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Mastery Practice 2.1 pg 27
1. AB
Motion : Constant acceleration
Displacement : 2.4 + 4.0 + 5.6 = 12.0 cm
Average velocity : 12.0 cm = 40 cm s-1
3 x 0.1 s
Acceleration : u = 2.4 = 24 cm s-1
0.1
v = 5.6 = 56 cm s-1
0.1
A = 56 24 = 160 cm s-2
2 x 0.1
BC
Motion : Constant velocity
Displacement : 7.2 x 3 = 21.6 cm
Average velocity : 21.6 cm = 72 cm s-1
3 x 0.1 s
Acceleration : 0
_____________________________________________
2. u = 30 m s-1 , v= 0 , t = 6 s, s = ?
Using s = ( u + v)t
s = (30 + 0)6
= 90 m
The obstacle is 10 m from the car.
Mastery
1. Pave the soft ground with wooden planks of large
surface area.
This will increase the area of contact.
When the area of contact increases, pressure
decreases.
Hence pressure on the ground is reduced while the
wheelbarrow is pushed over the planks.
Mastery Practice 3.2 pg 84
1 (a) The tea level is high above the tap. The tea exert
a high pressure at the tap.
(b) When there is very little tea, the pressure exerted
on the tap by the tea is very small. If the
container is tilted, the tea level above the tap
increases and hence the liquid pressure
increases.
2 (a) Pressure = h g
= 8000 x 1025 x 10
= 8.2 x 107 Pa
(b) Force = Pressure x area
= 8.2 x 107 x 0.16
= 1.312 x 107 N
Challenge Yourself pg 85
1.
(i) Water flows in from the top.
(ii) The water column above the sprinkler exerts
pressure to the sprinkles.
(iii) The pressure forces the water sprinkling
out .
(iv) The higher the water column, the bigger the
pressure.
2.1 ANALYSING LINEAR MOTION
2.1
2.2
3.2
3.2
2.2 UNDERSTANDING PRESSURE
IN LIQUID
Water from tap
sprinkler
3.2
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Mastery Practice 3.3 pg 91
1. (i) The air molecules in the tyre are in random
motion.
(ii) They collide all over the inner surface of the
tyre and among themselves.
(iii) The collisions result in a change of momentum
and produce force.
(iv) The force hence creates the gas pressure.
2. Pressure of gas X
= Atmospheric pressure + 10 cm Hg
= 76 cm Hg +10 cm Hg
= 86 cm Hg
Challenge Yourself pg 91
1. (a) (i) Air inside the hose is being pumped out.
(ii) Pressure inside the hose is reduced.
(iii) Atmospheric pressure outside is higher.
(iv) Atmospheric pressure then forces the
water to rise up in the hose and flow out
of the well.
(b) Atmospheric pressure can only support a
maximum water height of 10m. If the water
level drops until it is more than 10 m below
the well, water cannot rise to reach the top of
the hose. Hence water cannot be pumped out
anymore.
Mastery Practice 3.4 pg 95
1. (a) Pascals Principle states that pressure applied to
an enclosed liquid is transmitted equally to
every part of the liquid.
(b)2 1
2 1
W W
A A= therefore
1 1
2 2
A W
A W=
=2
2
1
3W
W
=1
3
A1 : A2 = 1 : 3
2. Pressure at the slave cylinder
=Pressure at the master cylinder
= 2100
(0.4)
=1625
N cm
Force at each slave cylinder = pressure x area
=625
( )(1.6)2
= 1600 N
Challenge Yourself pg 96
1. (a) (i) An input force is applied to compress the
air.
(ii) The pressure exerted is transmitted to the
output piston.
(iii) The output piston has a larger cross-
sectional area.
(iv) Thus a very large output force is
produced at the output piston.
(b) Open the release valve to let out the
compressed air.
Oil will flow back to the oil reservoir and
platform will be lowered down.
3.3 UNDERSTANDING GAS
PRESSURE & ATMOSPHERIC
PRESSUR
3.3
3.3
3.4 APPLYING
PACALS PRINCIPLE
3.4
3.4
