Edexcel Paper 3 2021 October — Question 2 12 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
Year2021
SessionOctober
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on rough incline, particle hanging
DifficultyStandard +0.3 This is a standard A-level mechanics pulley problem with routine application of Newton's second law. Given tan α = 3/4, students find sin α = 3/5, apply F = ma to both particles with friction μR = (1/6)(3mg cos α) = mg/10, and solve simultaneous equations. The sketch in (c) is straightforward (linear from origin), and part (d) tests conceptual understanding. Slightly easier than average due to the helpful tan α value and clear structure.
Spec3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03v Motion on rough surface: including inclined planes
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{63363c3e-13fc-49a1-8cef-951e6e97e801-04_396_993_246_536} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A small stone \(A\) of mass \(3 m\) is attached to one end of a string.
A small stone \(B\) of mass \(m\) is attached to the other end of the string.
Initially \(A\) is held at rest on a fixed rough plane.
The plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\) The string passes over a pulley \(P\) that is fixed at the top of the plane.
The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane.
Stone \(B\) hangs freely below \(P\), as shown in Figure 1.
The coefficient of friction between \(A\) and the plane is \(\frac { 1 } { 6 }\) Stone \(A\) is released from rest and begins to move down the plane.
The stones are modelled as particles.
The pulley is modelled as being small and smooth.
The string is modelled as being light and inextensible. Using the model for the motion of the system before \(B\) reaches the pulley,
  1. write down an equation of motion for \(A\)
  2. show that the acceleration of \(A\) is \(\frac { 1 } { 10 } \mathrm {~g}\)
  3. sketch a velocity-time graph for the motion of \(B\), from the instant when \(A\) is released from rest to the instant just before \(B\) reaches the pulley, explaining your answer. In reality, the string is not light.
  4. State how this would affect the working in part (b).
Question 2:
Part 2(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Equation of motion for \(A\)M1 Equation in \(T\) and \(a\) with correct no. of terms, condone sign errors and sin/cos confusion. If one of the 3's is missing, allow M1. Treat \(\sin(3/5)\) etc as an A error but allow recovery
\(3mg\sin\alpha - F - T = 3ma\)A1 Correct equation (allow \((-a)\) instead of \(a\) in both equations)
Part 2(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolve perpendicular to the planeM1 Correct no. of terms, condone sign errors and sin/cos confusion. Allow if appears in (a)
\(R = 3mg\cos\alpha\)A1 Correct equation
\(F = \frac{1}{6}R\)B1 Seen anywhere in (a) or (b), including on a diagram
Equation of motion for \(B\) OR for whole systemM1 Equation (for \(B\)) in \(T\) and \(a\) with correct no. of terms, condone sign errors and sin/cos confusion. OR Whole system equation with correct no. of terms, condone sign errors and sin/cos confusion
\(T - mg = ma\) OR \(3mg\sin\alpha - F - mg = 3ma + ma\)A1 Correct equation
Complete method to solve for \(a\)DM1 Complete method (trig may not be substituted), dependent on M1 in (a) and second M1 in (b) if two equations used, or second M1 in (b) if one equation used
\(a = \frac{1}{10}g\)A1* Correct answer correctly obtained
Part 2(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Straight line graph starting at origin, with positive gradientB1 Straight line starting at origin (could be reflected in the \(t\)-axis). B0 if continuous vertical line at the end
e.g. acceleration (of \(B\)) is constantDB1 Dependent on first B1, for any equivalent statement
Part 2(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
e.g. the tensions in the two equations of motion would be different. Tension on \(A\) would be different to tension on \(B\)B1 B0 if incorrect extras 
## Question 2:

### Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $A$ | M1 | Equation in $T$ and $a$ with correct no. of terms, condone sign errors and sin/cos confusion. If one of the 3's is missing, allow M1. Treat $\sin(3/5)$ etc as an A error but allow recovery |
| $3mg\sin\alpha - F - T = 3ma$ | A1 | Correct equation (allow $(-a)$ instead of $a$ in both equations) |

### Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to the plane | M1 | Correct no. of terms, condone sign errors and sin/cos confusion. Allow if appears in (a) |
| $R = 3mg\cos\alpha$ | A1 | Correct equation |
| $F = \frac{1}{6}R$ | B1 | Seen anywhere in (a) or (b), including on a diagram |
| Equation of motion for $B$ **OR** for whole system | M1 | Equation (for $B$) in $T$ and $a$ with correct no. of terms, condone sign errors and sin/cos confusion. **OR** Whole system equation with correct no. of terms, condone sign errors and sin/cos confusion |
| $T - mg = ma$ **OR** $3mg\sin\alpha - F - mg = 3ma + ma$ | A1 | Correct equation |
| Complete method to solve for $a$ | DM1 | Complete method (trig may not be substituted), dependent on M1 in (a) and second M1 in (b) if two equations used, or second M1 in (b) if one equation used |
| $a = \frac{1}{10}g$ | A1* | Correct answer correctly obtained |

### Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Straight line graph starting at origin, with positive gradient | B1 | Straight line starting at origin (could be reflected in the $t$-axis). B0 if continuous vertical line at the end |
| e.g. acceleration (of $B$) is constant | DB1 | Dependent on first B1, for any equivalent statement |

### Part 2(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. the tensions in the two equations of motion would be different. Tension on $A$ would be different to tension on $B$ | B1 | B0 if incorrect extras |

---
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{63363c3e-13fc-49a1-8cef-951e6e97e801-04_396_993_246_536}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A small stone $A$ of mass $3 m$ is attached to one end of a string.\\
A small stone $B$ of mass $m$ is attached to the other end of the string.\\
Initially $A$ is held at rest on a fixed rough plane.\\
The plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$\\
The string passes over a pulley $P$ that is fixed at the top of the plane.\\
The part of the string from $A$ to $P$ is parallel to a line of greatest slope of the plane.\\
Stone $B$ hangs freely below $P$, as shown in Figure 1.\\
The coefficient of friction between $A$ and the plane is $\frac { 1 } { 6 }$\\
Stone $A$ is released from rest and begins to move down the plane.\\
The stones are modelled as particles.\\
The pulley is modelled as being small and smooth.\\
The string is modelled as being light and inextensible.

Using the model for the motion of the system before $B$ reaches the pulley,
\begin{enumerate}[label=(\alph*)]
\item write down an equation of motion for $A$
\item show that the acceleration of $A$ is $\frac { 1 } { 10 } \mathrm {~g}$
\item sketch a velocity-time graph for the motion of $B$, from the instant when $A$ is released from rest to the instant just before $B$ reaches the pulley, explaining your answer.

In reality, the string is not light.
\item State how this would affect the working in part (b).
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 3 2021 Q2 [12]}}
This paper (5 questions)
View full paper
Q1 4 Q2 12 Q3 10 Q4 10 Q5 14