**2021 Physics paper 1**

**SECTION A**

**Answer all the questions in this section in the spaces provided.**

**1. Figure 1 shows part of the thimble scale of a screw gauge with 50 divisions.**

**On the diagram, draw the sleeve scale to show a reading of 3.87 mm. (1 mark)**

**2. Figure 2 shows a siphon used to empty a tank. In order to start the siphon, state why:**

**a. it must be full of liquid. (1 mark)**

- To overcome atmospheric pressure inside the siphon by expelling trapped air inside the siphon.
**b. end X must be below the level of the liquid in the tank. (1 mark)** - To create pressure difference.
**3. Figure 3(a) shows a horizontal tube containing air trapped by a mercury thread of length 5 cm. The****length of the enclosed air column is 7.5cm. The atmospheric pressure is 76 cmHg.****The tube is then turned vertically with its mouth facing down as shown in Figure 3(b).****a. Determine the length 1 of the air column. (3 marks)**P_{1}V_{1}= P_{2}V_{2}76 x 7.5 = (76 – 5)l

l = 8.03cm (2 dp)

**b. State the reason why the mercury thread did not fall out in Figure 3(b). (1 mark)****4. In a Physics experiment, a student filled a burette with water up to a level of 15 ml. The student ran out 3 drops of water each of volume 2 cm³ from the burette into a beaker. Determine the final reading of the burette. (2 marks)**Initial burette reading = 15ml

Volume of water dropped out = 3 drops x 2cm³

= 6cm³

New burette reading = 15cm³ + 6cm³

= 21cm

**5. State two factors that affect the angular velocity of a body moving in a circular path (marks)**- The instantaneous linear velocity of the moving body
- The radius of the circular path
**6. Figure 4 shows two capillary tubes X and Y of different diameters dipped in mercury.****Complete the diagram to show the meniscus in Y. (1 mark)** **7. In an experiment, a drop of black ink is introduced at the bottom of a container filled with water. It is observed that the water gradually turns black. State the effect on the observation when the experiment is carried out using water at a lower temperature. (1 mark)****8. Figure 5 shows two identical springs arranged side by side and supporting a weight of 50 N.****When the same weight is supported by one of the springs above, it produces an extension of 1 cm.****Determine the effective spring constant of the arrangement in Figure 5. (3 marks)**K = F/e = 50N/1cm

= 50N/cm

KT = 2x 50N/cm

= 100N/cm or 1000N/m

**9. On the axes provided, sketch a graph of density against temperature for water between 0°C and 10 °C. (1 mark)****10. State the reason why a student climbing a hill tends to bend forward. (1 mark)**- To shift the position of the centre of gravity to the front part to maintain equilibrium.
**11. Figure 6 shows a graph of temperature against time for a pure molten substance undergoing cooling.** **Explain what happens to the substance in region BC. (2 marks)**- The substance undergoes change of state from molten to solid without change in temperature.
**12. Figure 7 shows a uniform rod AB 2 m long and of mass 1 kg. It is pivoted 0.5 m from end A and balanced horizontally by a string attached 0.1 m from end B.** **Determine the tension in the string. (take g = 10 Nkg) (2 marks)**Sum of clockwise moments = sum of anti-clockwise moments10N × 0.5m = 1.4m × T

T= 3.57N

**13. Figure 8 shows two pieces of ice A and B trapped using a wire gauze in a large beaker containing water. Heat is supplied at the centre of the base of the beaker as shown. State the reason why B melted earlier than A (1 mark)**- Heated water at the bottom becomes less dense which rises to the top. Hence ice B melts earlier than A.
**14. Figure 9 shows a folded piece of paper. A stream of air is blown underneath the paper.** **Explain why the paper collapsed. (2 marks)**Air blown underneath the paper reduces pressure acting on the paper. Atmospheric pressure acting from top becomes higher. Hence the paper collapses.**SECTION B (55 marks)****Answer all the questions in this section in the spaces provided.****15.a. Figure 10 shows a wooden block of volume 90 cm floating with 1/3 of its body submerged inwater of density 1gcm-3 (g = 10 Nkg-1)****Determine:****i. The weight of the block**weight of block = vpg

1/3 x 90 x 1000 x 10

100000

= 0.3N

Alternatively

W = mg

m = 1/3 x 90 x 1

= 30g

W = 30 x 10

1000

= 0.3N

**ii. The weight of a metal block that can be placed onto the block so that its top surface is on the same level as the water surface.(3 marks)**Volume of remaining part = 90 – 30 = 60cm3 U → W = vpg

= 60/1000000 x 1000 x 10

= 0.6N

w = mg but m = v x p

60 x 1 = 60g

w = 60/1000 x 10

= 0.6N

**b. Figure 11 shows a solid metal suspended in oil using a thread.****i. Other than upthrust, list two other forces acting on the sphere. (2 marks)**- Tension force
- Weight, mg
**ii. The oil is carefully and gradually drawn from the beaker. State the effect on each of the two forces in 15(b)(i). (2 marks)** - Tension force will increase
- Weight, mg, will remain constant
**16.a. Define the term “specific latent heat of fusion” (1 mark)**Quantity of heat required to change a unit mass of the material from solid state to liquid without change in temperature.

**b. Ice of mass 5g at a temperature of -10°C is immersed into 10.5g of hot water at 100°C in a container of negligible heat capacity.****All the ice melts and the final temperature of the mixture is 40°C. Assuming there are no heat losses to the surrounding and taking the specific latent heat of fusion for ice as L (Cwar = 4200 Jkg-K and C = 2100 Jkg ‘K).****Determine the:****i. heat lost by the hot water. (3 marks)**heat lost by the water. (3 marks)

Heat lost by the water = m<subw< sub=””>c<subw< sub=””>Δθ</subw<></subw<>

0.0105 x 4200 x (100 – 40)

= 2646J

**ii. heat gained by ice from -10°C to 0°C. (2 marks)**heat gained by ice from −10°C to 0°C

heat gained by ice upto 0oC = m

_{ice}c_{ice}Δθ**iii. heat required to melt the ice in terms of Lf**(1 mark)heat required to melt the ice in terms of Lf (1 mark)

mLf

0.005Lf

**iv. heat gained by the melted ice. (2 marks)**heat gained by the melted ice. (2 marks)

heat gained by melted ice upto = m

_{ice}c_{ice}Δθ0.005 x 4200 x 40

= 840J

**v. specific latent heat of fusion of ice. (3 marks)**heat lost by hot water= heat gained by ice(-10oC to 0oC) + melting ice + melted ice up to 40oC

2646J = 105J + 0.05Lf + 840J

**17. Figure 12 shows a hydraulic lift system. The radius of the small piston is 5.64 cm while that of the large piston is 14.24 cm. The small piston is operated using a lever. A force of 100N is applied to the lever** **Determine the:****a. pressure exerted by the smaller piston. (5 marks)**w x 0.4 = 100N x 2.0m

w = 500N

P = F/A but A = πr2

22/7 x 5.642

= 99.97cm2

= 99.97 x 10-4m2

P = 500N

99.97 x 10-4m2

=5.0015 x 10

^{4}Pa**b. load that can be lifted. (3 marks)**L = P x A

_{larger piston}5.0015 x 10

^{4}x 22/7 x 14.24^{2}x 10^{-4}m²**c. mechanical advantage of the system. (3 marks)**M.A = Load/Effort

3185.22N/500N

= 6.37

**18.a. A bus moving initially at a velocity of 20ms-1 decelerates uniformly at 2 ms-2****i. Determine the time taken for the bus to come to a stop. (3 marks)**t =v – u OVER a

= 0 – 20 OVER 2

= 10sec

**ii. Sketch the velocity – time graph for the motion of the bus up to the time it stopped. (2 marks)****iii. Use the graph to determine the distance moved by the bus before stopping. (1 mark)**Distance = Area under the curve

1⁄2 x 20 x 10

= 100m

**b. A car of mass 1000 kg travelling at a constant velocity of 40ms-1 collides with a stationary metal block of mass 800 kg. The impact takes 3 seconds before the two move together.****Determine the impulsive force. (4 marks)**m1v1 + m2v2 = v(m1 + m2)

(1000 x 40) + (800 x 0) = v(1000 + 800)

= 22.22m/s

v = u + at

22.22 = 40 + 3a

a = -5.93m/s²

(decelerating)

F = ma

1800 x -5.93

= 10674N

**19.a. State two conditions necessary for a body to be in equilibrium. (2 marks)**- Sum of clockwise moments about a point must be equal to the sum of anti- clockwise moments about the same point.
- For a system of parallel forces in equilibrium, sum of forces in either direction is equal.
**b. Figure 13 shows a non-uniform log of wood AB of length 4m. The log is held horizontally by applying forces of 80N at end A and 120N at end B.** **Determine:****i. the value of R. (1 mark)**the value of R. (1 mark)

R = 80 + 120

= 200N

**ii. the position of the centre of gravity of the log from end B. (3 marks)**Let x be the distance from the pivot to point B

80(4 – x) = 120x

320 = 200x

x = 1.6m

**c. You are provided with a metre rule, a knife edge and a mass m1****i. Describe how the position of the centre of gravity of the metre rule can be determined using the knife edge. (2 marks)**Place the metre rule horizontally on knife edge. The position where it balances on the knife edge is the centre of gravity.

**ii. Using the position of the centre of gravity determined in 19(c)(i) and the mass M,describe how the mass M of the metre rule can be determined. (4 marks)**- Move the knife edge away from the centre of gravity to a new position. Note the distance from the knife edge and the centre of gravity as d1
- Place the mass m1 on one side of the metre rule and adjust it until the rule balances as in 19 c (i). Note the distance from the knife edge and the mass m
_{1}as d_{2}. - Using principle of moment;Md1 = m1d2
M = m1d2/d1

Where M is the mass of the metre rule