AQA Further Paper 2 2023 June — Question 16 16 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeProve SHM and find period: given force or equation of motion directly
DifficultyHard +2.3 This is a challenging Further Maths mechanics question requiring: (1) deriving a second-order differential equation from Newton's second law, (2) solving it to obtain the given damped harmonic motion solution with particular integral, (3) applying initial conditions to find constants involving exponential and trigonometric functions, and (4) solving a simpler undamped case. The multi-step nature, differential equations content, and algebraic complexity place this well above average difficulty, though it follows a standard damped oscillator framework that Further Maths students should recognize.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x4.10g Damped oscillations: model and interpret6.02e Calculate KE and PE: using formulae6.03a Linear momentum: p = mv
A bungee jumper of mass \(m\) kg is attached to an elastic rope. The other end of the rope is attached to a fixed point. The bungee jumper falls vertically from the fixed point. At time \(t\) seconds after the rope first becomes taut, the extension of the rope is \(x\) metres and the speed of the bungee jumper is \(v\) m s\(^{-1}\)
  1. A model for the motion while the rope remains taut assumes that the forces acting on the bungee jumper are • the weight of the bungee jumper • a tension in the rope of magnitude \(kx\) newtons • an air resistance force of magnitude \(Rv\) newtons where \(k\) and \(R\) are constants such that \(4km > R^2\)
    1. Show that this model gives the result $$x = e^{-\frac{Rt}{2m}} \left( A \cos \frac{\sqrt{4km - R^2}}{2m} t + B \sin \frac{\sqrt{4km - R^2}}{2m} t \right) + \frac{mg}{k}$$ where \(A\) and \(B\) are constants, and \(g\) m s\(^{-2}\) is the acceleration due to gravity. You do not need to find the value of \(A\) or the value of \(B\) [6 marks]
    2. It is also given that: \(k = 16\) \(R = 20\) \(m = 62.5\) \(g = 9.8\) m s\(^{-2}\) and that the speed of the bungee jumper when the rope becomes taut is 14 m s\(^{-1}\) Show that, to the nearest integer, \(A = -38\) and \(B = 16\) [6 marks]
  2. A second, simpler model assumes that the air resistance is zero. The values of \(k\), \(m\) and \(g\) remain the same. Find an expression for \(x\) in terms of \(t\) according to this simpler model, giving the values of all constants to two significant figures. [4 marks]
Question 16:
AnswerMarks
16(a)(i)Forms an equation of
motion with at least
three terms correct.
AnswerMarks Guidance
Accept a and/or v.3.3 M1
mx+Rx+kx = mg
CF:
mλ2 +Rλ+k =0
−R± R2 −4km R  4km−R2 
λ= =− ±i 
2m 2m  2m 
 
CF: − R t   4km−R2   4km−R2  
x=e 2 m Acos  t+Bsin  t 
  2m   2m  
     
PI:
mg
x= p, x =0, x=0⇒kp=mg ⇒ p=
k
So
− Rt   4km−R2   4km−R2   mg
x=e 2mAcos t+Bsin t+
  2m   2m   k
     
Obtains fully correct
AnswerMarks Guidance
differential equation.1.1b A1
Obtains solutions of
their auxiliary
AnswerMarks Guidance
equation.1.1a M1
Obtains a
complementary
function consistent
with their solutions to
their auxiliary
equation.
Must justify the choice
of complementary
function either from
the roots of their
auxiliary equation or
AnswerMarks Guidance
by a clear explanation.1.1b A1
Uses a correct method
to obtain the correct
AnswerMarks Guidance
particular integral.1.1b B1
Completes a fully
correct reasoned
argument to obtain the
required result.
Condone incorrect
AnswerMarks Guidance
brackets.2.1 R1
Subtotal6
QMarking Instructions AO
AnswerMarks
16(a)(ii)Correctly substitutes
all relevant values into
AnswerMarks Guidance
the general equation.3.1b B1
x=e Acos 0.48t +Bsin 0.48t +38.3
When t =0, x=0
0= A+38.3
A= −38.3
x=e −0.16t( −38.3cos ( 0.48t ) +Bsin ( 0.48t )) +38.3
x = −0.16e −0.16t( −38.3cos ( 0.48t ) +Bsin ( 0.48t ))
−0.16t( ( ) ( ))
+e 18.4sin 0.48t +0.48Bcos 0.48t
When t =0, x =14
14= −0.16 (−38.3 ) +0.48B
B=16.4
To the nearest integer, A=−38 and B =16
Substitutes t =0 and
x=0 into the general
equation, either with or
without other values
AnswerMarks Guidance
substituted.3.4 M1
Obtains the correct
value for A
Must have -38.3 or
AnswerMarks Guidance
better.1.1b A1
Uses the product rule
to differentiate the
expression for
AnswerMarks Guidance
displacement.1.1a M1
Substitutes t =0 and
v=14into their
equation for v, either
with or without other
AnswerMarks Guidance
values substituted.3.4 M1
Completes a fully
correct argument to
show that to the
nearest integer,
AnswerMarks Guidance
A= −38 and B=161.1b A1
Subtotal6
QMarking Instructions AO
AnswerMarks
16(b)Deduces that R=0 in the
equation from part (a)(i)
OR
Obtains a correct solution to
the differential equation
formed from an equation of
AnswerMarks Guidance
motion using R=02.2a B1
  4km  4km  mg
x=Acos t+Bsin t+
     
  2m   2m   k
( ( ) ( ))
x= Acos 0.506t +Bsin 0.506t +38.3
When t =0, x=0⇒ A=−38.3
x= (−38.3cos ( 0.506t)+Bsin ( 0.506t)) +38.3
x=19.4sin ( 0.506t)+0.506Bcos ( 0.506t)
When t =0, x =14
14
B= =27.7
0.506
x= ( −38cos ( 0.51t ) +28sin ( 0.51t )) +38
AnswerMarks Guidance
Differentiates displacement1.1a M1
Substitutes t =0, x=0 and
v=14into their equations
AnswerMarks Guidance
for x and v3.4 M1
Obtains a completely
correct expression for x,
with values to 2 significant
AnswerMarks Guidance
figures or better.1.1b A1
Subtotal4
Question total16
Paper total100 
Question 16:
--- 16(a)(i) ---
16(a)(i) | Forms an equation of
motion with at least
three terms correct.
Accept a and/or v. | 3.3 | M1 | mx= mg−kx−Rx
mx+Rx+kx = mg
CF:
mλ2 +Rλ+k =0
−R± R2 −4km R  4km−R2 
λ= =− ±i 
2m 2m  2m 
 
CF: − R t   4km−R2   4km−R2  
x=e 2 m Acos  t+Bsin  t 
  2m   2m  
     
PI:
mg
x= p, x =0, x=0⇒kp=mg ⇒ p=
k
So
− Rt   4km−R2   4km−R2   mg
x=e 2mAcos t+Bsin t+
  2m   2m   k
     
Obtains fully correct
differential equation. | 1.1b | A1
Obtains solutions of
their auxiliary
equation. | 1.1a | M1
Obtains a
complementary
function consistent
with their solutions to
their auxiliary
equation.
Must justify the choice
of complementary
function either from
the roots of their
auxiliary equation or
by a clear explanation. | 1.1b | A1
Uses a correct method
to obtain the correct
particular integral. | 1.1b | B1
Completes a fully
correct reasoned
argument to obtain the
required result.
Condone incorrect
brackets. | 2.1 | R1
Subtotal | 6
Q | Marking Instructions | AO | Marks | Typical Solution
--- 16(a)(ii) ---
16(a)(ii) | Correctly substitutes
all relevant values into
the general equation. | 3.1b | B1 | −0.16t( ( ) ( ))
x=e Acos 0.48t +Bsin 0.48t +38.3
When t =0, x=0
0= A+38.3
A= −38.3
x=e −0.16t( −38.3cos ( 0.48t ) +Bsin ( 0.48t )) +38.3
x = −0.16e −0.16t( −38.3cos ( 0.48t ) +Bsin ( 0.48t ))
−0.16t( ( ) ( ))
+e 18.4sin 0.48t +0.48Bcos 0.48t
When t =0, x =14
14= −0.16 (−38.3 ) +0.48B
B=16.4
To the nearest integer, A=−38 and B =16
Substitutes t =0 and
x=0 into the general
equation, either with or
without other values
substituted. | 3.4 | M1
Obtains the correct
value for A
Must have -38.3 or
better. | 1.1b | A1
Uses the product rule
to differentiate the
expression for
displacement. | 1.1a | M1
Substitutes t =0 and
v=14into their
equation for v, either
with or without other
values substituted. | 3.4 | M1
Completes a fully
correct argument to
show that to the
nearest integer,
A= −38 and B=16 | 1.1b | A1
Subtotal | 6
Q | Marking Instructions | AO | Marks | Typical solution
--- 16(b) ---
16(b) | Deduces that R=0 in the
equation from part (a)(i)
OR
Obtains a correct solution to
the differential equation
formed from an equation of
motion using R=0 | 2.2a | B1 | Setting R=0
  4km  4km  mg
x=Acos t+Bsin t+
     
  2m   2m   k
( ( ) ( ))
x= Acos 0.506t +Bsin 0.506t +38.3
When t =0, x=0⇒ A=−38.3
x= (−38.3cos ( 0.506t)+Bsin ( 0.506t)) +38.3
x=19.4sin ( 0.506t)+0.506Bcos ( 0.506t)
When t =0, x =14
14
B= =27.7
0.506
x= ( −38cos ( 0.51t ) +28sin ( 0.51t )) +38
Differentiates displacement | 1.1a | M1
Substitutes t =0, x=0 and
v=14into their equations
for x and v | 3.4 | M1
Obtains a completely
correct expression for x,
with values to 2 significant
figures or better. | 1.1b | A1
Subtotal | 4
Question total | 16
Paper total | 100
A bungee jumper of mass $m$ kg is attached to an elastic rope.
The other end of the rope is attached to a fixed point.

The bungee jumper falls vertically from the fixed point.

At time $t$ seconds after the rope first becomes taut, the extension of the rope is $x$ metres and the speed of the bungee jumper is $v$ m s$^{-1}$

\begin{enumerate}[label=(\alph*)]
\item A model for the motion while the rope remains taut assumes that the forces acting on the bungee jumper are

• the weight of the bungee jumper
• a tension in the rope of magnitude $kx$ newtons
• an air resistance force of magnitude $Rv$ newtons

where $k$ and $R$ are constants such that $4km > R^2$

\begin{enumerate}[label=(\roman*)]
\item Show that this model gives the result
$$x = e^{-\frac{Rt}{2m}} \left( A \cos \frac{\sqrt{4km - R^2}}{2m} t + B \sin \frac{\sqrt{4km - R^2}}{2m} t \right) + \frac{mg}{k}$$

where $A$ and $B$ are constants, and $g$ m s$^{-2}$ is the acceleration due to gravity.

You do not need to find the value of $A$ or the value of $B$
[6 marks]

\item It is also given that:
$k = 16$
$R = 20$
$m = 62.5$
$g = 9.8$ m s$^{-2}$

and that the speed of the bungee jumper when the rope becomes taut is 14 m s$^{-1}$

Show that, to the nearest integer, $A = -38$ and $B = 16$
[6 marks]
\end{enumerate}

\item A second, simpler model assumes that the air resistance is zero.

The values of $k$, $m$ and $g$ remain the same.

Find an expression for $x$ in terms of $t$ according to this simpler model, giving the values of all constants to two significant figures.
[4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2023 Q16 [16]}}
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