Question 5:
5 | (a) | 2 v d v = − ( v 2 + 1 )
d x | M1 | 3.3 | Using F = ma to derive a
differential equation in v and x
only. Allow a single sign error | Condone m seen in equation (not
𝑑𝑥
2). Do not allow
𝑑𝑣
22 v
 d v = − x + c
v + 1 | M1 | 1.1 | Correctly separating the
variables and integrating at least
one side | Condone m seen in equation (not
2). Can be given if m substituted
later
ln(v2+1)=−x+c | A1 | 1.1 | Correct relationship between v
and x (condone missing + c). | Must see brackets for this A mark,
and correct numerical m.
x = 0 , v = 3  c = ln ( u 2 + 1 ) = ln 1 0 | M1 | 3.4 | Using initial condition to find c
for an expression of correct
format ln(𝑓(𝑣)) = ±𝑥 +𝑐 | May see use of limits
[ln (𝑣2 +1)]𝑣 = [−x]𝑥
3 0
⇒ln(𝑣2 +1)−ln(10) = −𝑥
 v 2 + 1 = e − x + c = 1 0 e − x
 v = 1 0 e − x − 1 | A1 | 1.1 | The –1 must be clearly under the
root sign oe
[5]
(b) | v = 2 => ln(22 + 1) = –x + ln10 | M1 | 3.4 | Substituting v = 2 into a solution
of the correct form, which
includes a numerical value for
the constant of integration | Or integrating between the correct
limits.
[ln (𝑣2 +1)]2 = [−x]𝑥
3 0
ln(5)−ln(10) = −𝑥
x = ln10 – ln5 = ln2 so distance is ln2 m or
awrt 0.693 m | A1 | 1.1 | Award SC1 if answer seen with no
working (need to see v=2)
[2]
(c) | For Q, F = –1 => a = –0.5 => v = u – 0.5t’ or
v2 = u2 – s | M1 | 1.1 | Calculating the (constant)
acceleration and using it in a
suvat equation | ALT method 1: (using F = ma to
set up a differential equation):
𝑑𝑣 𝑑𝑣
𝐹 = −1 = 𝑚 = 2
𝑑𝑡 𝑑𝑡
⟹ −𝑡+𝑐 = 2𝑣
At 𝑡 = 0,𝑣 = 3 ⟹ 𝑐 = 6
1
Hence 𝑣 = (6−𝑡)
2
ALT method 2:
𝑑𝑣
2𝑣 = −1 ⇒𝑣2 = −𝑥 +𝑐
𝑑𝑥
At 𝑥 = 0,𝑣 = 3 ⇒𝑐 = 9
Hence 𝑣2 = 9−𝑥 oe
Q stops when v = 0.
Q
v = u + at’ => –0.5t’ = –3 => t’ = 6
v2 = u2 + 2as => s = 9 | A1 | 2.1 | AG Need to see v = 0 used. | ALT method 1
v = 0 when t’ = 6
Q
𝑑𝑥
−𝑡’+6 = 2
𝑑𝑡’
𝑡’2
⟹ − +6𝑡’(+𝑘) = 2𝑥
2
1 𝑡’2
(and k = 0) so 𝑥 = (6𝑡’− )
2 2
At t’ = 6, x = 9
ALT method 2
Q stops when v=0, therefore 9-x =
0, therefore x = 9.
They will also need to use one of
the other listed methods to justify
the limit on t for this method mark
But for P, F = 1 + v2 > 1 (so a  > 0.5 for all
P
v) so P will slow down more quickly than Q
(and be travelling more slowly at any
comparable time) so it must stop before time
t = 6 and it cannot reach x = 9.
 t < 6 and x < 9 | A1 | 2.2a | AG Include:
• F = 1 + v2 > 1 (explicit
comparison)
• A reference to P coming
to rest more quickly than
Q (comparison). Needs
to make the link – a
larger resistive force will
decelerate P more
quickly or a larger
resistive force will mean
P is travelling more
slowly than Q (from
starting point of 3)
• Final conclusion: So it
comes to rest when t < 6
and x < 9 | Be generous with t and t’ confusion
[3]
(d) | B1 | 3.1b | Graph in 1st quadrant only.
Strictly decreasing from positive
value on vertical (v) axis to the
horizontal (t) axis. Negative
gradient at vertical axis
intercept. Allow concave. Do
not allow linear
No values required. | Should come to rest
[1]
(e) | v = 0 => ln1 = –x + ln10 | M1 | 3.4 | Substituting v = 0 into a solution
of the DE needs to be correct
form | Or considering 10e–x – 1 ≥ 0
x = ln10 so maximum displacement is ln10 m
or awrt 2.30 m | A1 | 1.1 | Allow -ln(1/10) | Award SC1 if answer seen with no
working (need to see v=0)
[2]