How Forces Affect Motion Class 9 Notes | CBSE Science Chapter 6

Important Definitions

• Force: a push or pull that can make an object move from rest, change the speed or direction of a moving object, or change its shape. It has magnitude and direction.
• Newton (N): the SI unit of force. One newton is the force that produces an acceleration of 1 m s⁻² on an object of mass 1 kg.
• Weight: the gravitational force with which the Earth pulls an object, F = mg.
• Balanced forces: forces on an object that are equal in magnitude but opposite in direction, so the net force is zero.
• Net force: the overall force on an object when all the forces on it are combined.
• Force of friction: the force between two surfaces in contact that acts opposite to the direction of motion (or attempted motion). It depends on the nature of the surfaces.
• Normal force: the force exerted by a surface on an object placed on it, acting perpendicular to the surface.
• Inertia: the tendency of objects to resist a change in their state of rest or uniform motion.
• Newton’s first law: An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.
• Newton’s second law: When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object.
• Newton’s third law: Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object.
• Acceleration due to gravity (g): the acceleration of an object due to the gravitational force by the Earth, about 9.8 m s⁻² near the surface, independent of the object’s mass.
• Tension: the force transmitted through a string that connects two objects.
• Internal force: a force between parts of a system, such as tension in the connecting string.
• External force: a force that acts on the system from outside it.

Solved Examples (NCERT-Based)

Example 1: Pushing a block against friction (NCERT Example 6.5)

Question: A student pushes a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing the motion is 50 N. Find the displacement of the block in 2 s if the push is (i) 50 N, (ii) 55 N, in the forward direction.

Solution (i): The push equals the maximum friction, so the forces are balanced and the net force is zero. The block stays stationary: displacement = 0.

Solution (ii): Net force = 55 N − 50 N = 5 N forwards.

a = F/m = 5 N / 25 kg = 0.2 m s⁻².

u = 0, t = 2 s. s = ut + ½at² = 0 + ½ × 0.2 × (2)² = 0.1 × 4 = 0.4 m in the forward direction.

Example 2: Force on a sports car from its motion (NCERT Example 6.6)

Question: A sports car of mass 1500 kg moves towards the east. From its velocity-time graph: from 0 s to 5 s its velocity rises uniformly from 0 to 10 m s⁻¹; from 5 s to 10 s it stays at 10 m s⁻¹; from 10 s to 15 s it falls uniformly from 10 m s⁻¹ to 0. Find the force on the car in each interval.

(i) 0 s to 5 s: u = 0, v = 10 m s⁻¹, t = 5 s. From v = u + at: 10 = 0 + a × 5, so a = 2 m s⁻². F = ma = 1500 × 2 = 3000 N towards the east.

(ii) 5 s to 10 s: the graph is parallel to the time axis, so velocity is constant and a = 0. No net force acts on the car.

(iii) 10 s to 15 s: u = 10 m s⁻¹, v = 0, t = 5 s. 0 = 10 + a × 5, so a = −2 m s⁻². F = 1500 × (−2) = −3000 N. The minus sign means the force is opposite to the motion: 3000 N towards the west.

Example 3: Recoil of a gun (NCERT Example 6.8)

Question: When a 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, the gun recoils. What are the magnitudes of initial accelerations of the bullet and the gun?

Solution: By the third law, the recoil force on the gun is also 2 N.

Acceleration of the gun = force / mass of gun = 2 N / 5 kg = 0.4 m s⁻².

Acceleration of the bullet = force / mass of bullet = 2 N / 0.1 kg = 20 m s⁻².

The forces are equal, but the accelerations differ because the masses differ. The bullet is 50 times lighter, so its acceleration is 50 times larger.

Example 4: The snake boat race (NCERT Exercise 4)

Question: While practising for the snake boat race (Vallum kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Solution: Forward force = 95 × 200 N = 19000 N. Backward force = 5 × 200 N = 1000 N.

Net force = 19000 N − 1000 N = 18000 N in the forward direction.

Example 5: Stopping a bullet in a wooden block (NCERT Exercise 12)

Question: A bullet of mass 50 g moving with a speed of 100 m s⁻¹ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Solution: Convert units: m = 50 g = 0.05 kg, s = 50 cm = 0.5 m. u = 100 m s⁻¹, v = 0.

v² = u² + 2as gives 0 = (100)² + 2 × a × 0.5, so 0 = 10000 + a, and a = −10000 m s⁻².

F = ma = 0.05 × (−10000) = −500 N. The stopping force is 500 N, opposite to the bullet’s motion.

Example 6: Time of contact in a penalty kick (NCERT Exercise 13)

Question: An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻¹. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.

Solution: For a penalty the ball starts at rest, so u = 0. Convert the speed: v = 108 × 5/18 = 30 m s⁻¹.

a = F/m = 800 N / 0.4 kg = 2000 m s⁻².

v = u + at gives 30 = 0 + 2000 × t, so t = 30/2000 = 0.015 s (15 milliseconds).

Example 7: Object on a rough patch (NCERT Exercise 14)

Question: An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Solution: Both forces oppose motion, so the net force = 7 N + 3 N = 10 N against the motion.

a = −10 N / 2 kg = −5 m s⁻². u = 10 m s⁻¹, v = 0.

v² = u² + 2as gives 0 = 100 + 2 × (−5) × s, so 10s = 100 and s = 10 m.

Example 8: Tractor, harrow and trolley (NCERT Exercise 15)

Question: A tractor pulls a harrow (a ploughing tool) of mass m₁ with a net force F resulting in an acceleration of a₁. The same tractor pulls a trolley of mass m₂ with a force F producing an acceleration of a₂. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a₁ and a₂. Ignore friction.

Solution: From F = ma: m₁ = F/a₁ and m₂ = F/a₂.

The harrow on the trolley moves as one system of mass m₁ + m₂. So a = F / (m₁ + m₂) = F / (F/a₁ + F/a₂).

Cancel F: a = 1 / (1/a₁ + 1/a₂) = a₁a₂ / (a₁ + a₂).

Check: if a₁ = a₂, the total mass doubles and the formula gives a = a₁/2, as expected.

Competency-Based Questions (with answers)

1. Case-based: Pushing a cupboard

Riya pushes a 40 kg cupboard across her room floor. At first she pushes with 60 N and the cupboard does not move. She increases her push to 100 N and it starts moving. While it moves, the force of friction on it is 80 N.

(a) What is the friction on the cupboard while she pushes with 60 N and it stays at rest? (b) What is the acceleration of the cupboard while she pushes with 100 N? (c) What push should she apply to keep it moving at constant velocity?

Answer: (a) The cupboard is at rest, so the net force is zero. Friction balances the push: 60 N, opposite to the push. (b) Net force = 100 − 80 = 20 N; a = 20/40 = 0.5 m s⁻² in the direction of the push. (c) Constant velocity needs zero net force, so the push must equal friction: 80 N.

2. Case-based: Airbags and seat belts

In a collision a car stops abruptly. An airbag inflates quickly into a soft, compressible cushion. The passenger’s head and chest push into the bag instead of hitting the hard steering wheel or dashboard.

(a) Which quantity does the airbag increase? (b) Use Newton’s second law to explain how this reduces injury. (c) Name one other everyday use of the same idea.

Answer: (a) The time over which the passenger’s velocity drops to zero. (b) The same change in velocity over a longer time means a smaller acceleration. By F = ma, a smaller acceleration means a smaller force on the person, which lowers the risk of serious injury. (c) A fielder pulling the hands back while catching a fast ball, bubble wrap around glass items, or a landing mat in the high jump.

3. Source-based: Activity 6.3 data

A class pulls a cart over the same distance from rest using a hanging cup. With the first cup mass the cart takes T₁ = 2.0 s. With double the cup mass it takes T₂ = 1.5 s.

(a) Find the ratio a₂/a₁. (b) The ideal value is 2. Give two reasons why the measured value differs.

Answer: (a) From s = ½aT² for the same s, a₂/a₁ = T₁²/T₂² = 4.0/2.25 ≈ 1.78. (b) Friction between the cart’s wheels and the table, and errors in reading the times from the video.

4. Assertion-Reason

Assertion (A): An object moving with constant velocity on a straight road has no net force acting on it. Reason (R): A net force acting on an object always produces an acceleration in its 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.

Answer: (a). Constant velocity means zero acceleration. By R (the second law, a = F/m), any net force would produce an acceleration, so the net force must be zero. R correctly explains A.

5. Assertion-Reason

Assertion (A): The force exerted by a paddle on the water and the force exerted by the water on the paddle cancel each other. Reason (R): The two forces are equal in magnitude and opposite in direction.

Options as in Question 4.

Answer: (d). A is false: the two forces act on different objects (water and paddle), so they cannot cancel. R is true, as stated by the third law.

6. Assertion-Reason

Assertion (A): When a fruit falls, the Earth does not seem to move towards the fruit. Reason (R): The force exerted by the fruit on the Earth is much smaller than the force exerted by the Earth on the fruit.

Options as in Question 4.

Answer: (c). A is true. R is false: the two gravitational forces are equal in magnitude. The Earth’s acceleration (a = F/m) is extremely small because its mass is so large.

7. Reasoning: A spacecraft far from any planet

A spacecraft is moving in a region of space where the gravitational force on it is negligible. (a) If its engines are switched off, how will it move? (b) Suggest how it can change its velocity.

Answer: (a) With no net force it continues with constant velocity, by the first law. (b) It fires its engine to expel gas. To speed up, gas is expelled backwards, so the gas pushes the craft forwards. To slow down, the engine fires in the direction of motion, as the Vikram lander did before landing on the Moon. To change its direction, gas is expelled to one side, and the gas pushes the craft towards the opposite side.

8. Competency MCQ: Hand carts

A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

(i) the loaded cart exerts a force of larger magnitude on the empty cart. (ii) the empty cart exerts a force of larger magnitude on the loaded cart. (iii) neither cart exerts a force on the other. (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Answer: (iv). By the third law the forces are equal and opposite. The empty cart, having less mass, gets the larger acceleration.

Important Questions for Exams

1-Mark Questions

1. What is the SI unit of force? The newton, symbol N (1 N = 1 kg m s⁻²).
2. An object is moving with a constant velocity. Is there a net force acting upon it? No. Constant velocity means zero acceleration, so the net force is zero.
3. A toy car of mass 100 g is moving with a constant velocity of 0.5 m s⁻¹. What is the net force acting on the toy car? Zero, because its velocity is constant.
4. When a net force acts on an object, it accelerates: (i) opposite to the force, with acceleration proportional to the force; (ii) opposite to the force, with acceleration proportional to the mass; (iii) in the direction of the force, with acceleration inversely proportional to the force; (iv) in the direction of the force, with acceleration proportional to the force. (iv).
5. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table? F, acting opposite to the applied force.

2-Mark Questions

6. Distinguish between balanced forces and a third-law action-reaction pair. Both are equal and opposite. Balanced forces act on the same object and cancel, giving zero net force. A third-law pair acts on two different objects, so the forces never cancel each other.
7. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why. The heavier child. From F = ma, for the same acceleration the force is proportional to the mass.
8. Why does a fielder pull the hands back while catching a fast cricket ball? Pulling back increases the time in which the ball stops. The ball’s acceleration is then smaller, so by F = ma the force on the hands is smaller and injury is less likely.
9. Why does a fireperson sometimes struggle when holding the pipe issuing water? The pipe pushes a large amount of water forwards at high speed. By the third law the water pushes the pipe backwards with an equal force, so the fireperson must hold it firmly.

3-Mark Questions

10. A sailor jumps out from a small boat to the shore. As the sailor jumps forward, will the boat move? If yes, in which direction and why. Yes, the boat moves backwards, away from the shore. To jump, the sailor pushes the boat backwards with the feet. By the third law the boat pushes the sailor forwards with an equal force. The backward force on the boat, which floats on water, sets it moving backwards.
11. Define one newton and find the force needed to give a 1200 kg car an acceleration of 1.5 m s⁻². One newton is the force that produces an acceleration of 1 m s⁻² on an object of mass 1 kg. F = ma = 1200 × 1.5 = 1800 N.
12. A net force of 12 N acts on a 4 kg trolley at rest for 5 s. Find its acceleration, its final velocity and the distance covered. a = 12/4 = 3 m s⁻². v = u + at = 0 + 3 × 5 = 15 m s⁻¹. s = ut + ½at² = 0 + ½ × 3 × 25 = 37.5 m.
13. When the pole of a bar magnet is brought close to a magnetic compass, the magnet and the compass needle exert equal and opposite forces on each other. However, the compass needle moves, whereas the bar magnet does not move. Explain why. The forces are equal in magnitude, but equal forces do not produce equal accelerations when the masses differ (a = F/m). The compass needle has a very small mass and turns freely on its pivot, so the force gives it a noticeable acceleration and it swings. The bar magnet has a much larger mass, so the same force gives it only a very small acceleration. It is also usually held in the hand or rests on a table, where the hand or friction balances this small force. So the bar magnet shows no noticeable motion.

5-Mark Questions

14. State Newton’s second law of motion. Describe an activity to show that, for a fixed mass, a larger force produces a larger acceleration. Statement: When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and is inversely proportional to the mass of the object. Activity: a wheeled cart is pulled along a table by a thread passing over a pipe at the far end to a hanging paper cup with coins. Measure the time T₁ to cover the table from rest using a slow-motion video. Double the mass in the cup and measure T₂. Since s = ½aT² for the same s, a₁/a₂ = T₂²/T₁². T₂ is shorter, so a₂ is larger: more force, more acceleration.
15. State Newton’s third law of motion. Explain with examples how it accounts for (a) walking, (b) rowing a canoe, (c) the launch of a rocket. Why don’t the two forces cancel? Statement: Whenever one object is exerting a force on a second object, the second object is simultaneously exerting an equal and opposite force on the first object. (a) The foot pushes the ground backwards; friction from the ground pushes the foot forwards. (b) The paddle pushes water backwards; the water pushes the paddle and canoe forwards with an equal force. Pushing harder gives a larger forward force. (c) The engine expels gas downwards; the gas pushes the rocket upwards. When this force exceeds the rocket’s weight, the net force is upwards and it lifts off. The two forces of a pair act on different objects, so they cannot cancel; only forces on the same object can balance.
16. A 1000 kg car moving at 20 m s⁻¹ brakes uniformly and stops in 4 s. Find (a) its deceleration, (b) the braking force, (c) the distance it covers while braking. (d) If the car carried extra load so that its mass became 1250 kg, what braking force would stop it in the same time? (a) a = (v − u)/t = (0 − 20)/4 = −5 m s⁻², a deceleration of 5 m s⁻². (b) F = ma = 1000 × 5 = 5000 N, opposite to the motion. (c) s = ut + ½at² = 20 × 4 + ½ × (−5) × 16 = 80 − 40 = 40 m. (d) The same deceleration is needed, so F = 1250 × 5 = 6250 N.

Common Mistakes and Examiner Tips

1. Mistake: saying a moving object needs a force to keep moving. Fix: a force is needed only to overcome friction or to change velocity. With zero net force, a moving object keeps a constant velocity (first law).
2. Mistake: writing “no force acts” when the velocity is constant. Fix: write “no net force acts”. Friction, weight and the normal force may all be acting and cancelling.
3. Mistake: adding forces that act in opposite directions. Fix: subtract them and give the direction of the larger force. Add only forces in the same direction.
4. Mistake: saying action and reaction cancel. Fix: they act on two different objects, so they never cancel. Only forces on the same object can balance.
5. Mistake: thinking the heavier object exerts a larger force in a collision or in gravity. Fix: the forces in a pair are always equal. The lighter object has the larger acceleration because a = F/m.
6. Mistake: using the applied force in F = ma when friction also acts. Fix: find the net force first (applied force minus friction), then divide by mass.
7. Mistake: forgetting unit conversion. Fix: change g to kg, cm to m and km h⁻¹ to m s⁻¹ (multiply by 5/18) before substituting. 50 g = 0.05 kg; 108 km h⁻¹ = 30 m s⁻¹.
8. Mistake: dropping the minus sign of a retarding acceleration and then losing track of direction. Fix: keep the sign in the working, then state the answer as a magnitude with a direction (for example, “3000 N towards the west”).
9. Mistake: writing the unit as Newton or n. Fix: the full name is newton (small n) and the symbol is N (capital).
10. Mistake: believing g depends on the mass of the object. Fix: g is about 9.8 m s⁻² for every object near the Earth’s surface. The weight mg changes with mass; g does not.
11. Mistake: including tension when finding the acceleration of connected boxes treated as one system. Fix: tension is internal. Use a = F / (m₁ + m₂) with external forces only.
12. Tip: write the three laws in the exact textbook wording. Key phrases such as “unless a net force acts”, “in the direction of the net force” and “simultaneously” are the parts examiners look for.

Quick Revision Points

• A force can start motion, change speed or direction, or change shape.
• Force has magnitude and direction. SI unit: newton (N). Measured with a spring balance.
• Equal and opposite forces on the same object are balanced: net force zero.
• Same direction: add forces. Opposite directions: subtract, direction of the larger force.
• An object’s motion depends only on the net force.
• Friction acts opposite to motion and depends on the nature of the surfaces.
• Smaller friction: velocity falls more slowly and the object travels farther.
• Galileo: with all impediments removed, a body on a horizontal plane moves indefinitely.
• First law: at rest stays at rest, moving keeps constant velocity, unless a net force acts.
• Inertia: the tendency to resist a change in the state of rest or uniform motion.
• Zero net force: zero acceleration. Position-time graph is a straight line.
• Second law: acceleration is along the net force, proportional to it and inversely proportional to mass.
• a = F/m, F = ma, 1 N = 1 kg m s⁻².
• Weight F = mg, g = 9.8 m s⁻², independent of mass.
• Longer stopping time means smaller acceleration and smaller force: catching, airbags, mats, bubble wrap.
• Very short stopping time means a very large force: cracking a coconut.
• Third law: forces come in equal and opposite pairs acting on two different objects.
• A third-law pair never cancels.
• Walking, pushing a bicycle along with your feet, tree climbing, rowing, balloons and rockets are third-law examples.
• The third law holds for contact and non-contact forces (magnetic, electrostatic, gravitational).
• Equal forces, unequal accelerations: gun and bullet, Earth and fruit.
• Connected objects: a = F / (m₁ + m₂); ignore internal forces such as tension.

Weightage and Exam Pattern

Class 9 has no board exam. This chapter is assessed in school exams (periodic tests, half-yearly and annual papers), and the NCERT source gives no fixed mark allocation for it. It combines statements of laws, reasoning about everyday situations and numericals built on F = ma and the kinematic equations of Chapter 4.

This chapter is the foundation for Class 10 physics. Finding the net force first, converting units before substituting and naming the direction of every force are habits used in every later chapter on force and motion.

Related chapters: read our Work, Energy, and Simple Machines Class 9 notes and System of Particles and Rotational Motion Class 11 notes.