| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle in circular tube or on wire |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem combining energy conservation with circular motion dynamics. While it requires multiple steps (energy equation with elastic PE, then force resolution at point B), the approach is methodical and follows well-established techniques taught in M3. The elastic string component adds moderate complexity beyond basic circular motion, but the 'show that' format guides students to the answer, making it slightly above average difficulty for A-level. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 - mgl\) or \(mgl - \frac{1}{2}mv^2\) seen or implied | B1 | Difference between KE and GPE, seen either way round |
| Use of EPE | M1 | Use of EPE formula at top or at \(B\) |
| \(\frac{mg}{2l}l^2\) | A1 | Correct EPE at top |
| \(\frac{mg}{2l}(l\sqrt{2}-l)^2\) | A1 | Correct EPE at \(B\) |
| \(\frac{1}{2}mv^2 + \frac{mg}{2l}(l\sqrt{2}-l)^2 = mgl + \frac{mg}{2l}l^2\) | M1 | Use of conservation of energy, with 1 GPE, 1 KE and 2 EPE terms, condone sign errors |
| Solve for \(v^2\) | dM1 | Solve for \(v^2\), dependent on previous M |
| \(v^2 = 2gl\sqrt{2}\) * | A1* | Exact given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \frac{mg(l\sqrt{2}-l)}{l} = mg(\sqrt{2}-1)\) | M1 A1 | Use of Hooke's Law at \(B\) – may appear in an attempted equation of motion. Correct unsimplified tension at \(B\) |
| \(\pm N + T\cos 45° = \frac{mv^2}{l}\) | M1 A1 A1 | Equation of motion at \(B\) horizontally with correct terms, condone sign errors. Correct equation with at most one error. Correct equation |
| \(\pm N + mg(\sqrt{2}-1) \times \frac{\sqrt{2}}{2} = \frac{m}{l} \times 2gl\sqrt{2}\) | dM1 | Sub for \(T\) and \(v^2\), dependent on both previous M marks |
| \(N = \frac{1}{2}mg(5\sqrt{2}-2)\) * | A1* | Given answer correctly obtained (exactly). If \(N = -\frac{1}{2}mg(5\sqrt{2}-2)\) then clear justification required using 'magnitude' or modulus signs |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - mgl$ or $mgl - \frac{1}{2}mv^2$ seen or implied | B1 | Difference between KE and GPE, seen either way round |
| Use of EPE | M1 | Use of EPE formula at top or at $B$ |
| $\frac{mg}{2l}l^2$ | A1 | Correct EPE at top |
| $\frac{mg}{2l}(l\sqrt{2}-l)^2$ | A1 | Correct EPE at $B$ |
| $\frac{1}{2}mv^2 + \frac{mg}{2l}(l\sqrt{2}-l)^2 = mgl + \frac{mg}{2l}l^2$ | M1 | Use of conservation of energy, with 1 GPE, 1 KE and 2 EPE terms, condone sign errors |
| Solve for $v^2$ | dM1 | Solve for $v^2$, dependent on previous M |
| $v^2 = 2gl\sqrt{2}$ * | A1* | Exact given answer correctly obtained |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{mg(l\sqrt{2}-l)}{l} = mg(\sqrt{2}-1)$ | M1 A1 | Use of Hooke's Law at $B$ – may appear in an attempted equation of motion. Correct unsimplified tension at $B$ |
| $\pm N + T\cos 45° = \frac{mv^2}{l}$ | M1 A1 A1 | Equation of motion at $B$ horizontally with correct terms, condone sign errors. Correct equation with at most one error. Correct equation |
| $\pm N + mg(\sqrt{2}-1) \times \frac{\sqrt{2}}{2} = \frac{m}{l} \times 2gl\sqrt{2}$ | dM1 | Sub for $T$ and $v^2$, dependent on both previous M marks |
| $N = \frac{1}{2}mg(5\sqrt{2}-2)$ * | A1* | Given answer correctly obtained (exactly). If $N = -\frac{1}{2}mg(5\sqrt{2}-2)$ then clear justification required using 'magnitude' or modulus signs |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-16_574_506_283_776}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
A small smooth ring $R$ of mass $m$ is threaded on to a smooth wire in the shape of a circle with centre 0 and radius $I$. The wire is fixed in a vertical plane. The ring $R$ is attached to one end of a light elastic string of natural length I and modulus of elasticity mg . The other end of the elastic string is attached to A , the lowest point of the wire. The point B is on the wire and $O B$ is horizontal.
The ring $R$ is at rest at the highest point of the wire, as shown in Figure 6.\\
The ring $R$ is slightly disturbed from rest and slides along the wire.\\
At the instant when $R$ reaches the point $B$, the speed of $R$ is $v$ and the magnitude of the force exerted on R by the wire is N .
\begin{enumerate}[label=(\alph*)]
\item Show that
$$v ^ { 2 } = 2 g l \sqrt { 2 }$$
\item Show that
$$N = \frac { 1 } { 2 } m g ( 5 \sqrt { 2 } - 2 )$$
\begin{center}
\end{center}
$\_\_\_\_$ VIAV SIHI NI JIIHM ION OC\\
VILU SIHIL NI GLIUM ION OC\\
VEYV SIHI NI ELIUM ION OC
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q6 [14]}}