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Class 9 Science: How Forces Affect Motion — Chapter Question Set

Chapter-wise questions and answers for Class 9 Science: How Forces Affect Motion — Chapter Question Set in Class 9 Science, with a clear summary, key points, and exam-ready model answers.

What you will find on this page: a clear chapter summary, revision notes, word meanings, important questions, MCQs, and exam-ready question answers for Class 9 Science.

After reviewing the chapter explanation, answer the short-answer, long-answer, extract-based, and objective questions below.

About 62 min readChapter 6: How Forces Affect MotionUpdated 15 Sep 2026
148 questions

Question 1: The SI unit of force is: (a) newton (b) dyne (c) joule (d) watt

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(a)

Question 2: The symbol used for the SI unit of force is written as: (a) n (b) N (c) Nt (d) NT

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(b)

Question 3: A force can change all of the following EXCEPT: (a) speed of a moving object (b) direction of motion (c) mass of an object (d) shape of an object

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(c)

Question 4: A spring balance is used to measure the: (a) magnitude of a force (b) mass of an object only (c) speed of an object (d) density of an object

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(a)

Question 5: The weight of an object is the: (a) gravitational force by Earth on the object (b) normal force on the object (c) buoyant force on the object (d) mass of the object

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(a)

Question 6: When two teams in a tug of war pull with equal and opposite forces, the rope: (a) does not move (b) moves toward the larger team (c) moves toward the smaller team (d) snaps immediately

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(a)

Question 7: Forces equal in magnitude but opposite in direction acting on an object are called: (a) net forces (b) frictional forces (c) unbalanced forces (d) balanced forces

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(d)

Question 8: When forces on an object are unbalanced, the object experiences: (a) a non-zero net force (b) no acceleration (c) constant velocity forever (d) zero net force

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(a)

Question 9: Two forces of 10 N and 6 N act on a block in the SAME direction. The net force is: (a) 4 N (b) 60 N (c) 16 N (d) 0 N

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(c)

Question 10: Two forces of 10 N and 6 N act on a block in OPPOSITE directions. The magnitude of the net force is: (a) 0 N (b) 60 N (c) 16 N (d) 4 N

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(d)

Question 11: When two unequal, opposite forces act on an object, the net force acts along: (a) the smaller force (b) the larger force (c) no particular direction (d) perpendicular to both

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(b)

Question 12: The force of friction between two surfaces acts: (a) opposite to the direction of motion (b) perpendicular to the surfaces (c) in the direction of motion (d) vertically downward only

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(a)

Question 13: A box starts moving when the applied force becomes: (a) smaller than the frictional force (b) equal to the frictional force (c) zero (d) larger than the frictional force

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(d)

Question 14: When you stop pedalling a bicycle, it eventually stops because of: (a) the force of friction (b) balanced forces (c) Newton's first law (d) zero net force

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(a)

Question 15: The force exerted by a surface on an object placed on it, in the upward perpendicular direction, is called the: (a) normal force (b) gravitational force (c) frictional force (d) buoyant force

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(a)

Question 16: Galileo argued through thought experiments that on a frictionless horizontal plane, a moving body would: (a) move indefinitely (b) slow down gradually (c) move faster and faster (d) stop after some distance

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(a)

Question 17: The tendency of objects to resist a change in their state of rest or uniform motion is called: (a) momentum (b) acceleration (c) gravity (d) inertia

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(d)

Question 18: Newton presented his three laws of motion in the year: (a) 1808 (b) 1787 (c) 1687 (d) 1911

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(c)

Question 19: According to Newton's first law, an object at rest remains at rest unless acted upon by: (a) a balanced force (b) a net force (c) gravity only (d) friction only

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(b)

Question 20: If the net force acting on a moving object is zero, the object: (a) stops immediately (b) accelerates uniformly (c) moves in a circle (d) continues with constant velocity

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(d)

Question 21: A person pushes a box forward with a force exactly equal to the opposing friction. The box will: (a) continue moving with constant velocity (b) accelerate forward (c) come to rest (d) reverse direction

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(a)

Question 22: The velocity-time graph of an object moving with constant velocity is a straight line: (a) inclined to the time axis (b) parallel to the time axis (c) parallel to the velocity axis (d) curving upward

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(b)

Question 23: In Newton's second law, the mathematical relation between force (F), mass (m) and acceleration (a) is: (a) F = ma (b) F = a/m (c) F = m/a (d) a = Fm

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(a)

Question 24: One newton is the force that produces an acceleration of 1 m s⁻² in an object of mass: (a) 10 kg (b) 1 g (c) 1 kg (d) 100 g

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(c)

Question 25: The acceleration produced by a net force is inversely proportional to the: (a) mass of the object (b) magnitude of the force (c) displacement (d) time taken

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(a)

Question 26: If the net force on an object doubles (mass constant), its acceleration: (a) halves (b) remains same (c) becomes zero (d) doubles

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(d)

Question 27: If the mass of an object doubles (force constant), its acceleration: (a) becomes four times (b) halves (c) doubles (d) remains same

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(b)

Question 28: The value of acceleration due to gravity (g) near the Earth's surface is approximately: (a) 98 m s⁻² (b) 9.8 m s⁻² (c) 1 m s⁻² (d) 0.98 m s⁻²

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(b)

Question 29: The gravitational force acting on an object of mass m is given by: (a) F = mg (b) F = m + g (c) F = m/g (d) F = g/m

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(a)

Question 30: While catching a fast-moving cricket ball, a fielder pulls the hands backwards to: (a) increase the stopping time and reduce force (b) increase the force on hands (c) apply Newton's first law (d) make the ball travel further

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(a)

Question 31: Airbags in vehicles reduce injury during collisions by: (a) increasing acceleration (b) reducing mass of passenger (c) eliminating net force (d) increasing the time over which the impact occurs

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(d)

Question 32: A coconut breaks when it hits the ground at high velocity because: (a) its mass increases (b) of Newton's first law (c) the ground exerts a very large force to stop it quickly (d) of balanced forces

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(c)

Question 33: A weightlifter holds a 30 kg barbell steady. The upward force she applies is about (g = 9.8 m s⁻²): (a) 29.4 N (b) 9.8 N (c) 294 N (d) 30 N

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(c)

Question 34: A block of mass 25 kg experiences a net force of 5 N. Its acceleration is: (a) 125 m s⁻² (b) 0.5 m s⁻² (c) 5 m s⁻² (d) 0.2 m s⁻²

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(d)

Question 35: Newton's third law states that when object A exerts a force on object B, object B exerts on A a force that is: (a) equal and opposite (b) equal and same direction (c) larger and same direction (d) smaller and opposite

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(a)

Question 36: The pair of forces in Newton's third law act on: (a) only the larger object (b) the same object (c) two different objects (d) only the smaller object

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(c)

Question 37: When you walk, you move forward because the ground applies a force on your feet: (a) opposite to your push, via friction (b) equal to your weight (c) in the direction of gravity (d) vertically upward

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(a)

Question 38: A canoe moves forward when the canoeist pushes water backwards. This is explained by: (a) Newton's first law (b) Newton's second law (c) Newton's third law (d) law of inertia

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(c)

Question 39: The equal and opposite forces of Newton's third law do NOT cancel each other because they: (a) act in the same direction (b) act on two different objects (c) are imaginary forces (d) have different magnitudes

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(b)

Question 40: A rocket lifts off because the exhaust gases pushed downwards exert: (a) an upward force on the rocket (b) no force on the rocket (c) a sideways force on the rocket (d) a downward force on the rocket

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(a)

Question 41: In the cannon-ball pair (gun and bullet), the accelerations produced are different because: (a) their masses are different (b) Newton's third law does not apply (c) the forces are unequal (d) friction differs

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(a)

Question 42: A 0.1 kg bullet and a 5 kg gun experience equal recoil forces of 2 N. The bullet's acceleration is: (a) 10 m s⁻² (b) 0.4 m s⁻² (c) 20 m s⁻² (d) 2 m s⁻²

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(c)

Question 43: The Earth does not noticeably move towards a falling fruit because: (a) the Earth's huge mass makes its acceleration extremely small (b) the force on Earth is zero (c) the fruit has no mass (d) Newton's third law fails here

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(a)

Question 44: When two boxes connected by a string are pulled by force F on a frictionless surface, the system accelerates like a single object of mass: (a) m1 × m2 (b) m1 + m2 (c) m1 − m2 (d) m1 / m2

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(b)

Question 45: In the two-box system pulled by an external force F, the tension in the connecting string is a(n): (a) gravitational force (b) normal force (c) internal force (d) external force

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(c)

Question 46: The force of friction depends upon the: (a) colour of the surfaces (b) volume of the objects (c) mass of the Earth (d) nature of the surfaces in contact

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(d)

Question 47: On which surface would a sliding stack of coins travel the FARTHEST before stopping? (a) rough wooden table (b) polished marble floor (c) cemented floor (d) laminated table top

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(b)

Question 48: Grooves are made on the soles of footwear to: (a) reduce weight (b) increase friction and prevent slipping (c) decrease friction (d) increase the normal force

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(b)

Question 49: It is difficult to walk on wet polished floors or ice because: (a) gravity is weaker there (b) friction is too high (c) normal force is zero (d) friction is too low

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(d)

Question 50: While walking, the friction between your feet and the ground acts: (a) vertically downward (b) in no direction (c) against your motion (d) in the forward direction helping you move

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(d)

Question 51: A toy car of mass 100 g moves with a constant velocity of 0.5 m s⁻¹. The net force on it is: (a) 50 N (b) 0 N (c) 0.5 N (d) 0.05 N

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(b)

Question 52: Two children of different masses sit on identical swings. To give the SAME initial acceleration, the larger force is needed for the child with: (a) no force needed at all (b) either child (same force) (c) larger mass (d) smaller mass

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(c)

Question 53: If a spacecraft far from any gravitational field fires its engine in the direction of its motion, it will: (a) stop immediately (b) slow down (c) speed up (d) change direction only

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(b)

Question 54: A falling object approaching the Earth has an acceleration that is: (a) dependent on its mass (b) zero (c) independent of its mass (d) always increasing with mass

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(c)

Question 55: A bullet of mass 50 g moving at 100 m s⁻¹ stops in 0.5 m inside a wooden block. The stopping force is approximately: (a) 500 N (b) 50 N (c) 1000 N (d) 100 N

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(a) ---

Question 56: The SI unit of force is the ______ and its symbol is ______.

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newton; N

Question 57: Two forces equal in magnitude but opposite in direction are called ______ forces.

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balanced

Question 58: When unbalanced forces act on an object, a ______ force acts on it.

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non-zero net

Question 59: The force of friction always acts in a direction ______ to the direction of motion.

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opposite

Question 60: An object at rest remains at rest, and an object in motion continues with constant velocity, unless a ______ force acts upon it.

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net

Question 61: Newton's second law of motion is mathematically expressed as F = ______.

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ma

Question 62: One newton is defined as the force that produces an acceleration of ______ on an object of mass ______.

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1 m s⁻²; 1 kg

Question 63: The value of g near the Earth's surface is approximately ______ m s⁻².

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9.8

Question 64: Whenever one object exerts a force on a second object, the second object simultaneously exerts an ______ and ______ force on the first.

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equal; opposite

Question 65: The upward force exerted by a surface on an object placed on it is called the ______ force.

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normal

Question 66: The tendency of objects to resist a change in their state of rest or uniform motion is called ______.

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inertia

Question 67: A person climbing a coconut tree is pushed upward by the force of ______ between the trunk and their legs.

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friction

Question 68: A rocket works on the basis of Newton's ______ law of motion.

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third

Question 69: When two connected boxes of masses m₁ and m₂ are pulled with force F on a frictionless surface, the acceleration of the system is a = F/(______).

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m₁ + m₂

Question 70: The reading of a spring balance pulled horizontally gives an approximate measure of the ______ acting between the block and the surface.

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force of friction

Question 71: Force is a physical quantity for which both magnitude and direction need to be specified.

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True

Question 72: If either the magnitude or the direction of a force changes, the effect of the force also changes.

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True

Question 73: Balanced forces can change the velocity of a moving object.

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False

Question 74: The net force of two opposite, unequal forces equals the sum of their magnitudes.

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False

Question 75: The force of friction acts in the same direction as the motion of the object.

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False

Question 76: A box being pushed with a force equal to the frictional force remains stationary if initially at rest.

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True

Question 77: According to Newton's first law, a net force is required to keep an object moving with constant velocity.

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False

Question 78: An object moving with constant velocity has zero acceleration.

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True

Question 79: Acceleration due to gravity (g) depends on the mass of the falling object.

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False

Question 80: For the same force, a lighter object acquires a larger acceleration than a heavier one.

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True

Question 81: The forces in a Newton's third law pair act on the same object and hence balance each other.

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False

Question 82: The equal and opposite forces of the third law can produce different accelerations on the two interacting objects.

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True

Question 83: Friction always opposes motion and can never help an object move.

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False

Question 84: Newton's third law applies only to contact forces.

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False

Question 85: Treating connected objects as a single system simplifies the analysis of their motion.

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True

Question 86: Assertion (A): A rope in a tug of war does not move when both teams pull with equal force. Reason (R): Equal and opposite forces acting on the same object are balanced and produce zero net force. (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.

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(a)

Question 87: Assertion (A): A cyclist must keep pedalling on a rough road to maintain constant speed. Reason (R): The force of friction continuously acts opposite to the motion and must be counteracted. (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.

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(a)

Question 88: Assertion (A): A ball released on an ideal frictionless horizontal surface keeps moving forever. Reason (R): Newton's first law states that an object in motion continues with constant velocity if no net force acts on it. (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.

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(a)

Question 89: Assertion (A): A cricketer pulls his hands back while catching a fast ball. Reason (R): Increasing the time of contact reduces the acceleration and hence the force on the hands. (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.

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(a)

Question 90: Assertion (A): A truck and a cricket ball experience the same magnitude of force when the ball hits the truck's windshield. Reason (R): Newton's third law states that interacting objects exert equal and opposite forces on each other. (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.

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(a)

Question 91: Assertion (A): The Earth does not visibly accelerate towards a falling apple. Reason (R): The force exerted by the apple on the Earth is much smaller than the force exerted by the Earth on the apple. (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.

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(c)

Question 92: Assertion (A): A gun recoils when a bullet is fired from it. Reason (R): The bullet exerts an equal and opposite force on the gun, in accordance with Newton's third law. (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.

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(a)

Question 93: Assertion (A): It is easier to walk on a rough road than on a wet polished floor. Reason (R): The normal force on a wet floor is smaller than on a rough road. (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.

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(c) ---

Question 94: Define balanced and unbalanced forces with one example of each.

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Balanced forces are equal in magnitude and act in opposite directions, so their net force is zero — for example, the two equal pulls on a tug-of-war rope when the teams are evenly matched. Unbalanced forces leave a non-zero net force — for example, when one team pulls harder, the rope moves in that direction.

Question 95: State Newton's first law of motion and explain the term inertia.

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Newton's first law states that an object at rest stays at rest, and an object in motion continues to move with constant velocity in a straight line, unless acted upon by a net external force. Inertia is the natural tendency of an object to resist any change in its state of rest or of uniform motion.

Question 96: State Newton's second law of motion and write its mathematical form.

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Newton's second law states that when a net force acts on an object, the object accelerates in the direction of the net force; the acceleration is directly proportional to the net force and inversely proportional to the mass of the object. Mathematically, a = F/m, or equivalently F = ma.

Question 97: State Newton's third law of motion with one daily-life example.

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Newton's third law states that whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first. For example, while walking we push the ground backwards, and in return the ground pushes us forward, enabling us to move.

Question 98: Why does a fielder pull his hands backward while catching a fast cricket ball?

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By pulling his hands backward, the fielder increases the time in which the ball's velocity falls to zero. This reduces the ball's rate of change of velocity (a = Δv/Δt), so a much smaller force acts on his hands, preventing injury.

Question 99: Why are grooves made on the soles of footwear and treads on tyres?

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Grooves and treads increase the friction between the sole or tyre and the ground by providing a better interlocking grip. This increased friction prevents slipping and gives a firm hold on the road while walking or driving.

Question 100: Explain why it is difficult to walk on a wet polished floor or ice.

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On a wet polished floor or on ice, the friction between our feet and the surface is extremely small. When we push the ground backwards, our foot slips instead of getting a firm grip, and the forward frictional push is too small to move us forward, making walking difficult.

Question 101: A box moving on a floor with friction stops when no one pushes it. Does this violate Newton's first law? Explain.

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No, this does not violate Newton's first law, because the first law applies only when the net force on an object is zero. Here friction is an unbalanced force acting opposite to the motion, which decelerates the box and brings it to rest.

Question 102: Two forces of 8 N and 3 N act on an object in opposite directions. Find the net force and its direction.

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Since the forces act in opposite directions, the net force is their difference: 8 N − 3 N = 5 N. It acts in the direction of the larger force, i.e. along the 8 N force.

Question 103: A car of mass 1000 kg accelerates at 2 m s⁻². Find the net force acting on it.

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By Newton's second law, F = ma = 1000 kg × 2 m s⁻² = 2000 N. The net force of 2000 N acts in the direction of the acceleration of the car.

Question 104: What is the weight of an object of mass 5 kg? (Take g = 10 m s⁻²)

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Weight is the gravitational force on a body, W = mg. Therefore W = 5 kg × 10 m s⁻² = 50 N, acting vertically downward towards the centre of the Earth.

Question 105: Define the newton (SI unit of force).

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One newton is defined as the force that produces an acceleration of 1 m s⁻² in an object of mass 1 kg. Hence 1 N = 1 kg m s⁻².

Question 106: Why does a fireperson struggle to hold a pipe issuing water at high speed?

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The pipe pushes the water forward with a large force to eject it at high speed. By Newton's third law, the water pushes the pipe backward with an equal and opposite force, which the fireperson has to struggle to hold against.

Question 107: Explain how a rocket moving in space (negligible gravity) can change its velocity.

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A rocket changes its velocity by firing its engine and expelling gases. When the gases are expelled backward, they push the rocket forward, speeding it up; expelling gases forward exerts a backward reaction force that slows it down, in accordance with Newton's third law.

Question 108: How does a person climb a vertical coconut tree using friction?

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While climbing, a person grips the trunk and pushes his legs downward against it. The friction between the trunk and the legs then pushes the person upward with an equal force, allowing the climb — this is harder on smooth, low-friction trunks.

Question 109: Explain why the pair of action-reaction forces never cancel each other.

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Action and reaction forces are equal in magnitude and opposite in direction, but they always act on two different objects. Only equal and opposite forces acting on the same object can balance or cancel each other, so an action-reaction pair never cancels out.

Question 110: A bullet of 0.1 kg is fired from a 5 kg gun. Both experience a force of 2 N. Compare their initial accelerations.

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For the bullet, a = F/m = 2/0.1 = 20 m s⁻²; for the gun, a = 2/5 = 0.4 m s⁻². The same force gives very different accelerations because their masses are very different — the bullet accelerates 50 times more than the gun.

Question 111: What is a thought experiment? Why did Galileo use one to study motion?

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A thought experiment is an imagined experiment whose idealised conditions are difficult or impossible to create in reality. Galileo imagined a surface with all friction (all impediments) removed, and concluded that on such a frictionless horizontal plane a moving body would continue moving forever with constant velocity.

Question 112: Two people push a stalled car in the same direction with forces 300 N and 200 N. What is the net force on the car?

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Since both forces act in the same direction, they add up: net force = 300 N + 200 N = 500 N. This net force acts on the car in the direction in which both people are pushing.

Question 113: Distinguish between internal and external forces for a system of two boxes joined by a string and pulled by force F.

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Internal forces act between the parts of the system themselves, such as the tension in the string joining the two boxes. External forces act on the system from outside it, such as the applied force F, gravity and the normal reaction from the floor.

Question 114: Explain balanced and unbalanced forces with the help of a tug-of-war example.

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Introduction: Forces decide whether an object moves or stays still, as seen in a tug of war between two teams.
Key Points:
1. Balanced forces — Both teams pull with equal magnitude in opposite directions, so the net force on the rope is zero.
2. Unbalanced forces — When one team pulls harder, forces become unequal and the rope moves toward the larger force.
3. Role of net force — Only a non-zero net force can change the state of rest or motion of the rope or any object.
Conclusion: Thus, the motion of an object depends only on the net force acting upon it, not on individual forces.

Question 115: State Newton's laws of motion and give one example of each.

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Introduction: Newton formulated three laws in 1687 that connect force and motion for all objects.
Key Points:
1. First law — An object stays at rest or in uniform motion unless a net force acts; e.g. passengers jerk forward when a bus suddenly brakes.
2. Second law — Acceleration is proportional to net force and inversely proportional to mass; e.g. a harder hit sends a ball farther.
3. Third law — Every action has an equal and opposite reaction; e.g. a rocket expels gas downward and moves upward.
Conclusion: Together, the three laws allow us to predict how any object moves under given forces.

Question 116: Describe the coin-stack activity showing that friction depends on the surfaces in contact.

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Introduction: Activities with a rubber band and a coin stack demonstrate how friction changes with surface type.
Key Points:
1. Setup — A stretched rubber band launches a taped stack of four coins across different horizontal surfaces.
2. Observation — The coins travel the shortest distance on a rough wooden top and farthest on polished marble.
3. Conclusion — Smaller friction slows the coins more gradually, so they travel farther before coming to rest.
Conclusion: Hence, the force of friction depends on the nature of the surfaces in contact.

Question 117: Prove with an example that equal forces on different masses produce different accelerations.

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Introduction: Newton's second law links force, mass and acceleration through a = F/m for any object.
Key Points:
1. Law relation — For a fixed net force, acceleration is inversely proportional to the mass of the object.
2. Gun-bullet example — A 2 N recoil force acts on a 5 kg gun and a 0.1 kg bullet simultaneously.
3. Calculation — Gun: a = 2/5 = 0.4 m s⁻²; bullet: a = 2/0.1 = 20 m s⁻² — very different accelerations.
Conclusion: Equal action-reaction forces need not give equal accelerations because masses of interacting objects differ.

Question 118: Explain how airbags and backward hand-pulls reduce injury, using Newton's second law.

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Introduction: Safety devices work by stretching the time of an impact, which reduces the force experienced.
Key Points:
1. Second law insight — For a given velocity change, smaller acceleration needs a smaller force on the body.
2. Airbags — The inflated soft cushion increases the stopping time of the passenger's head during a collision.
3. Fielder's technique — Pulling hands back with the moving ball increases stopping time, reducing the required force.
Conclusion: By increasing collision time, both devices lower the force on the human body and prevent serious injuries.

Question 119: A 25 kg block is pushed with 55 N against a maximum friction of 50 N. Find its acceleration and displacement in 2 s.

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Introduction: Newton's second law and kinematics together describe the block's motion after the net force is found.
Key Points:
1. Net force — F(net) = 55 N − 50 N = 5 N, acting in the forward direction of the push.
2. Acceleration — a = F/m = 5 N / 25 kg = 0.2 m s⁻² in the forward direction.
3. Displacement — s = ut + ½at² = 0 + ½ × 0.2 × (2)² = 0.4 m forward.
Conclusion: The block accelerates at 0.2 m s⁻² and covers 0.4 m in the first two seconds.

Question 120: Explain the working of a rocket on the basis of Newton's third law.

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Introduction: A rocket moves by expelling gas, in accordance with the action-reaction principle.
Key Points:
1. Action — The rocket engine expels hot gas downwards with a large force.
2. Reaction — The gas exerts an equal and opposite upward force on the rocket.
3. Lift-off — When this upward force exceeds the rocket's weight, the net force is upward and it lifts off.
Conclusion: The same principle was used by the Vikram lander of Chandrayaan-3 to slow down for a soft Moon landing.

Question 121: Explain how walking is possible only because of friction.

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Introduction: Walking forward relies on friction pushing the foot, not on friction opposing it.
Key Points:
1. Push on ground — While walking, your foot pushes the ground backwards along the surface.
2. Friction's direction — The ground exerts an equal forward frictional force on your foot.
3. Consequence — This forward friction moves you ahead; without it the foot would slip and you would fall.
Conclusion: Thus friction, usually seen as a hindrance, actually enables walking, running and driving.

Question 122: A car moves along a straight road; its v-t graph rises linearly from 0 to 10 m s⁻¹ in 5 s, stays flat for 5 s, then falls to 0 in 5 s. Describe the forces on the car in each phase (mass 1500 kg).

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Introduction: The velocity-time graph reveals the acceleration, from which Newton's second law gives the force.
Key Points:
1. 0–5 s — Constant acceleration a = (10−0)/5 = 2 m s⁻²; F = ma = 1500 × 2 = 3000 N forward.
2. 5–10 s — Flat line means constant velocity; acceleration and hence net force are zero.
3. 10–15 s — a = (0−10)/5 = −2 m s⁻²; F = −3000 N, i.e. 3000 N opposing the motion.
Conclusion: The graph's slope gives acceleration, and F = ma converts it into the force for every phase.

Question 123: Explain the concept of treating connected objects as a system with the two-box example.

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Introduction: Newton's laws can be applied to several connected objects treated as one system.
Key Points:
1. System idea — Two boxes joined by a string on a frictionless surface are pulled by external force F.
2. Internal forces — The tension in the string acts within the system and is not considered.
3. Result — The system accelerates as a single object: a = F/(m₁ + m₂).
Conclusion: Treating connected objects as one system greatly simplifies the analysis of their motion.

Question 124: A boy kicks a 0.4 kg football with a speed of 108 km h⁻¹ with force 800 N. Find the contact time between foot and ball.

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Introduction: Newton's second law links the force, the velocity change and the contact time.
Key Points:
1. Convert speed — 108 km h⁻¹ = 108 × 1000/3600 = 30 m s⁻¹; the ball starts from rest.
2. Acceleration — a = F/m = 800 N / 0.4 kg = 2000 m s⁻² during the kick.
3. Time — t = v/a = 30/2000 = 0.015 s of contact between foot and ball.
Conclusion: The foot remains in contact with the ball for about 0.015 s during the penalty kick.

Question 125: Distinguish between mass and inertia with examples.

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Introduction: Mass measures the quantity of matter, while inertia describes resistance to change of motion.
Key Points:
1. Mass — Mass is the amount of matter in an object and its SI unit is the kilogram.
2. Inertia — Inertia is the tendency of an object to resist any change in its state of rest or motion.
3. Link — A larger mass means larger inertia — a loaded truck is harder to start or stop than a bicycle.
Conclusion: Therefore, mass is the measure of inertia: heavier objects resist changes to motion more strongly.

Question 126: Explain why the canoe moves forward and moves faster when pushed harder, using Newton's laws.

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Introduction: The motion of a canoe follows directly from Newton's third law and second law.
Key Points:
1. Third law — The paddle pushes water backwards; the water pushes the paddle and canoe forward equally.
2. Second law — When the canoeist pushes harder, the larger forward force gives the canoe a larger acceleration.
3. Mass effect — With a passenger aboard, the same paddle force produces a smaller acceleration of the canoe.
Conclusion: The canoe's forward motion and its response to harder paddling follow Newton's second and third laws.

Question 127: An object of 2 kg moving at 10 m s⁻± meets a rough patch with friction 7 N and an opposing force 3 N. Find the distance travelled before it stops.

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Introduction: The combined opposing forces decelerate the object until it comes to rest.
Key Points:
1. Total opposing force — F = 7 N + 3 N = 10 N opposite to the motion of the object.
2. Deceleration — a = F/m = 10 N / 2 kg = 5 m s⁻² against the direction of motion.
3. Stopping distance — v² = u² − 2as → 0 = 10² − 2×5×s, giving s = 10 m.
Conclusion: The object travels 10 m along the rough patch before it finally comes to rest.

Question 128: Explain why a sailor jumping forward from a small boat observes the boat moving backward.

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Introduction: The jump demonstrates Newton's third law between the sailor and the boat.
Key Points:
1. Action — The sailor's legs exert a backward force on the boat while jumping toward the shore.
2. Reaction — The boat exerts an equal and opposite forward force on the sailor, launching them ahead.
3. Boat's motion — The backward force on the light boat gives it a noticeable backward acceleration.
Conclusion: The boat moves backward because the action force acts on it while the reaction propels the sailor forward.

Question 129: A group of students performed Activity 6.1: they stacked four ₹10 coins, taped them, and launched the stack with a stretched rubber band on three surfaces — a wooden table top, a laminated table top, and a polished marble floor. The rubber band was stretched to the same mark C each time. (a) On which surface did the coins travel the largest distance, and why? (b) What does the gradual decrease in the velocity of the coins after release indicate about the force acting on them? (c) What would happen if the experiment could be done on a perfectly frictionless floor? (d) Name the instrument the students could use next to quantify the friction on each surface.

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(a) The coins travelled the largest distance on the polished marble floor, because it offers the least friction among the three surfaces.
(b) It indicates that after release only the frictional force acts on the coins, opposing their motion and gradually decelerating them to rest.
(c) On a perfectly frictionless floor the coins would continue moving with constant velocity forever, as stated by Newton's first law.
(d) They could use a spring balance, pulled horizontally with a wooden block on each surface, to measure the force of friction.

Question 130: Rahul tries to push a 25 kg stationary block on a horizontal floor. The maximum friction opposing the motion is 50 N. (a) What happens when Rahul pushes with exactly 50 N? (b) Calculate the acceleration when he pushes with 55 N. (c) Calculate the displacement of the block in 2 s in case (b). (d) Which law of motion did you use for parts (b) and (c)?

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(a) The applied force exactly balances the maximum friction, so the net force is zero and the block remains stationary.
(b) Net force = 55 − 50 = 5 N; acceleration a = F/m = 5/25 = 0.2 m s⁻².
(c) Displacement s = ½at² = ½ × 0.2 × (2)² = 0.4 m.
(d) Newton's second law of motion (F = ma) was used, along with a kinematic equation for displacement.

Question 131: During a cricket match, a fielder catches a fast-moving ball. The commentator remarks, 'Notice how gracefully he pulls his hands back with the ball!' (a) Why does the fielder pull his hands backwards? (b) Which law of physics is involved in reducing the impact? (c) Explain how increasing the stopping time reduces the force. (d) Name one safety device in vehicles based on the same principle.

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(a) He pulls his hands backwards to increase the time in which the ball's velocity becomes zero, which reduces the hurt or injury.
(b) Newton's second law of motion is involved, since the force depends on the rate of change of velocity.
(c) Force is proportional to the rate of change of velocity; by stretching the stopping time, the acceleration of the ball is reduced and hence the force on the hands becomes smaller.
(d) Airbags in vehicles work on the same principle — they inflate to cushion the passenger and increase the stopping time during a collision.

Question 132: A sports car of mass 1500 kg moves east. Its velocity-time graph shows the velocity rising uniformly from 0 to 10 m s⁻¹ during 0–5 s, staying constant at 10 m s⁻¹ during 5–10 s, and falling uniformly to zero during 10–15 s. (a) Calculate the acceleration in each interval. (b) Find the net force on the car in each interval. (c) In which interval is the net force zero, and why? (d) In which direction does the force act during 10–15 s?

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(a) Acceleration during 0–5 s = 10/5 = +2 m s⁻²; during 5–10 s = 0; during 10–15 s = (0 − 10)/5 = −2 m s⁻².
(b) Net force = ma: during 0–5 s it is 1500 × 2 = 3000 N east; during 5–10 s it is 0 N; during 10–15 s it is 1500 × (−2) = −3000 N, i.e. 3000 N west.
(c) The net force is zero during 5–10 s, because the car moves with constant velocity, so its acceleration and hence F = ma are zero.
(d) During 10–15 s the force acts opposite to the direction of motion, i.e. towards the west.

Question 133: India's Chandrayaan-3 Vikram lander had to slow down for a soft landing near the south pole of the Moon. (a) Which law of motion explains how rockets change their velocity in space? (b) Explain how firing the engine in the direction of motion slows a spacecraft. (c) What happens when a rocket's engine expels gas downward during lift-off? (d) Why does the equal and opposite force pair not cancel out?

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(a) Newton's third law of motion explains how rockets change their velocity in space.
(b) The engine throws the exhaust gases forward, and the gases exert an equal and opposite reaction force backward on the spacecraft, reducing its velocity.
(c) The expelled gas pushes the rocket upward with an equal force; when this upward force exceeds the rocket's weight, the net force lifts the rocket off the ground.
(d) The equal and opposite forces act on two different objects — the gas and the rocket — so they cannot cancel each other.

Question 134: Two spring balances are connected hook-to-hook on a table. The free end of one is fixed; a student pulls the free end of the other with different forces and observes both readings. (a) What relation is observed between the two readings every time? (b) Which law of motion does this experiment verify? (c) What would happen if the two equal and opposite forces acted on the same object instead? (d) State one daily-life example of the same law.

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(a) The two spring balances always show equal readings, whatever force is applied.
(b) The experiment verifies Newton's third law of motion — the mutual forces of action and reaction are equal and opposite.
(c) The two forces would then balance each other, giving zero net force and producing no change in the object's state of motion.
(d) Rowing a canoe is an example — the paddle pushes the water backwards, and the water pushes the canoe forward with an equal force.

Question 135: A gun of mass 5 kg fires a bullet of mass 0.1 kg with a force of 2 N acting during the shot. (a) What is the recoil force on the gun? (b) Calculate the initial accelerations of the bullet and the gun. (c) Why are the two accelerations different even though the forces are equal? (d) Why does a heavy gun hurt the shoulder less than a light gun firing the same bullet?

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(a) By Newton's third law, the gun experiences an equal and opposite recoil force of 2 N backward.
(b) Bullet: a = F/m = 2/0.1 = 20 m s⁻²; gun: a = 2/5 = 0.4 m s⁻².
(c) The accelerations differ because acceleration a = F/m is inversely proportional to mass, and the masses of the bullet and the gun differ greatly.
(d) The heavier gun's larger mass gives it a much smaller recoil acceleration and hence a smaller recoil velocity, so it hurts the shoulder less.

Question 136: A weightlifter holds a barbell steady overhead. The bar weighs 10 kg and each fixed plate weighs 10 kg. (Take g = 9.8 m s⁻²) (a) What is the total mass of the barbell system? (b) Calculate the gravitational force on the barbell. (c) How much force and in which direction must the weightlifter apply to hold it steady? (d) Is she doing any work while holding the barbell steady? Why or why not?

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(a) Total mass = 10 kg (bar) + 10 kg + 10 kg (plates) = 30 kg.
(b) Gravitational force F = mg = 30 × 9.8 = 294 N, directed vertically downward.
(c) To hold the barbell steady she must apply an equal force of 294 N in the upward direction, so that the net force is zero.
(d) No, she does no work on the barbell because there is no displacement of the barbell while she holds it steady, although her muscles tire internally.

Question 137: Two children, Riya (20 kg) and Kabir (40 kg), sit on identical swings. Their father applies forces to give both swings the same initial acceleration. (a) For which child is a larger force required? (b) Give the reason using Newton's second law. (c) If the father applies 60 N to Kabir's swing, what is the initial acceleration? (d) What force would give Riya's swing the same acceleration?

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(a) A larger force is required for Kabir, the heavier child.
(b) By Newton's second law, a = F/m; for the same acceleration, the required force F = ma is directly proportional to the mass.
(c) a = F/m = 60/40 = 1.5 m s⁻².
(d) F = ma = 20 × 1.5 = 30 N would give Riya's swing the same acceleration.

Question 138: 100 oarsmen row a snake boat (Vallam Kalli) in Kerala. 95 row backwards to propel the boat forward, but 5 mistakenly row in the opposite direction. Each oarsman applies 200 N (ignore drag). (a) What is the total forward propelling force from the 95 oarsmen? (b) What is the force from the 5 mistaken oarsmen? (c) Calculate the net horizontal force on the boat. (d) Which laws of motion are used in answering (a)–(c)?

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(a) The 95 oarsmen together apply 95 × 200 N on the water backwards, and by reaction the water pushes the boat forward with 19000 N.
(b) The 5 mistaken oarsmen apply 5 × 200 N = 1000 N in the reverse direction, i.e. backward on the boat.
(c) Net horizontal force = 19000 − 1000 = 18000 N in the forward direction.
(d) Newton's third law (the water pushes the boat forward as the oars push the water back) and Newton's second law (the net force accelerates the boat) are used.

Question 139: State and explain Newton's three laws of motion with one example of each.

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Introduction: Newton's three laws, presented in 1687, form the foundation of classical mechanics.
Key Points:
1. First law — An object stays at rest or in uniform motion unless acted upon by a net external force.
2. Second law — The acceleration produced is proportional to the net force and inversely proportional to mass (F = ma).
3. Third law — For every action, there is a simultaneous equal and opposite reaction on the other object.
4. Example — first law — A book lying on a table remains at rest until someone pushes it with a net force.
5. Examples — second and third laws — A harder kick sends a ball farther; the recoil of a gun when a bullet is fired.
Conclusion: Together, these laws explain and predict the motion of objects from everyday life to planets.

Question 140: Explain the concept of balanced and unbalanced forces with examples and their effects on motion.

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Introduction: Real objects usually experience several forces at once, whose combined effect decides the motion.
Key Points:
1. Balanced forces — Equal and opposite forces on the same object give zero net force and no change of motion.
2. Example — balanced — A tug-of-war rope pulled equally by both teams does not move at all.
3. Unbalanced forces — Unequal or non-cancelling forces leave a net force that accelerates the object.
4. Example — unbalanced — A box pushed with 55 N against 50 N friction accelerates due to a 5 N net force.
5. Key idea — Only the net force matters for motion; individual forces may act but cancel out.
Conclusion: Hence, whether an object starts moving, stops, or maintains its motion depends solely on the net force.

Question 141: Describe the gold-foil style friction activities (6.1 and 6.2) and explain what they reveal about the force of friction.

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Introduction: Two classroom activities together show how friction varies and how it can be measured.
Key Points:
1. Activity 6.1 — A stretched rubber band launches a coin stack on different surfaces — wood, laminate, marble.
2. Observation — The stack travels farthest on marble and shortest on wood, i.e. least friction on marble.
3. Activity 6.2 — A spring balance dragged with a wooden block measures the friction at the moment of sliding.
4. Result — The spring balance reading is smallest for marble and largest for the wooden table top.
5. Conclusion — Smaller friction means slower deceleration and larger travel distance before the coins stop.
Conclusion: These activities establish that friction depends on the nature of the surfaces in contact.

Question 142: A block of mass m slides on a floor and stops due to friction alone. Using Newton's laws, explain why continuous force is needed in real life to maintain uniform motion, and reconcile this with Newton's first law.

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Introduction: The need for continuous pedalling or pushing in real life seems to contradict the first law but does not.
Key Points:
1. Real situation — Friction always acts on moving objects, opposing their motion on any surface.
2. Effect of friction — Friction is an unbalanced force that decelerates the object until it stops.
3. Why push continuously — A continuous applied force merely cancels friction, making the net force zero.
4. First law reconciliation — The first law speaks of zero NET force; with friction balanced, velocity stays constant.
5. Ideal case — On a frictionless surface, an object would need no force to keep moving uniformly forever.
Conclusion: Thus Newton's first law holds perfectly once all forces, including friction, are properly accounted for.

Question 143: A sports car (1500 kg) travels east; draw and interpret its velocity-time graph in three phases (acceleration 0–5 s, constant velocity 5–10 s, deceleration 10–15 s, peak speed 10 m s⁻¹). Calculate the force in each phase and explain the significance of the graph's slope.

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Introduction: A velocity-time graph encodes acceleration through its slope, which Newton's second law converts to force.
Key Points:
1. Phase 0–5 s — Straight inclined line: a = 2 m s⁻²; F = 1500 × 2 = 3000 N towards the east.
2. Phase 5–10 s — Line parallel to time axis: a = 0; net force is zero though the car keeps moving.
3. Phase 10–15 s — Inclined falling line: a = −2 m s⁻²; F = −3000 N, i.e. 3000 N towards the west.
4. Slope meaning — Positive slope means accelerating force; zero slope means zero net force.
5. Negative slope — A negative slope indicates a force opposite to the direction of motion.
Conclusion: The slope of the v-t graph, combined with F = ma, gives the magnitude and direction of the force.

Question 144: Explain the concept of inertia with everyday examples, and describe Galileo's thought experiment that led to Newton's first law.

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Introduction: Inertia is the resistance of matter to changes in its state of motion, an idea rooted in Galileo's work.
Key Points:
1. Meaning of inertia — Inertia is the tendency of an object to remain at rest or continue in uniform motion.
2. Everyday examples — Passengers lurch backward when a bus starts and forward when it brakes suddenly.
3. Galileo's thought experiment — Galileo imagined a body moving on a horizontal plane with all impediments removed.
4. Conclusion drawn — Such an ideal body would continue moving indefinitely without any force.
5. Newton's contribution — Newton coined the word 'inertia' and used Galileo's idea to frame his first law.
Conclusion: Galileo's reasoning and Newton's formulation together established that uniform motion needs no cause.

Question 145: State Newton's second law. Derive/explain F = ma and use it to solve: a force acts on 5 kg producing 2 m s⁻², and the same force acts on 10 kg — find the new acceleration.

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Introduction: Newton's second law quantitatively relates the net force on an object to the acceleration it produces.
Key Points:
1. Statement — The object accelerates along the net force; a ∝ F and a ∝ 1/m, giving a = F/m.
2. Definition of newton — F = ma with m = 1 kg and a = 1 m s⁻² defines 1 N = 1 kg m s⁻².
3. First calculation — F = ma = 5 kg × 2 m s⁻² = 10 N is the applied force.
4. Second calculation — Same 10 N on 10 kg gives a = F/m = 10/10 = 1 m s⁻².
5. Insight — For a fixed force, doubling the mass halves the acceleration produced.
Conclusion: The relation F = ma lets us predict accelerations, forces or masses whenever two are known.

Question 146: Explain Newton's third law with the examples of walking, rowing a canoe, and rocket propulsion. Why do the action-reaction forces not cancel?

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Introduction: Newton's third law appears in many everyday phenomena wherever two objects interact.
Key Points:
1. Statement — When object A pushes B, B simultaneously pushes A with equal force in the opposite direction.
2. Walking — Feet push the ground backward; the ground's forward friction pushes the walker ahead.
3. Rowing a canoe — The paddle pushes water backward; the water pushes the paddle and canoe forward.
4. Rocket propulsion — The engine expels gas downward; the gas pushes the rocket upward with equal force.
5. No cancellation — Action and reaction act on two different objects, so they never balance each other.
Conclusion: Every interaction involves a pair of equal, opposite forces acting on the two bodies involved.

Question 147: Two boxes (m₁ = 2 kg, m₂ = 3 kg) connected by a light string rest on a frictionless table. A force of 10 N pulls box 1. Find the acceleration of the system, and explain the internal vs. external forces. Also compute the system's acceleration if the two boxes were glued together.

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Introduction: Connected objects can be analysed simply by treating them as one system.
Key Points:
1. System view — The two boxes plus the string form one system of total mass m₁ + m₂ = 5 kg.
2. Internal force — The tension in the string acts between the boxes and is internal to the system.
3. External force — The applied 10 N pull is the only external horizontal force on the system.
4. Acceleration — a = F/(m₁ + m₂) = 10/5 = 2 m s⁻² for both boxes together.
5. Glued case — Gluing changes nothing — the system still accelerates at 2 m s⁻² as one object.
Conclusion: This shows the power of Newton's laws: complicated systems behave like single objects of total mass.

Question 148: Describe how Newton's laws explain the following safety and sporting situations: (i) airbags in cars, (ii) a high-jumper landing on a mat, (iii) packing glass in bubble wrap, and (iv) cracking a coconut in one go.

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Introduction: Newton's laws explain why impact time determines the damage in collisions and strikes.
Key Points:
1. Airbags — The soft cushion lengthens the stopping time of the passenger, reducing the force on the body.
2. High-jump mat — The sand bed/mat deforms, increasing stopping time so the athlete lands gently without injury.
3. Bubble wrap — The wrapping increases the impact time for glass items, so a much smaller force reaches them.
4. Coconut cracking — The coconut stopped in a very short time by hard ground experiences a huge breaking force.
5. Common principle — Force = mass × acceleration; longer stopping times mean smaller accelerations and forces.
Conclusion: All four situations show that stretching or shortening the collision time controls the force experienced.

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