Where Mindful Learning Begins
Where Mindful Learning Begins
Complete NCERT Solutions for Chapter 6 of the new Class 9 Science Exploration textbook (CBSE 2026-27) — every Think It Over, Activity, Pause & Ponder, Worked Example, Think as a Scientist, Ready to Go Beyond, Revise Reflect Refine, and Journey Beyond question on this one page, with full step-by-step working for every numerical.
This chapter builds directly on the motion concepts from Chapter 4, now asking why objects move the way they do. You'll work through Newton's three laws — inertia, the F = ma relationship, and action-reaction pairs — along with friction (why it slows things down, and why we couldn't walk without it) and the law of conservation of momentum. These ideas turn up constantly in board exams, from braking-distance numericals to recoil and collision problems, so a solid grip on this chapter pays off well beyond just this unit.
How Forces Affect Motion is the heart of Class 9 mechanics — it moves from simply describing motion (Chapter 4) to explaining what causes it. The chapter builds up balanced and unbalanced forces, friction, and then all three of Newton's Laws of Motion in sequence: inertia, F = ma, and action-reaction pairs — finishing with how to treat connected objects as a single system. Every question is solved here, section by section, exactly as the textbook presents them, with full working for every numerical.
Net force, why friction opposes motion, and how the surfaces in contact change how far an object travels.
Inertia, F = ma, and equal-and-opposite reaction pairs — applied to everything from barbells to rocket launches.
Treating connected boxes, carts, or even your own body as a single system to simplify force calculations.
Newton's three laws of motion
| Law | Statement | Key idea |
|---|---|---|
| First Law (Inertia) | An object stays at rest, or moves with constant velocity in a straight line, unless acted upon by an unbalanced (net) force | Explains why we lurch forward when a bus suddenly brakes |
| Second Law | The rate of change of momentum of an object is directly proportional to the applied unbalanced force, and takes place in the direction of the force | Gives us \(F = ma\) |
| Third Law | For every action, there is an equal and opposite reaction, acting on two different bodies | Explains recoil of a gun, walking, swimming |
Key formulae
where \(m\) = mass, \(u\) = initial velocity, \(v\) = final velocity, \(a\) = acceleration, \(F\) = force, \(t\) = time.
Friction
| Type | When it acts | Relative magnitude |
|---|---|---|
| Static friction | Opposes the start of motion between surfaces at rest relative to each other | Highest (up to a maximum limiting value) |
| Sliding (kinetic) friction | Opposes relative motion once an object is already sliding | Less than maximum static friction |
| Rolling friction | Opposes motion of a rolling object (e.g., wheel, ball) | Least of the three — why wheels reduce effort needed to move things |
Answer: when the canoeist pushes water backwards with the paddle, by Newton's Third Law of Motion, the water exerts an equal and opposite force on the paddle in the forward direction. This forward reaction force propels the canoe forward. When the canoeist pushes harder, the force on the water increases, and the reaction force on the paddle is also larger. A larger net force on the canoe means greater acceleration (Newton's Second Law: F = ma), so the canoe moves faster.
Answer: the empty canoe will move faster. By Newton's Second Law, acceleration \(a = F/m\). For the same applied force F, a greater mass m produces a smaller acceleration. The canoe carrying a passenger has more total mass, so it accelerates less and moves more slowly — the empty canoe has less mass and attains a greater velocity.
This question previews both Newton's Second Law and Third Law, covered later in Sections 6.5 and 6.6 of this chapter.
Observation: the stack of coins travels the largest distance on polished marble, a smaller distance on the laminated table, and the smallest distance on the wooden table.
Explanation: before release, the forces on the coins are balanced (stationary). Upon release, the rubber band exerts a forward force, accelerating the coins. Once the coins lose contact with the rubber band, only friction acts — opposing motion, slowing them down, and bringing them to rest. Since the rubber band is stretched by the same amount each time, the initial energy given to the coins is the same in all cases — a smaller friction force decelerates the coins less rapidly, so they travel further before stopping.
Conclusion: the force of friction depends on the nature of the surfaces in contact. Smoother surfaces (polished marble) have less friction; rougher surfaces (wood) have more. A smaller friction force results in a smaller deceleration and a longer distance travelled.
This activity sets up the intuition for Newton's First Law — if friction were zero, the coins would continue moving indefinitely.
Answer: the spring balance reading when the block just starts to move gives the approximate magnitude of the force of friction. At this instant, the block is on the verge of moving but its velocity is neither increasing nor decreasing, meaning the net force is zero — so the force applied by the spring equals the force of friction.
Expected observations: polished marble gives the smallest reading (least friction); the laminated table gives an intermediate reading; the wooden table gives the largest reading (most friction).
Yes — the surface with the smallest spring balance reading (polished marble) is the same surface on which the stack of coins travelled the greatest distance in Activity 6.1: lower friction means a longer distance before stopping.
Conclusion (from Activities 6.1 and 6.2): when the force of friction is smaller, the velocity of the object decreases more slowly and it travels a larger distance before coming to rest.
Answer: both trials start with the cart at rest (\(u=0\)) and travel the same distance \(s\). Using \(s = \tfrac12at^2\): Trial 1: \(s = \tfrac12a_1T_1^2\); Trial 2: \(s = \tfrac12a_2T_2^2\). Equating gives \(a_1/a_2 = T_2^2/T_1^2\).
Since the mass in the cup is doubled in Trial 2, the force pulling the cart is approximately doubled. If \(T_2 < T_1\), then \(a_2 > a_1\), confirming that doubling the force increases the acceleration.
Conclusion: for an object of fixed mass, a larger net force produces a larger acceleration — acceleration is proportional to the applied force, directly supporting Newton's Second Law.
In practice, the increase in acceleration may be slightly less than double, due to friction between the wheels and the surface, and because the hanging mass is also being accelerated.
Answer: using the same analysis, \(a_1/a_2 = T_2^2/T_1^2\), but now the force is kept the same while the mass of the cart is doubled. If doubling the mass causes \(T_2 > T_1\) (the cart takes longer to travel the same distance), then \(a_2 < a_1\).
Expected result: when the cart's mass is doubled and the force is kept constant, the acceleration is approximately halved.
Conclusion: for a given net force, acceleration is inversely proportional to mass — confirming Newton's Second Law, \(F=ma\).
(i) Pushing the table away: when you apply a forward force on the table, it exerts an equal and opposite force back on you — pushing you (and the chair) backward.
(ii) Pulling the table towards you: when you apply a backward force on the table, it exerts a forward reaction force on your hands — the chair moves forward.
Conclusion: every force you apply on the table is accompanied by an equal and opposite force applied by the table on you — a direct demonstration of Newton's Third Law of Motion: every action has an equal and opposite reaction, acting on different objects.
Prediction: since the two spring balances are connected and stationary, by Newton's Third Law, the force Balance 1 exerts on Balance 2 must be equal and opposite to the force Balance 2 exerts on Balance 1 — both balances should show the same reading.
Observation: the readings on both spring balances are identical every time, regardless of the magnitude of the applied force.
Conclusion: the forces that two objects exert on each other are always equal in magnitude and opposite in direction — this experimentally verifies Newton's Third Law of Motion.
Observation: when the neck is released, air rushes out of the balloon in one direction and the balloon (with straw) moves in the opposite direction along the string.
Explanation: the stretched elastic material of the balloon exerts a force on the air molecules inside, pushing them out through the neck. By Newton's Third Law, the expelled air exerts an equal and opposite force on the balloon, pushing it in the opposite direction to the airflow.
This is exactly how a rocket works: the engine pushes exhaust gas downward at high speed, and the reaction force pushes the rocket upward. When this upward thrust exceeds the weight of the rocket, the net force is upward and the rocket lifts off.
This same principle was used by the Vikram lander of Chandrayaan-3: by firing its engines in the direction of motion (downward toward the Moon), it generated an upward retro-thrust that slowed the lander for a soft landing near the Moon's south pole.
Two forces acting on the barbell: the gravitational force (weight) acting downward, equal to \(mg\); and the force applied by the weightlifter acting upward.
When the barbell is held steady (not accelerating), the net force on it is zero. By Newton's First Law, the two forces must be balanced — equal in magnitude (both = \(mg\)) and opposite in direction.
Answer: no, the forces are not balanced at this instant. If the arms are tilting towards R, a net force acts in R's direction — meaning R is exerting a larger force than S. The forces are unbalanced, with the net force in the direction of R's push.
Answer: No. By Newton's First Law, an object moving with constant velocity has zero acceleration. Zero acceleration means the net force acting on the object is zero — the individual forces may or may not be zero, but they must sum to zero (balanced forces).
(i) Possible — an object at rest with zero net force has zero acceleration and remains at rest (Newton's First Law).
(ii) Possible — an object already in motion with zero net force continues moving with the same constant velocity (also Newton's First Law).
(iii) Not possible — a constant, non-zero acceleration requires a non-zero net force (Newton's Second Law, \(F=ma\)). If \(F=0\), then \(a=0\) — constant acceleration with zero force contradicts this law.
Example 1: a book resting on a table. Two forces act on it — gravity pulling it downward and the normal force from the table pushing it upward. These are equal and opposite, so the net force is zero and the book remains at rest.
Example 2: a car travelling at constant velocity on a highway. The engine provides a forward driving force, and friction/air resistance opposes the motion. When these are equal in magnitude, the net force is zero and the car continues at constant velocity without accelerating or decelerating.
Answer: the toy car moves with constant velocity, so its acceleration is zero. By Newton's Second Law: \(F = ma = 0.1 \times 0 = 0\text{ N}\). The net force is zero.
Answer: a larger force is required for the heavier child. By Newton's Second Law, \(F=ma\) — for the same acceleration \(a\), force \(F\) is directly proportional to mass \(m\). The heavier child has greater mass, so a greater force must be applied to produce the same initial acceleration.
Answer: bubble wrap and hay act as cushioning materials. When the package is jolted or dropped, the glass item's velocity changes from a certain value to zero. Without cushioning, this change happens in a very short time, producing a very large deceleration and a very large force (\(F=ma\)) on the glass, which can shatter it. With cushioning, the material compresses gradually, increasing the time over which the glass decelerates — a longer stopping time means a smaller deceleration and therefore a smaller force, preventing breakage.
Answer: by Newton's Third Law, when the hose expels water forward at high speed, the water exerts an equal and opposite reaction force on the hose (and the firefighter) in the backward direction. The faster and more forcefully the water is expelled, the larger this backward reaction force — if the flow rate and pressure are high, this can be strong enough to make it difficult to hold the hose steady without bracing against it.
Answer: the spacecraft can fire its rocket engines (thrusters). By expelling exhaust gas in one direction at high velocity, Newton's Third Law ensures the spacecraft receives an equal and opposite force in the other direction. By choosing the direction and duration of the engine firing, the spacecraft can increase speed (fire engines opposite to motion), slow down (fire in the direction of motion), or change direction (fire thrusters sideways). No external medium is needed — rockets work in the vacuum of space.
(a) Both forces in the same direction (rightward): net force = 10 + 6 = 16 N, towards the right.
(b) 10 N rightward, 6 N leftward: net force = 10 − 6 = 4 N, towards the right (direction of the larger force).
(c) 6 N rightward, 10 N leftward: net force = 10 − 6 = 4 N, towards the left (direction of the larger force).
When two forces act in opposite directions, the net force equals the difference of their magnitudes, in the direction of the larger force.
Answer: the friction force acts backward while the applied force acts forward, and both are equal in magnitude — they balance each other, giving a net force of zero. By Newton's First Law, the box will continue moving with constant velocity, neither accelerating nor decelerating.
Case 1 — object at rest: no net force → zero acceleration → object stays at rest. Position-time graph: horizontal straight line. Velocity-time graph: horizontal line at \(v=0\).
Case 2 — object moving with constant velocity: no net force → zero acceleration → velocity unchanged. Position-time graph: straight line with positive slope. Velocity-time graph: horizontal line at constant, non-zero \(v\).
A straight-line position-time graph always means constant velocity. A curved position-time graph means the object is accelerating (non-zero net force).
Total mass of barbell = 10 + 10 + 10 = 30 kg.
To keep the barbell steady, the weightlifter must apply an equal and opposite force — 294 N upward.
(i) Applied force = 50 N = friction force → net force = 0 N. By Newton's First Law, the block remains stationary — displacement = 0 m.
(ii) Applied force = 55 N, friction = 50 N. Net force = 5 N (forward).
Velocity is constant → \(a=0\) → \(F=0\) N (no net force).
10 to 15 s — deceleration phase (10 to 0 m/s)The negative sign means the force acts opposite to the direction of motion — 3000 N acting towards the west.
Answer: by Newton's Third Law, the forces are equal in magnitude. However, by Newton's Second Law, acceleration \(a=F/m\). The fruit has a very small mass, so the same force produces a large, visible acceleration. The Earth has an enormous mass (6 × 10²⁴ kg), so its acceleration due to the same force is ≈10⁻²² m/s² — negligibly small and undetectable. This is why we observe only the fruit moving.
By Newton's Third Law, the recoil force on the gun is also 2 N (equal and opposite to the force on the bullet).
Acceleration of bullet = \(2/0.1 = 20\text{ m/s}^2\) (forward). Acceleration of gun = \(2/5 = 0.4\text{ m/s}^2\) (backward — recoil).
The forces are equal (Newton's Third Law), but the accelerations differ because the masses differ (Newton's Second Law) — the light bullet accelerates much more than the heavy gun.
Answer: if friction is zero, once the object is given an initial push, no force acts on it in the direction of (or opposing) its motion. By Newton's First Law, the object will continue moving with constant velocity indefinitely — it will never come to rest on its own. This is the idealised situation Galileo described in the 17th century through thought experiments, leading directly to the concept of inertia.
In the real world, some friction is always present, but experiments on air tracks (where the object rides on a cushion of air, eliminating surface friction) closely approximate this ideal.
Hypothesis: for the same mass, \(a \propto F\) (larger force → larger acceleration).
Test: use the cart-pulley setup from Activity 6.3. Keep the cart's mass constant, apply different forces by changing the mass of the hanging cup, and measure the time \(T\) for the cart to travel the same fixed distance from rest. Using \(s=\tfrac12at^2\), we get \(a = 2s/T^2\) — comparing \(a_1\) and \(a_2\) for different forces confirms whether \(a\) increases proportionally with \(F\).
Hypothesis: for the same force, \(a \propto 1/m\) (larger mass → smaller acceleration).
Test (Activity 6.4): keep the mass of the hanging cup constant (same force) but add objects to the cart to double its mass. Measure the time for the same fixed distance, and compare \(a=2s/T^2\) for the two trials. If doubling the mass approximately halves the acceleration, the hypothesis is confirmed.
Answer: when forces act at angles to each other, they must be added as vectors, not simply added or subtracted as numbers. The net force is found using vector addition — typically the parallelogram law, or by resolving each force into horizontal and vertical components, adding the components separately, and finding the resultant magnitude and direction using the Pythagorean theorem and trigonometry. This is covered in detail in higher grades.
Answer: when equal and opposite forces are applied at the two ends of an extended object (not at the same point), the object doesn't translate but instead rotates. This pair of forces acting on different points of the same object is called a couple or torque. The turning effect depends on the magnitude of each force and the distance between their lines of action — this is distinct from Newton's Third Law, where the forces act on two different objects.
Answer: momentum (\(p\)) of an object is the product of its mass and velocity: \(p=mv\), in the same direction as the velocity.
The more complete form of Newton's Second Law states that the net force on an object equals the rate of change of its momentum:
For constant mass, this reduces to \(F=m(\Delta v/\Delta t) = ma\), the familiar form. However, the momentum form is more general and also applies where the mass of the object is changing — e.g., a rocket burning fuel and losing mass.
Answer: when two connected objects are treated as a single system, internal forces (like the tension in the string) need not be considered — they're equal and opposite by Newton's Third Law and cancel within the system. Only the external force \(F\) matters:
The entire system accelerates as though it were a single object of mass \(m_1+m_2\). This approach of treating connected objects as a system greatly simplifies the analysis.
In addition to F, the system also has external forces: gravity \((m_1+m_2)g\) downward and normal force \((N_1+N_2)\) upward from the surface — these are balanced and don't affect the horizontal motion.
Answer: Newton's laws work extremely well for everyday objects and speeds. They require modification in three extreme situations:
For all everyday situations — from throwing a ball to launching a satellite — Newton's laws give accurate and reliable predictions.
Answer: since the table moves at constant velocity, its acceleration is zero, and by Newton's First Law the net force is zero. The applied force F (forward) and the frictional force (backward) must be equal and opposite, so frictional force = F.
(i) If no net force is applied on the ball, the velocity will remain the same (Newton's First Law).
(ii) If a net force is applied in the direction of motion, the magnitude of velocity will increase (force in the direction of motion → acceleration → speed increases).
(iii) If a net force is applied opposite to the direction of motion, the magnitude of velocity will decrease (force opposing motion → deceleration).
Answer: Block P has a net force of 5 − 4 = 1 N (unbalanced), so P experiences a net force. Block Q moves at constant velocity, so \(a=0\) and the net force is zero — Q does not experience a net force. The correct statement is: P experiences a net force and Q does not.
Forward force (95 oarsmen) = 95 × 200 = 19,000 N. Backward force (5 oarsmen) = 5 × 200 = 1,000 N.
Answer: the object accelerates in the direction of the net force, with the acceleration's magnitude proportional to the magnitude of the net force (and inversely proportional to the mass). This is Newton's Second Law: \(a=F/m\).
A net force causes acceleration, which appears as a curved position-time graph (non-uniform velocity).
Net force acts on Objects C and D.
Answer: Yes, the boat will move. By Newton's Third Law, when the sailor pushes forward off the boat (exerting a backward force on it), the boat exerts an equal and opposite reaction force on the sailor in the forward direction. This reaction causes the boat to move backward (away from the shore) at the moment the sailor jumps — since the boat's mass is small, it may move noticeably.
Answer: when an athlete falls onto a hard surface, their velocity changes from a large value to zero in a very short time — a short stopping time means a very large deceleration and a very large force on the body (\(F=ma\)), which can cause serious injury. A landing mat or sand bed is soft and deformable, increasing the time over which the athlete decelerates. A longer stopping time means a smaller deceleration and a smaller impact force, protecting the athlete from injury.
Answer: both the loaded cart and the empty cart exert equal magnitude forces on each other. By Newton's Third Law, the force exerted by the loaded cart on the empty cart is equal in magnitude and opposite in direction to the force exerted by the empty cart on the loaded cart — the masses differ, so the accelerations will differ, but the forces are always equal.
From Newton's Second Law, \(F=ma\). Reading values from the graph: at \(m=1\text{ kg}\), \(a≈10\text{ m/s}^2 → F=10\text{ N}\); at \(m=2\text{ kg}\), \(a≈5\text{ m/s}^2 → F=10\text{ N}\); at \(m=4\text{ kg}\), \(a≈2.5\text{ m/s}^2 → F=10\text{ N}\).
Since \(F=ma\) is constant for all masses (as the graph is a hyperbola, \(a = \text{constant}/m\)), the force-mass graph is a horizontal straight line at F = 10 N — the force is constant regardless of mass, confirming the acceleration-mass graph represents a fixed applied force.
The slope of the velocity-time graph gives acceleration — always read the slope carefully from the graph.
Given: \(m=0.05\text{ kg}\), \(u=100\text{ m/s}\), \(v=0\), \(s=0.5\text{ m}\).
Stopping force = \(ma = 0.05 \times 10{,}000 = \) 500 N (opposing motion).
The stopping force is enormous (500 N) despite the bullet's small mass, because the deceleration is extremely large due to the very short stopping distance.
Speed = 108 km/h = 30 m/s. Using impulse = change in momentum:
The contact time is only 0.015 seconds (15 milliseconds) — far too short to perceive consciously. In this brief interval, the ball changes from rest to 30 m/s.
Total opposing force = 7 + 3 = 10 N. Deceleration: \(a = -10/2 = -5\text{ m/s}^2\).
From Newton's Second Law: \(F=m_1a_1 \Rightarrow m_1 = F/a_1\), and \(F=m_2a_2 \Rightarrow m_2 = F/a_2\).
This is the harmonic mean formula. The combined acceleration is always less than both a₁ and a₂ individually, as expected when the total mass increases.
Answer: by Newton's Third Law, the magnetic force on the compass needle is equal and opposite to the force on the bar magnet. However, by Newton's Second Law, \(a=F/m\). The compass needle is extremely light, so the same force produces a large, easily visible acceleration — it swings noticeably. The bar magnet is held in the hand (a much larger effective mass, including the holder's hand and body), so the force produces negligible acceleration. This is exactly analogous to Example 6.7 (Earth and fruit) and Example 6.8 (gun and bullet).
Yes, friction depends on how hard the surfaces press together (the normal force). The force of friction is given by \(f = \mu N\), where N is the normal force and \(\mu\) is the coefficient of friction. A heavier object presses harder on the surface, increasing friction proportionally.
2. Static vs kinetic frictionYes, static friction (just before the object starts moving) is slightly larger than kinetic friction (while it's sliding) — this is why it takes more force to start moving an object than to keep it moving at constant velocity.
3. Rolling vs sliding frictionRolling friction is significantly less than sliding friction. A ball or wheel rolling on a surface deforms less than a surface sliding against another, so rolling friction can be 100–1000 times smaller than sliding friction — this is why the invention of the wheel was a transformative milestone, replacing sliding with rolling and dramatically reducing the force needed to transport loads.
Three side-by-side panels: (a) heavier vs lighter block showing f = μN with arrows; (b) static vs kinetic showing a force-displacement graph with a peak (static) before a plateau (kinetic); (c) a rolling wheel vs a sliding block, comparing distances travelled from the same initial push.
Expected observation: both cars should travel equal distances in opposite directions. By Newton's Third Law, the magnetic repulsion forces on the two cars are equal and opposite. Since their masses are equal, they experience equal accelerations (\(a=F/m\)) and should travel equal distances in equal times.
With added equal masses: the total mass of each car increases by the same amount. The repulsion force stays the same (same poles at the same initial distance), so each car's acceleration decreases, and each travels less distance before stopping — but both still travel equal distances (symmetric).
Distance vs mass graph: as mass increases (with the same magnetic force), acceleration decreases and the stopping distance decreases — the graph should show a decreasing curve, approximately \(d \propto 1/m\) for the same force.
If the cars travel unequal distances, it may indicate different friction on the two sides, unequal masses, or asymmetric magnets — a good opportunity to test experimental precision.
Answer: the wrapping of a rope around a cylindrical post is described by the capstan equation (Euler's formula): \(T_{hold} = T_{load} \times e^{-\mu\theta}\), where \(\mu\) is the coefficient of friction and \(\theta\) is the total wrap angle in radians.
Why exponential? each small segment of rope presses against the post with a normal force proportional to the tension at that point, generating friction proportional to that tension. As friction reduces tension, the next segment presses less, generating slightly less friction — this cascading effect leads to an exponential (not linear) decay in tension around the post.
This is why sailors can control massive sails and ships with surprisingly thin ropes — a few wraps around a capstan (bollard) multiplies the holding force exponentially, a principle also used in cranes, lifts, and rope-braking systems.
Answer: Isaac Newton published his laws of motion in 1687 in a landmark work titled Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), commonly called the Principia. Written in Latin, it is considered one of the most important scientific books ever written.
In the Principia, Newton stated his three laws of motion as Axiomata sive Leges Motus (Axioms or Laws of Motion):
Newton built on the earlier work of Galileo Galilei (who established the concept of inertia through thought experiments) and Johannes Kepler (whose laws of planetary motion Newton explained using his theory of gravitation). The formulation of these three laws was a defining moment in the history of physics — it unified terrestrial and celestial mechanics into a single mathematical framework.
Both the original Latin text of the Principia and English translations with commentaries are freely available online — reading even a few pages gives insight into how Newton organised his thinking.
Motion doesn't need a cause to continue, only to change — Newton's First Law says an object keeps its velocity until a net force acts, the Second Law says that force produces acceleration in proportion to itself and inversely to mass, and the Third Law says every force comes paired with an equal and opposite one acting on a different object, from a canoe paddle pushing water to a rocket expelling exhaust gas.
Pair these solutions with our free Class 9 Science Notes PDF — quick, chapter-wise revision notes covering every unit, perfect for last-minute recall before a test.
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