Question 9:
--- 9(a) ---
9(a) | dv
5g−0.05v2 =5v
dx
dv  100v dv
980−v2 =100v ⇒  =1
dx 980−v2 dx | M1
A1
[2] | 3.3
1.1 | Applying N2L with correct number of terms (allow sign
dv dv
errors but dimensionally correct) – allow v or only
dx dt
for a
AG – www (so any errors seen is A0)
--- 9(b) ---
9(b) | When x = 0, v2 =980 ( 1−e−0.02(0)) =980(1−1)=0
dv 980
2v = e −0.02x
dx 50
980 
−0.02x
100 e 
100 
980−(980−980e −0.02x)
−0.02x
980e
=1 (as required)
−0.02x
980e | B1
M1*
M1dep*
A1
[4] | 3.4
1.1
1.1
1.1 | Confirming that v = 0 when x = 0 – as a minimum must
show at least one step of working e.g. v2 =980(1−e0)=0
but B0 for just ‘when x = 0, v = 0’ – showing that v2 =0
without explicitly mentioning that v = 0 is B1
Attempt to differentiate either v or v2 - must be of the
dv
form 2v =ke −0.02x where k ≠−980(or equivalent if
dx
dv
differentiating v) – if implying 2v =ke −0.02x then M0
dt
Substitute their derivative and correct expression for v2
 100v dv
into
 
980−v2 dx
AG so sufficient working must be shown (but no
conclusion after arriving at ‘= 1’ is required) – allow
 100v dv
other complete methods to show that   =1
980−v2 dx
e.g. using v2 =980(1−e−0.02x)to derive given de
Alternative Solution
⌠ 100v
x= dv=kln(980−v2)(+c)
⌡ 980−v2 | M1* | Solve the given differential equation by separating
variables and obtaining x=kln(980−v2)or
x=kln 980−v2 with k ≠0
x=−50ln(980−v2)(+c) | A1 | or x=−50ln 980−v2 - condone lack of constant of
integration
x=0,v=0⇒0=−50ln980+c∴c=50ln980 | M1dep* | Using correct initial conditions to find c
x=−50ln(980−v2)+50ln980 | A1 | AG – sufficient working must be shown – at least one
line of additional working from
x=−50ln(980−v2)+50ln980to given answer required –
any errors is A0
 980 
x=50ln

980−v2 
980 x − x
=e50 ⇒980−v2 =980e 50
980−v2
v2 =980(1−e−0.02x)
⌠ 100v
x= dv=kln(980−v2)(+c)
⌡ 980−v2
variables and obtaining x=kln(980−v2)or
x=kln | 980−v2 | with k ≠0
or x=−50ln 980−v2 - condone lack of constant of
integration
M1*
x=−50ln(980−v2)(+c)
A1
AG – sufficient working must be shown – at least one
line of additional working from
x=−50ln(980−v2)+50ln980to given answer required –
any errors is A0
--- 9(c) ---
9(c) | 980−v2 =100(8.36)⇒v=...
 980 
x = 50ln  (=7.946…)
980−122 
Loss in PE = 5g(7.946...)
Gain in KE = 1(5)(12)2 (= 360)
2
Work done against resistance = 389.363… – 360 =
29.4 (J) | M1*
M1dep*
B1FT
B1FT
A1
[5] | 3.1b
3.1b
1.1
1.1
2.2a | Use N2L equation from part (a) (or re-start) with a = 8.36
to find v (if correct then v = 12) or v2– allow minor slips
when solving for v only
Using the equation from part (b) to find the distance P
falls from O to given point – dependent on the previous
M mark – allow minor slips when solving for x only
Dependent on both previous M marks – FT their x
Using their value of v – dependent on the first M mark
only – FT their v
awrt 29.4 or awrt 29.6 (from using x=7.95) or allow 29
www or allow 30 www (from using x = 7.95)
Alternative Solution
980−v2 =100(8.36)⇒v=... | M1* | Use N2L equation from part (a) (or re-start) with a = 8.36
to find v (if correct then v = 12) – allow minor slips when
solving for v only
to find v (if correct then v = 12) – allow minor slips when
solving for v only
 980 
x = 50ln  (=7.946…)
980−122  | M1dep* | Using the equation from part (b) to find the distance P
falls from O to given point – allow minor slips when
solving for x only
Work done is ∫0.05×980(1−e−0.02x)dx | B1* | Setting up correct integral – ignore limits for this mark
=49(x+50e−0.02x) | M1dep* | Integrate to get an expression of the form k x+k e−0.02x
1 2
where k ,k are both non-zero – this mark is dependent
1 2
on the previous B mark only – ignore limits for this
mark – a correct answer implies this mark (as can be done
BC)
29.4 (J) | A1 | awrt 29.4 or allow 29 www
Integrate to get an expression of the form k x+k e−0.02x
1 2
where k ,k are both non-zero – this mark is dependent
1 2
on the previous B mark only – ignore limits for this
mark – a correct answer implies this mark (as can be done