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How Forces Affect Motion, Class 9 Science Chapter 6: NCERT question answers for every Think It Over, Activity, Pause & Ponder, Worked Example, Think as a Scientist, Revise Reflect Refine and Journey Beyond question (CBSE 2026-27), with full working for every numerical. Go through your Class 9 Science Ch 6 notes first, then check your answers here.
This chapter moves from describing motion to explaining its cause, and these Class 9 Science Chapter 6 How Forces Affect Motion NCERT solutions work through Newton's three laws of motion numericals, F = ma problems, friction, inertia examples and conservation of momentum with the working shown for every step. Use them for connected-body problems, recoil of a gun and braking-force questions.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
How Forces Affect Motion is the heart of Class 9 mechanics. It moves from describing motion (Chapter 4) to explaining what causes it, building 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, with full working for every numerical, as complete Class 9 Science Chapter 6 NCERT solutions with answers.
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.
When the canoeist pushes the water backwards with the paddle, the paddle exerts a force on the water in the backward direction.
By Newton’s third law of motion, the water exerts an equal and opposite force on the paddle in the forward direction.
This force on the paddle makes the paddle and the canoe move forward.
When the canoeist pushes harder on the water, the forward force on the paddle is larger.
By Newton’s second law, \(a = F/m\), a larger force produces a larger acceleration, so the velocity of the canoe increases and it moves faster.
The force applied by the canoeist is the same in both cases.
By Newton’s second law, \(a = F/m\). For the same force \(F\), the acceleration \(a\) is smaller when the mass \(m\) is larger.
The canoe carrying a passenger has a larger mass, so its acceleration is smaller.
The empty canoe has a smaller mass, so its acceleration is larger.
Step 4: Yes. After losing contact with the rubber band, the velocity of the stack of coins decreases gradually and it comes to rest after travelling some distance.
Step 5: Yes. The stack travels a larger distance than on the wooden table top, and its velocity decreases more slowly.
Step 6: Yes. The stack travels an even larger distance on the polished marble or tile floor, and its velocity decreases even more slowly.
ExplanationBefore the stack is released, it is stationary. So the forces acting on it are balanced.
Upon release, the force applied by the stretched rubber band acts on the stack in the forward direction. The coins start moving, so this force is larger than the force of friction and a net force acts on them in the forward direction.
Due to the net force, the velocity changes from zero to a certain value, i.e., the net force gives an acceleration in the forward direction.
When the stack loses contact with the rubber band, only the force of friction acts on it, in the direction opposite to its motion. It gradually decreases the velocity and finally brings the stack to rest.
The rubber band is stretched by the same amount each time, but the distance travelled is different on different surfaces. So the force of friction is different on different surfaces.
Step 3: The reading shows the force applied by the spring on the block.
If the velocity of the block is neither increasing nor decreasing, the net force acting on the block is zero.
So the force applied by the spring is equal to the force of friction. Yes, the reading of the spring balance indicates the magnitude of the force of friction acting on the block.
Step 5: Yes, the readings are different for different surfaces.
Yes, the reading is smallest for the surface on which the stack of coins travelled the largest distance (polished marble or tile floor).
Yes, the reading is largest for the surface on which the distance travelled was the smallest.
In both cases the cart starts with zero velocity, \(u = 0\), and travels the same distance \(s\).
Let \(a_1\) and \(a_2\) be the accelerations in the two cases. Using the kinematic equation \(s = ut + \tfrac12 at^2\) with \(u=0\):
Equating the two equations:
On substituting the values of \(T_1\) and \(T_2\), we find that when the force is increased for the same mass of the cart, the acceleration increases.
The force is the same in both cases (the mass of the cup and objects is constant). The cart starts from rest and travels the same distance \(s\).
As in Activity 6.3, using \(s = \tfrac12 at^2\):
When the mass of the cart is doubled, the cart takes more time to cover the same distance (\(T_2 > T_1\)), so \(a_2 < a_1\).
Yes, for the same force, when the mass of the cart is increased, the acceleration decreases.
Step 2: When I push the table forward, I move backwards along with the chair. Yes, the chair moves in the direction opposite to the force applied by me on the table.
Step 3: When I pull the table towards me, the chair moves forward, i.e., in the direction opposite to that in step 2.
Prediction: Both spring balances will show the same reading, because each pulls the other with a force of the same magnitude.
Observation: The readings of the scales of the two spring balances are the same every time, even when the force applied by me is changed.
Yes, my observation is the same as my prediction.
Observation: The air rushes out of the balloon in one direction, and the straw and the balloon move in the opposite direction along the thread.
Reason: The stretched material of the balloon applies a force on the air molecules inside to expel them as it shrinks in size.
The air rushing out exerts an equal force on the balloon material in the opposite direction (Newton’s third law).
This force makes the balloon move in a direction opposite to the direction in which the air is rushing out.
A rocket moves in a similar manner. Its engine expels gas in the downward direction, and the gas exerts an equal and opposite force on the rocket in the upward direction.
The two forces acting on the barbell are:
(i) the gravitational force (weight) due to the Earth, acting in the downward direction, and
(ii) the force applied by the weightlifter, acting in the upward direction.
When the weightlifter keeps the barbell steady, its velocity does not change, so its acceleration is zero and the net force on it is zero.
So the two forces are equal in magnitude and opposite in direction.
No, the forces exerted by the two players are not balanced at this instant.
If the forces were balanced, the net force would be zero and the arms would not move. Here the arms are tilting, so a net force acts on them.
The net force acts in the direction of the force of larger magnitude, so the player whose force is in the direction in which the arms tilt exerted the larger force.
No. An object moving with a constant velocity has zero acceleration, because its velocity does not change.
If there were a net force, the object would accelerate (Newton’s laws of motion).
Many forces may act on the object, but they balance each other.
When the net force is zero, an object at rest remains at rest (Newton’s first law).
When the net force is zero, an object in motion continues to move with a constant velocity (Newton’s first law).
When the net force is zero, the acceleration is zero.
So the object cannot move with a constant acceleration.
Example: A person pushes a box moving on a floor.
The force of friction acts on the box in the backward direction. If we apply an additional force on the box in the forward direction equal to the force of friction, the two forces are equal and opposite.
So they balance each other, and the net force on the box is zero.
As per Newton’s first law, the box continues to move with a constant velocity.
Another example: A ball floating on water. The gravitational force acts downwards and the buoyant force of the liquid acts upwards. They are equal and opposite, so the net force is zero.
Given: m = 100 g = 0.1 kg, constant velocity \(v = 0.5\text{ m s}^{-1}\)
To find: net force \(F\)
The velocity is constant, so the acceleration \(a = 0\).
By Newton’s second law:
By Newton’s second law, \(F = ma\).
For the same acceleration \(a\), the force \(F\) is directly proportional to the mass \(m\).
So a larger force is required for the child having the larger mass.
When a package is dropped or jolted, the velocity of the glass item changes from some value to zero.
If the time taken for this change is very short, the acceleration is very large, and so is the force (\(F = ma\)). The glass can break.
Bubble wrap or hay is soft and increases the time over which the velocity of the glass item reduces to zero.
So the acceleration is smaller and a smaller force acts on the glass item.
The pipe throws water out in the forward direction with a large force.
By Newton’s third law, the water exerts an equal and opposite force on the pipe, in the backward direction.
The fireperson has to hold the pipe against this large backward force, so sometimes they struggle.
The spacecraft can fire its engine and expel gas in one direction.
By Newton’s third law, the exhaust gas exerts an equal and opposite force on the spacecraft in the opposite direction.
This force produces an acceleration, so the velocity changes.
If the engine fires in the direction of motion, the force on the spacecraft is opposite to its motion and it slows down (as done by the Vikram lander of Chandrayaan-3).
Net force = 10 N + 6 N = 16 N
Net force = 10 N – 6 N = 4 N
Net force = 10 N – 6 N = 4 N
The force of friction acts on the box in the backward direction.
The two forces acting on the box are equal and opposite, and thus they balance each other.
The net force acting on the box is zero.
As per Newton’s first law of motion, the box will continue moving with constant velocity.
When no net force is acting on an object, there are two possibilities — either the object is at rest or the object is moving with a constant velocity.
If the object is at rest, its position will not change with time, so its position-time graph is a horizontal straight line (Fig. 6.15a).
Its velocity will remain zero, so its velocity-time graph is a straight line along the time axis (Fig. 6.15b).
If the object is moving with a constant velocity and no net force is acting upon it, it will continue moving with the same velocity.
So the velocity-time graph is a horizontal straight line (Fig. 6.16b).
The position changes uniformly with time, so the position-time graph is a straight line inclined to the time axis (Fig. 6.16a).
Given: mass of the barbell = 10 kg + 10 kg + 10 kg = 30 kg, \(g = 9.8\text{ m s}^{-2}\)
The total mass of the barbell is 30 kg.
The gravitational force acting on the barbell in the downward direction is (using Eq. 6.3):
To keep the barbell steady, the weightlifter has to apply an equal force in the opposite direction.
The force applied by the student is equal to the opposing force of friction.
Thus, the two forces are balanced and the net force acting on the block is zero.
So, the block will remain stationary.
The net force on the block is 55 N – 50 N = 5 N.
The mass of the block is 25 kg. Using Newton’s second law of motion, the acceleration of the block is
Using the kinematic equation, the displacement of the block in 2 seconds is
The velocity-time graph is a straight line inclined to the time axis, so the sports car is moving with a constant acceleration, with \(u = 0\text{ m s}^{-1}\), \(v = 10\text{ m s}^{-1}\) and \(t = 5\text{ s}\).
Using \(v = u + at\):
Using \(F = ma\):
The velocity-time graph is a straight line parallel to the time axis, so the sports car is moving with a constant velocity.
The acceleration is zero, so \(F = ma = 0\).
The velocity-time graph is a straight line inclined to the time axis, so the sports car is moving with a constant acceleration, with \(u = 10\text{ m s}^{-1}\), \(v = 0\text{ m s}^{-1}\) and \(t = 5\text{ s}\).
Using \(v = u + at\):
Using \(F = ma\):
The negative sign shows that the force is acting in a direction opposite to the direction of motion.
The forces acting on both the Earth and the fruit are equal in magnitude.
The mass of the Earth is so large as compared to the fruit that the acceleration of the Earth caused by the force is extremely small, as per \(a = F/m\).
Thus, its effect on the Earth is too small to be noticed.
The fruit has a very small mass, so the same force gives it a large acceleration and it moves towards the Earth.
Given: mass of the bullet = 0.1 kg, mass of the gun = 5 kg, force on the bullet = 2 N
To find: initial accelerations of the gun and the bullet
From Newton’s third law of motion, the recoil force on the gun is also 2 N.
From Newton’s second law of motion, the initial magnitude of acceleration of the gun is:
The initial acceleration of the bullet is:
Even though the pair of forces are equal in magnitude, the magnitudes of accelerations are not equal because their masses are different.
If the force of friction is zero, no force acts on the object opposing its motion.
So the velocity of the object will not decrease.
By Newton’s first law, the object will continue to move with a constant velocity. It will never come to rest; it will keep moving forever.
Galileo argued through thought experiments that if a body moves along a horizontal plane and all impediments to its motion are removed, it will continue to move indefinitely.
I can test the hypothesis by Activity 6.3, using weights of different magnitudes to apply forces of different magnitudes.
Keep the mass of the cart the same. Pull it with a hanging cup, and measure the time taken to travel the same distance from rest.
Then double the mass in the cup, so that the force is doubled, and measure the time again.
Using \(s = \tfrac12 at^2\), the accelerations are compared as \(a_1/a_2 = T_2^2/T_1^2\).
I can test the second hypothesis by Activity 6.4.
Keep the mass of the cup and objects inside it constant, so that the force is the same.
Double the mass of the cart by adding more objects in it, and measure the time taken to travel the same distance from rest.
Using \(s = \tfrac12 at^2\), find the ratio of the accelerations in the two cases.
The table moves at a constant velocity, so its acceleration is zero.
Therefore the net force on the table is zero (Newton’s first law).
The applied force F (forward) and the force of friction (backward) are equal and opposite.
No net force acts on the ball, so by Newton’s first law its velocity does not change.
The net force is in the direction of motion, so the ball accelerates in the direction of motion.
The net force is opposite to the direction of motion, so the acceleration is opposite to the velocity.
Block P: Two forces of 4 N and 5 N act in opposite directions. Net force = 5 N – 4 N = 1 N. So P experiences a net force.
Block Q: It is moving with a constant velocity, so its acceleration is zero and the net force is zero. So Q does not experience a net force.
Given: number of oarsmen rowing backwards = 95, in the opposite direction = 5, force by each = 200 N
Force in the forward direction = 95 × 200 N = 19000 N
Force in the backward direction = 5 × 200 N = 1000 N
Net force = 19000 N – 1000 N = 18000 N
By 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 (\(a = F/m\)).
The position-time graph is a straight line inclined to the time axis.
The velocity is constant, so the acceleration is zero.
The position does not change with time, so the object is at rest.
The acceleration is zero.
The position-time graph is a curve; its slope keeps increasing.
So the velocity changes with time, i.e., the object accelerates.
The position-time graph is a straight line, sloping downwards.
The velocity is constant (in the backward direction), so the acceleration is zero.
Yes, the boat will move.
While jumping forward, the sailor pushes the boat backwards with his feet (the sailor applies a force on the boat).
By Newton’s third law, the boat applies an equal and opposite force on the sailor, in the forward direction. This pushes the sailor forward.
The force applied by the sailor on the boat is in the backward direction, so the boat moves backwards, away from the shore.
When the athlete falls on the ground, the velocity of the athlete reduces to zero. On a hard surface this happens in a very short time.
So the acceleration is very large and a very large force acts on the athlete (\(F = ma\)), which can cause injury.
A landing mat or sand bed is soft. It increases the time during which the velocity reduces to zero.
So the acceleration is smaller and the force exerted on the athlete is smaller.
According to Newton’s third law of motion, whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first.
So the two carts exert forces of equal magnitude on each other. Their accelerations are different because their masses are different.
From the graph (Fig. 6.40): when \(m = 1\text{ kg}\), \(a = 10\text{ m s}^{-2}\); \(m = 2\text{ kg}\), \(a = 5\text{ m s}^{-2}\); \(m = 4\text{ kg}\), \(a = 2.5\text{ m s}^{-2}\); \(m = 5\text{ kg}\), \(a = 2\text{ m s}^{-2}\).
Using \(F = ma\):
The force is 10 N for every mass, so the force-mass graph is a straight line parallel to the mass axis.
Given: m = 10 kg; from the graph, \(u = 10\text{ m s}^{-1}\) at \(t = 0\), \(v = 30\text{ m s}^{-1}\) at \(t = 8\text{ s}\)
The graph is a straight line, so the acceleration is constant. Using \(v = u + at\):
Using Newton’s second law:
Given: m = 50 g = 0.05 kg, \(u = 100\text{ m s}^{-1}\), \(v = 0\), \(s = 50\text{ cm} = 0.5\text{ m}\)
To find: stopping force \(F\)
Using the kinematic equation \(v^2 = u^2 + 2as\):
Using Newton’s second law:
The negative sign shows that the force is opposite to the direction of motion.
Given: \(v = 108\text{ km h}^{-1}\), F = 800 N, m = 0.4 kg, \(u = 0\)
To find: time of contact \(t\)
Speed of the ball:
Using Newton’s second law:
Using \(v = u + at\):
Given: m = 2 kg, \(u = 10\text{ m s}^{-1}\), force of friction = 7 N, additional force = 3 N (both opposing the motion), \(v = 0\)
To find: distance \(s\)
Total opposing force = 7 N + 3 N = 10 N
Acceleration:
Using \(v^2 = u^2 + 2as\):
For the harrow: \(F = m_1 a_1\), so \(m_1 = \dfrac{F}{a_1}\).
For the trolley: \(F = m_2 a_2\), so \(m_2 = \dfrac{F}{a_2}\).
When the harrow is placed on the trolley, the total mass is
The acceleration with the same force F is
The magnetic forces on the bar magnet and on the compass needle are equal in magnitude and opposite in direction (Newton’s third law).
But the acceleration produced depends on the mass, \(a = F/m\).
The compass needle has a very small mass, so the force produces a large acceleration and the needle moves.
The mass of the bar magnet is much larger, so the same force produces a very small acceleration and it does not appear to move.
Yes. When the surfaces press each other harder, the force of friction between them is larger (for example, it is harder to push a heavily loaded box than an empty one).
Yes. A larger force is needed to start moving an object than to keep it moving.
Yes. Rolling friction is much less than sliding friction. This is why the invention of the wheel was a major milestone: loads could be moved with much less force.
Did the cars travel equal distances in opposite directions? Yes. The magnetic forces on the two cars are equal and opposite (Newton’s third law). The masses are equal, so the accelerations are equal (\(a = F/m\)) and the cars travel equal distances in opposite directions.
After adding equal masses: the force is the same but the mass of each car is larger, so the acceleration is smaller and each car travels a smaller distance. Both cars still travel equal distances.
Observation: With one turn of rope, a large force is needed to hold the bucket. With each extra turn, the force required becomes much smaller.
Reason: Each extra turn increases the contact between the rope and the branch. This increases the grip and the force of friction, so a much smaller force is needed to hold the same load.
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.
Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.
Two forces act on the book: the force of gravity (weight) downwards, and the force exerted by the table on the book upwards.
These two forces are equal and opposite, so they balance each other.
The net force on the book is zero.
By Newton’s first law, an object at rest remains at rest if the net force is zero.
Given: F = 15 N, m = 3 kg, \(u = 0\), t = 4 s
To find: acceleration \(a\) and velocity \(v\)
No, it does not violate Newton’s third law.
The force exerted by the car on the fly is equal and opposite to the force exerted by the fly on the car.
But the masses are very different. By \(a = F/m\), the fly, having a very small mass, gets a very large acceleration.
The car, having a very large mass, gets an extremely small acceleration which cannot be noticed.
Given: m = 1200 kg; car 1: \(u_1 = 20\text{ m s}^{-1}\), \(s_1 = 40\text{ m}\); car 2: \(u_2 = 40\text{ m s}^{-1}\); same braking force; final velocity = 0
Car 1: using \(v^2 = u^2 + 2as\):
Car 2: the braking force is the same and the mass is the same, so the deceleration is the same, \(a = -5\text{ m s}^{-2}\).
Since \(a = F/m\) is the same, \(s = u^2/2a\) depends on \(u^2\). When the velocity doubles, the stopping distance becomes 4 times.
Both blocks are pushed with the same force.
The rubber block has a rougher underside, so the force of friction on it is larger.
Net force = applied force – force of friction, so the net force on the rubber block is smaller.
Acceleration \(a = F_{net}/m\), so the rubber block has a smaller acceleration.
Given: m₁ = 3 kg (pulled by F = 10 N), m₂ = 2 kg, frictionless surface
To find: acceleration \(a\) and tension \(T\)
Taking both the boxes and the string as one system, the tension is an internal force. Only the external force \(F\) matters:
The tension is the only horizontal force on the 2 kg box:
Check on the 3 kg box: \(F - T = m_1 a\), i.e., \(10 - 4 = 3 \times 2 = 6\text{ N}\), which is correct.
The force \(F\) has to accelerate both the boxes, so only a part of it (4 N) is transmitted through the string to the 2 kg box. That is why \(T\) is less than 10 N.
The rocket fires its engine and expels exhaust gas in one direction.
By Newton’s third law, the gas exerts an equal and opposite force on the rocket in the opposite direction.
This force changes the velocity of the rocket, so it can change its direction without anything to push against.
Next up is Chapter 7 — Work, Energy and Simple Machines. Or explore the full chapter list, browse the Class 9 Science hub, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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