# Physics Paper 1 Questions and Answers - Bunamfan Cluster Pre Mock Exam 2022

INSTRUCTIONS TO THE CANDIDATES:

• Answer all the questions both in section A and B in the spaces provided below each question
• All workings must be clearly shown; marks may be awarded for correct steps even if the answers are wrong.
• Let g = 10m/s2

Question 1

SECTION A 25 MARKS: Attempt all the questions in this section

1. The figure below shows a piece of metal stuck in a hollow glass pipe. .Explain how temperature change may be used to separate them (2mark)
2. Form four students were playing football game during which the ball got deflated. Explain what happened to its density (2marks)
3. Micrometer screw gauge A has a zero error of – x mm. Micrometer screw gauge B has a zero error of x mm When used to measure the diameter of a tube the difference between their readings is 0.04mm. If the actual diameter of the tube is 5.56mm determinex hence state the reading of micrometer screw gauge A(3 marks)
4. A car of mass 1000kg travelling at a constant velocity of 40m/s collides with a stationary metal block of mass 800kg. The impact takes 3s before the two move together. Determine the impulsive force (3marks)
5. The figure below shows a drop of water about to fall from a pipette and after falling. Explain why the shapes of the drop are different (2 marks)
6. Figure shows a liquid manometer. The gas pressure is 755mmHg and that of the surround is 760mmHg. The height h is 80mm. Determine the density of the liquid. (Take density of mercury = 13600kgm-3 and g =10Nkg-1) (3 marks)
7. A student balances a V - shaped uniform wire on a tight string as shown in A and B. With reason state the one which is easier to do (2marks)
8. The figure below shows a Bunsen burner. Explain how air is drawn into the burner when the gas tap is opened. (2marks)
9. The figure shows a uniform metal bar of length 10m and weight W = 200N held at equilibrium by a light chain fixed at the cog and tethered on the floor using a light chain. Determine the tension of the chain (3marks)
10. A student set up the apparatus as shown below. The boiling tube was heated in the middle as shown

1. State the role of the lead shot in the experiment (1mark)
2. With reason, state the wax that will melt first (2marks)

SECTION B 55 MARKS: Attempt all the questions in this section

1. Marble A is projected horizontally from the top of a cliff at a velocity of 50m/s. The height of the cliff from its foot is 31.25m. At the same time another marble B is projected horizontally from the same point. The figure below shows the trajectories taken by the marbles.

Determine
1. The distance of marble A from the foot of the cliff as it hits the ground (3marks)
2. Vertical velocity of marble A as it hits the ground (2marks)
3. Horizontal velocity of marble B as it hits the ground (2marks)
4. The shortest distance between the marbles upon hitting the ground (2marks)
2. The figure below shows two identical light springsand other apparatus used in an experiment

After the data was collected the following graph was obtained

1. State two measurements taken in the experiment (2mark)
2. Explain how the measurements can be used to come up with the graph (2marks)
3. Explain the graph in sections
1. AB (2marks)
2. CD (2marks)
4. Determine the spring constant of each spring (3marks)
5. Determine the work done in section CD (2marks)
6. On the same axes sketch the graph expected when the experiment is repeated using one of the springs only (1mark)
3. The figure below shows an inclined plane on which a trolley of mass 30kg is pulled up a slope by a force of 100N, parallel to the slope. The trolley moves so that its centre of mass travels from points A to B.

1. Determined the work done on the trolley against the gravitational force in moving from A to B. (2 marks)
2. Determine the work done by the force in moving the trolley from A to B. (3 marks)
3. Determine the percentage of the work input that goes to waste (3 marks)
4. Determine the frictional force. (1 mark)
5. Determine the mechanical advantage of the system. (1 mark)
6. Find the velocity ratio (1 mark)
4.
1. The figure below shows a set-up that can be used to determine the specific heat capacity of a metal block.

1. Other than temperature and current, state two measurements that should be taken in the experiment to determine the specific heat capacity of the block.(2marks)
2. Describe how the method can be used to determine the specific heat capacity of the metal block. (3marks)
3. State the purpose of oil in the set-up. (1mark)
2. A well lagged copper can together with a copper stirrer of total heat capacity 60JK-1 contains 200g of water at 200C. Dry steam at 1000C is passed in while the water is stirred until the content reach a temperature of 500C. Determine the mass of condensed steam. (Specific latent heat of vaporization of water is 2.26 X 106 J/kg and specific heat capacity of water is 4200 J/kgK)(4marks)
5.
1. A uniform metal strip is 3.0cm wide 0.6cm thick and 100cm long. The density of the metal is 2.7g/cm3.
1. Determine the weight of the metal strip. (2marks)
The strip is used to support two masses in equilibrium by applying force F as shown below.
2. Determine the value of F (3 marks)
3. Determine reaction R due to the pivot (2 marks)
2. The Figure belowshows a set up that may be used to verify a gas law.
1. State the law being verified (1mark)
2. State two functions of the concentrated Sulphuric acid in the experiment (2marks)?
3. State one assumption in the experiment (1mark)

Marking Scheme

1. Cool the joint.
The metal contracts at a higher rate than glass hence the separation.
2. The density increases since the volume reduce due to the exit of air and ρ=m/v
3. A= R-E
R=A+E
RA= A+X
RB= A+X
RB - R
(A+X)-(A-X)= 0.04
A+X - A+X= 0.04
2X= 0.04
X= 0.02
RA= A-X
5.56 - 0.02
= 5.54mm
4. M1V1 + M2V2 = (M1 + M2)
1000 x 40 + 0 = 1800
V= 22.22m/s
F= Mv - Mv = 1000 x 22.22 - 1000 x 40
6                   3
OR
800 x 22.22 - 0 = +5925N
3
5. Adhessive force hold part of the drop against the pipette
Cohesion force holds the water molecules together as a sphere.
6. Atm = GP + CP
760 = 755 + CP
CP = 760 - 755 = 55mmHg
mervury      liquid
ρgh            ρgh
13600 x `0 x 5/100 = ρ x 10 x 80/100
680 = 0.8ρ
7. in A the cog is lower than the point of support unlike B
8. Gas passes the nozzle at a high speed creates a region of low pressure at A. Atm pushes air through the hole
9. Sin 30 =d/5
d= 5Sin 30= 2.5m
F1d1 = F2d2
F1 x 2.
10.
1. Hold the wax to prevent it from floating
2. Floating wax, heat reaches it by convection of heat (hot water rises)
11.
1. h= 1/2gt2
31.25 = 1/2 x 10x t2
t= 2.55
R= uxt
= 50 x 2.5 =125
2. V=gt
v= 10 x 2.5
Ux= 8m/s
3. R= Uxt
20= Ux x 2.5
Ux = 8m/s
4. √(1252 + 202) = 126.6m
12.
1.
• Mass on the springs
• Length of the compressing springs
2.
• Gradually add the masses each time getting the corresponding length x
• Draw the graph of x against force (x- axis)
3.
1.
• Length x reduces as the weight in creases due to compression
2.
• NO change in length
• Since all the turns had come into contact with each other/ one another.
4. Combined spring contact
Grad =    19 - 09       = 31.25
0.32 - 0
KT = 1/grod = 0.032N/cm
each spring
K= 0.032/2
= 0.016N/cm
5. Area under graph = work
length = 9  x 0.04 = 0.36N
Width = 8/100 = 0.08mwork = 0.36 x 0.08
= 0.0288J
13.
1.
1. W= mgh
30 x 10 x 10 = 3000J
2. Cos 75 = 10/AB
AB= 10/Cos75
AB= 38.64m
W= F.d
100 x 38.86
= 3864J
3. Wastage = 3864 - 3000 = 864J
864   x 100 = 22.36%
3864
4. MA= L/E = 300/100 = 3
5. VR= Ed/Cx = 38.64/10 = 3.864
14.
1.
1.
• Time of heating
• P.d accross the heating coil
• Mass of the copper block.
2.
• Determine the mass of the block.
• Record the initial temp.
• Put on the switch for some time.
• Note the voltmeter reading
• Electrical energy = Heat . Record final temp
VIE = McΔT
C=   VIt
MDr
3.
• To create thermal contact between the thermometer/ heater and the block
2. Heat lost = heat gained.
MW + MCDT = HC x DT + MCDT
M x 2.26 x 106 + M + 4200(50) = 60 x 30 + 0.2 x 4200 x 30
2,470,000 = 1800 + 25200
2, 470, 000 = 27, 000
M= 0.01093kg or 10.93g
15.
1.
1. m= ρV
= 2.7 x 3 x 0.6 x 100 = 486g
W= mg
= 4.86N
2. moment anout the pivot
50 x 4.86 + 0.15 x 60 x 0.25 x 80 = 100F
243 + 9 + 20 = 100F
272 = 100F
F= 2.72N
3. Upward force = Downward force
2.72 + R = 4.86 + 0.15 + 0.25
2.72 + R = 5.26
R= 5.26 - 2.72
= 2.54 N
2.
1.
• The volume of a fixed mass of a gas is directly proportional to its absolute temperature provided pressure remains constant.
2.
• Server as an index
• Drying agent to keep the air dry
3. The temperature of water in the name as the temeprature od the air in the tube.

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