| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2021 |
| 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 | Standard +0.8 This is a Further Maths FM1 question combining elastic strings, inclined planes, friction, and energy methods. Part (a) requires resolving forces with the spring compressed, handling the trigonometry (tan α = 3/4 means sin α = 3/5, cos α = 4/5), and accounting for friction direction. Part (b) requires energy conservation with work done against friction over distance 4l. While systematic, it demands careful bookkeeping of multiple energy terms and force components across several steps, placing it moderately above average difficulty. |
| 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 | Mark | Guidance |
| Thrust in spring \(= \frac{3mg \cdot 2l}{3l}\) \((= 2mg)\) | B1 | Correct unsimplified expression for the thrust |
| Equation of motion: \(2mg - mg\sin\alpha - \frac{1}{3}mg\cos\alpha = ma\) | M1 | Equation of motion. All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| \(\left(2mg - \frac{3mg}{5} - \frac{4mg}{15} = ma\right)\) | A1ft | Unsimplified equation with at most one error (in \(T\) or their \(T\)) |
| A1ft | Correct unsimplified equation (in \(T\) or their \(T\)) | |
| \(a = \frac{17g}{15}\) | A1* | Obtain given result from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Initial EPE \(= \frac{3mg \cdot 4l^2}{2 \times 3l}\) \((= 2mgl)\) | B1 | Use model to obtain one correct term |
| Gain in GPE \(= mg \cdot 2l\sin\alpha\) \(\left(= \frac{6}{5}mgl\right)\) | B1 | Use model to obtain two correct terms |
| Work done against friction \(= \frac{1}{3}mg\cos\alpha \times 2l\) \(\left(= \frac{8}{15}mgl\right)\) | B1 | Use model to obtain three correct terms |
| Work-energy equation: \(\frac{1}{2}mv^2 + \frac{2}{3}mgl\cos\alpha + 2mgl\sin\alpha = 2mgl\) | M1 | All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| A1 | Correct unsimplified equation | |
| \(v = \sqrt{\frac{8gl}{15}}\) | A1 | Accept \(0.73\sqrt{gl}\) |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Thrust in spring $= \frac{3mg \cdot 2l}{3l}$ $(= 2mg)$ | B1 | Correct unsimplified expression for the thrust |
| Equation of motion: $2mg - mg\sin\alpha - \frac{1}{3}mg\cos\alpha = ma$ | M1 | Equation of motion. All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| $\left(2mg - \frac{3mg}{5} - \frac{4mg}{15} = ma\right)$ | A1ft | Unsimplified equation with at most one error (in $T$ or their $T$) |
| | A1ft | Correct unsimplified equation (in $T$ or their $T$) |
| $a = \frac{17g}{15}$ | A1* | Obtain given result from correct working |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Initial EPE $= \frac{3mg \cdot 4l^2}{2 \times 3l}$ $(= 2mgl)$ | B1 | Use model to obtain one correct term |
| Gain in GPE $= mg \cdot 2l\sin\alpha$ $\left(= \frac{6}{5}mgl\right)$ | B1 | Use model to obtain two correct terms |
| Work done against friction $= \frac{1}{3}mg\cos\alpha \times 2l$ $\left(= \frac{8}{15}mgl\right)$ | B1 | Use model to obtain three correct terms |
| Work-energy equation: $\frac{1}{2}mv^2 + \frac{2}{3}mgl\cos\alpha + 2mgl\sin\alpha = 2mgl$ | M1 | All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion |
| | A1 | Correct unsimplified equation |
| $v = \sqrt{\frac{8gl}{15}}$ | A1 | Accept $0.73\sqrt{gl}$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7f077b82-6b39-4cb5-8574-bfa308c88df3-20_401_814_246_628}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light elastic spring has natural length $3 l$ and modulus of elasticity $3 m g$.\\
One end of the spring is attached to a fixed point $X$ on a rough inclined plane.\\
The other end of the spring is attached to a package $P$ of mass $m$.\\
The plane is inclined to the horizontal at an angle $\alpha$ where $\tan \alpha = \frac { 3 } { 4 }$\\
The package is initially held at the point $Y$ on the plane, where $X Y = l$. The point $Y$ is higher than $X$ and $X Y$ is a line of greatest slope of the plane, as shown in Figure 2.
The package is released from rest at $Y$ and moves up the plane.\\
The coefficient of friction between $P$ and the plane is $\frac { 1 } { 3 }$\\
By modelling $P$ as a particle,
\begin{enumerate}[label=(\alph*)]
\item show that the acceleration of $P$ at the instant when $P$ is released from rest is $\frac { 17 } { 15 } \mathrm {~g}$
\item find, in terms of $g$ and $l$, the speed of $P$ at the instant when the spring first reaches its natural length of 31 .
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 2021 Q6 [11]}}