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
- 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)
- Form four students were playing football game during which the ball got deflated. Explain what happened to its density (2marks)
- 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)
- 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)
- 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)
- 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)
- 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)
- The figure below shows a Bunsen burner. Explain how air is drawn into the burner when the gas tap is opened. (2marks)
- 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)
- A student set up the apparatus as shown below. The boiling tube was heated in the middle as shown
- State the role of the lead shot in the experiment (1mark)
- With reason, state the wax that will melt first (2marks)
SECTION B 55 MARKS: Attempt all the questions in this section
- 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- The distance of marble A from the foot of the cliff as it hits the ground (3marks)
- Vertical velocity of marble A as it hits the ground (2marks)
- Horizontal velocity of marble B as it hits the ground (2marks)
- The shortest distance between the marbles upon hitting the ground (2marks)
- The figure below shows two identical light springsand other apparatus used in an experiment
After the data was collected the following graph was obtained- State two measurements taken in the experiment (2mark)
- Explain how the measurements can be used to come up with the graph (2marks)
- Explain the graph in sections
- AB (2marks)
- CD (2marks)
- Determine the spring constant of each spring (3marks)
- Determine the work done in section CD (2marks)
- On the same axes sketch the graph expected when the experiment is repeated using one of the springs only (1mark)
- 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.
- Determined the work done on the trolley against the gravitational force in moving from A to B. (2 marks)
- Determine the work done by the force in moving the trolley from A to B. (3 marks)
- Determine the percentage of the work input that goes to waste (3 marks)
- Determine the frictional force. (1 mark)
- Determine the mechanical advantage of the system. (1 mark)
- Find the velocity ratio (1 mark)
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- The figure below shows a set-up that can be used to determine the specific heat capacity of a metal block.
- 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)
- Describe how the method can be used to determine the specific heat capacity of the metal block. (3marks)
- State the purpose of oil in the set-up. (1mark)
- 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)
- The figure below shows a set-up that can be used to determine the specific heat capacity of a metal block.
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- A uniform metal strip is 3.0cm wide 0.6cm thick and 100cm long. The density of the metal is 2.7g/cm3.
- 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. - Determine the value of F (3 marks)
- Determine reaction R due to the pivot (2 marks)
- Determine the weight of the metal strip. (2marks)
- The Figure belowshows a set up that may be used to verify a gas law.
- State the law being verified (1mark)
- State two functions of the concentrated Sulphuric acid in the experiment (2marks)?
- State one assumption in the experiment (1mark)
- A uniform metal strip is 3.0cm wide 0.6cm thick and 100cm long. The density of the metal is 2.7g/cm3.
Marking Scheme
- Cool the joint.
The metal contracts at a higher rate than glass hence the separation. - The density increases since the volume reduce due to the exit of air and ρ=m/v
- A= R-E
R=A+E
RA= A+X
RB= A+X
RB - RA
(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 - 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 - Adhessive force hold part of the drop against the pipette
Cohesion force holds the water molecules together as a sphere. - 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 - in A the cog is lower than the point of support unlike B
- Gas passes the nozzle at a high speed creates a region of low pressure at A. Atm pushes air through the hole
- Sin 30 =d/5
d= 5Sin 30= 2.5m
F1d1 = F2d2
F1 x 2. -
- Hold the wax to prevent it from floating
- Floating wax, heat reaches it by convection of heat (hot water rises)
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- h= 1/2gt2
31.25 = 1/2 x 10x t2
t= 2.55
R= uxt
= 50 x 2.5 =125 - V=gt
v= 10 x 2.5
Ux= 8m/s - R= Uxt
20= Ux x 2.5
Ux = 8m/s - √(1252 + 202) = 126.6m
- h= 1/2gt2
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- Mass on the springs
- Length of the compressing springs
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- Gradually add the masses each time getting the corresponding length x
- Draw the graph of x against force (x- axis)
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- Length x reduces as the weight in creases due to compression
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- NO change in length
- Since all the turns had come into contact with each other/ one another.
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- 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 - Area under graph = work
length = 9 x 0.04 = 0.36N
Width = 8/100 = 0.08mwork = 0.36 x 0.08
= 0.0288J
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- W= mgh
30 x 10 x 10 = 3000J - Cos 75 = 10/AB
AB= 10/Cos75
AB= 38.64m
W= F.d
100 x 38.86
= 3864J - Wastage = 3864 - 3000 = 864J
864 x 100 = 22.36%
3864 - MA= L/E = 300/100 = 3
- VR= Ed/Cx = 38.64/10 = 3.864
- W= mgh
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- Time of heating
- P.d accross the heating coil
- Mass of the copper block.
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- 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
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- To create thermal contact between the thermometer/ heater and the block
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- 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
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- m= ρV
= 2.7 x 3 x 0.6 x 100 = 486g
W= mg
= 4.86N - 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 - 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
- m= ρV
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- The volume of a fixed mass of a gas is directly proportional to its absolute temperature provided pressure remains constant.
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- Server as an index
- Drying agent to keep the air dry
- The temperature of water in the name as the temeprature od the air in the tube.
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