| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Challenging +1.2 This is a multi-step Further Mechanics problem requiring energy methods and force resolution on an inclined plane with friction and elastic strings. While it involves several components (friction, elasticity, inclined plane), the approach is systematic: part (a) uses work-energy principle with given answer to verify, part (b) requires standard force resolution. The calculations are involved but follow standard FM1 techniques without requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work done against friction \(= 3l \times \mu mg\cos\theta \left(= \frac{9mgl}{13}\right)\) | B1 | Use model to obtain one correct term |
| Gain in EPE \(= \frac{kmg \times 4l^2}{2l}\ (= 2kmgl)\) | B1 | Use model to obtain two correct terms |
| Gain in GPE \(= mg \times 3l\sin\theta \left(= \frac{15mgl}{13}\right)\) | B1 | Use model to obtain three correct terms |
| Work-energy equation: \(\frac{1}{2}m \times 6gl = \frac{9mgl}{13} + 2kmgl + \frac{15mgl}{13}\) | M1 | Need all terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| \(2k = 3 - \frac{24}{13} - \frac{15}{13},\quad k = \frac{15}{26}\) | A1* | Obtain given result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tension in string at \(B\): \(\dfrac{\frac{15}{26}mg \times 2l}{l} \left(= \frac{15mg}{13}\right)\) | B1 | Correct unsimplified expression for tension in string |
| Equation of motion: tension + component of weight \(-\) friction \(= ma\) | M1 | Need all terms and no extras. Condone sign errors and sin/cos confusion. Allow with \(T\) or their \(T\) |
| \(\dfrac{15mg}{13} + mg\sin\theta - \dfrac{1}{4}mg\cos\theta = ma\) | A1 | Unsimplified equation with at most one error |
| \(mg\left(\dfrac{15}{13} + \dfrac{5}{13} - \dfrac{3}{13}\right) = ma\) | A1 | Correct unsimplified equation |
| \(a = \dfrac{17g}{13}\) | A1 | Exact answer or accept 12.8 or 13 \((\text{ms}^{-2})\) |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work done against friction $= 3l \times \mu mg\cos\theta \left(= \frac{9mgl}{13}\right)$ | B1 | Use model to obtain one correct term |
| Gain in EPE $= \frac{kmg \times 4l^2}{2l}\ (= 2kmgl)$ | B1 | Use model to obtain two correct terms |
| Gain in GPE $= mg \times 3l\sin\theta \left(= \frac{15mgl}{13}\right)$ | B1 | Use model to obtain three correct terms |
| Work-energy equation: $\frac{1}{2}m \times 6gl = \frac{9mgl}{13} + 2kmgl + \frac{15mgl}{13}$ | M1 | Need all terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| $2k = 3 - \frac{24}{13} - \frac{15}{13},\quad k = \frac{15}{26}$ | A1* | Obtain given result from correct working |
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## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension in string at $B$: $\dfrac{\frac{15}{26}mg \times 2l}{l} \left(= \frac{15mg}{13}\right)$ | B1 | Correct unsimplified expression for tension in string |
| Equation of motion: tension + component of weight $-$ friction $= ma$ | M1 | Need all terms and no extras. Condone sign errors and sin/cos confusion. Allow with $T$ or their $T$ |
| $\dfrac{15mg}{13} + mg\sin\theta - \dfrac{1}{4}mg\cos\theta = ma$ | A1 | Unsimplified equation with at most one error |
| $mg\left(\dfrac{15}{13} + \dfrac{5}{13} - \dfrac{3}{13}\right) = ma$ | A1 | Correct unsimplified equation |
| $a = \dfrac{17g}{13}$ | A1 | Exact answer or accept 12.8 or 13 $(\text{ms}^{-2})$ |
---\begin{enumerate}
\item A light elastic string with natural length $l$ and modulus of elasticity $k m g$ has one end attached to a fixed point $A$ on a rough inclined plane. The other end of the string is attached to a package of mass $m$.
\end{enumerate}
The plane is inclined at an angle $\theta$ to the horizontal, where $\tan \theta = \frac { 5 } { 12 }$\\
The package is initially held at $A$. The package is then projected with speed $\sqrt { 6 g l }$ up a line of greatest slope of the plane and first comes to rest at the point $B$, where $A B = 31$.\\
The coefficient of friction between the package and the plane is $\frac { 1 } { 4 }$\\
By modelling the package as a particle,\\
(a) show that $k = \frac { 15 } { 26 }$\\
(b) find the acceleration of the package at the instant it starts to move back down the plane from the point $B$.
\hfill \mbox{\textit{Edexcel FM1 2020 Q6 [11]}}