Class 9 Science Chapter 6 Question Bank CBSE Board Pattern

Section A — MCQs (10 questions, 1 mark each)

1. The SI unit of force is:
(a) dyne (b) newton (c) kg m s⁻¹ (d) pascal

2. When two forces of equal magnitude act on an object in opposite directions, the net force is:
(a) equal to sum of forces (b) equal to difference of forces (c) zero (d) twice the magnitude of one force

3. According to Newton’s first law of motion, an object at rest remains at rest if:
(a) no force acts on it (b) net force acting on it is zero (c) friction acts on it (d) gravitational force acts on it

4. The force of friction always acts:
(a) in the direction of motion (b) opposite to the direction of motion (c) perpendicular to the surface (d) upwards

5. Newton’s third law of motion states that the forces of action and reaction:
(a) act on the same object (b) act on two different objects (c) are unequal in magnitude (d) are in the same direction

6. Assertion (A): When a net force acts on an object, the object accelerates in the direction of the net force.
Reason (R): The magnitude of acceleration is proportional to the magnitude of the net force and inversely proportional to the mass of the object.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

7. Assertion (A): Balanced forces acting on an object produce zero acceleration.
Reason (R): Balanced forces are equal in magnitude and opposite in direction.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

8. In Activity 6.1, the stack of coins travels different distances on different surfaces because:
(a) gravitational force changes (b) force of friction is different (c) applied force changes (d) normal force is absent

9. The gravitational force acting on an object of mass m near the Earth’s surface is given by:
(a) F = ma (b) F = mg (c) F = m/v (d) F = ma/g

10. When a person pushes the ground backwards while walking, the ground exerts:
(a) a force of smaller magnitude forward (b) an equal and opposite force forward (c) no force (d) a force in the same direction

Section B — Very Short Answer (6 questions, 2 marks each)

11. Define balanced forces with one example from the chapter.
12. What is the magnitude and direction of net force when two forces of 10 N and 6 N act on a block in the same direction?
13. State Newton’s first law of motion.
14. Why does a moving object come to rest when the applied force is removed?
15. What does the reading of the spring balance indicate in Activity 6.2?
16. Give one example each of contact and non-contact force from the chapter.

Section C — Short Answer (5 questions, 3 marks each)

17. Distinguish between balanced and unbalanced forces with the help of a tug-of-war example.
18. Explain, with a labelled diagram, the forces acting on a box being pushed on a horizontal surface (applied force, friction, weight and normal force).
19. A person is exerting a force on a moving box equal to the force of friction. Will the box continue moving with constant velocity? Justify using Newton’s first law.
20. State Newton’s second law of motion. Derive the relation F = ma from the definition of force and acceleration.
21. Why does a fielder pull his hands backwards while catching a fast-moving cricket ball? Explain using Newton’s second law.

Section D — Long Answer (3 questions, 5 marks each)

22. State Newton’s third law of motion. Explain its application in the following situations with diagrams: (i) rowing a canoe, (ii) launching of a rocket. (Labelled diagram description required)
23. Two forces of 10 N and 6 N act on a block in three different ways as shown in the chapter (parallel same direction, opposite, and at an angle). Calculate net force and direction in each case. Further, if the block has mass 2 kg and net force is 4 N, find its acceleration. (Multi-step numerical)
24. Describe Activity 6.3 (cart and pulley system) to show how acceleration depends on net force. Derive the conclusion that acceleration is directly proportional to net force for fixed mass, using the kinematic equations given in the chapter.

Section E — Case/Source-Based (2 questions, 4 marks each)

25. Case: While catching a fast-moving cricket ball, a fielder gradually pulls his hands backwards. Airbags are provided in vehicles for similar reasons. In the event of collision, the airbag increases the time over which the velocity of the passenger’s head reduces to zero.
(a) Name the law of motion used to explain this situation.
(b) How does increasing time affect the force experienced?
(c) What would happen if the time of contact were very small?
(d) Relate this to the definition of acceleration and force.

26. Case: A stack of four coins is released after stretching a rubber band on different surfaces (wooden table, laminated top, polished marble). The distance travelled before coming to rest varies with the surface.
(a) Which force brings the stack to rest after losing contact with the rubber band?
(b) Why does the stack travel different distances on different surfaces?
(c) How is this force measured approximately using a spring balance?
(d) What would happen if friction were completely absent (thought experiment from the chapter)?

Answer Key Attempt all questions first,
then tap to reveal

1. (b) newton — 1 mark
2. (c) zero — 1 mark
3. (b) net force acting on it is zero — 1 mark
4. (b) opposite to the direction of motion — 1 mark
5. (b) act on two different objects — 1 mark
6. (a) Both true and R explains A — 1 mark
7. (a) Both true and R explains A — 1 mark
8. (b) force of friction is different — 1 mark
9. (b) F = mg — 1 mark
10. (b) an equal and opposite force forward — 1 mark

11. Balanced forces: equal magnitude, opposite direction → no change in motion (tug-of-war example) — definition 1 mark, example 1 mark
12. Net force = 16 N in the direction of larger force — 2 marks
13. Object at rest remains at rest; object in motion continues with constant velocity unless net force acts — 2 marks
14. Force of friction acts opposite to motion and brings it to rest (no continuous force applied) — 2 marks
15. Reading gives approximate magnitude of force of friction — 2 marks
16. Contact: friction; Non-contact: gravitational/magnetic/electrostatic — 1 mark each

17. Balanced: equal & opposite → no motion (tug-of-war); Unbalanced: unequal → motion in direction of larger force — 3 marks (definition + example + effect)
18. Diagram showing applied force forward, friction backward, weight downward, normal upward — diagram 2 marks, explanation 1 mark
19. Forces balanced → net force zero → constant velocity (Newton’s first law) — 3 marks
20. Statement of law + F ∝ a, F ∝ 1/m → F = ma — 3 marks
21. Increases time → decreases acceleration → smaller force (F = ma) — 3 marks

22. Statement of third law + canoe (paddle pushes water back, water pushes canoe forward) + rocket (gases expelled down, rocket pushed up) with labelled diagrams — law 1 mark, each example 2 marks
23. (a) 16 N right; (b) 4 N right; (c) 4 N left; acceleration = 2 m s⁻² — each case 1 mark, acceleration 2 marks (working shown)
24. Description of cart-pulley activity + kinematic equations s = ½aT² → conclusion a ∝ F — activity 2 marks, derivation 3 marks

25. (a) Newton’s second law — 1 mark; (b) larger time → smaller acceleration → smaller force — 1 mark; (c) very large force/injury — 1 mark; (d) a = Δv/Δt, F = ma — 1 mark
26. (a) Force of friction — 1 mark; (b) different surfaces → different friction — 1 mark; (c) spring balance reading when block just moves — 1 mark; (d) object continues moving with constant velocity forever — 1 mark

All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.