Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | dv
m mg1kv
dt | B1 | Mass must be seen at this point or earlier.
[SUVAT does not apply.]
1
 ln1kvgt A
k | M1 | Separate variables and integrate to logarithm.
A1 | Correct, with constant of integration.
t0, v0 [A0] | M1 | Use initial condition to evaluate their constant.
1 1ekgt
v
k | A1 | Any correct form with v as subject.
Final A0 if numerical value of g present.
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | dx  1e0.5t
k 0.05 and so 20
dt | *M1 | Attempt to integrate if expression contains a term of
the form bect .
 t2e0.5t
Integrate: x20 B | A1 | 1 1 
x t ekgt B
k gk 
t0, x0 B40 | DM1 | Use initial condition to evaluate their constant.
When v12, from part (a), e0.5t 10.05120.4, t2ln0.4 | M1 | 1.83…
x40ln0.4400.440 12.7 | A1 | 5
40ln 24
2
Alternative method for question 6(b)
dv  1 
v g1kv leading to  1  dvkgdx
dx  1kv | *M1 | Separate variables and write in integrable form
1
v ln1kvkgxB
k | DM1 A1 | Dependent on previous M1. Attempt to integrate.
v0, x0 B0 and k 0.05, v12 1 220ln0.40.5x | M1 | Dependent on both previous M1s.
Use initial condition to evaluate their constant and
use v12
x12.7 | A1 | 5
40ln 24
2
5
Question | Answer | Marks | Guidance