Question 8:
8 | (a) | 25000 25000
Driving force = or
v 7
800gsin5
25000
750800gsin5800a
v
v7 a2.67m s–2
a0 v17.4m s–1 | B1
B1
M1
A1
A1
[5] | 1.2
1.1
3.3
1.1
1.1 | Not for mgsin5- award when value for
m substituted
N2L with either 3 or 4 terms – allow
for the M mark
D750800gsin5800aor
D750800gsin50
Answer to least 3 sf
Answer to least 3 sf | Or implied by their
answer
Condone sign errors
and sin/cos confusion –
lhs must be a weight
component and the rhs
must be mass only
2.67265943…
17.442253…
8 | (b) | WD by car = 25000(10.4)
WD against reistance = 750(131)
800g 
Change in PE = 131sin5
Change in KE =  1  800  v272 
2
1  800  v2 72  800g 131sin5 
2
  750  
25000 10.4 131
v15.2m s–1 | B1
B1
B1
B1
M1
A1
[6] | 1.1
1.1
1.1
1.1
3.3
2.2a | 260000
98250
89512.4340…
Use of correct formula for KE
Use of work-energy principle, all
terms present – all values (condone g)
substituted or implied by later working
(so not in terms of m but award when
m substituted or implied)
Answer to least 3 sf | mg 
Not for 131sin5 -
however, can be
awarded if m implied or
substituted later
As above - must use
value for m at some pt.
Allow sign errors but
must be a change in KE
(so must imply
subtraction of the two
KE terms)
15.1523567…
8 | (b) | ALT | Using diff. equations
800v
x dvor
25000
750800gsin5
v
800
t dv
25000
750800gsin5
v | Award B1 for either of these two
integrals (allow numerical equivalents)
Then the remaining 5 marks are for the
correct answer of 15.2 (no interim
marks after the first B mark) | So may have attempted
to simplified the
constant terms
1
At angle , PE = mglcos, KE = mv2
2
1 d 2
ml2 mgl coscos
 
2  dt 
d 2 2g 2g
 cos cos
 
 dt  l l | B1
M1
A1
[4] | 1.1
3.3
1.1 | 1
Or B1 for KE = mv2 and B1 for
2
PE = mglcoscos
Note that 1mv2 mglcoscos
2
implies both B marks
Conservation of energy and use of
d
vl
dt
2g
k  cos
1 l | Must be three terms but
allow sign errors and
sin/cos confusion