AQA Further AS Paper 2 Mechanics 2024 June — Question 4 8 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done against air resistance - vertical motion
DifficultyStandard +0.3 This is a straightforward mechanics question testing standard energy methods. Part (a) is direct KE formula application, part (b) is routine conservation of energy, parts (c)(i-ii) involve simple work-energy calculations with resistance, and part (c)(iii) requires only basic commentary on model validity. All techniques are standard A-level mechanics with no novel problem-solving required, making it slightly easier than average.
Spec6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle
In this question use \(g = 9.8 \text{ m s}^{-2}\) A ball of mass 0.5 kg is projected vertically upwards with a speed of \(10 \text{ m s}^{-1}\)
  1. Calculate the initial kinetic energy of the ball. [1 mark]
  2. Assuming that the weight is the only force acting on the ball, use an energy method to show that the maximum height reached by the ball is approximately 5.1 m above the point of projection. [2 marks]
    1. A student conducts an experiment to verify the accuracy of the result obtained in part (b). They observe that the ball rises to a height of 4.4 m above the point of projection and concludes that this height difference is due to a resistance force, \(R\) newtons. Find the total work done against \(R\) whilst the ball is moving upwards. [2 marks]
    2. Using a model that assumes \(R\) is constant, find the magnitude of \(R\) [2 marks]
    3. Comment on the validity of the model used in part (c)(ii). [1 mark]
Question 4:
AnswerMarks
4(a)Obtains 25.
Condone missing or incorrect
AnswerMarks Guidance
units.1.1b B1
KE = mv2
2
1
= (0.5)(10)2
2
= 25 J
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks
4(b)Uses conservation of energy to
form an equation in h with PE
AnswerMarks Guidance
and their KE from part (a).3.3 M1
25
h = =5.102...
(0.5)(9.8)
Therefore, height reached is
approx. 5.1 m above point of
projection.
Solves the energy equation to
obtain
h = 5.1 m
Obtains correct value (5.10) to
25
at least 3 sf or oe
4.9
Condone missing units.
AnswerMarks Guidance
AG1.1b A1
Subtotal2
QMarking instructions AO
AnswerMarks
4(c)(i)Translates problem into finding
difference between two energy
AnswerMarks Guidance
terms.3.3 M1
= 25 – 21.56
= 3.44 J
Obtains AWRT 3.4 J
Condone missing units.
AnswerMarks Guidance
Do not ISW1.1b A1
Subtotal2
QMarking instructions AO
AnswerMarks
4(c)(ii)Forms an equation using
work done from part (c)(i) =
AnswerMarks Guidance
R times 4.43.4 M1
43
R = = 3.44
55
R = 0.78
Obtains AWRT 0.78
or AWRT 0.77
Follow through their answer
AnswerMarks Guidance
from (c)(i).1.1b A1F
Subtotal2
QMarking instructions AO
AnswerMarks
4(c)(iii)Explains that resistance is
unlikely to be constant (model
not valid) as it would vary
AnswerMarks Guidance
according to speed.3.5b E1
realistic as resistance would
decrease as the speed decreases.
AnswerMarks Guidance
Subtotal1
Question total8
QMarking instructions AO 
Question 4:
--- 4(a) ---
4(a) | Obtains 25.
Condone missing or incorrect
units. | 1.1b | B1 | 1
KE = mv2
2
1
= (0.5)(10)2
2
= 25 J
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 4(b) ---
4(b) | Uses conservation of energy to
form an equation in h with PE
and their KE from part (a). | 3.3 | M1 | mgh = 25
25
h = =5.102...
(0.5)(9.8)
Therefore, height reached is
approx. 5.1 m above point of
projection.
Solves the energy equation to
obtain
h = 5.1 m
Obtains correct value (5.10) to
25
at least 3 sf or oe
4.9
Condone missing units.
AG | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 4(c)(i) ---
4(c)(i) | Translates problem into finding
difference between two energy
terms. | 3.3 | M1 | Work done = 25 – (0.5)(9.8)(4.4)
= 25 – 21.56
= 3.44 J
Obtains AWRT 3.4 J
Condone missing units.
Do not ISW | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 4(c)(ii) ---
4(c)(ii) | Forms an equation using
work done from part (c)(i) =
R times 4.4 | 3.4 | M1 | R×4.4 = 3.44
43
R = = 3.44
55
R = 0.78
Obtains AWRT 0.78
or AWRT 0.77
Follow through their answer
from (c)(i). | 1.1b | A1F
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 4(c)(iii) ---
4(c)(iii) | Explains that resistance is
unlikely to be constant (model
not valid) as it would vary
according to speed. | 3.5b | E1 | A constant resistance force is not
realistic as resistance would
decrease as the speed decreases.
Subtotal | 1
Question total | 8
Q | Marking instructions | AO | Marks | Typical solution
In this question use $g = 9.8 \text{ m s}^{-2}$

A ball of mass 0.5 kg is projected vertically upwards with a speed of $10 \text{ m s}^{-1}$

\begin{enumerate}[label=(\alph*)]
\item Calculate the initial kinetic energy of the ball. [1 mark]

\item Assuming that the weight is the only force acting on the ball, use an energy method to show that the maximum height reached by the ball is approximately 5.1 m above the point of projection. [2 marks]

\item \begin{enumerate}[label=(\roman*)]
\item A student conducts an experiment to verify the accuracy of the result obtained in part (b).

They observe that the ball rises to a height of 4.4 m above the point of projection and concludes that this height difference is due to a resistance force, $R$ newtons.

Find the total work done against $R$ whilst the ball is moving upwards. [2 marks]

\item Using a model that assumes $R$ is constant, find the magnitude of $R$ [2 marks]

\item Comment on the validity of the model used in part (c)(ii). [1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q4 [8]}}
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