NCERT Solutions Class 9 Science Ch 6 Forces 2026-27
NCERT Solutions Class 9 Science Chapter 6 How Forces Affect Motion 2026-27. All 36 questions solved - Newton's laws, force and graphs.
This free NCERT Solutions for CBSE Class IX Science, Chapter 6: How Forces Affect Motion, contains step-by-step solutions to every NCERT textbook question and exercise. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.
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These are the NCERT Solutions for Class 9 Science Chapter 6, How Forces Affect Motion, from the new NCERT Science textbook Exploration (session 2026-27). All 36 questions are solved in simple, exam-ready language, covering Newton’s three laws, balanced and unbalanced forces, momentum, force and acceleration numericals, and graphs.
- Class
- 9 (IX)
- Subject
- Science
- Textbook
- Exploration (NCERT, 2026-27)
- Chapter
- 6 – How Forces Affect Motion
- Questions solved
- 36
Jump to a question (36 questions)
NCERT Class 9 Science Chapter 6 Questions and Answers
Question 1: Why does a canoe move forward when the paddle pushes water backwards?
Why does a canoe move forward when the canoeist pushes water backwards with their paddle and why does it move faster when they push harder?
When the paddle pushes the water backward, the water pushes the canoe forward with an equal and opposite force (Newton’s third law). This reaction force moves the canoe ahead.
When the canoeist pushes harder, the backward force on the water is larger, so the forward reaction force is also larger. A larger force gives a larger acceleration (F = ma), so the canoe moves faster.
Question 2: Same paddle force on an empty canoe and a loaded canoe
Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?
The empty canoe will move faster. For the same force, the acceleration is a = F ÷ m, so a smaller mass gives a larger acceleration. The canoe with the extra passenger has more mass, so it accelerates less and moves slower.
Question 3: Forces on a barbell held by a weightlifter
A weightlifter lifts a barbell. List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?
Two forces on the barbell:
- The downward force of gravity (the weight of the barbell).
- The upward force applied by the weightlifter.
Yes, when the barbell is held steady the two forces are equal in size and opposite in direction, so the net force is zero and the forces are balanced.
Question 4: Forces in an arm-wrestling match
Two players R and S are participating in an arm-wrestling match. At the instant when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?
No, the forces are not balanced at that instant. The arms are moving (tilting), so there is a net force on them.
The player who is pushing the arms in the direction of the tilt exerts the larger force and is winning the match at that instant.
Question 5: Is there a net force on an object moving with constant velocity?
An object is moving with a constant velocity. Is there a net force acting upon it?
No, there is no net force on it. A constant velocity means the acceleration is zero. By Newton’s second law, F = ma = m × 0 = 0. By Newton’s first law, the object continues to move uniformly unless an unbalanced external force acts on it.
Question 6: What can happen when no net force acts on an object?
Suppose no net force is acting on an object. Which of the following situations are possible?
- Object remains at rest if at rest.
- Object keeps moving with a constant velocity if already moving.
- Object is moving with a constant acceleration.
With no net force the acceleration is zero (Newton’s first law).
- (i) Possible. An object at rest stays at rest.
- (ii) Possible. A moving object keeps moving with constant velocity.
- (iii) Not possible. A constant acceleration needs a net force.
So situations (i) and (ii) are possible.
Question 7: Getting a net force of zero with additional forces
In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.
In real life several forces act on an object. If they balance each other so that the resultant force is zero, the object behaves as if no force acts on it.
Example: a book kept on a table. Two forces act on it: its weight downward (gravity) and the normal reaction of the table upward. They are equal and opposite, so they cancel. The net force on the book is zero and it stays at rest.
Question 8: Net force on a toy car moving with constant velocity
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?
- Givenm = 100 g = 0.1 kg, velocity constant, so acceleration a = 0
- Newton’s second lawF = ma = 0.1 × 0 = 0 N
The net force on the toy car is 0 N.
Question 9: Force needed to swing two children of different masses
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.
By Newton’s second law, F = ma. For the same acceleration, the force is directly proportional to the mass.
So a larger force is needed for the heavier child.
Question 10: How packing material protects glass items
How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Bubble wrap or hay increases the time over which the momentum of the glass becomes zero during a jerk or collision. The force is F = Δp ÷ Δt, so when Δt increases, the force F on the glass decreases.
The packing acts as a cushion and reduces the force of impact, so the glass items are protected from damage.
Question 11: Why does a fireperson struggle to hold the water pipe?
Why does a fireperson sometimes struggle when holding the pipe issuing water?
Water leaves the pipe at high speed and carries a large momentum. By Newton’s third law, the water exerts an equal and opposite reaction force on the pipe. This backward force acts on the fireperson holding the pipe, which makes it hard to control.
Question 12: How can a spacecraft change its velocity in space?
Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.
With no external force the spacecraft keeps moving with constant velocity (Newton’s first law). To change its velocity, it can fire its thrusters and expel gases backwards. By Newton’s third law, the gases exert an equal and opposite force on the spacecraft, which makes it accelerate.
Question 13: Accelerations of a bullet and a gun
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?
- Recoil force on the gunBy Newton’s third law, the force on the gun is also 2 N.
- Acceleration of the guna = F ÷ m = 2 ÷ 5 = 0.4 m/s2
- Acceleration of the bulleta = F ÷ m = 2 ÷ 0.1 = 20 m/s2
Gun: 0.4 m/s2 Bullet: 20 m/s2
The forces are equal, but the accelerations are different because the masses are different.
Question 14: Net force on a block in three cases
Two forces of 10 N and 6 N are acting on a block lying on the table as shown in the figure. What is the magnitude and the direction of the net force acting on the block in each case?
| Case | Calculation | Net force |
|---|---|---|
| (a) both forces to the right | 10 + 6 | 16 N towards the right |
| (b) 10 N right, 6 N left | 10 − 6 | 4 N towards the right |
| (c) 6 N right, 10 N left | 10 − 6 | 4 N towards the left |
Question 15: Force of the person equals the friction on a moving box
A person is exerting a force on a moving box in the forward direction which is equal to the force of friction acting between the bottom surface of the box and the floor. Will the box continue moving or will it come to rest after some time?
Friction acts on the box in the backward direction. The two forces are equal and opposite, so they balance each other and the net force is zero.
By Newton’s first law, the box will continue moving with constant velocity. It will not come to rest.
Question 16: Graphs for an object on which no net force acts
Draw (i) position-time, and (ii) velocity-time graphs for an object on which no net force is acting.
With no net force, the object is either at rest or moving with constant velocity.
- Object at rest: the position-time graph is a horizontal line (Fig. 1a) and the velocity-time graph is a horizontal line on the time axis (Fig. 1b).
- Object moving with constant velocity: the position-time graph is a straight sloping line (Fig. 2a) and the velocity-time graph is a horizontal line above the time axis (Fig. 2b).
Question 17: Force applied by a weightlifter to hold a barbell
A weightlifter is holding a barbell with mass of 10 kg fixed on each side of the bar. The mass of the bar itself is 10 kg. How much force is she applying to keep the barbell steady?
- Total mass10 + 10 + 10 = 30 kg
- Weight (gravitational force)F = mg = 30 × 9.8 = 294 N
- To keep it steadyShe must apply an equal force in the opposite (upward) direction.
She applies 294 N upward.
Question 18: Displacement of a block pushed with 50 N and 55 N
A student is trying to push a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing this motion is 50 N. Determine the displacement of the block in 2 seconds if Rahul pushes it with a constant force of (i) 50 N and (ii) 55 N in the forward direction.
- (i) Force of 50 NThe push equals the maximum friction, so the net force is zero and the block stays at rest.Displacement = 0
- (ii) Force of 55 NNet force = 55 − 50 = 5 Na = F ÷ m = 5 ÷ 25 = 0.2 m/s2s = ut + ½at2 = 0 + ½ × 0.2 × 22 = 0.4 m
(i) 0 m (ii) 0.4 m in the forward direction
Question 19: Force on a sports car from its velocity-time graph
A sports car of mass 1500 kg is moving towards the east and its velocity-time graph is shown in the figure. Calculate the force acting on the car during (i) 0 s to 5 s, (ii) 5 s to 10 s, (iii) 10 s to 15 s.
- (i) 0 s to 5 sThe velocity rises from 0 to 10 m/s in 5 s.a = (10 − 0) ÷ 5 = 2 m/s2F = ma = 1500 × 2 = 3000 N towards the east
- (ii) 5 s to 10 sThe velocity is constant (10 m/s), so the acceleration is zero.F = 0 N
- (iii) 10 s to 15 sThe velocity falls from 10 m/s to 0 in 5 s.a = (0 − 10) ÷ 5 = −2 m/s2F = 1500 × (−2) = −3000 NThe minus sign shows that the force acts opposite to the motion, that is towards the west.
(i) 3000 N east (ii) 0 (iii) 3000 N west
Question 20: Why does the fruit move towards the Earth and not the Earth towards the fruit?
As shown in the figure, the Earth and the fruit apply equal and opposite gravitational forces on each other. Then why does the fruit move towards the Earth while the Earth doesn’t seem to move towards the fruit?
The forces on the Earth and the fruit are equal in size. But the mass of the Earth is extremely large compared with the fruit, and the acceleration is a = F ÷ m. So the acceleration of the Earth is so tiny that it cannot be noticed, while the fruit, with its small mass, accelerates visibly.
Question 21: Friction on a table moved with constant velocity
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?
The table moves with constant velocity, so its acceleration is zero and the net force is zero (Newton’s first law). So the applied force F is balanced by the friction.
Frictional force = F
It is equal in size to F and opposite in direction.
Question 22: Ball on a smooth frictionless surface
For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct. (i) If no net force is applied on the ball, the velocity of the ball will remain the same / increase / decrease. (ii) If a net force is applied in the direction of its motion, the magnitude of the velocity will remain the same / increase / decrease. (iii) If a net force is applied opposite to the direction of its motion, the magnitude of the velocity will remain the same / increase / decrease.
| Statement | Correct word | Reason |
|---|---|---|
| (i) No net force | remain the same | The ball continues in uniform motion |
| (ii) Net force along the motion | increase | The acceleration is in the direction of motion |
| (iii) Net force opposite to the motion | decrease | The acceleration is opposite to the motion, so the ball slows down |
Question 23: Net force on blocks P and Q
Two blocks P and Q on a smooth horizontal surface are shown in figure (a) and figure (b). Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity. Which of the following statements is correct?
- P experiences a net force and Q does not experience a net force.
- P does not experience a net force and Q experiences a net force.
- Neither P nor Q experiences a net force.
- Both P and Q experience a net force.
Correct option: (a)
- Block P: the forces are 5 N and 4 N in opposite directions, so the net force = 5 − 4 = 1 N, which is not zero.
- Block Q: it moves with constant velocity, so its acceleration is zero and the net force is zero.
Question 24: Net force on a snake boat
While practising for the snake boat race (Vallam Kali 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.)
- Forward force95 × 200 = 19000 N
- Backward force5 × 200 = 1000 N
- Net force19000 − 1000 = 18000 N
The net force is 18000 N in the forward direction.
Question 25: Direction and size of acceleration produced by a net force
When a net force acts on an object, we observe that the object accelerates:
- Opposite to force; proportional to force.
- Opposite to force; proportional to mass.
- Along the force; inversely proportional to force.
- Along the force; proportional to force.
Correct option: (d)
By Newton’s second law, the acceleration is in the direction of the net force and is directly proportional to the force (a = F ÷ m).
Question 26: On which object does a net force act?
The position-time graphs for four objects A, B, C and D moving along a straight line are given in the figure. A net force acts on:
- Object D
- Object B
- Object C
- Object A
Correct option: (c) Object C
A net force acts on an object only when its velocity changes, that is, when it accelerates. In a position-time graph:
- A straight sloping line means constant velocity, so there is no net force (Objects A and D).
- A horizontal line means the object is at rest, so there is no net force (Object B).
- A curved line means the velocity is changing, so a net force acts (Object C).
Question 27: A sailor jumps from a boat to the shore
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 will move backward (opposite to the direction of the jump).
When the sailor jumps forward he pushes the boat backward. By Newton’s third law, every action has an equal and opposite reaction, so the boat is pushed backward with an equal force.
Question 28: Why is a landing mat used in high jump?
During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason behind it.
A landing mat or sand bed increases the time of impact when the athlete lands. Since F = Δp ÷ Δt, a longer time of contact means a smaller force on the athlete’s body.
So the mat reduces the impact force and prevents injury.
Question 29: Collision of a loaded and an empty hand cart
A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
- The empty cart exerts more force.
- No force acts between them.
- Both exert equal forces.
- The loaded cart exerts more force.
Correct option: (c)
By Newton’s third law, the two carts exert forces on each other that are equal in size and opposite in direction, whatever their masses are.
Question 30: Force-mass graph from an acceleration-mass graph
The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in the figure. Plot the force-mass graph for this case.
From the graph, the acceleration falls as the mass rises (a ∝ 1/m). For example, 1 kg gives 10 m/s2 and 5 kg gives 2 m/s2.
F = ma = 1 × 10 = 5 × 2 = 10 NSo the force is the same for every mass. The force-mass graph is a horizontal straight line at F = 10 N, parallel to the mass axis. The force does not depend on the mass.
Question 31: Force on a 10 kg object from its velocity-time graph
The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in the figure. Calculate the force acting on the object by using the graph.
- Read the graphThe velocity is 10 m/s at t = 0, 20 m/s at t = 4 s and 30 m/s at t = 8 s.
- Acceleration (slope of the graph)a = (30 − 10) ÷ 8 = 2.5 m/s2
- ForceF = ma = 10 × 2.5 = 25 N
The force acting on the object is 25 N.
Question 32: Stopping force on a bullet in a wooden block
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).
- Givenm = 0.05 kg, u = 100 m/s, v = 0, s = 0.5 m
- AccelerationUsing v2 = u2 + 2as0 = 1002 + 2 × a × 0.5 ⇒ a = −10000 m/s2
- ForceF = ma = 0.05 × (−10000) = −500 N
The stopping force is 500 N, opposite to the motion.
Question 33: Time of contact of a footballer’s kick
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.
- Givenv = 108 km/h = 30 m/s, u = 0, m = 0.4 kg, F = 800 N
- FormulaF = m(v − u) ÷ t, so t = m(v − u) ÷ F
- Calculationt = 0.4 × 30 ÷ 800 = 12 ÷ 800 = 0.015 s
The time of contact is 0.015 s.
Question 34: Distance covered on a rough patch
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?
- Total opposing force7 + 3 = 10 N
- Accelerationa = −10 ÷ 2 = −5 m/s2
- DistanceUsing v2 = u2 + 2as0 = 102 + 2(−5)s ⇒ s = 10 m
The object travels 10 m before coming to rest.
Question 35: Acceleration of a tractor pulling a harrow on a trolley
A tractor pulls a harrow 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), obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.
- Masses from the two casesm1 = F ÷ a1 m2 = F ÷ a2
- Combined massm1 + m2 = F(1/a1 + 1/a2)
- Resulting acceleration aUsing F = (m1 + m2)aF = F(1/a1 + 1/a2) a ⇒ a = 1 ÷ (1/a1 + 1/a2)
a = a1a2 ÷ (a1 + a2)
Question 36: Why does the compass needle move but the bar magnet does not?
When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move. Explain why.
The forces are equal, but motion depends on the acceleration, a = F ÷ m.
- The compass needle has a very small mass, so even a small force gives it a large acceleration. It moves easily.
- The bar magnet has a much larger mass (and is usually held fixed), so the same force gives it a negligible acceleration. It does not move.
Quick revision: Class 9 Science Chapter 6
- Newton’s first law: an object stays at rest or moves with constant velocity unless an unbalanced force acts on it.
- Newton’s second law: F = ma; the acceleration is along the force and is proportional to it (and inversely proportional to the mass).
- Newton’s third law: forces come in equal and opposite pairs acting on two different objects.
- Constant velocity means zero acceleration, so the net force is zero.
- F = Δp ÷ Δt: increasing the time of impact reduces the force (bubble wrap, landing mats).
- Equal forces give different accelerations when the masses are different (the Earth and a fruit, a gun and a bullet).
- In a position-time graph a curved line means a net force; in a velocity-time graph the slope gives the acceleration.
- Net force on an object = sum of forces in one direction minus the sum in the opposite direction.
Frequently asked questions
How many questions are solved in Class 9 Science Chapter 6?
All 36 questions of Chapter 6, How Forces Affect Motion, from the new NCERT Exploration textbook (2026-27) are solved on this page.
What are Newton’s three laws of motion?
The first law says an object keeps its state of rest or uniform motion unless a net force acts on it, the second law says F = ma, and the third law says that every action has an equal and opposite reaction.
Why do equal action and reaction forces not cancel each other?
They act on two different objects, so they cannot cancel. Each force produces an acceleration in the object it acts on.
How do you find the force from a velocity-time graph?
Find the acceleration from the slope of the graph, a = (v − u) ÷ t, and then use F = ma.
More free CBSE study material
Notes, worksheets, PPTs and practice papers for Classes 6 to 10 are available at uniquestudyonline.com.
Solutions prepared by Unique Study Point by Sumeet Sahu. Figures are from the NCERT Science textbook Exploration (Class 9).
📋 Details
| Class | Class IX (CBSE / NCERT) |
| Subject | Science |
| Chapter | Chapter 6: How Forces Affect Motion |
| Resource Type | NCERT Solutions |
| Last Updated | 26 October 2026 |
| Session | 2026-27 (Latest NCERT Syllabus) |
| Downloads | 0+ |
| Prepared by | Sumeet Sahu, Unique Study Point, Indore |
| Cost | Free |