Edexcel M1 2020 January — Question 4 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2020
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeConnected particles with friction
DifficultyStandard +0.3 This is a standard M1 connected particles problem with friction. Part (a) requires simple vertical resolution at particle C, part (b) needs vertical resolution at ring A, and part (c) applies limiting friction with horizontal resolution. All steps are routine applications of equilibrium equations with no novel insight required, making it slightly easier than average.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{916543cb-14f7-486c-ba3c-eda9be134045-10_633_1237_258_356} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Two identical small rings, \(A\) and \(B\), each of mass \(m\), are threaded onto a rough horizontal wire. The rings are connected by a light inextensible string. A particle \(C\) of mass \(3 m\) is attached to the midpoint of the string. The particle \(C\) hangs in equilibrium below the wire with angle \(B A C = \beta\), as shown in Figure 2. The tension in each of the parts, \(A C\) and \(B C\), of the string is \(T\)
  1. By considering particle \(C\), find \(T\) in terms of \(m , g\) and \(\beta\)
  2. Find, in terms of \(m\) and \(g\), the magnitude of the normal reaction between the wire and \(A\). The coefficient of friction between each ring and the wire is \(\frac { 4 } { 5 }\) The two rings, \(A\) and \(B\), are on the point of sliding along the wire towards each other.
  3. Find the value of \(\tan \beta\) \includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-11_2255_50_314_34}
    VIXV SIHIANI III IM IONOOVIAV SIHI NI JYHAM ION OOVI4V SIHI NI JLIYM ION OO

Question 4:
Part (a)
AnswerMarks Guidance
Working/AnswerMark Guidance
\(2T\sin\beta = 3mg\) OR \(\frac{T}{\sin(90°-\beta)} = \frac{3mg}{\sin 2\beta}\)M1 Resolve vertically for \(C\) with usual rules or use triangle of forces
\(T = \frac{3mg}{2\sin\beta}\) OR \(T = \frac{3mg\cos\beta}{\sin 2\beta}\)A1 (2) Allow \(\cos(90°-\beta)\) for \(\sin\beta\) or \(\sin(90°-\beta)\) for \(\cos\beta\)
Part (b)
AnswerMarks Guidance
Working/AnswerMark Guidance
For \(A\) or \(B\): \((\uparrow)\ R = mg + T\sin\beta\) OR whole system: \((\uparrow)\ 2R = 3mg + mg + mg\) OR for \(AC\) or \(BC\): \((\uparrow)\ R + T\sin\beta = mg + 3mg\)M1 A1 M1 resolve vertically for \(A\) or \(B\), for whole system or for \(AC\) or \(BC\) with usual rules. A1 correct equation
\(R = 2.5mg\)A1 (3)
Part (c)
AnswerMarks Guidance
Working/AnswerMark Guidance
\(F = T\cos\beta\)M1A1 M1 resolve horizontally for \(A\) with usual rules. A1 correct equation
\(F = \frac{4}{5} \times 2.5mg\)B1 ft B1 ft for \(F = \frac{4}{5} \times\) their \(R\) (allow magnitude if \(R < 0\)). B0 for just \(F = 4/5\ R\)
Eliminate \(T\) and solve for \(\tan\beta\)M1 Eliminate \(T\) and solve for \(\tan\beta\) correctly
\(\tan\beta = \frac{3}{4}\)A1 (5) oe
## Question 4:

### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2T\sin\beta = 3mg$ **OR** $\frac{T}{\sin(90°-\beta)} = \frac{3mg}{\sin 2\beta}$ | M1 | Resolve vertically for $C$ with usual rules or use triangle of forces |
| $T = \frac{3mg}{2\sin\beta}$ **OR** $T = \frac{3mg\cos\beta}{\sin 2\beta}$ | A1 (2) | Allow $\cos(90°-\beta)$ for $\sin\beta$ or $\sin(90°-\beta)$ for $\cos\beta$ |

### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| For $A$ or $B$: $(\uparrow)\ R = mg + T\sin\beta$ **OR** whole system: $(\uparrow)\ 2R = 3mg + mg + mg$ **OR** for $AC$ or $BC$: $(\uparrow)\ R + T\sin\beta = mg + 3mg$ | M1 A1 | M1 resolve vertically for $A$ or $B$, for whole system or for $AC$ or $BC$ with usual rules. A1 correct equation |
| $R = 2.5mg$ | A1 (3) | |

### Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F = T\cos\beta$ | M1A1 | M1 resolve horizontally for $A$ with usual rules. A1 correct equation |
| $F = \frac{4}{5} \times 2.5mg$ | B1 ft | B1 ft for $F = \frac{4}{5} \times$ their $R$ (allow magnitude if $R < 0$). B0 for just $F = 4/5\ R$ |
| Eliminate $T$ and solve for $\tan\beta$ | M1 | Eliminate $T$ and solve for $\tan\beta$ correctly |
| $\tan\beta = \frac{3}{4}$ | A1 (5) | oe |
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{916543cb-14f7-486c-ba3c-eda9be134045-10_633_1237_258_356}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

Two identical small rings, $A$ and $B$, each of mass $m$, are threaded onto a rough horizontal wire. The rings are connected by a light inextensible string. A particle $C$ of mass $3 m$ is attached to the midpoint of the string. The particle $C$ hangs in equilibrium below the wire with angle $B A C = \beta$, as shown in Figure 2.

The tension in each of the parts, $A C$ and $B C$, of the string is $T$
\begin{enumerate}[label=(\alph*)]
\item By considering particle $C$, find $T$ in terms of $m , g$ and $\beta$
\item Find, in terms of $m$ and $g$, the magnitude of the normal reaction between the wire and $A$.

The coefficient of friction between each ring and the wire is $\frac { 4 } { 5 }$\\
The two rings, $A$ and $B$, are on the point of sliding along the wire towards each other.
\item Find the value of $\tan \beta$\\

\includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-11_2255_50_314_34}\\

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
\hline
\end{tabular}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2020 Q4 [10]}}