + T_B \sin PBA = mg\) | M1 | Vertical equilibrium; 3 forces; Both tensions resolved |
| \(\frac{2\lambda_A}{2} \times \frac{3}{5} + \frac{3}{2}\lambda_A \times \frac{4}{5} = mg\) | M1 | Using tensions and \(\lambda_B = \frac{3}{8}\lambda_A\) |
| \(\lambda_A = \frac{5}{8}mg\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(EPE = \frac{\lambda_A \times 2^2}{2 \times 2} + \frac{\lambda_B \times 1^2}{2 \times 2} = mg\) | M1 | soi |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T_A' = 0.3mg = \frac{\frac{5}{8}mg x_A}{2}\) | M1 | Considering equilibrium of \(Q\) and using Hooke's law; (\(x_A = 1\)) |
| \(T_B'' = mg + T_A'\) | M1 | Considering equilibrium of \(P\); must involve 3 terms; Or \(T_B'' = mg + 0.3mg\) soi |
| \(1.3mg = \frac{8 \times \frac{5}{8}mg \times x_B}{2}\) | M1 | Use of Hooke's law for upper string, with \(T_A'\) eliminated; (\(x_B = \frac{13}{8}\)) |
| Distance is \(1 + 2 + \frac{13}{8} + 2 = 6\frac{5}{8}\) m | A1 | oe, e.g. awrt 6.63 m |
## (i)(a)
$T_A = \frac{\lambda_A \times 2}{2}$ and $T_B = \frac{\lambda_B \times 1}{2}$ | B1 | Hooke's law used for both strings
$\cos PAB = \frac{4}{5}$ and $\cos PBA = \frac{3}{5}$ oe | B1 | Correct trig for both angles used
$T_A \cos PAB = T_B \cos PBA$ | M1 | Equating horizontal components
$\frac{2\lambda_A}{2} \times \frac{4}{5} = \frac{\lambda_B}{2} \times \frac{3}{5} \Rightarrow \lambda_p = \frac{8}{3}\lambda_A$ | A1 | AG
## (i)(b)
$T_A \sin PAB + T_B \sin PBA = mg$ | M1 | Vertical equilibrium; 3 forces; Both tensions resolved
$\frac{2\lambda_A}{2} \times \frac{3}{5} + \frac{3}{2}\lambda_A \times \frac{4}{5} = mg$ | M1 | Using tensions and $\lambda_B = \frac{3}{8}\lambda_A$
$\lambda_A = \frac{5}{8}mg$ | A1 |
## (ii)
$EPE = \frac{\lambda_A \times 2^2}{2 \times 2} + \frac{\lambda_B \times 1^2}{2 \times 2} = mg$ | M1 | soi
| A1 |
## (iii)
$T_A' = 0.3mg = \frac{\frac{5}{8}mg x_A}{2}$ | M1 | Considering equilibrium of $Q$ and using Hooke's law; ($x_A = 1$)
$T_B'' = mg + T_A'$ | M1 | Considering equilibrium of $P$; must involve 3 terms; Or $T_B'' = mg + 0.3mg$ soi
$1.3mg = \frac{8 \times \frac{5}{8}mg \times x_B}{2}$ | M1 | Use of Hooke's law for upper string, with $T_A'$ eliminated; ($x_B = \frac{13}{8}$)
Distance is $1 + 2 + \frac{13}{8} + 2 = 6\frac{5}{8}$ m | A1 | oe, e.g. awrt 6.63 m
---$A$ and $B$ are two points a distance of 5 m apart on a horizontal ceiling. A particle $P$ of mass $m$ kg is attached to $A$ and $B$ by light elastic strings. The particle hangs in equilibrium at a distance of 4 m from $A$ and 3 m from $B$ so that angle $APB = 90°$ (see diagram).
\includegraphics{figure_4}
The string joining $P$ to $A$ has natural length 2 m and modulus of elasticity $\lambda_A$ N. The string joining $P$ to $B$ also has natural length 2 m but has modulus of elasticity $\lambda_B$ N.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Show that $\lambda_B = \frac{3}{4}\lambda_A$. [4]
\item Find an expression for $\lambda_A$ in terms of $m$ and $g$. [3]
\end{enumerate}
\item Find, in terms of $m$ and $g$, the total elastic potential energy stored in the strings. [2]
\end{enumerate}
The string joining $P$ to $A$ is detached from $A$ and a second particle, $Q$, of mass $0.3m$ kg is attached to the free end of the string. $Q$ is then gently lowered into a position where the system hangs vertically in equilibrium.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the distance of $Q$ below $B$ in this equilibrium position. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q4 [13]}}