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AQA PAST PAPER: NEWTON’S LAWS OF MOTION – EXTRA QUESTIONS - ANSWERS
Q1 FoP Jan 2009 Q10
(a)
9.0 tan 50° 
10.7 ms-1 
(b) (i)
(ii)
(c)
12/9 = 1.33 s 
use of s = ut + ½at2 
substitution correct [10.7 × 1.33 – ½ x 9.8 × 1.332] 
s = 5.6 m 
maximum four from:
at A only vertical force is weight downwards 
initial vertical speed/ke zero at A then starts to accelerate down 
air resistance/drag acts upwards 
drag increases as speed increases 
resultant downward force decreases so acceleration decreases 
(until) drag = weight/equilibrium/balanced 
acceleration is zero not ‘stop accelerating’ 
constant speed reached/remains at terminal speed 
Q2 FoP June 2004 Q2
(a) (i) 2.4 ms−2 
(ii)
(b)
(c)
(2)
(4)
max 4
(1)
F = ma 
= 132 000 N (ecf from (i)) 
(2)
final speed = (8902 + 602)½ 
= 892 ms−1
(allow 890 m s−1 as final answer but 892 must be seen in working) 
(2)
tan−1(60/890) or sin−1(60/892) = 3.9° (3.86)° or cos−1(890/892) = 3.8 (4)°
or sin−1(60/890) =3.9° (3.86)° if ecf from (b) 
(1)
Q3 FoP June 2004 Q7
(a)
s = ½at2 
a = 9.6 ms−2 
(2)
(b)
lower value obtained 
air resistance has a greater effect on the ball 
resistive force is same/higher (for given speed) and downward force/weight/gravitational
force lower or resultant downward force will be lower for the lower mass/ball or ball reaches
terminal velocity quicker/sooner 
(3)
(c)
curve drawn with correct shape always below the given line
(line may or may not be shown reaching terminal velocity) 
curve showing a time greater than given line; initial slope as given line 
(2)
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Q4 FoP June 2009 Q8
(a) (i) 3.6 s 
(ii)
(b)
correct construction of sensible size > one-third area of page 
61 ± 1 (ms-1) 
55 ± 1 to vertical/35 to horizontal (°) 
2 sf from construction 
(5)
using area under graph, calculates speed change for 1 square (0.5 ms-1) 
estimates time from zero to attain 70 squares 
7.5 ± 1.0 (s) 
(3)
weight force downwards constant and air resistance/drag force increases with speed 
net vertical force = weight − drag or drag opposes weight 
net vertical acceleration = (weight - drag)/mass 
acceleration tends to 0/acceleration = 0 when weight = drag
allow terminal speed reached for acceleration = 0 
(4)
(c)
Q5 FoP June 2005 Q7
(a) (i) 1.5 x 104 sin 35 or 1.5 x 104 cos 55 seen = 8603 (N)
(to minimum 4 sf to see difference from 8600) 
(ii)
(b) (i)
(ii)
(1)
11 600 N or 12 000 N 
(1)
accelerating force = 5600 N 
mass of cable car = 1530 kg (or 15 000/9.8 seen) 
a = 3.7 ms−2 
(3)
v2 = u2 + 2as 
v = 30 (29.6) ms−1 (ecf for acceleration) 
(2)
(iIi) any equation of uniformly accelerated motion that includes t 
t = 8.1 s (ecf for v or a) 
(2)
Q6 FoP June 2008 Q10
musket/cannon ball
acceleration the same/falls at the same speed 
effects of different masses and weights cancel out (reference to F = ma) 
air resistance insignificant/affects both in the same way 
musket/table tennis ball
table tennis ball is (significantly) affected by air resistance 
which reduces acceleration/table tennis ball lands second/has slower speed 
appropriate comment about air resistance with reference to size, weight, density or
surface quality 
max 5