Question 3:
3 | Attempt to resolve either direction | M1 | Correct number of terms. Allow sin/cos mix. Allow
sign errors. Allow g missing.
0.3g + T cos  – 4 sin 60 = 0 (T cos α° = 0.464…) | A1 | OE
T sin  – 4 cos 60 = 0 (T sin α° = 2) | A1 | OE If the two Ts are different, award maximum
A1A0 unless subsequently stated that the two Ts are
the same.
 4 cos 60   2 
=tan−1 =tan−1
  
4 sin 60−0.3g 0.464 | M1 | Attempt to solve for α. No missing/extra terms. Allow
g missing. Must get to ‘α =’.
4cos60
T = = (4 cos 60 )2 +(4 sin 60−0.3g)2 = 22 +(0.464)2
sin(their) | M1 | OE Attempt to solve for T. No missing/extra terms.
Allow g missing. Must get to ‘T =’.
Tension = 2.05 N α = 76.9 | A1 | For both AWRT 2.05, 76.9
(Tension = 2.05314… N α = 76.9356…)
Alternative method for Q3 using triangle of forces
Attempt at cosine rule from triangle of forces | M1 | Must use lengths 4 and 0.3g with a suitable angle.
Allow g missing.
T2 =42 +(0.3g)2 −24(0.3g)cos30 | A1
Tension = 2.05 | A1 | Tension = 2.05314… AWRT 2.05
Attempt at sin rule | M1 | Must have angle 30° and another angle in terms of α
with correct numerators, but allow g missing.
Their T 4 Their T 0.3g
= or =
sin30 sin(180−) sin30 sin(−30) | A1 | Correct. Allow sin instead of sin(180−) .
α = 76.9 | A1 | α = 76.9356… AWRT 76.9
Question | Answer | Marks | Guidance
3 | Alternative method for Q3 using Lami’s theorem
Attempt at Lami’s theorem | M1 | Must have numerators correct and at least one angle
correct. Allow g missing.
4 0.3g T
= =
sin sin(210−) sin(150) | A1 A1 | A1 for two parts second A1 for all three.
 4sin210 
=tan−1
 
0.3g+4cos210 | M1 | For solving for α using compound angle formula. Must
be correct for their angles. Allow g missing.
4sin(150) 0.3gsin(150)
T = or T =
sin sin(210−) | M1 | For solving for T using their α. Allow g missing.
Tension = 2.05 N α = 76.9 | A1 | For both AWRT 2.05, 76.9
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SC: Tension and the 4 N force considered in the wrong directions
Attempt to resolve either direction | M1 | Correct number of terms. Allow sin/cos mix. Allow
sign errors. Allow g missing.
T cos 60 – 4 sin  = 0
And: T sin 60 – 4 cos −0.3g= 0 | A1 | For both
OE If the two Ts are different, they get SC A0 unless
they subsequently state that the two Ts are the same.
T cos 60 2 T sin 60 −0.3g 2 1 3
+ =1 T2 + T2 −33T +9=16
   
 4   4  4 4
T2−33T−7=0T =6.31 (or−1.11)
3
OR: 4 3sin−4cos=3 ⇒ 8sin(−30)=3 ⇒ =sin−1 +30
8 | M1 | OE Attempt to solve for T or α. No missing/extra
terms. Allow g missing. Must get to ‘T =’ or ‘α=’.
Question | Answer | Marks | Guidance
3 | T =6.31 N  =52.0 | A1 | (T = 6.30617…, α=52.0243…)
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Question | Answer | Marks | Guidance