Question 17 - A-Level Maths

AQA AS Paper 1 2022 June — Question 17 8 marks

Exam Board	AQA
Module	AS Paper 1 (AS Paper 1)
Year	2022
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Newton's laws and connected particles
Type	Car towing trailer, horizontal
Difficulty	Moderate -0.8 This is a straightforward AS-level mechanics question requiring basic differentiation to find acceleration, then standard application of Newton's second law to connected particles. All steps are routine: differentiate v to get a, apply F=ma to the system and individual components, with clearly signposted values. No problem-solving insight needed, just methodical application of standard techniques.
Spec	3.02f Non-uniform acceleration: using differentiation and integration 3.03d Newton's second law: 2D vectors 3.03k Connected particles: pulleys and equilibrium

\includegraphics{figure_17} A car and caravan, connected by a tow bar, move forward together along a horizontal road. Their velocity $v$ m s$^{-1}$ at time $t$ seconds, for $0 \leq t < 20$, is given by $$v = 0.5t + 0.01t^2$$

Show that when $t = 15$ their acceleration is 0.8 m s$^{-2}$ [2 marks]
The car has a mass of 1500 kg The caravan has a mass of 850 kg When $t = 15$ the tension in the tow bar is 800 N and the car experiences a resistance force of 100 N
1. Find the total resistance force experienced by the caravan when $t = 15$ [2 marks]
2. Find the driving force being applied by the car when $t = 15$ [3 marks]
State one assumption you have made about the tow bar. [1 mark]

Show mark scheme Show mark scheme source

Question 17:

Answer	Marks
17(a)	Differentiates to find expression

for acceleration with at least one

Answer	Marks	Guidance
term correct.	3.4	M1

𝑎𝑎 = 0.5+0.02×15 = 0.8

Substitutes t = 15 to obtain

Or

𝑎𝑎Sh=ow0 .s8 that when a = 0.8, t = 15

Answer	Marks	Guidance
AG	1.1b	A1
Subtotal	2
Q	Marking instructions	AO

Answer	Marks
17(b)(i)	Uses to form an

equation with three-terms

Answer	Marks	Guidance
model𝐹𝐹lin=g 𝑚𝑚th𝑎𝑎e caravan.	3.3	M1

N

𝑅𝑅 = 120

Obtains 120 N

Answer	Marks	Guidance
Condone missing units.	1.1b	A1
Subtotal	2
Q	Marking instructions	AO

Answer	Marks
17(b)(ii)	Uses to form equation

modelling the car or the whole

system𝐹𝐹 w=it𝑚𝑚h 𝑎𝑎one side correct.

Answer	Marks	Guidance
Whole system equation is	3.3	M1

N

𝐷𝐷 = 2 100

𝐷𝐷−220 = 2350(0.8)

Answer	Marks	Guidance
Forms fully correct equation.	1.1b	A1

Obtains 2100 N

Answer	Marks	Guidance
Condone missing units.	1.1b	A1
Subtotal	3
Q	Marking instructions	AO

Answer	Marks
17(c)	States a valid assumption about

the tow bar.

Accept

• Inelastic/inextensible OE

• Light OE

• Rigid

• Tension is constant

throughout the tow bar

• Tow bar is in the

direction of the car and

caravan

• Tow bar is horizontal

Do not accept ‘the tow bar

Answer	Marks	Guidance
breaks’ or ‘uniform’.	3.5b	B1
Subtotal	1
Question 17 Total	8
Question Paper Total	80

Question 17:
--- 17(a) ---
17(a) | Differentiates to find expression
for acceleration with at least one
term correct. | 3.4 | M1 | 𝑎𝑎 = 0.5+0.02𝑡𝑡
𝑎𝑎 = 0.5+0.02×15 = 0.8
Substitutes t = 15 to obtain
Or
𝑎𝑎Sh=ow0 .s8 that when a = 0.8, t = 15
AG | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b)(i) ---
17(b)(i) | Uses to form an
equation with three-terms
model𝐹𝐹lin=g 𝑚𝑚th𝑎𝑎e caravan. | 3.3 | M1 | 800−𝑅𝑅 = 850×0.8
N
𝑅𝑅 = 120
Obtains 120 N
Condone missing units. | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b)(ii) ---
17(b)(ii) | Uses to form equation
modelling the car or the whole
system𝐹𝐹 w=it𝑚𝑚h 𝑎𝑎one side correct.
Whole system equation is | 3.3 | M1 | 𝐷𝐷−800−100= 1500×0.8
N
𝐷𝐷 = 2 100
𝐷𝐷−220 = 2350(0.8)
Forms fully correct equation. | 1.1b | A1
Obtains 2100 N
Condone missing units. | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 17(c) ---
17(c) | States a valid assumption about
the tow bar.
Accept
• Inelastic/inextensible OE
• Light OE
• Rigid
• Tension is constant
throughout the tow bar
• Tow bar is in the
direction of the car and
caravan
• Tow bar is horizontal
Do not accept ‘the tow bar
breaks’ or ‘uniform’. | 3.5b | B1 | Tow bar is light
Subtotal | 1
Question 17 Total | 8
Question Paper Total | 80

Show LaTeX source

\includegraphics{figure_17}

A car and caravan, connected by a tow bar, move forward together along a horizontal road.

Their velocity $v$ m s$^{-1}$ at time $t$ seconds, for $0 \leq t < 20$, is given by
$$v = 0.5t + 0.01t^2$$

\begin{enumerate}[label=(\alph*)]
\item Show that when $t = 15$ their acceleration is 0.8 m s$^{-2}$
[2 marks]

\item The car has a mass of 1500 kg

The caravan has a mass of 850 kg

When $t = 15$ the tension in the tow bar is 800 N and the car experiences a resistance force of 100 N

\begin{enumerate}[label=(\roman*)]
\item Find the total resistance force experienced by the caravan when $t = 15$
[2 marks]

\item Find the driving force being applied by the car when $t = 15$
[3 marks]
\end{enumerate}

\item State one assumption you have made about the tow bar.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2022 Q17 [8]}}

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