| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough incline, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with connected particles. While it involves multiple components (incline, friction, pulley), the setup is routine: write two equations of motion using F=ma, solve simultaneously for acceleration, then find tension. The trig is given (tan α = 3/4 means sin α = 3/5, cos α = 4/5), and all steps follow a well-practiced algorithm. Slightly above average difficulty only due to the multi-part nature and the final pulley force calculation requiring vector resolution. |
| Spec | 3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 2mg\sin\alpha - F = 2ma\) | M1A1 | Correct no. of terms, condone sin/cos confusion and sign errors; correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(3mg - T = 3ma\) | M1A1 | Correct no. of terms, condone sign errors; correct equation. N.B. Could have \(a\) replaced by \((-a)\) in both. Ignore labelling (i) and (ii). (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 2mg\cos\alpha\) | M1A1 | Allow if this appears in (a). Correct no. of terms, condone sin/cos confusion and sign errors; correct equation |
| \(F = \frac{1}{2}R\) | B1 | Seen, possibly on a diagram or in (a) |
| Substitute for trig. and solve for \(a\) | DM1 | Dependent on the two M's in (a), for solving 2 simultaneous equations or using a whole system equation to find \(a\) |
| \(a = \frac{1}{5}g\) | A1 | cao (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \frac{12mg}{5}\) | DM1 | Dependent on relevant 1st or 2nd M1 in (a), for attempt to find \(T\); must be of form \(km\) or \(kmg\). Apply isw if they 'cancel' \(m\)'s. |
| \(2T\cos\left(\frac{90°-\alpha}{2}\right)\) OR \(\sqrt{T^2+T^2-2T^2\cos(90°+\alpha)}\) OR \(\sqrt{(T\cos\alpha)^2+(T+T\sin\alpha)^2}\) | M1 | For a correct expression in terms of \(T\) and \(\alpha\) only; \(\alpha\) does not need to be substituted |
| Substitute for trig and \(T\) to obtain expression in \(m\) or \(mg\) | DM1 | Dependent on previous M, for substituting their \(T\) and for trig, to give an expression of form \(km\) or \(kmg\) |
| \(\frac{48\sqrt{5}mg}{25}\); Accept \(4.3mg\) or better, \(42m\) or \(42.1m\) | A1 | cao (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Tension is the same on either side of the pulley, tension across the pulley is the same. | B1 | For any equivalent statement. B0 for incorrect extras. B0 for tension is same for \(A\) and \(B\) or is the same for both strings etc (1 mark) |
# Question 7:
## Part 7(a)(i):
$T - 2mg\sin\alpha - F = 2ma$ | M1A1 | Correct no. of terms, condone sin/cos confusion and sign errors; correct equation
## Part 7(a)(ii):
$3mg - T = 3ma$ | M1A1 | Correct no. of terms, condone sign errors; correct equation. N.B. Could have $a$ replaced by $(-a)$ in both. Ignore labelling (i) and (ii). **(4 marks)**
## Part 7(b):
$R = 2mg\cos\alpha$ | M1A1 | Allow if this appears in (a). Correct no. of terms, condone sin/cos confusion and sign errors; correct equation
$F = \frac{1}{2}R$ | B1 | Seen, possibly on a diagram or in (a)
Substitute for trig. and solve for $a$ | DM1 | Dependent on the two M's in (a), for solving 2 simultaneous equations or using a whole system equation to find $a$
$a = \frac{1}{5}g$ | A1 | cao **(5 marks)**
## Part 7(c):
$T = \frac{12mg}{5}$ | DM1 | Dependent on relevant 1st or 2nd M1 in (a), for attempt to find $T$; must be of form $km$ or $kmg$. Apply isw if they 'cancel' $m$'s.
$2T\cos\left(\frac{90°-\alpha}{2}\right)$ **OR** $\sqrt{T^2+T^2-2T^2\cos(90°+\alpha)}$ **OR** $\sqrt{(T\cos\alpha)^2+(T+T\sin\alpha)^2}$ | M1 | For a **correct** expression in terms of $T$ and $\alpha$ only; $\alpha$ does not need to be substituted
Substitute for trig and $T$ to obtain expression in $m$ or $mg$ | DM1 | Dependent on previous M, for substituting their $T$ **and** for trig, to give an expression of form $km$ or $kmg$
$\frac{48\sqrt{5}mg}{25}$; Accept $4.3mg$ or better, $42m$ or $42.1m$ | A1 | cao **(4 marks)**
## Part 7(d):
Tension is the same on **either side of the pulley**, tension across the pulley is the same. | B1 | For any equivalent statement. B0 for incorrect extras. B0 for tension is same for $A$ and $B$ or is the same for both strings etc **(1 mark)**
**Total: 14 marks**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{151d9232-5a78-4bc1-a57e-6c9cae80e473-24_446_624_260_708}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
One end of a light inextensible string is attached to a particle $A$ of mass $2 m$. The other end of the string is attached to a particle $B$ of mass $3 m$. The string passes over a small, smooth, light pulley $P$ which is fixed at the top of a rough inclined plane. The plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$
Particle $A$ is held at rest on the plane with the string taut and $B$ hanging freely below $P$, as shown in Figure 4. The section of the string $A P$ is parallel to a line of greatest slope of the plane.
The coefficient of friction between $A$ and the plane is $\frac { 1 } { 2 }$\\
Particle $A$ is released and begins to move up the plane.\\
For the motion before $A$ reaches the pulley,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item write down an equation of motion for $A$,
\item write down an equation of motion for $B$,
\end{enumerate}\item find, in terms of $g$, the acceleration of $A$,
\item find the magnitude of the force exerted on the pulley by the string.
\item State how you have used the information that $P$ is a smooth pulley.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q7 [14]}}