| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Challenging +1.2 This is a standard M3 energy conservation problem with a spring-pulley system. Part (a) requires setting up energy equations at maximum extension (where KE=0) - a routine technique. Part (b) applies energy conservation with given values. While it involves multiple energy forms (GPE, EPE, KE) and careful bookkeeping, the method is well-practiced in M3 and follows directly from textbook examples. The 9 total marks reflect moderate length rather than exceptional difficulty. |
| Spec | 6.02g 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 |
|---|---|---|
| (a) Loss in P.E. = gain in E.P.E.: \(Mge = \frac{2l}{2l}e\) | \(e = \frac{2Mg\ell}{\lambda}\) | M1 A1 A1 |
| (b) Loss in P.E. of \(A\) = gain in K.E. of \((A \& B)\) + gain in E.P.E. | M1 M1 | |
| \(3 \times 9 \cdot 8 \times 0 \cdot 5 = \frac{1}{2}(4 \cdot 5)v^2 + 17 \cdot 5(0 \cdot 25)\) | \(v^2 = 4 \cdot 589\) | \(v = 2 \cdot 14 \text{ ms}^{-1}\) |
(a) Loss in P.E. = gain in E.P.E.: $Mge = \frac{2l}{2l}e$ | $e = \frac{2Mg\ell}{\lambda}$ | M1 A1 A1
(b) Loss in P.E. of $A$ = gain in K.E. of $(A \& B)$ + gain in E.P.E. | M1 M1
$3 \times 9 \cdot 8 \times 0 \cdot 5 = \frac{1}{2}(4 \cdot 5)v^2 + 17 \cdot 5(0 \cdot 25)$ | $v^2 = 4 \cdot 589$ | $v = 2 \cdot 14 \text{ ms}^{-1}$ | A1 A1 M1 A1
**Total: 9 marks**Two particles $A$ and $B$, of masses $M$ kg and $m$ kg respectively, are connected by a light inextensible string passing over a smooth fixed pulley. $B$ is placed on a smooth horizontal table and $A$ hangs freely, as shown. $B$ is attached to a spring of natural length $l$ m and modulus of elasticity $\lambda$ N, whose other end is fixed to a vertical wall.
\includegraphics{figure_3}
The system starts to move from rest when the string is taut and the spring neither extended nor compressed. $A$ does not reach the ground, nor does $B$ reach the pulley, during the motion.
\begin{enumerate}[label=(\alph*)]
\item Show that the maximum extension of the spring is $\frac{2Mgl}{\lambda}$ m. [3 marks]
\item If $M = 3$, $m = 1.5$ and $\lambda = 35l$, find the speed of $A$ when the extension in the spring is $0.5$ m. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [9]}}