| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Heavier particle hits ground, lighter continues upward - vertical strings |
| Difficulty | Moderate -0.3 This is a standard two-stage pulley problem requiring Newton's second law for connected particles, then kinematics after one particle hits the ground. The setup is straightforward with clear modeling assumptions, and the multi-part structure guides students through the solution. While it requires careful bookkeeping across two phases of motion, it involves routine application of mechanics techniques without novel insight or complex problem-solving, making it slightly easier than average. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution |
| VIAV SIHI NI III IM ION OC | VIIN SIHI NI III M M O N OO | VIIV SIHI NI IIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) Equation of motion for \(P\) | M1 | Translate situation into model and set up equation of motion for \(P\) (must contain \(T\) and \(a\)) |
| \(T - 2mg = 2ma\) | A1 | Correct equation |
| (ii) Equation of motion for \(Q\) | M1 | Translate situation into model and set up equation of motion for \(Q\) (must contain \(T\) and \(a\)) |
| \(5mg - T = 5ma\) | A1 | Correct equation. N.B. allow \((-a)\) in both equations. Allow above 4 marks if equations appear in (b). If \(m\)'s omitted consistently, max (a) M1A0M1A0 (b) M1A0M1A1M1A1M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solve equations for \(a\), or use whole system equation and solve for \(a\) | M1 | Solve for \(a\) |
| \(a = \dfrac{3g}{7} = 4.2\) | A1 | Allow \(4.2\ \text{ms}^{-2}\) or must be in terms of \(g\) only. N.B. allow above 2 marks if they appear in (a) |
| \(v = \sqrt{2 \times \dfrac{3g}{7} \times h} = \sqrt{8.4h}\) or \(v^2 = 2 \times \dfrac{3g}{7} \times h\ (= 8.4h)\) | M1 | Complete method to produce an expression for \(v\) or \(v^2\) in terms of \(h\), using their \(a\) |
| \(0 = \dfrac{6gh}{7} - 2gH\) | M1 | Complete method to produce an expression for \(H\) in terms of \(h\), using \(a = -g\) and \(v = 0\) |
| \(H = \dfrac{3h}{7}\) | A1 | Correct expression for \(H\) |
| Total height \(= 2h + h + H\) | M1 | Complete method to find total distance |
| Total height \(= \dfrac{24h}{7}\) | A1 | cao but allow \(3.4h\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. The distance that \(Q\) falls to the ground would not be exactly \(h\) | B1 | B0 if any incorrect extras are given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. The accelerations of the balls would not have equal magnitude (allow 'wouldn't be the same' oe) | B1 | B0 if any incorrect extras are given, or for an incorrect statement e.g. tension is not constant so accelerations will be different. B0 if they say 'inextensible \(\Rightarrow\) acceleration same' |
## Question 3:
### Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) Equation of motion for $P$ | M1 | Translate situation into model and set up equation of motion for $P$ (must contain $T$ and $a$) |
| $T - 2mg = 2ma$ | A1 | Correct equation |
| (ii) Equation of motion for $Q$ | M1 | Translate situation into model and set up equation of motion for $Q$ (must contain $T$ and $a$) |
| $5mg - T = 5ma$ | A1 | Correct equation. N.B. allow $(-a)$ in both equations. Allow above 4 marks if equations appear in (b). If $m$'s omitted consistently, max (a) M1A0M1A0 (b) M1A0M1A1M1A1M1A0 |
**(4 marks)**
---
### Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve equations for $a$, or use whole system equation and solve for $a$ | M1 | Solve for $a$ |
| $a = \dfrac{3g}{7} = 4.2$ | A1 | Allow $4.2\ \text{ms}^{-2}$ or must be in terms of $g$ only. N.B. allow above 2 marks if they appear in (a) |
| $v = \sqrt{2 \times \dfrac{3g}{7} \times h} = \sqrt{8.4h}$ or $v^2 = 2 \times \dfrac{3g}{7} \times h\ (= 8.4h)$ | M1 | Complete method to produce an expression for $v$ or $v^2$ in terms of $h$, using their $a$ |
| $0 = \dfrac{6gh}{7} - 2gH$ | M1 | Complete method to produce an expression for $H$ in terms of $h$, using $a = -g$ and $v = 0$ |
| $H = \dfrac{3h}{7}$ | A1 | Correct expression for $H$ |
| Total height $= 2h + h + H$ | M1 | Complete method to find total distance |
| Total height $= \dfrac{24h}{7}$ | A1 | cao but allow $3.4h$ or better |
**(7 marks)**
---
### Part 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. The distance that $Q$ falls to the ground would not be exactly $h$ | B1 | B0 if any incorrect extras are given |
**(1 mark)**
---
### Part 3(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. The accelerations of the balls would not have equal magnitude (allow 'wouldn't be the same' oe) | B1 | B0 if any incorrect extras are given, or for an incorrect statement e.g. tension is not constant so accelerations will be different. B0 if they say 'inextensible $\Rightarrow$ acceleration same' |
**(1 mark)**
---
**(13 marks total)**3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4a022ec0-7640-4664-87a6-1963309cad6a-08_761_595_210_735}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A ball $P$ of mass $2 m$ is attached to one end of a string.\\
The other end of the string is attached to a ball $Q$ of mass $5 m$.\\
The string passes over a fixed pulley.\\
The system is held at rest with the balls hanging freely and the string taut.\\
The hanging parts of the string are vertical with $P$ at a height $2 h$ above horizontal ground and with $Q$ at a height $h$ above the ground, as shown in Figure 1.
The system is released from rest.\\
In the subsequent motion, $Q$ does not rebound when it hits the ground and $P$ does not hit the pulley.
The balls are modelled as particles.\\
The string is modelled as being light and inextensible.\\
The pulley is modelled as being small and smooth.\\
Air resistance is modelled as being negligible.\\
Using this model,
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item write down an equation of motion for $P$,
\item write down an equation of motion for $Q$,
\end{enumerate}\item find, in terms of $h$ only, the height above the ground at which $P$ first comes to instantaneous rest.
\item State one limitation of modelling the balls as particles that could affect your answer to part (b).
In reality, the string will not be inextensible.
\item State how this would affect the accelerations of the particles.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIAV SIHI NI III IM ION OC & VIIN SIHI NI III M M O N OO & VIIV SIHI NI IIIYM ION OC \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2021 Q3 [13]}}