Question 4 - A-Level Maths

Edexcel AS Paper 2 2023 June — Question 4 7 marks

Exam Board	Edexcel
Module	AS Paper 2 (AS Paper 2)
Year	2023
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Newton's laws and connected particles
Type	Car towing trailer, horizontal
Difficulty	Moderate -0.3 This is a standard connected particles problem requiring straightforward application of Newton's second law to two bodies separately. Part (a) involves setting up two equations (F=ma for car and trailer) and solving simultaneously for R and a—routine mechanics with no conceptual challenges. Parts (b) and (c) test basic understanding of modeling assumptions through simple qualitative reasoning. Slightly easier than average due to being a textbook-style question with clear setup and standard method.
Spec	3.03d Newton's second law: 2D vectors 3.03k Connected particles: pulleys and equilibrium 3.03l Newton's third law: extend to situations requiring force resolution

4. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d9615c4f-d8fa-4e44-978a-cf34b2b1c0b5-10_211_1527_294_269} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} A car of mass 1200 kg is towing a trailer of mass 400 kg along a straight horizontal road using a tow rope, as shown in Figure 2.
The rope is horizontal and parallel to the direction of motion of the car.

The resistance to motion of the car is modelled as a constant force of magnitude \(2 R\) newtons
The resistance to motion of the trailer is modelled as a constant force of magnitude \(R\) newtons
The rope is modelled as being light and inextensible
The acceleration of the car is modelled as \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\)

The driving force of the engine of the car is 7400 N and the tension in the tow rope is 2400 N . Using the model,

find the value of \(a\) In a refined model, the rope is modelled as having mass and the acceleration of the car is found to be \(a _ { 1 } \mathrm {~ms} ^ { - 2 }\)
State how the value of \(a _ { 1 }\) compares with the value of \(a\)
State one limitation of the model used for the resistance to motion of the car.

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Question 4:

Part 4(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Equation of motion for the car	M1	AO 3.3
\(7400 - 2R - 2400 = 1200a\)	A1	AO 1.1b
Equation of motion for the trailer	M1	AO 3.4
\(2400 - R = 400a\)	A1	AO 1.1b
\(a = 0.5\)	A1	AO 1.1b

Notes: Either equation may be replaced by equation of motion for whole system: \(7400 - 3R = 1600a\). M1 requires correct number of terms, condone sign errors, driving force as 7400, tension as 2400.

Part 4(b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
The value of \(a_1\) would be less than the value of \(a\). Allow '\(a_1\) would be slower than \(a\)'. Allow 'it would be less than \(a\)'	B1	AO 3.5a

Part 4(c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
The resistance won't be constant / 'it won't be constant'. Allow negative: the resistance is constant / 'it is constant'	B1	AO 3.5b

Notes: B0 for 'it doesn't take account of air resistance'. B0 if any incorrect extras given or for an incorrect statement.

## Question 4:

### Part 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for the car | M1 | AO 3.3 |
| $7400 - 2R - 2400 = 1200a$ | A1 | AO 1.1b |
| Equation of motion for the trailer | M1 | AO 3.4 |
| $2400 - R = 400a$ | A1 | AO 1.1b |
| $a = 0.5$ | A1 | AO 1.1b |

Notes: Either equation may be replaced by equation of motion for whole system: $7400 - 3R = 1600a$. M1 requires correct number of terms, condone sign errors, driving force as 7400, tension as 2400.

### Part 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The value of $a_1$ would be less than the value of $a$. Allow '$a_1$ would be slower than $a$'. Allow 'it would be less than $a$' | B1 | AO 3.5a |

### Part 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The resistance won't be constant / 'it won't be constant'. Allow negative: the resistance is constant / 'it is constant' | B1 | AO 3.5b |

Notes: B0 for 'it doesn't take account of air resistance'. B0 if any incorrect extras given or for an incorrect statement.

Show LaTeX source

4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{d9615c4f-d8fa-4e44-978a-cf34b2b1c0b5-10_211_1527_294_269}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A car of mass 1200 kg is towing a trailer of mass 400 kg along a straight horizontal road using a tow rope, as shown in Figure 2.\\
The rope is horizontal and parallel to the direction of motion of the car.

\begin{itemize}
  \item The resistance to motion of the car is modelled as a constant force of magnitude $2 R$ newtons
  \item The resistance to motion of the trailer is modelled as a constant force of magnitude $R$ newtons
  \item The rope is modelled as being light and inextensible
  \item The acceleration of the car is modelled as $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$
\end{itemize}

The driving force of the engine of the car is 7400 N and the tension in the tow rope is 2400 N .

Using the model,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$

In a refined model, the rope is modelled as having mass and the acceleration of the car is found to be $a _ { 1 } \mathrm {~ms} ^ { - 2 }$
\item State how the value of $a _ { 1 }$ compares with the value of $a$
\item State one limitation of the model used for the resistance to motion of the car.
\end{enumerate}

\hfill \mbox{\textit{Edexcel AS Paper 2 2023 Q4 [7]}}

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