| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Ring or bead on wire/rod, equilibrium |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem with friction. Part (a) requires resolving forces in two directions with a given thrust value—straightforward application of Newton's first law. Part (b) involves finding the coefficient of friction at limiting equilibrium, which is a routine extension. The tan α = 4/3 setup is a standard exam trick to avoid messy trigonometry. Slightly above average due to the two-part structure and need to understand limiting friction, but still a textbook exercise requiring no novel insight. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\uparrow)\ \pm F = 0.2g - 2.5\cos\alpha\) | M1 A1 | Allow use of \((\mu R)\) for \(F\). Correct no. of terms, condone sin/cos confusion and sign errors, allow if they have \(T\) instead of 2.5 |
| \(F = 0.46\ \text{(N)}\) oe including fractions, upwards | A1 | Need both magnitude (must be positive) and direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\uparrow)\ F + 0.2g = 6.125\cos\alpha\) | M1A1 | Correct terms, condone sin/cos confusion and sign errors, allow if they have \(T\) instead of 6.125 (but M0 if using \(T = 2.5\)) |
| \((\rightarrow)\ R = 6.125\sin\alpha\ \ (= 4.9)\) | M1A1 | Correct terms, condone sin/cos confusion and sign errors, allow if they have \(T\) instead of 6.125 (but M0 if using \(T = 2.5\)) |
| \(F = \mu R\) | B1 | Seen, but B0 if they use a value for \(R\) found in (a) |
| Solve for \(\mu\) | DM1 | Dependent on both M's |
| \(\mu = 0.35\) oe including fractions | A1 | cao |
| N.B. If \(F\) and \(R\) are interchanged, max B1 can be scored |
# Question 5:
## Part 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\uparrow)\ \pm F = 0.2g - 2.5\cos\alpha$ | M1 A1 | Allow use of $(\mu R)$ for $F$. Correct no. of terms, condone sin/cos confusion and sign errors, allow if they have $T$ instead of 2.5 |
| $F = 0.46\ \text{(N)}$ oe including fractions, upwards | A1 | Need both magnitude (must be positive) and direction |
## Part 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\uparrow)\ F + 0.2g = 6.125\cos\alpha$ | M1A1 | Correct terms, condone sin/cos confusion and sign errors, allow if they have $T$ instead of 6.125 (but M0 if using $T = 2.5$) |
| $(\rightarrow)\ R = 6.125\sin\alpha\ \ (= 4.9)$ | M1A1 | Correct terms, condone sin/cos confusion and sign errors, allow if they have $T$ instead of 6.125 (but M0 if using $T = 2.5$) |
| $F = \mu R$ | B1 | Seen, but B0 if they use a value for $R$ found in (a) |
| Solve for $\mu$ | DM1 | Dependent on both M's |
| $\mu = 0.35$ oe including fractions | A1 | cao |
| **N.B.** If $F$ and $R$ are interchanged, max B1 can be scored | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{151d9232-5a78-4bc1-a57e-6c9cae80e473-18_440_230_248_856}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small bead of mass 0.2 kg is attached to the end $P$ of a light rod $P Q$. The bead is threaded onto a fixed vertical rough wire.
The bead is held in equilibrium with the $\operatorname { rod } P Q$ inclined to the wire at an angle $\alpha$, where $\tan \alpha = \frac { 4 } { 3 }$, as shown in Figure 2.
The thrust in the rod is $T$ newtons.\\
The bead is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude and direction of the friction force acting on the bead when $T = 2.5$
The coefficient of friction between the bead and the wire is $\mu$.\\
Given that the greatest possible value of $T$ is 6.125
\item find the value of $\mu$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q5 [10]}}