| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.3 This is a standard Hooke's law equilibrium problem requiring resolution of forces and Pythagoras to find extensions. The setup is straightforward with symmetry, and the elastic energy formula is direct application. Slightly easier than average due to the methodical nature and clear geometric setup, though the multi-step calculation and need to handle two string sections adds modest complexity. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| Resolving vertically: \(27\cos\theta = W\) | M1A2, 1.0 |
| Hooke's Law: \(T = \frac{80 \times 3.5}{4}\) | M1A1 |
| \(W = 84N\) | A1 |
| Answer | Marks |
|---|---|
| \(EPE = 2 \times \frac{80 \times 3.5^2}{2 \times 4} = 245\) (or awrt 245) | M1A1ft, A1 |
| Alternative: \(\frac{80 \times 7^2}{16} = 245\) |
## Part (a)
Resolving vertically: $27\cos\theta = W$ | M1A2, 1.0 |
Hooke's Law: $T = \frac{80 \times 3.5}{4}$ | M1A1 |
$W = 84N$ | A1 |
## Part (b)
$EPE = 2 \times \frac{80 \times 3.5^2}{2 \times 4} = 245$ (or awrt 245) | M1A1ft, A1 |
Alternative: $\frac{80 \times 7^2}{16} = 245$ | |
---A light elastic string has natural length $8$ m and modulus of elasticity $80$ N.
The ends of the string are attached to fixed points $P$ and $Q$ which are on the same horizontal level and $12$ m apart. A particle is attached to the mid-point of the string and hangs in equilibrium at a point $4.5$ m below $PQ$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the weight of the particle. [6]
\item Calculate the elastic energy in the string when the particle is in this position. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q1 [9]}}