| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string on inclined plane |
| Difficulty | Challenging +1.8 This is a multi-stage work-energy problem requiring careful tracking of energy through three phases (string slack, string taut, and equilibrium analysis). It involves impulse-momentum, work done against friction and gravity, elastic potential energy, and force resolution to check equilibrium. The calculation is lengthy with multiple components but follows standard M3 techniques without requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02h Elastic PE: 1/2 k x^26.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.5u = 4 \rightarrow u = 8\) | B1 | |
| \(F_{max} = \frac{\sqrt{5}}{5} \times 0.5g \times \frac{\sqrt{45}}{7} \left(= \frac{3g}{14} = 2.1\right)\) | B1 | |
| \(\frac{1}{2} \times 0.5 \times 8^2 = 0.5g(x+2)\sin\theta + F_r(x+2) + \frac{3x^2}{2 \times 2}\) | M1A1A1 | |
| \(64 = 14(x+2) + 3x^2\) | ||
| \(3x^2 + 14x - 36 = 0\) | M1 | |
| \(x = 1.8\text{m}\) \((1.84\text{m})\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T = \frac{3 \times 1.84}{2} (= 2.76)\) | B1ft | |
| \(0.5a = 2.76 + 1.4 - 2.1\) \((= 2.06)\) Acceleration down slope, so particle does not remain at \(A\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.5u = 4 \rightarrow u = 8\) | B1 | |
| \(F_{max} = \frac{\sqrt{5}}{5} \times 0.5g \times \frac{\sqrt{45}}{7} (= 2.1)\) | B1 | |
| \(\frac{1}{2} \times 0.5 \times 8^2 = 0.5gd\sin\theta + F_r d + \frac{3(d-2)^2}{2 \times 2}\) | M1A1A1 | |
| \(64 = 14d + 3(d-2)^2\) | ||
| \(3d^2 + 2d - 52 = 0\) | M1 | |
| \(d = 3.8 \rightarrow x = 1.8\text{m}\) \((1.84\text{m})\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T = \frac{3 \times 1.84}{2} (= 2.76)\) | B1ft | |
| Upslope: \(F_{max}(= 2.1)\), Downslope: \(0.5g\sin\theta + T (= 4.16)\) | M1 | |
| \(4.16 > 2.1\) so there is a resultant force down slope, so the particle does not remain at \(A\) | A1 |
# Question 7:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.5u = 4 \rightarrow u = 8$ | B1 | |
| $F_{max} = \frac{\sqrt{5}}{5} \times 0.5g \times \frac{\sqrt{45}}{7} \left(= \frac{3g}{14} = 2.1\right)$ | B1 | |
| $\frac{1}{2} \times 0.5 \times 8^2 = 0.5g(x+2)\sin\theta + F_r(x+2) + \frac{3x^2}{2 \times 2}$ | M1A1A1 | |
| $64 = 14(x+2) + 3x^2$ | | |
| $3x^2 + 14x - 36 = 0$ | M1 | |
| $x = 1.8\text{m}$ $(1.84\text{m})$ | M1A1 | |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T = \frac{3 \times 1.84}{2} (= 2.76)$ | B1ft | |
| $0.5a = 2.76 + 1.4 - 2.1$ $(= 2.06)$ Acceleration down slope, so particle does not remain at $A$ | M1A1 | |
### ALT (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.5u = 4 \rightarrow u = 8$ | B1 | |
| $F_{max} = \frac{\sqrt{5}}{5} \times 0.5g \times \frac{\sqrt{45}}{7} (= 2.1)$ | B1 | |
| $\frac{1}{2} \times 0.5 \times 8^2 = 0.5gd\sin\theta + F_r d + \frac{3(d-2)^2}{2 \times 2}$ | M1A1A1 | |
| $64 = 14d + 3(d-2)^2$ | | |
| $3d^2 + 2d - 52 = 0$ | M1 | |
| $d = 3.8 \rightarrow x = 1.8\text{m}$ $(1.84\text{m})$ | M1A1 | |
### ALT (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T = \frac{3 \times 1.84}{2} (= 2.76)$ | B1ft | |
| Upslope: $F_{max}(= 2.1)$, Downslope: $0.5g\sin\theta + T (= 4.16)$ | M1 | |
| $4.16 > 2.1$ so there is a resultant force down slope, so the particle does not remain at $A$ | A1 | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-24_394_1027_248_461}
\captionsetup{labelformat=empty}
\caption{Figure 8}
\end{center}
\end{figure}
A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 3 N . The other end of the string is attached to a fixed point $O$ on a rough plane. The plane is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 2 } { 7 }$
The coefficient of friction between $P$ and the plane is $\frac { \sqrt { 5 } } { 5 }$
The particle $P$ is initially at rest at the point $O$, as shown in Figure 8.
The particle $P$ then receives an impulse of magnitude 4 Ns, directed up a line of greatest slope of the plane.
The particle $P$ moves up the plane and comes to rest at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the extension of the elastic string when $P$ is at $A$.
\item Show that the particle does not remain at rest at $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q7 [11]}}