Question 10 - A-Level Maths

OCR Further Pure Core 1 2024 June — Question 10 10 marks

Exam Board	OCR
Module	Further Pure Core 1 (Further Pure Core 1)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Applied/modelling contexts
Difficulty	Challenging +1.2 This is a standard integrating factor question with a mechanics context. Part (a) requires applying F=ma to set up the differential equation (routine). Part (b) involves finding the integrating factor sec²(2t), integrating sin(4t)sec²(2t) using a standard identity, and solving v=0. While it requires multiple steps and careful trigonometric manipulation, it follows the textbook template for integrating factor problems without requiring novel insight or particularly challenging integration.
Spec	1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 4.10c Integrating factor: first order equations

10 A particle \(B\), of mass 3 kg , moves in a straight line and has velocity \(v \mathrm {~ms} ^ { - 1 }\).
At time \(t\) seconds, where \(0 \leqslant t < \frac { 1 } { 4 } \pi\), a variable force of \(- ( 15 \sin 4 \mathrm { t } + 6 \mathrm { v } \tan 2 \mathrm { t } )\) Newtons is applied to \(B\). There are no other forces acting on \(B\). Initially, when \(t = 0 , B\) has velocity \(4.5 \mathrm {~ms} ^ { - 1 }\). The motion of \(B\) can be modelled by the differential equation \(\frac { d v } { d t } + P ( t ) v = Q ( t )\) where \(P ( t )\) and \(\mathrm { Q } ( \mathrm { t } )\) are functions of \(t\).

Find the functions \(\mathrm { P } ( \mathrm { t } )\) and \(\mathrm { Q } ( \mathrm { t } )\).
Using an integrating factor, determine the first time at which \(B\) is stationary according to the model.

Show mark scheme Show mark scheme source

Question 10:

Answer	Marks	Guidance
10	(a)	F =ma so

dv

So −(15sin4t+6vtan2t)=3

dt

dv+2vtan2t

=−5sin4t (so P ( t ) = 2 t a n 2 t and Q ( t ) = − 5 s i n 4 t )

Answer	Marks
dt	M1

A1

Answer	Marks
[2]	3.3
1.1	Using F = 3a and a = dd vt to form a

differential equation - allow minor slips or

sign errors but intention must be clear.

Their F must be two terms only.

The correct differential equation in the

form dd vt + P ( t ) v = Q ( t ) can imply this

mark. P(t) and Q(t) do not need to be

explicitly stated. ISW once correct form

seen. If not written in the form,

dd vt + P ( t ) v = Q ( t ) then P(t) and Q(t) must

be explicitly stated.

Answer	Marks
(b)	 2 ta n 2 t d t

I ( t ) = e

= e − ln (c o s 2 t ) ( = s e c 2 t )

( dd vt + 2 v t a n 2 t )  I ( t ) = − 5 s i n 4 t  I ( t )  dd ( v  I ( t ) ) = − 5 s i n 4 t  I ( t )

t

v  s e c 2 t = − 5  ( s i n 4 t  s e c 2 t ) d t

 s i n t 2 c o s 2 t

( v s e c 2 t = ) − 1 0 d t = − 1 0  s i n 2 t d t

c o s 2 t

v s e c 2 t = 5 c o s 2 t ( + c )

v ( 0 ) = 4 .5  c = − 0 .5

0 = 5 c o s 2 t − 0 . 5  c o s 2 t = 0 . 1

Answer	Marks
So, B stationary after 0.735 seconds.	M1*

M1

M1dep*

M1

A1

M1

M1dep*

A1

Answer	Marks
[8]	1.1

1.1

3.3

3.4

Answer	Marks
2.2a	P(t)dt

For I(t)=e for their P(t)

For I ( t ) = e  k ln (c o s 2 t ) or e  k ln (s e c 2 t ) or

 k c o s 2 t or ksec2tor  a c o s k 2 t or

aseck 2t for a,k 0.

For v  I ( t ) = k  s i n 4 t  I ( t ) d t with their

1 I ( t ) (in any form) and k  0 .

1 For simplifying RHS to k sin2tdt for

2 any k  0 - dependent on all previous M

2 marks.

For correct general solution (any

equivalent form). Condone lack of +c.

Using v ( 0 ) = 4 .5 to find constant term –

dependent on first three M marks and an

attempt at integration.

For a two-term equation of the form

cos2t =k where k 1 and k 0 -

3 3 3

dependent on all previous M marks.

Ignore if any other solution(s) found.

Allow awrt 0.735 – for reference

0.7353144…

42.1 seconds (calculated in degrees) scores

A0.

Question 10:
10 | (a) | F =ma so
dv
So −(15sin4t+6vtan2t)=3
dt
dv+2vtan2t
=−5sin4t (so P ( t ) = 2 t a n 2 t and Q ( t ) = − 5 s i n 4 t )
dt | M1
A1
[2] | 3.3
1.1 | Using F = 3a and a = dd vt to form a
differential equation - allow minor slips or
sign errors but intention must be clear.
Their F must be two terms only.
The correct differential equation in the
form dd vt + P ( t ) v = Q ( t ) can imply this
mark. P(t) and Q(t) do not need to be
explicitly stated. ISW once correct form
seen. If not written in the form,
dd vt + P ( t ) v = Q ( t ) then P(t) and Q(t) must
be explicitly stated.
(b) |  2 ta n 2 t d t
I ( t ) = e
= e − ln (c o s 2 t ) ( = s e c 2 t )
( dd vt + 2 v t a n 2 t )  I ( t ) = − 5 s i n 4 t  I ( t )  dd ( v  I ( t ) ) = − 5 s i n 4 t  I ( t )
t
v  s e c 2 t = − 5  ( s i n 4 t  s e c 2 t ) d t
 s i n t 2 c o s 2 t
( v s e c 2 t = ) − 1 0 d t = − 1 0  s i n 2 t d t
c o s 2 t
v s e c 2 t = 5 c o s 2 t ( + c )
v ( 0 ) = 4 .5  c = − 0 .5
0 = 5 c o s 2 t − 0 . 5  c o s 2 t = 0 . 1
So, B stationary after 0.735 seconds. | M1*
M1
M1dep*
M1
A1
M1
M1dep*
A1
[8] | 1.1
1.1
1.1
1.1
1.1
3.3
3.4
2.2a | P(t)dt
For I(t)=e for their P(t)
For I ( t ) = e  k ln (c o s 2 t ) or e  k ln (s e c 2 t ) or
 k c o s 2 t or ksec2tor  a c o s k 2 t or
aseck 2t for a,k 0.
For v  I ( t ) = k  s i n 4 t  I ( t ) d t with their
1
I ( t ) (in any form) and k  0 .
1
For simplifying RHS to k sin2tdt for
2
any k  0 - dependent on all previous M
2
marks.
For correct general solution (any
equivalent form). Condone lack of +c.
Using v ( 0 ) = 4 .5 to find constant term –
dependent on first three M marks and an
attempt at integration.
For a two-term equation of the form
cos2t =k where k 1 and k 0 -
3 3 3
dependent on all previous M marks.
Ignore if any other solution(s) found.
Allow awrt 0.735 – for reference
0.7353144…
42.1 seconds (calculated in degrees) scores
A0.

Show LaTeX source

10 A particle $B$, of mass 3 kg , moves in a straight line and has velocity $v \mathrm {~ms} ^ { - 1 }$.\\
At time $t$ seconds, where $0 \leqslant t < \frac { 1 } { 4 } \pi$, a variable force of $- ( 15 \sin 4 \mathrm { t } + 6 \mathrm { v } \tan 2 \mathrm { t } )$ Newtons is applied to $B$. There are no other forces acting on $B$. Initially, when $t = 0 , B$ has velocity $4.5 \mathrm {~ms} ^ { - 1 }$. The motion of $B$ can be modelled by the differential equation $\frac { d v } { d t } + P ( t ) v = Q ( t )$ where $P ( t )$ and $\mathrm { Q } ( \mathrm { t } )$ are functions of $t$.
\begin{enumerate}[label=(\alph*)]
\item Find the functions $\mathrm { P } ( \mathrm { t } )$ and $\mathrm { Q } ( \mathrm { t } )$.
\item Using an integrating factor, determine the first time at which $B$ is stationary according to the model.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2024 Q10 [10]}}

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