OCR Further Pure Core 2 2021 November — Question 8 16 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2021
SessionNovember
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeApplied/modelling contexts
DifficultyChallenging +1.2 This is a structured multi-part mechanics question using integrating factor method. While it requires setting up Newton's second law, finding the proportionality constant, solving a standard first-order linear ODE, and interpreting the solution physically, each step is clearly signposted. The integrating factor technique is routine for Further Maths students, and the conceptual demands (understanding maximum speed occurs when dv/dt=0, recognizing model validity ends at boundary) are moderate. More challenging than a basic C3 question due to the applied context and multiple parts, but less demanding than questions requiring novel insight or complex proof.
Spec4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations
8 A particle \(P\) of mass 2 kg can only move along the straight line segment \(O A\), where \(O A\) is on a rough horizontal surface. The particle is initially at rest at \(O\) and the distance \(O A\) is 0.9 m . When the time is \(t\) seconds the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). \(P\) is subject to a force of magnitude \(4 \mathrm { e } ^ { - 2 t } \mathrm {~N}\) in the direction of \(A\) for any \(t \geqslant 0\). The resistance to the motion of \(P\) is modelled as being proportional to \(v\). At the instant when \(t = \ln 2 , v = 0.5\) and the resultant force on \(P\) is 0 N .
  1. Show that, according to the model, \(\frac { d v } { d t } + v = 2 e ^ { - 2 t }\).
  2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
  3. By considering the behaviour of \(v\) as \(t\) becomes large explain why, according to the model, \(P\) 's speed must reach a maximum value for some \(t > 0\).
  4. Determine the maximum speed considered in part (c).
  5. Determine the greatest value of \(t\) for which the model is valid.
Question 8:
AnswerMarks Guidance
8(a) dv
F =ma=2 =4e−2t −kv
dt
t = ln2, v = 0.5, F = 0 => 0 = 1 – 0.5k
dv dv
k = 2 =>2 =4e−2t −2v=> +v=2e−2t
AnswerMarks
dt dtM1
M1
A1
AnswerMarks
[3]3.3
2.2a
AnswerMarks
1.1Use of NII with m and a
replaced and with 2 forces, the
given force and kv
Use of given conditions to
derive an equation in k
AnswerMarks
AGF=ma can be implicit here
Can be done first
Complete argument including
F=ma
AnswerMarks
(b)∫1dt
IF = e =et
et dv +etv= d ( etv ) =et ×2e−2t
dt dt
etv=∫2e−tdt =−2e−t +c
t = 0, v = 0 => c = 2
AnswerMarks
v=2e−t −2e−2t*B1
*M1
A1
dep*M1
A1
AnswerMarks
[5]1.1
1.1
1.1
3.4
AnswerMarks
3.4Multiplying by IF and writing
LHS as an exact derivative
“+ c” required
Use of initial conditions to
AnswerMarks
derive a value for cOr CF
Or subst correct PI into DE
Or GS
−𝑡𝑡 −2𝑡𝑡
Or using𝑣𝑣 a=lte𝐴𝐴rn𝑒𝑒ativ−e b2o𝑒𝑒undary
condition
AnswerMarks
(c)As t→∞, v→ 0
So speed starts at 0 and ends at 0 (and is
continuous and positive between) so m ust
AnswerMarks
reach a maximum somewhere in t > 0M1
A1
AnswerMarks
[2]3.4
2.4
AnswerMarks
(d)dv
v is max when =0 so t = ln2
dt
So v = 0.5 (given) (or
max
2 1
v =2e−ln2 −2e−2ln2 =1− = )
AnswerMarks
max 4 2M1
A1
AnswerMarks
[2]2.2a
3.4Deducing time when v is
maximumdv
Or by finding expression for
dt
dv
and solving =0
dt
AnswerMarks Guidance
8(e) dx
v= =2e−t −2e−2t ⇒x=−2e−t +e−2t +d
dt
t =0, x=0⇒0=−2+1+d ⇒d =1
0.9=−2e−t +e−2t +1
( e−t)2 −2e−t +0.1=0⇒e−t = 10±3 10
10
 10 
⇒t =ln  =2.97 (3 sf)
AnswerMarks
10−3 10M1
M1
M1
A1
AnswerMarks
[4]3.3
3.3
3.5a
AnswerMarks
2.3Integrating to find expression
for x
Using initial conditions to find
value of (new) constant
Recognising that the model is
only valid when x lies between 0
and 0.9
 10 
Rejecting t =ln  <0
10+3 10
AnswerMarks
(can be implicit)Or definite integral with correct
lower limit..
…and upper limit
Question 8:
8 | (a) | dv
F =ma=2 =4e−2t −kv
dt
t = ln2, v = 0.5, F = 0 => 0 = 1 – 0.5k
dv dv
k = 2 =>2 =4e−2t −2v=> +v=2e−2t
dt dt | M1
M1
A1
[3] | 3.3
2.2a
1.1 | Use of NII with m and a
replaced and with 2 forces, the
given force and kv
Use of given conditions to
derive an equation in k
AG | F=ma can be implicit here
Can be done first
Complete argument including
F=ma
(b) | ∫1dt
IF = e =et
et dv +etv= d ( etv ) =et ×2e−2t
dt dt
etv=∫2e−tdt =−2e−t +c
t = 0, v = 0 => c = 2
v=2e−t −2e−2t | *B1
*M1
A1
dep*M1
A1
[5] | 1.1
1.1
1.1
3.4
3.4 | Multiplying by IF and writing
LHS as an exact derivative
“+ c” required
Use of initial conditions to
derive a value for c | Or CF
Or subst correct PI into DE
Or GS
−𝑡𝑡 −2𝑡𝑡
Or using𝑣𝑣 a=lte𝐴𝐴rn𝑒𝑒ativ−e b2o𝑒𝑒undary
condition
(c) | As t→∞, v→ 0
So speed starts at 0 and ends at 0 (and is
continuous and positive between) so m ust
reach a maximum somewhere in t > 0 | M1
A1
[2] | 3.4
2.4
(d) | dv
v is max when =0 so t = ln2
dt
So v = 0.5 (given) (or
max
2 1
v =2e−ln2 −2e−2ln2 =1− = )
max 4 2 | M1
A1
[2] | 2.2a
3.4 | Deducing time when v is
maximum | dv
Or by finding expression for
dt
dv
and solving =0
dt
8 | (e) | dx
v= =2e−t −2e−2t ⇒x=−2e−t +e−2t +d
dt
t =0, x=0⇒0=−2+1+d ⇒d =1
0.9=−2e−t +e−2t +1
( e−t)2 −2e−t +0.1=0⇒e−t = 10±3 10
10
 10 
⇒t =ln  =2.97 (3 sf)
10−3 10 | M1
M1
M1
A1
[4] | 3.3
3.3
3.5a
2.3 | Integrating to find expression
for x
Using initial conditions to find
value of (new) constant
Recognising that the model is
only valid when x lies between 0
and 0.9
 10 
Rejecting t =ln  <0
10+3 10
(can be implicit) | Or definite integral with correct
lower limit..
…and upper limit
8 A particle $P$ of mass 2 kg can only move along the straight line segment $O A$, where $O A$ is on a rough horizontal surface. The particle is initially at rest at $O$ and the distance $O A$ is 0.9 m .

When the time is $t$ seconds the displacement of $P$ from $O$ is $x \mathrm {~m}$ and the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$. $P$ is subject to a force of magnitude $4 \mathrm { e } ^ { - 2 t } \mathrm {~N}$ in the direction of $A$ for any $t \geqslant 0$. The resistance to the motion of $P$ is modelled as being proportional to $v$.

At the instant when $t = \ln 2 , v = 0.5$ and the resultant force on $P$ is 0 N .
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, $\frac { d v } { d t } + v = 2 e ^ { - 2 t }$.
\item Find an expression for $v$ in terms of $t$ for $t \geqslant 0$.
\item By considering the behaviour of $v$ as $t$ becomes large explain why, according to the model, $P$ 's speed must reach a maximum value for some $t > 0$.
\item Determine the maximum speed considered in part (c).
\item Determine the greatest value of $t$ for which the model is valid.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q8 [16]}}
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