| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.8 This is a multi-step energy conservation problem involving elastic potential energy in a spring. Part (a) requires setting up and solving a quadratic equation from energy conservation when the spring extends beyond natural length. Part (b) requires similar analysis with changed geometry. While the method is standard for M3, the algebraic manipulation and careful tracking of spring extension vs natural length makes it moderately challenging, above average difficulty. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy: \(\frac{1}{2} \times 2 \times 10^2 = 2 \times 9.8 \times h + \frac{1}{2} \times \frac{120 \times h^2}{3}\) | M1 A1 A1 | |
| \(20h^2 + 19.6h - 100 = 0\) | M1 | |
| \(h = \frac{-19.6 \pm \sqrt{19.6^2 + 4 \times 20 \times 100}}{40}\) | M1 | |
| \(= 1.7991\ldots \approx 1.8\) (or 1.80) m | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2} \times 2 \times V^2 = 2 \times 9.8 \times 1.8 + \frac{1}{2} \times \frac{120 \times 2.3^2}{3} - \frac{1}{2} \times \frac{120 \times 0.5^2}{3}\) | M1 A1 A1 | |
| \(V = 11.7\) (3 s.f.) or \(12\) (2 s.f.) m s\(^{-1}\) | M1 A1 | (5) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy: $\frac{1}{2} \times 2 \times 10^2 = 2 \times 9.8 \times h + \frac{1}{2} \times \frac{120 \times h^2}{3}$ | M1 A1 A1 | |
| $20h^2 + 19.6h - 100 = 0$ | M1 | |
| $h = \frac{-19.6 \pm \sqrt{19.6^2 + 4 \times 20 \times 100}}{40}$ | M1 | |
| $= 1.7991\ldots \approx 1.8$ (or 1.80) m | A1 | (6) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2} \times 2 \times V^2 = 2 \times 9.8 \times 1.8 + \frac{1}{2} \times \frac{120 \times 2.3^2}{3} - \frac{1}{2} \times \frac{120 \times 0.5^2}{3}$ | M1 A1 A1 | |
| $V = 11.7$ (3 s.f.) or $12$ (2 s.f.) m s$^{-1}$ | M1 A1 | (5) |
**(11 marks)**
---5. In a "test your strength" game at an amusement park, competitors hit one end of a small lever with a hammer, causing the other end of the lever to strike a ball which then moves in a vertical tube whose total height is adjustable. The ball is attached to one end of an elastic spring of natural length 3 m and modulus of elasticity 120 N . The mass of the ball is 2 kg . The other end of the spring is attached to the top of the tube. The ball is modelled as a particle, the spring as light and the tube is assumed to be smooth.
The height of the tube is first set at 3 m . A competitor gives the ball an initial speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the height to which the ball rises before coming to rest.
The tube is now adjusted by reducing its height to 2.5 m . The spring and the ball remain unchanged.
\item Find the initial speed which the ball must now have if it is to rise by the same distance as in part (a).\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [11]}}