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Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 5

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TuitionGoWhere Practice Paper - Combined Science Secondary 3

TuitionGoWhere Exam Practice (AI)
Subject: Combined Science (Physics Component)
Level: Secondary 3
Paper: SA2 Practice Paper (Version 5 of 5)
Duration: 1 hour 15 minutes
Total Marks: 65

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates:

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.
Take $g = 10 \, \text{m/s}^2$ and density of water $= 1000 \, \text{kg/m}^3$ unless otherwise stated.

Section A: Multiple Choice & Short Structured Questions (20 Marks)

1. A student measures the length of a metal rod using a ruler. The rod starts at the 2.0 cm mark and ends at the 14.5 cm mark. What is the length of the rod? A. 12.5 cm B. 14.5 cm C. 16.5 cm D. 12.50 cm

Answer: ____________________ [1]

2. Which of the following is a vector quantity? A. Speed B. Distance C. Mass D. Displacement

Answer: ____________________ [1]

3. A car travels at a constant speed of $20 \, \text{m/s}$ for 10 seconds. It then decelerates uniformly to rest in 5 seconds. Calculate the total distance travelled by the car.

Answer space

Answer: ____________________ m [3]

4. State Newton’s First Law of Motion.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

5. A box of mass 50 kg is pushed across a horizontal floor with a constant force of 200 N. The frictional force acting on the box is 50 N. Calculate the acceleration of the box.

Answer space

Answer: ____________________ $\text{m/s}^2$ [2]

6. Define the term density.

Answer: _________________________________________________________________________ [1]

7. A block of wood has a volume of $0.02 \, \text{m}^3$ and a mass of 10 kg. Calculate its density.

Answer space

Answer: ____________________ $\text{kg/m}^3$ [2]

8. A force of 10 N is applied perpendicular to a door at a distance of 0.8 m from the hinge. Calculate the moment of this force about the hinge.

Answer space

Answer: ____________________ Nm [2]

9. Explain why it is easier to open a door by pushing near the handle rather than near the hinge.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

10. A diver is swimming at a depth of 10 m in sea water (density $1030 \, \text{kg/m}^3$ ). Calculate the pressure due to the water at this depth. ( $g = 10 \, \text{m/s}^2$ )

Answer space

Answer: ____________________ Pa [2]

Section B: Structured Questions (30 Marks)

11. Fig. 11.1 shows a speed-time graph for a cyclist.

Image pending generation for this question.

(a) Describe the motion of the cyclist during the first 5 seconds.

Answer space

Answer: _________________________________________________________________________ [1]

(b) Calculate the acceleration of the cyclist during the first 5 seconds.

Answer space

Answer: ____________________ $\text{m/s}^2$ [2]

Answer space

Answer: ____________________ m [3]

12. A student investigates the principle of moments using a uniform metre rule pivoted at the 50 cm mark. A weight of 2.0 N is hung at the 20 cm mark. A weight of $W$ is hung at the 80 cm mark to balance the rule.

(a) State the Principle of Moments.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

(b) Calculate the value of weight $W$ .

Answer space

Answer: ____________________ N [3]

(c) The pivot is moved to the 40 cm mark. The 2.0 N weight remains at the 20 cm mark. Explain whether the weight $W$ needs to be moved closer to or further from the new pivot to maintain equilibrium, assuming $W$ remains at the 80 cm mark position relative to the rule end (i.e., its distance from the new pivot changes). Note: Assume W is adjusted to a new position to balance. Actually, let's rephrase for clarity: The pivot is moved to the 40 cm mark. The 2.0 N weight is still at the 20 cm mark. The weight $W$ is moved to a new position $X$ to balance the rule. If $W$ is increased, does the distance of $W$ from the pivot need to increase or decrease to maintain balance? Explain.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

13. Fig. 13.1 shows a hydraulic press. Piston A has an area of $0.01 \, \text{m}^2$ . Piston B has an area of $0.5 \, \text{m}^2$ . A force of 100 N is applied to Piston A.

(a) Calculate the pressure transmitted through the liquid.

Answer space

Answer: ____________________ Pa [2]

(b) Calculate the force exerted by Piston B.

Answer space

Answer: ____________________ N [2]

Answer space

Answer: _________________________________________________________________________ [1]

14. A crane lifts a load of mass 500 kg to a height of 20 m in 10 seconds.

(a) Calculate the work done by the crane. ( $g = 10 \, \text{m/s}^2$ )

Answer space

Answer: ____________________ J [3]

(b) Calculate the power developed by the crane.

Answer space

Answer: ____________________ W [2]

Answer space

Answer: ____________________ J [2]

15. A gas is trapped in a cylinder by a movable piston. The volume of the gas is $0.05 \, \text{m}^3$ and the pressure is $200,000 \, \text{Pa}$ . The piston is pushed in until the volume is $0.02 \, \text{m}^3$ . The temperature remains constant.

(a) State the law that applies to this situation.

Answer space

Answer: _________________________________________________________________________ [1]

(b) Calculate the new pressure of the gas.

Answer space

Answer: ____________________ Pa [3]

Section C: Free Response & Application (15 Marks)

16. A student investigates the efficiency of an inclined plane. She pulls a block of weight 10 N up a slope of length 2.0 m and height 0.5 m. The force applied parallel to the slope is 4.0 N.

(a) Calculate the useful work output (gain in gravitational potential energy).

Answer space

Answer: ____________________ J [2]

(b) Calculate the total work input.

Answer space

Answer: ____________________ J [2]

Answer space

Answer: ____________________ % [2]

(d) Suggest one reason why the efficiency is less than 100%.

Answer space

Answer: _________________________________________________________________________ [1]

17. Fig. 17.1 shows a simple mercury barometer.

(a) Explain what supports the column of mercury in the tube.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

(b) What is the approximate value of atmospheric pressure in Pa?

Answer space

Answer: ____________________ Pa [1]

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

18. A ball of mass 0.5 kg is dropped from a height of 20 m.

(a) Calculate the gravitational potential energy of the ball at the start.

Answer space

Answer: ____________________ J [2]

(b) Assuming no air resistance, calculate the speed of the ball just before it hits the ground.

Answer space

Answer: ____________________ m/s [3]

19. Two forces act on an object: 3 N to the East and 4 N to the North.

(a) Draw a vector diagram to show the resultant force.

Answer space

[Space for diagram] [2]

(b) Calculate the magnitude of the resultant force.

Answer space

Answer: ____________________ N [2]

20. A submarine is submerged in water.

(a) Explain why the pressure on the submarine increases as it dives deeper.

Answer space

Answer: _________________________________________________________________________ ___________________________________________________________________________________ [2]

(b) The submarine has a window of area $0.1 \, \text{m}^2$ . At a certain depth, the pressure difference between the outside and inside is $500,000 \, \text{Pa}$ . Calculate the force acting on the window.

Answer space

Answer: ____________________ N [2]

End of Paper

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Answers

TuitionGoWhere Practice Paper - Combined Science Secondary 3

Answer Key & Marking Scheme (Version 5)

Subject: Combined Science (Physics Component)
Level: Secondary 3
Paper: SA2 Practice Paper

Section A: Multiple Choice & Short Structured Questions

1. A
Reasoning: Length = End reading - Start reading = $14.5 - 2.0 = 12.5$ cm. [1]

2. D
Reasoning: Displacement has both magnitude and direction. Speed, distance, and mass are scalars. [1]

3. 250 m
Working:
Distance 1 (constant speed) = $20 \times 10 = 200$ m.
Distance 2 (deceleration) = Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 20 = 50$ m.
Total distance = $200 + 50 = 250$ m.
[Marks: 1 for dist 1, 1 for dist 2, 1 for total]

4. An object remains at rest or continues to move at a constant velocity in a straight line unless acted upon by a resultant external force.
[Marks: 1 for "rest or constant velocity", 1 for "unless acted on by resultant force"]

5. 3.0 $\text{m/s}^2$
Working:
Resultant Force $F = 200 - 50 = 150$ N.
$F = ma \Rightarrow 150 = 50 \times a$ .
$a = 150 / 50 = 3.0 \, \text{m/s}^2$ .
[Marks: 1 for resultant force, 1 for correct answer]

6. Mass per unit volume.
[Marks: 1]

7. 500 $\text{kg/m}^3$
Working:
$\text{Density} = \text{Mass} / \text{Volume} = 10 / 0.02 = 500 \, \text{kg/m}^3$ .
[Marks: 1 for formula/substitution, 1 for answer]

8. 8.0 Nm
Working:
$\text{Moment} = \text{Force} \times \text{perpendicular distance} = 10 \times 0.8 = 8.0$ Nm.
[Marks: 1 for formula/substitution, 1 for answer]

9. Pushing near the handle increases the perpendicular distance from the pivot (hinge). Since $\text{Moment} = \text{Force} \times \text{Distance}$ , a larger distance produces a larger moment for the same force, making it easier to rotate the door.
[Marks: 1 for mentioning increased distance, 1 for linking to larger moment/easier rotation]

10. 103,000 Pa
Working:
$P = h \rho g = 10 \times 1030 \times 10 = 103,000$ Pa.
[Marks: 1 for formula/substitution, 1 for answer]

Section B: Structured Questions

11.
(a) The cyclist accelerates uniformly (or constant acceleration). [1]
(b) $a = \Delta v / \Delta t = (10 - 0) / 5 = 2.0 \, \text{m/s}^2$ . [2]
(c) Total distance = Area under graph.
Area 1 (triangle) = $\frac{1}{2} \times 5 \times 10 = 25$ m.
Area 2 (rectangle) = $10 \times 10 = 100$ m.
Area 3 (triangle) = $\frac{1}{2} \times 5 \times 10 = 25$ m.
Total = $25 + 100 + 25 = 150$ m. [3]

12.
(a) For an object in equilibrium, the sum of clockwise moments about any pivot is equal to the sum of anticlockwise moments about that same pivot. [2]
(b)
Anticlockwise Moment = $2.0 \, \text{N} \times (50 - 20) \, \text{cm} = 2.0 \times 30 = 60$ Ncm.
Clockwise Moment = $W \times (80 - 50) \, \text{cm} = W \times 30$ .
$60 = 30 W \Rightarrow W = 2.0$ N. [3]
(c) Decrease.
Explanation: If $W$ is increased, the clockwise moment ( $W \times d$ ) increases for the same distance. To maintain equilibrium (balance the constant anticlockwise moment), the distance $d$ must decrease. [2]

13.
(a) $P = F / A = 100 / 0.01 = 10,000$ Pa. [2]
(b) $F = P \times A = 10,000 \times 0.5 = 5,000$ N. [2]
(c) Liquids are incompressible. [1]

14.
(a) $\text{Work} = mgh = 500 \times 10 \times 20 = 100,000$ J. [3]
(b) $\text{Power} = \text{Work} / \text{Time} = 100,000 / 10 = 10,000$ W. [2]
(c) $\text{Efficiency} = (\text{Useful Output} / \text{Total Input}) \times 100\%$ .
$40 = (100,000 / \text{Input}) \times 100$ .
$\text{Input} = 100,000 / 0.4 = 250,000$ J. [2]

15.
(a) Boyle’s Law. [1]
(b) $P_1 V_1 = P_2 V_2$ .
$200,000 \times 0.05 = P_2 \times 0.02$ .
$10,000 = 0.02 P_2$ .
$P_2 = 10,000 / 0.02 = 500,000$ Pa. [3]

Section C: Free Response & Application

16.
(a) $\text{GPE} = mgh = \text{Weight} \times h = 10 \times 0.5 = 5$ J. [2]
(b) $\text{Work Input} = \text{Force} \times \text{distance along slope} = 4.0 \times 2.0 = 8$ J. [2]
(c) $\text{Efficiency} = (5 / 8) \times 100\% = 62.5\%$ . [2]
(d) Energy is lost as heat due to friction between the block and the slope. [1]

17.
(a) Atmospheric pressure acting on the surface of the mercury in the reservoir supports the column. [2]
(b) $101,325$ Pa (or approx $1.01 \times 10^5$ Pa). [1]
(c) The vertical height remains the same. The atmospheric pressure supports a specific vertical column of mercury regardless of the tube's tilt. The length of the mercury column increases, but the vertical height is constant. [2]

18.
(a) $\text{GPE} = mgh = 0.5 \times 10 \times 20 = 100$ J. [2]
(b) $\text{GPE converted to KE}$ .
$100 = \frac{1}{2} mv^2$ .
$100 = 0.5 \times 0.5 \times v^2$ .
$100 = 0.25 v^2$ .
$v^2 = 400$ .
$v = 20$ m/s. [3]

19.
(a) Diagram should show two vectors at right angles (3 units East, 4 units North) and a resultant vector from the tail of the first to the head of the second (hypotenuse). [2]
(b) $R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ N. [2]

20.
(a) Pressure in a liquid increases with depth because there is more weight of liquid above pushing down. $P = h \rho g$ . [2]
(b) $F = P \times A = 500,000 \times 0.1 = 50,000$ N. [2]