| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.8 This M3 question requires energy conservation with elastic potential energy in two scenarios, including finding extension at equilibrium and handling a collision scenario. It demands careful setup of energy equations and solving a quadratic, which is more challenging than routine mechanics but follows standard M3 methods without requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0.3g(x+0.4) = \frac{49x^2}{2 \times 0.4}\) OR \(0.3gy = \frac{49(y-0.4)^2}{2 \times 0.4}\) | M1A1A1 | M1: Use energy equation with \(x\) as extension at \(B\) or \(y\) as distance fallen. Must have PE term and EPE term of form \(k\frac{\lambda x^2}{l}\); A1A1: Deduct one mark per incorrect term |
| \(5x^2 - 0.24x - 0.096 = 0\) | dM1 | Simplify to 3-term quadratic. Depends on first M1 |
| \(x = \frac{0.24 \pm \sqrt{0.24^2 + 20 \times 0.096}}{10}\) | dM1 | Solve quadratic by formula or completing the square. Allow calculator solution only if \(x = 0.1646\) or final answer correct. Depends on both previous M marks |
| \(x = 0.1646\ldots\) (neg not needed) \ | \(y = 0.5646\ldots\) (0.24 need not be shown) | |
| \(AB = 0.56\) or \(0.565\) m \ | \(AB = y = 0.56\) or \(0.565\) | A1 (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{49 \times 0.2^2}{0.8} = \frac{1}{2} \times 0.3v^2 + 0.3g \times 0.6\) | M1A1A1 | Forming energy equation from release to \(A\). Must have 3 terms: initial EPE, PE and KE term. EPE term of form \(k\frac{\lambda x^2}{l}\) and extension \(\neq 0.165\) |
| \(v^2 = \frac{2}{0.3}\left(\frac{49 \times 0.2^2}{0.8} - 0.3 \times 9.8 \times 0.6\right)\) | ||
| \(v = 2.1\) or \(2.14\) m s\(^{-1}\) | dM1A1 (5) | dM1: Solve to \(v^2 = (4.5733\ldots)\) or \(v = \ldots\); A1: Speed = 2.1 or 2.14 m s\(^{-1}\) |
# Question 4(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0.3g(x+0.4) = \frac{49x^2}{2 \times 0.4}$ OR $0.3gy = \frac{49(y-0.4)^2}{2 \times 0.4}$ | M1A1A1 | M1: Use energy equation with $x$ as extension at $B$ or $y$ as distance fallen. Must have PE term and EPE term of form $k\frac{\lambda x^2}{l}$; A1A1: Deduct one mark per incorrect term |
| $5x^2 - 0.24x - 0.096 = 0$ | dM1 | Simplify to 3-term quadratic. Depends on first M1 |
| $x = \frac{0.24 \pm \sqrt{0.24^2 + 20 \times 0.096}}{10}$ | dM1 | Solve quadratic by formula or completing the square. Allow calculator solution only if $x = 0.1646$ or final answer correct. Depends on both previous M marks |
| $x = 0.1646\ldots$ (neg not needed) \| $y = 0.5646\ldots$ (0.24 need not be shown) | | |
| $AB = 0.56$ or $0.565$ m \| $AB = y = 0.56$ or $0.565$ | A1 (6) | Correct length of $AB$. Must be 2 or 3 sf |
---
# Question 4(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{49 \times 0.2^2}{0.8} = \frac{1}{2} \times 0.3v^2 + 0.3g \times 0.6$ | M1A1A1 | Forming energy equation from release to $A$. Must have 3 terms: initial EPE, PE and KE term. EPE term of form $k\frac{\lambda x^2}{l}$ and extension $\neq 0.165$ |
| $v^2 = \frac{2}{0.3}\left(\frac{49 \times 0.2^2}{0.8} - 0.3 \times 9.8 \times 0.6\right)$ | | |
| $v = 2.1$ or $2.14$ m s$^{-1}$ | dM1A1 (5) | dM1: Solve to $v^2 = (4.5733\ldots)$ or $v = \ldots$; A1: Speed = 2.1 or 2.14 m s$^{-1}$ |
---4. A light elastic string has natural length 0.4 m and modulus of elasticity 49 N . A particle $P$ of mass 0.3 kg is attached to one end of the string. The other end of the string is attached to a fixed point $A$ on a ceiling. The particle is released from rest at $A$ and falls vertically. The particle first comes to instantaneous rest at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $A B$.
The particle is now held at the point 0.6 m vertically below $A$ and released from rest.
\item Find the speed of $P$ immediately before it hits the ceiling.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q4 [11]}}