Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion Mamun Sir - 1 - Mamun sir 5 m s –1 0 Time t Speed 0 t = 5s Time t Speed Describing Motion Answers 1) 12.5 m s–1 2) a) 0.018 s b) 0.036 m 3) a) b) Distance and Displacement Answers 1) a) 380 km b) 34 km North 2) –2.5 m s –2
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Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
Mamun Sir - 1 -
Mamun sir 5 m s–1
0 Time t
Sp
eed
0 t = 5s Time t
Sp
eed
Describing Motion Answers
1) 12.5 m s–1
2)
a) 0.018 s
b) 0.036 m
3)
a)
b)
Distance and Displacement Answers
1)
a) 380 km
b) 34 km North
2) –2.5 m s–2
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
Mamun Sir - 2 -
Mamun sir
20
10
0 200 me/s
–10
–20
Sp
eed
/m s–1
50 100 150
Ti
More Information from Graphs of Motion Answers
1)
a)
b)
i. 1800 m forwards
ii. 1400 m forwards
iii. 1 m s–2; –0.5 m s–2; –1 m s–2; 1 m s–2
Equations of Motion Answers
1) Unless the pedestrian gets out of the way, there will be a collision.
2)
a) 10 m s–2
b)
i) 45 m
ii) 3 s
c) 15 s
d) 24 m
e) 120 m (to 2 s.f.)
Moving in More Than One Direction – Using Vectors Answers
1) The relative velocity against wind increases their wind speed for a comparatively
low ground speed. Thus they don’t have to hit the ground so fast but still get
enough lift from the wind passing over the wings.
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
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Mamun sir
reaction force
centre of gravity of block
weight of ruler
reaction force
forward force from engine
drag forces
weight of racing car
weight
2) 58 cm (to 2 s.f.) at an angle of 37° (2 s.f.) north of west.
3) 5.4 m s–1 with a bearing 30° east of north.
Causes of Motion Answers
1) Examples such as a ball that is kicked will stop rolling, a puck on ice will
eventually slow and stop, and a clock pendulum needs a weight or a spring to
keep it ticking. The scientific explanation is that a friction acts to oppose the
motion, so a force is needed to overcome friction.
Newton’s First Law of Motion Answers
1)
2)
Drag Forces Answers
1)
2) At first the only vertical force acting on the skydiver is their weight. As the
skydiver gains speed the air resistance increases until this drag force is equal to
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
Mamun Sir - 4 -
Mamun sir
1.00
0.50
0 0 0.5 1.0 1.5 2.0
1/mass /kg –1
Acc
ele
ratio
n a
/m s–2
1.20
1.00
0.80
0.60
0.40
0.20
0 0 0.1 0.2 0.3 0.4 0.5 0.6
Force F/N
Acc
ele
ratio
n a
/m s–2
the weight and the skydiver reaches a constant terminal velocity. When the
parachute is opened the air resistance increases hugely so there is a net force
upwards. This slows the skydiver down until once again the air resistance
balances the skydiver’s weight and the skydiver reaches a new, slower terminal
velocity.
Newton’s Second Law of Motion Answers
1)
a)
b)
c) For part a the gradient is 2.0 m s–2 N–1.
For part b the gradient is 0.5 m s–2 kg.
d) Acceleration is proportional to the applied force for constant mass, and also to the
reciprocal of mass for a constant force (it is inversely proportional to the mass).
2) Mass of locomotive (m) = 70 tonnes = 70 000 kg
Rate of acceleration of locomotive (a) = 1 m s–2
.
Force exerted by locomotive (F) = 70 000 kg × 1 m s–2 = 70 000 N
3)
a) Mass of woman = 60 kg
Acceleration = 14 m s–2
Time = 0.15 seconds
v = u + at (equation 1)
14 = 0 + a × 0.15
a = 14/0.15 = 93.3
Force = 60 kg × 93.3 m s–1 = 5600 N
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
Mamun Sir - 5 -
Mamun sir
b) Weight = 9.81 × 60 kg = 588.6 N. The force acting on the woman is
approximately 9.5 times as large as her weight.
Inertia, Mass and Weight Answers
1) When the bus accelerates, if the person is to accelerate with it a force must be applied. For
someone seated this comes from the reaction of the seat, but for someone standing it must
come from friction with the floor. This can result in the person being thrown forward or
backward, as they experience the force as if their feet were being pulled out from under
them, while their body remains in its original position.
2) 19.6 m
3) 1.6 N kg–1
4)
a) 6.9 s
b) Yes. Sound would only take 0.69 s to reach the ground.
Newton’s Third Law of Motion Answers
1)
a)
b)
Statics Answers
1) 30 N
Projectiles Answers
1)
a) 1.5 s (2 s.f.)
b) 19.9 m
c) 2.87 m
2) 2.4 m s–1 (2 s.f.)
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
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Mamun sir
3) 93.4 m s–1
4)
a) 6.4 s
b) 570 m
c) 110 m s–1
The Concept of Energy Answers
1) Energy is not lost, but is transferred to other forms such as heat and sound.
2) This method does not actually ‘save’ energy, but simply allows energy generated
while there is less demand to be used to provide a source of energy when demand is
high. The energy is stored as gravitational potential energy of the water in the higher
reservoir.
Energy Transformations Answers
1)
a) Boiling water in a kettle.
b) Putting a can of paint on a shelf; stretching a spring.
2) 96 J (2 s.f.)
Energy and Efficiency Answers
1)
a) 4.1 m s–1
b) 0.85 m
2)
a) 22.6 J
b) 14.4 m
c) 15.8 m s–1
Power Answers
1) 2.2 kW
2) 2710 kW
3) 27.5 kW
HSW The Mechanics of Hockey Answers
21 N
1.13 s
Eureka Answers
1) 915 kg m–3
2)
a) 0.82 g cm−3
b) 820 kg m−3
Edexcel AS Physics Advanced Institution of Physics 1.1.1 Describing Motion
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streamline flow
turbulent flow
3) Suitable estimates. For a room 3 m × 8 m × 8 m the mass of air would be 192 kg.
4) 0.54 N
5) The line for fresh water is higher on the hull because fresh water is less dense than salt
water. For a certain load, a ship will sink its lowest in fresh water, so this line needs to