Standard +0.8 This is a Further Mechanics question requiring multiple techniques: deriving a differential equation from Newton's second law with v dv/dx, solving a separable DE, interpreting physical behavior at infinity, calculating work done via integration, and connecting work-energy principles. While each individual step is methodical, the multi-part structure, requirement to interpret model limitations, and synthesis of calculus with mechanics concepts places this above average difficulty for Further Maths students.
Show that \(B\) 's motion can be modelled by the differential equation \(\frac { 1 } { \mathrm { v } } \frac { \mathrm { dv } } { \mathrm { dx } } = - 4\).
Solve the differential equation in part (a) to find the particular solution for \(v\) in terms of \(x\) and \(u\).
By considering the behaviour of \(v\) as \(x \longrightarrow \infty\) describe one feature of the model that is not realistic.
At the instant when \(B\) reaches the point \(A\), where \(\mathrm { x } = \mathrm { X }\), its speed is \(V \mathrm {~ms} ^ { - 1 }\). The work done by the resistance as \(B\) moves from \(O\) to \(A\) is denoted by \(W \mathrm {~J}\).
Use the formula \(\mathrm { W } = \int \mathrm { F } \mathrm { dx }\) to determine an expression for \(W\) in terms of \(X\) and \(u\).
Explain the relevance of the sign of your answer in part (c)(i).
By writing your answer to part (c)(i) in terms of \(V\) and \(u\) show how the quantity \(W\) relates to the energy of \(B\).
= 3 ( V 2 − u 2 ) = 3 ( u e − 4 X ) 2 − u 2
4 4
= 3 u 2 ( e − 8 X − 1 ) A1
4
Answer
Marks
Guidance
(c)
(ii)
Since X > 0, (e−8X< 1 so) W < 0
and
It is negative because work is being
done by B against the resistance (rather
than the resistance actually doing work
on B.) ie B is losing (mechanical)
Answer
Marks
Guidance
energy
B1
[1]
3.2a
A clear explanation of negative
sign from correct expression for
W
and
Accept valid alternatives, eg
“the particle is doing work
against the resistance” or “the
particle is losing (mechanical)
energy” or “the particle is losing
KE to do work against
resistance” etc.
Answer
Marks
or “KE is lost”
Do not condone:
force and displacement are in
opposite directions
Do not condone “slowing down”
with no comment on energy
the resistance does work on B
Do not condone “energy is being
removed/taken from B”
Do not accept “work done against
motion”
Do not allow additional incorrect
statements
Answer
Marks
Guidance
(c)
(iii)
Since V = u e − 4 X ,
3u2 3u2 3u2
W = ( e −4X )2 −1 = ( e −4X )2 −
4 4 4
3 3
= V2 − u2
4 4
1 .5 1 .5 1 1
= V 2 − u 2 = m V 2 − m u 2 = Δ𝐾𝐸
2 2 2 2
(m = 1.5)
(so the work done is the change in KE.
ie KE loss is equal to the energy lost to
resistance or work done against
Answer
Marks
resistance)
M1
A1
Answer
Marks
[2]
3.4
2.4
Substitute into correct W to
eliminate X and express W in
terms of V and u.
Need to see use of V = u e − 4 X or
V 2 = u 2 e − 8 X
Must state WD is Δ𝐾𝐸 of B.
Clear algebraic or stated
conclusion. Condone additional
mention of total energy.
Answer
Marks
WWW
SC1 if correct integration wrt v
using limit of V seen in part ci)
3 3 1 ( )
W = V 2 − u 2 = 1 .5 V 2 − u 2
4 4 2
= K E
stated in this part. (m=1.5 seen)
SC1 for:
W = − 3 u 2 ( e − 4 X ) 2 − 1 = − 3 u 2 ( e − 4 X ) 2 + 3 u 2
4 4 4
3 3
= u 2 − V 2
4 4
Question 7:
7 | (a) | R = –kv2 so v = 2, R = –24 => k = 6
d v
1 .5 a = 1 .5 v = − 6 v 2
d x
1 .5 v2 d v − 6 v 2
=
1 .5 v d x 1 .5 v 2
1 d v
= − 4
v d x | B1
M1
A1
[3] | 3.3
3.3
1.1 | DR
Minus sign could be missing
throughout here
Use of NII with correct form for
𝑑𝑣
a= 𝑣 and resistance term
𝑑𝑥
which must be negative here
AG so clear evidence of division
must be seen and www. If chain
𝑑𝑣 𝑑𝑣
rule used to convert = ×
𝑑𝑡 𝑑𝑥
𝑑𝑥
must see clear complete
𝑑𝑡
rearrangement. | Or R = kv2 leading to k = –6
Condone incorrect or missing k, or
“k” left in.
(b) | (i) | 1
d v = − 4 d x = − 4 x + c
v
lnv = –4x + c
x = 0, v = u => c = lnu
v=ue −4x | M1
A1
M1
A1
[4] | 1.1
1.1
3.4
1.1 | Separating the variables
correctly and integrating one
side (condone missing “+ c” or
incorrect letter (eg t for x))
Correct general solution, any
form (condone missing “+ c”)
Using correct initial conditions
in a solution to the DE
containing one arbitrary constant
(or as the lower limits in a
definite integral solution with x
and v as upper limits)
Condone v = e − 4 x + ln u | Do not condone missing dx’s but
recovery possible (eg once
integration correctly carried out).
Could also be a definite integral;
ignore limits for this M1A1.
This could be done after
exponentiation:
v = e − 4 x + c = A e − 4 x
x = 0 , v = u A = u
(b) | (ii) | As (x → ∞), v will tend to, but never
reach, 0.
and
So the model predicts B will continue
to move when in reality it will stop. | B1
[1] | 3.5b | Clear reference to the idea that
the model predicts that B never
comes to rest while in real life it
does. | Do not condone “unlikely/likely”
Do not allow “suggest v will never
reach 0 which is not realistic” as no
reference to v tending to 0.
(c) | (i) | X0 X0
W = − 6 v 2 d x = − 6 ( u e − 4 x ) 2 d x
X0 e − 8 x X
= − 6 u 2 e − 8 x d x = − 6 u 2
− 8
0
6 u 2
= e − 8 X − e 0
8
= 3 u 2 ( e − 8 X − 1 )
4 | M1FT
M1FT
A1
[3] | 3.1b
1.1
1.1 | DR
Correct form for integral and
limits for their v and k (may not
be numeric). Condone missing
dx or missing sign
If this is done as an indefinite
integral then this mark is
awarded when x = 0 and WD = 0
and x = X WD = W both used
Carrying out the integral for
their v of correct form 𝑣 =
𝑢𝑒−𝑐𝑥, c>0 only, and their k.
Condone x not X. Condone
missing sign.
Can ISW once final expression
seen. May not be factorised.
Allow 6/8 oe | ALT method
Changing variable to integrate
w.r.t. v, Must see V in limits from
next part.
𝑊 = ∫ 𝑋 −6𝑣2𝑑𝑥 = ∫ 𝑉 −6𝑣2 ×
0 𝑢
1
− dv M1FT
4𝑣
𝑉 𝑣2 𝑉
= ∫ 1.5𝑣dv=1.5[ ]
2
𝑢 𝑢
M1FT
= 3 ( V 2 − u 2 ) = 3 ( u e − 4 X ) 2 − u 2
4 4
= 3 u 2 ( e − 8 X − 1 ) A1
4
(c) | (ii) | Since X > 0, (e−8X< 1 so) W < 0
and
It is negative because work is being
done by B against the resistance (rather
than the resistance actually doing work
on B.) ie B is losing (mechanical)
energy | B1
[1] | 3.2a | A clear explanation of negative
sign from correct expression for
W
and
Accept valid alternatives, eg
“the particle is doing work
against the resistance” or “the
particle is losing (mechanical)
energy” or “the particle is losing
KE to do work against
resistance” etc.
or “KE is lost” | Do not condone:
force and displacement are in
opposite directions
Do not condone “slowing down”
with no comment on energy
the resistance does work on B
Do not condone “energy is being
removed/taken from B”
Do not accept “work done against
motion”
Do not allow additional incorrect
statements
(c) | (iii) | Since V = u e − 4 X ,
3u2 3u2 3u2
W = ( e −4X )2 −1 = ( e −4X )2 −
4 4 4
3 3
= V2 − u2
4 4
1 .5 1 .5 1 1
= V 2 − u 2 = m V 2 − m u 2 = Δ𝐾𝐸
2 2 2 2
(m = 1.5)
(so the work done is the change in KE.
ie KE loss is equal to the energy lost to
resistance or work done against
resistance) | M1
A1
[2] | 3.4
2.4 | Substitute into correct W to
eliminate X and express W in
terms of V and u.
Need to see use of V = u e − 4 X or
V 2 = u 2 e − 8 X
Must state WD is Δ𝐾𝐸 of B.
Clear algebraic or stated
conclusion. Condone additional
mention of total energy.
WWW | SC1 if correct integration wrt v
using limit of V seen in part ci)
3 3 1 ( )
W = V 2 − u 2 = 1 .5 V 2 − u 2
4 4 2
= K E
stated in this part. (m=1.5 seen)
SC1 for:
W = − 3 u 2 ( e − 4 X ) 2 − 1 = − 3 u 2 ( e − 4 X ) 2 + 3 u 2
4 4 4
3 3
= u 2 − V 2
4 4
\begin{enumerate}[label=(\alph*)]
\item Show that $B$ 's motion can be modelled by the differential equation $\frac { 1 } { \mathrm { v } } \frac { \mathrm { dv } } { \mathrm { dx } } = - 4$.
\item \begin{enumerate}[label=(\roman*)]
\item Solve the differential equation in part (a) to find the particular solution for $v$ in terms of $x$ and $u$.
\item By considering the behaviour of $v$ as $x \longrightarrow \infty$ describe one feature of the model that is not realistic.
At the instant when $B$ reaches the point $A$, where $\mathrm { x } = \mathrm { X }$, its speed is $V \mathrm {~ms} ^ { - 1 }$. The work done by the resistance as $B$ moves from $O$ to $A$ is denoted by $W \mathrm {~J}$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the formula $\mathrm { W } = \int \mathrm { F } \mathrm { dx }$ to determine an expression for $W$ in terms of $X$ and $u$.
\item Explain the relevance of the sign of your answer in part (c)(i).
\item By writing your answer to part (c)(i) in terms of $V$ and $u$ show how the quantity $W$ relates to the energy of $B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q7 [14]}}