5B Observing and recording motion
Practice questions Label axes [1 mark]
1 a) U Accurate points [1 mark]
Straight line [1 mark]
b) i) 67.5 km [1 mark]
Drag Forward push
ii) 2.3 hours [1 mark]
distance
c) speed = time [1 mark]
W
3 a) The train is slowing down. [1 mark]
The four forces are: OR The gradient of the graph is
• Forwards push from the water (shown) negative. [1 mark]
• Drag backwards [1 mark] b) The distance travelled is the area under the
• Weight (W) [1 mark] graph.
• Upthrust (U) [1 mark] [Areas A + B + C] [1 mark]
c)
Velocity
b) i) speed = distance
Train 1
time Train 2
= 1500
1200
= 1.25 m/s [1 mark]
distance
ii) speed =
time Time
51500 1 mark for a constant velocity.
= 1 mark for a velocity between 0.5 and 0.9 of
6800 [1 mark]
= 7.6 m/s [1 mark] the original velocity.
4 a) i) F is bigger because the lorry accelerates
Add all the distances, then divide by the in the direction of the resultant force.
sum of the times. (F−B). [1 mark]
c) The gradient of the graph is the ii) resultant force = mass × acceleration
speed. [1 mark] [1 mark]
For 700 s he went at a constant iii) F = ma
speed. [1 mark] 15 000 = 12 500 × a [1 mark]
Then he slowed down, then went 15000
quickly again, [1 mark] a =
12500
before slowing down towards = 1.2 m/s2 [1 mark] [1 mark]
the end. [1 mark] You must include the correct unit.
One mark for each point (up to 3). b) i) The driver is distracted. [1 mark]
2 a) Or The driver is under the influence of
Distance in km
100
alcohol or drugs.
Or
90
Some drivers are just slower than others.
80 ii) An icy road. [1 mark]
Or Worn tyres.
70 Or The road surface – water or mud.
67.5
Or Worn brakes.
60 Or The speed of the car.
Or Having a heavy load in the car.
50
c) The driver’s reaction time does not depend
40
on the speed. [1 mark]
35
The councillor should have said the
30 braking distance is less at 20 mph. [1 mark]
5 a) stopping distance = thinking distance +
20 braking distance [1 mark]
b) The graph shows:
10 the thinking distance is proportional
to the speed [1 mark]
0
0 1 2 2.3 3 4 4.5 5 6 the braking distance increases rapidly
Time in hours at high speeds. [1 mark]
131
c) About 30 m. [1 mark] A crumple zone increases the time
5B Motion
d) For the minimum stopping distance, you a car takes to stop. [1 mark]
need to take the smallest distance found in Therefore the force acting on the car and
the test. [1 mark] its passenger is lower. [1 mark]
e) i) There is no change to the thinking 9 a) i) The length of the card. [1 mark]
distance. [1 mark] ii) If the track is tilted, gravity will slow
This just depends on the reaction time of down or speed up the glider. Friction
the driver. [1 mark] would slow the glider down. [1 mark]
ii) The braking distance increases, because b) i) A vector has direction as well as size
there is a smaller braking force on the (or magnitude). [1 mark]
car, so its deceleration is less. [1 mark] ii) momentum = m × v
[1 mark] = 2.4 × 0.6 [1 mark]
change of speed = 1.44 kg m/s [1 mark]
6 a) acceleration =
time You must have the correct unit.
78 iii) Zero.
= [1 mark] c) i) Momentum is conserved,
60
= 1.3 m/s2 [1 mark] [1 mark] so combined momentum = 1.44 kg m/s
You must have the correct unit. [1 mark]
So the resultant force on the plane ii) momentum = mass × velocity
decreases. [1 mark] 1.44 = m × 0.4 [1 mark]
Acceleration decreases, because 1.44
=
resultant force = mass × acceleration. [1 mark] 0.4
c) Distance = area under the graph [1 mark] = 3.6 kg [1 mark]
The area is about 30 squares. [1 mark] mass of Q = 3.6 − mass of P
1 square = 10 m/s × 10 s = 100 m = 3.6 − 2.4
So distance = 30 × 100 m = 1.2 kg [1 mark]
= 3000 m [1 mark] iii) change of momentum = mv1 − mv2
change of speed = 2.4 × 0.6 − 2.4 × 0.4 [1 mark]
7 a) acceleration = = 1.44 − 0.96
time
4 = 0.48 kg m/s [1 mark]
= [1 mark] change of momentum
8 iv) force = [1 mark]
= 0.5 m/s2 [1 mark] [1 mark] time
You must have the correct unit. 0.48 [1 mark]
=
b) resultant force = mass × acceleration 0.05
60 − R = 80 × 0.5 [1 mark] = 9.6 N [1 mark]
60 − R = 40 [1 mark] v) 9.6 N [1 mark]
R = 20 N [1 mark] Working scientifically
8 a) momentum = mv 1 The 500 g mass has inertia; it is also an
= 68 × 6 example of Newton’s third law of motion.
= 408 kg m/s [1 mark] [1 mark] 2 a) Type of material used for the crumple zone.
You must have the correct unit. b) How far the 500 g mass moved forwards
change of momentum before stopping.
b) force = 3 The area and thickness of the materials used to
time
840 model the crumple zone.
= [1 mark] 4 From 8.0 cm (no material) to 14.8 cm (rubber
0.14
carpet underlay); the range could also be
= 6000 N [1 mark]
expressed as 6.8 (14.8 − 8.0).
c) i) Worn tyres.
5 Bar chart drawn, x-axis labelled type of
Worn brakes. material, y-axis labelled distance 500 g mass
Carrying a heavy load. moves before stopping.
Speed. 6 Type of material is a categoric variable.
Road conditions (e.g. ice, water, mud). 7 Yes, they are consistent; for each force, the
One mark for each correct answer, up to 2. order of the materials for increasing stopping
change of momentum distance is the same.
ii) force =
time
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