Questions
SECTION A: (25 MARKS)
Answer all questions in this section.
 A partially inflated balloon at sea level becomes fully inflated at higher altitudes. Explain this
observation (2mks)  Xcm^{3} of substance A of density 800kgm^{3} is mixed with 100cm^{3} of water of density 1000kgm^{3}. The
Density of the mixture is 960kgm^{3}. Determine the value of x. (3mks)  The figure below shows part of a vernier caliper when the jaws are closed without an object between the jaws
 What is the value of the zero error of the callipers? (1mk)
 A student used the same vernier calipers in (a) above to measure the diameter of a test tube whose actual diameter is 2.15cm. what was the reading shown by the calipers. (1mk)
 The figure below shows a bimetallic wheel whose diameter is not affected by changes in temperature. Briefly explain how the diameter of the wheel remain unchanged as the temperature increases (3mks)
 A uniform meter rule of mass 40g is pivoted at the 60cm mark and held horizontally with a vertical String as shown below. Determine the tension in the string. (3mks)
 The figure below shows a pith ball being lifted into the funnel by blowing air into the funnel.
Explain this observation (2mks)  Explain why a hole in a ship near the bottom is more dangerous than the one near the top (1mk)
 A bullet of mass log travelling at a speed of 400ms^{1} hits a tree trunk, it penetrates the tree trunk and stops inside the trunk after 4 cm.
 Calculate the average resistance force offered by the trunk to the bullet. (3mks)
 State the energy changes that takes place. (1mk)
 State the effect on motion of smoke particles when the temperature inside the smoke cell is lowered. (1mk)

 Explain the washing effect of soap. (1mk)
 State one way of making surface tension of water stronger. (1mk)
SECTION B (55 MKS)
Answer all questions in this section

 A modern car has a strengthened passenger cage but with front and back regions which can collapse in a crash. Explain how this collapsible regions should reduce injury in a car crash. (3mks)
 An object of mass 150kg moving at 20m/s collides with a stationery object of mass 90kg. They couple after collision. Determine;
 The total momentum before collision (2mks)
 Total momentum after collision (1mk)
 Their common velocity after collision (2mks)

 Distinguish between velocity and speed. (1mk)
 The velocity time graph in the figure below illustrates the motion of a ball which has been projected vertically upwards from the surface of a planet. The weight of the ball on earth is 30N
Determine the weight of a ball on the planet (3mks)  The figure below shows a section of a tape from a ten tick timer whose frequency is 50 Hz.
Calculate; The average velocity of the trolley between points
 WX (2mks)
 YZ (2mks)
 Find the acceleration of the trolley. (2mks)
 The average velocity of the trolley between points

 State Hooke’s Law. (1mk)
 The figure below shows the variation force with extension for a steel coil spring.
 On the same axes, sketch the variation of force with extension for a wire form which the spring is made. (1mk)
Explain the difference between the two lines drawn (2mks)
 On the same axes, sketch the variation of force with extension for a wire form which the spring is made. (1mk)
 A stone of mass 5g is released from a catapult. The catapult is stretched by 10cm. If the constant of elasticity is 100N/cm. Calculate;
 The horizontal velocity with which the stone is released (3mks)
 Sketch a graph of horizontal velocity against time from the time the stone is released to when it reaches the ground. (1mk)
 The following results were obtained in a experiment to verify Hooke’s law when a spring was extended by hanging various loads on it.
Load(N) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 Length of spring in cm 10.00 11.50 13.00 14.50 16.00 18.00 24.00 Extension 0.00  Complete the table for the extension e above. (1mk)
 Plot a graph of load (yaxis) against extension (3mks)
 From the graph determine the springs constant. (2mks)
 Calculate the energy stored when the spring is stretched to 16 cm. (3mks)

 State two advantages of mercury over alcohol as a thermometric liquid. (2 mks)
 When making the fixed points on a thermometer it is observed that at 00c the mercury thread is of length 2 cm and 8 cm at 1000c. What temperature correspond to a length of 6 cm. (3 mks)

 The figure below shows how volume of a given mass of water varies with temperature.
On the graph provided, sketch a graph of density against Temperature.  State and explain one effect of the anomalous expansion of water. (2 mks)
 The figure below shows how volume of a given mass of water varies with temperature.
 The figure below shows a bimetallic strip made of copper and iron at room temperature.
If copper expands more than iron. Identify A and B if the bimetallic strip is placed in a refrigerator whose temperature is 70^{0}c. (1 mk)
A ……………………………………………………………….
B ………………………………………………………………..
 The figure below shows a system used to draw water from a well. An effort is applied on the handlewhich turns on a radius of 70 cm. As the handle turns a rope is wound on the drum of diameter 28 cm, thus raising a bucket of water from a well.
 If an effort of 10N is needed to lift a bucket full of water of mass 4 kg.
Calculate: The energy gained by the water and bucket when the drum turns through one revolution. (3 mks)
 The work done by the effort during this one revolution. (3 mks)
 Calculate the velocity ratio of the system. (1 mk)
 Calculate the efficiency of the system. (3 mks)
 If an effort of 10N is needed to lift a bucket full of water of mass 4 kg.
Marking Scheme
 Atmospheric pressure reduces, Air pressure in the balloon becomes more than atmospheric pressure.
 Total mass = m1 + m2
M1 = 0.8x x m2= 1 x 100
= 0.8xg = 100g
Total = (0.8x + 100)g
Total volume = (x + 100) cm^{3}
Density of mixture =
0.8x + 100 = 0.96
x + 100
x = 4 = 25 cm^{3}
0.16 
 – 0.02 cm
 2.15 + 0.02
= 2.17 cm
 – 0.02 cm
 As the temperature increases the bimetallic strip expands with brass expanding more than iron , thus increases the curvature reducing the size of the gaps.
 T x 40 = 0.4 x 10
T = 0.4 x 10
40
= 0.1N  The velocity of air in narrow section is higher than in the wider section. The pressure in the narrow section is therefore lower than in the wider section. The higher pressure in the wider section raises the ball up
 Pressure at the bottom of the ship is greater than pressure near the top, more water will enter into the ship at the bottom then at the top.

 Work done = F x d = Initial K.E.
½ x ^{10}/_{1000} x (400)^{2} = F x ^{4}/_{100}
F = 20000 N  K.E. changes to heat and sound.
 Work done = F x d = Initial K.E.
 They slow down/their kinetic energy is reduced.

 Detergent lowers the surface tension of water making it stick to dirt and remove it. /Penetrate the space between dirt and fabric.
 – Lower the temperature
 Remove impurities.

 On collision they collapse increasing the time of impact . This reduces the rate of change of
momentum of the passengers, thus the force of the impact reduced. 
 Total momentum = m_{1}u_{1} + m_{2}u_{2} before collision
= 150 x 20 + 90 x 0
= 3000 kgms^{1}  Total momentum = 3000kgms^{1} after collision
 Momentum before collision = momentum after collision
(150 + 90)v = 3000
V = 3000
240
V = 12.5 ms^{1}
 Total momentum = m_{1}u_{1} + m_{2}u_{2} before collision
 On collision they collapse increasing the time of impact . This reduces the rate of change of

 Velocity is a vector quantity while speed is a scalar quantity.
 Acceleration on planet = 5 ms^{2}
3
Weight of object w = mg
30 = m x 1.0
M = 3 kg
Weight on planet w_{p} = mg_{p}
= 3 x 5/3
SN.  Velocity between w and x = 10
0.02 x 10
= 10 = 50 cm/s
02
Velocity between Y and z
= 30 = 30
0.02 x 10 02
= 150 cm/s
Acceleration = 150 – 50
20 x 0.02
= 150 – 50
0.4
= 100 = 250 cms^{2}
0.4
= 25ms^{2}

 For an elastic material, the extension is directly proportional to the force producing it provided the
elastic limit is not exceeded. 

 The wire has a greater constant of elasticity than coil of the same material hence greater gradient.

 K.E of stone = elastic potential energy of catapult
½ mv^{2} = ½ ke^{2}
½ x 5 x v2 = ½ 100N
1000 1/1000
M (10/100)^{2} 5v^{2} = ½ x 100 x 100 x 10 x 10
2000 100 x 100
5v^{2} = 100
100
v2= 100 x 1000 = 20,000
5
v = 141 ms^{2}
^{} 

Load(N) 0.00 1.00 2.00 3.00 4.00 5.00 6.00 Length of spring in cm 10.00 11.50 13.00 14.50 16.00 18.00 24.00 Extension 0.00 1.50 3.50 6.00 8.00 14.00  Suitable axes labelled
All points correct
Suitable line  Springs constant K = F
e
Use students graph
Correct units  Energy stored when the length is stretched by 16 cm
Area under the graph
Or E = ½ ke^{2}
Use k from graph and e = 16 cm.
K must be correct.
Correct substitution
Answer correct unit

 For an elastic material, the extension is directly proportional to the force producing it provided the


 Good thermal condudivity
 Expands regularly
 Wide range of temperature
 Clearly visible
 Range of length = (8 – 2) = 6 cm
Temperature = 1/6 x 100 x 4
= 66.67^{0}C 

 Freezing of lakes and ponds ice formed at 0^{0}c is less dense than water thus floats, water at 40c is denser thus remains at the bottom of the lake
Weathering of rocks. when water in the cracks of a rock freeze it expands and breaks the rocks.
Water pipes
Water pipes burst when water flowing through them freezes and expands.
Any one correct.

 A iron
B copper



 Energy gain = work done
= Load x distance
= 40 x 2 x 22/7 x 14/100
= 35.20 J  W.D. = effort x distance
= 10 x 2 x 22/7 x 70/100
= 44 J
 Energy gain = work done
 V.R. = R/r = 70/14 = 5
 A = Work output x 100 OR
Work input = M.A x 100
V.R.
= 35.20 x 100 4 x 100
44 5
= 80% = 80%

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