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NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion (PDF) – 2027

NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.

Chapter 4 told you how an object moves; this chapter tells you why. A force is a push or a pull with a direction, and only the net force matters. From that single idea come Newton’s three laws of motion — the reason a coin slides to a stop, a fielder pulls his hands back, an airbag saves a life and a rocket lifts off.

NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion PDF

All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.

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NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – Questions and Answers

All 62 questions of the chapter (textbook pages 94 – 115) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.

Think It Over — Page 94Chapter opener

Think It Over
Q1Why does a canoe move forward when the canoeist pushes water backwards with their paddle and why does it move faster when they push harder?
Answer

The canoe moves forward because of Newton’s third law of motion: the paddle pushes the water backwards, so the water pushes the paddle — and with it the canoe — forwards with an equal force. When the canoeist pushes harder, the water pushes forward harder too, and a larger force on the same canoe gives a larger acceleration (a = F/m), so the canoe moves faster.

Paddle pushes water backwards with force F
Water pushes paddle forwards with the same force F
The two forces act on different objects — water and paddle — so they do not cancel
Q2Suppose 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?
Answer

The empty canoe moves faster.

a = F ÷ m (Newton’s second law)
Same paddle force F in both cases
Empty canoe: smaller m → larger a
Loaded canoe: larger m → smaller a

In-text Questions — Page 956.2 Balanced and Unbalanced Forces

Q1In such cases, what is the effect of forces when more than one force is acting on an object at rest or in motion?
Answer

Only the net force matters — the object behaves exactly as if that single net force were the only force acting on it.

  • If the forces are balanced (equal magnitudes, opposite directions), the net force is zero. An object at rest stays at rest; an object already moving keeps moving with constant velocity. A ball floating on water is like this — the downward gravitational force and the upward buoyant force balance.
  • If the forces are unbalanced, a non-zero net force acts. Forces in the same direction add; forces in opposite directions subtract, and the net force points along the larger one. The object then accelerates in the direction of the net force.

Pause and Ponder — Page 976.2 Balanced and Unbalanced Forces

Pause and Ponder
Q1A weightlifter lifts a barbell (Fig. 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?
Barbell Weightlifter
Fig. 6.8, page 97 — a weightlifter holding a barbell steady above her head (redrawn sketch).
Answer

The two forces on the barbell are:

  1. The gravitational force (weight) of the barbell, exerted by the Earth, acting vertically downwards.
  2. The upward force applied by the weightlifter’s hands on the bar.

Yes — while she holds it steady, the two forces are balanced.

Q2Two players R and S are participating in an arm-wrestling match (Fig. 6.9). 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?
Table R S
Fig. 6.9, page 97 — the arm-wrestling match: player R on the left, player S on the right, elbows on the table and hands locked together (redrawn sketch).
Answer

No, the forces are not balanced. The locked hands move in the direction of the larger force. In Fig. 6.9 both players use their right hands: R pushes the hands into the page (towards R’s left) and S pushes them out of the page (towards S’s left). Since the arms tilt out of the page, player S exerted the larger force.

In-text Questions — Page 976.3 The Force of Friction: Often Overlooked but Always Present

Q1Will the force applied by you make the object move? Many a times, you might have experienced that on applying a force on an object, it did not move and you had to apply a larger force to move it. Why is it so?
Answer

Not necessarily. The object moves only if your push is larger than the force of friction.

Applied force Fapplied acts forwards
Force of friction Ffriction acts backwards, between the bottom of the object and the floor
Up to a limit, friction grows to match your push → net force = zero → the object stays at rest
Push larger than the largest friction the surfaces can give → net force forwards → the object starts to move
Q2Does it mean that you need to continuously apply a force to keep it moving?
Answer

Yes — but only because friction is present. The force is needed to cancel friction, not to keep the motion going.

To keep constant velocity: Fapplied = Ffriction
→ net force = 0 → no acceleration → velocity stays the same
Stop applying the force: net force = Ffriction backwards → the object decelerates and stops

What if … — Page 986.3 The Force of Friction: Often Overlooked but Always Present

What if …
Q1the force of friction disappears in the world? How will the motion of objects be impacted?
Answer

Nothing that is moving would ever slow down on its own — and almost nothing could be started, steered or stopped by us.

  • Moving objects would never stop. A ball rolled once, a cycle after you stop pedalling, a stack of coins released by a rubber band — each would keep moving with constant velocity for ever, exactly as Newton’s first law predicts when the net force is zero.
  • Walking and running would be impossible. You move forward because you push the ground backwards and friction pushes your foot forwards. With no friction your foot would just slide back and you would fall.
  • Vehicles could neither accelerate nor brake. Tyres would spin without gripping; brakes work purely by friction.

Activity 6.1: Let us investigate — Page 986.3 The Force of Friction: Often Overlooked but Always Present

Activity
Q1Do you find that 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?
Answer

Yes. The stack slows down steadily and stops after covering some distance from C.

While in contact with the band: force by band > friction → net force forwards → stack accelerates
After losing contact: only friction acts, backwards → net force backwards → stack decelerates
Velocity falls from its highest value to 0 m s–1, then the stack stops
Q2Does the stack of coins travel a larger distance than it did on the wooden table top before coming to rest? Does its velocity decrease more slowly now?
ABC(b) marks A, B and CABCstack of 4 coinsrubber band(c) band pulled back to C
Fig. 6.12 (b) and (c), page 98 — redrawn sketch of the Activity 6.1 set-up.
Answer

Yes to both. On the laminated top the stack goes further, and its velocity falls more gradually. The laminated top is smoother than wood, so the force of friction on the coins is smaller.

Q3Does the stack of coins travel an even larger distance and its velocity decrease even more slowly?
Answer

Yes. On polished marble or tiles the distance is the largest of the three and the slowing is the gentlest.

Ranking of friction: wooden top > laminated top > polished marble/tile
Ranking of deceleration (a = Ffriction ÷ m): wooden > laminated > marble
Ranking of distance travelled: marble > laminated > wooden
Q4What conclusion do you draw from your observations?
Answer

The force of friction depends on the nature of the surfaces in contact — and the smaller the friction, the slower the velocity falls and the further the object travels before stopping.

SurfaceNatureForce of frictionDistance travelled from C
Wooden table topRoughest of the threeLargestSmallest
Laminated table topSmootherSmallerLarger
Polished marble / tileSmoothestSmallestLargest

In-text Questions — Page 996.3 The Force of Friction: Often Overlooked but Always Present

Q1But how can you check if the force of friction is indeed different for different surfaces?
Answer

Measure it directly with a spring balance, as Activity 6.2 does.

  1. Lay the spring balance horizontally on the surface and check that it reads zero.
  2. Hook a wooden block to it and pull with a slowly increasing force.
  3. Note the reading at the instant the block just begins to move. That reading is an approximate measure of the force of friction between the block and that surface.
  4. Repeat on each of the four surfaces and compare the readings.

Activity 6.2: Let us measure — Page 996.3 The Force of Friction: Often Overlooked but Always Present

Activity
Q1Pull the spring balance with gradually increasing force and note down the reading on it when the block just starts moving. What does this reading indicate?
Answer

It indicates the magnitude of the force of friction acting on the block from that surface (an approximate value of the largest friction the surface can supply before sliding begins).

Forces on the block, horizontal: pull by spring (forwards), friction (backwards)
At the instant of “just starting to move” the velocity is not yet changing
→ acceleration ≈ 0 → net force ≈ 0
→ Fspring = Ffriction
So the scale reading = force of friction, in newtons
Q2If the velocity of the block is neither increasing nor decreasing, what can you say about the net force acting on the block?
wooden blockstringspring balancepull
Fig. 6.14, page 99 — redrawn sketch: a block being pulled by a spring balance.
Answer

The net force on the block is zero.

Velocity constant → change in velocity = 0
→ acceleration a = 0 m s–2
F = ma = m × 0 = 0 N
Q3Does the reading of the spring balance indicate the magnitude of the force of friction acting on the wooden block?
Answer

Yes — approximately. Because the net force is zero, the spring’s pull and friction must be equal in magnitude and opposite in direction, so the scale reading gives the frictional force.

Q4Compare the readings of the spring balance for all surfaces. Are the readings different? Is the reading smallest for the surface on which the stack of coins travelled the largest distance? Is the reading largest for which the distance travelled was the smallest?
Answer

Yes to all three. The readings differ from surface to surface, and they line up exactly with Activity 6.1.

SurfaceDistance in Activity 6.1Spring balance reading in Activity 6.2Force of friction
Polished marble / tileLargestSmallestSmallest
Laminated table topMediumMediumMedium
Wooden table topSmallestLargestLargest

Think as a Scientist — Page 1006.3 The Force of Friction: Often Overlooked but Always Present

Think as a Scientist
Q1Now, conduct a thought experiment. We do a thought experiment when the conditions required for the experiment are difficult to recreate in the real world. Suppose, you find an object and a horizontal floor having such smooth surfaces that the force of friction between them is zero. Imagine, what will happen if you repeat steps 3 and 4 of Activity 6.1 with such an object and a horizontal floor? Will the velocity of the object decrease? Will the object ever come to rest or continue moving forever?
Answer

The object would be set moving by the rubber band, and then it would keep moving for ever with constant velocity. Its velocity would not decrease, and it would never come to rest.

While in contact with the band: net force forwards → velocity rises from 0 to v
After losing contact: friction = 0 N, and no other horizontal force acts
→ net horizontal force = 0 N
→ a = F/m = 0 ÷ m = 0 m s–2
→ velocity stays at v for ever, in a straight line

Pause and Ponder — Page 1016.4 Newton’s First Law of Motion

Pause and Ponder
Q3An object is moving with a constant velocity. Is there a net force acting upon it?
Answer

No. The net force on it is zero.

Constant velocity → no change in magnitude or direction of velocity
→ acceleration a = 0 m s–2
Net F = ma = m × 0 m s–2 = 0 N
Q4Suppose, no net force is acting on an object. Which of the following situations are possible?
(i) Object remains at rest if at rest.
(ii) Object keeps moving with a constant velocity if already moving.
(iii) Object is moving with a constant acceleration.
Answer

(i) and (ii) are possible. (iii) is not possible.

SituationPossible?Reason
(i) Remains at rest if at restYesZero net force means zero acceleration; a body at rest has no reason to start moving
(ii) Keeps moving with constant velocityYesZero acceleration means the velocity cannot change in magnitude or direction
(iii) Moving with constant accelerationNoAny acceleration a ≠ 0 needs a net force F = ma ≠ 0 — a contradiction
Q5In 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.
Answer

You cannot remove forces, but you can balance them — add forces so that they cancel in pairs and the net force becomes zero.

Example — a book lying on a table.

Downward: weight of the book, F = mg
Upward: normal force N from the table
Net force = N – mg = 0 N, since N = mg
→ a = 0 m s–2 → the book stays at rest

Think as a Scientist — Page 1026.5 Newton’s Second Law of Motion

Think as a Scientist
Q1Now, how can you test your hypothesis? You will have to think of an activity where you can apply forces of different magnitudes on the same object and the same surface to find the resultant acceleration. How can you apply forces of different magnitudes?
Answer

Use weights of different magnitudes as the source of the force — the Earth pulls a heavier hanging load with a larger gravitational force.

The design of the test (Activity 6.3):

  1. Make a cart with ball-bearing wheels and tie a thread to it.
  2. Pass the thread over a pipe acting as a pulley at the edge of the table, and hang a paper cup from the free end.
  3. Put objects in the cup. As the cup falls, the thread pulls the cart with a constant force equal to the weight of the cup and its contents.
  4. Time the cart over a fixed distance s using a slow-motion video — call it T1.
  5. Double the mass in the cup (so the force doubles) and time it again — T2.

Activity 6.3: Let us experiment (Demonstration activity) — Page 1026.5 Newton’s Second Law of Motion

Activity
Q1Using the values of the time measured, let us do some analysis.
Answer

Analysis: both runs start from rest and cover the same distance s.

Kinematic equation, u = 0: s = ut + ½at2 = ½at2

Run 1 (force F): s = ½ a1T12
Run 2 (force 2F): s = ½ a2T22

Distance is the same in both runs, so
½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22

What the numbers show: with double the load in the cup the cart reaches the pipe sooner, so T2 < T1, which makes T12/T22 greater than 1 and therefore a2 > a1. So, for a fixed mass, the acceleration increases as the net force increases.

Think as a Scientist — Page 1036.5 Newton’s Second Law of Motion

Think as a Scientist
Q1Apart from force, does acceleration depends on any other factor? From everyday experiences, you know that with the same magnitude of force, it is easier to set lighter objects in motion than heavier ones. This leads to a second hypothesis, that for the same force, a smaller mass has a larger acceleration (or a larger mass has a smaller acceleration). Now how can you test your second hypothesis?
Answer

Run the same experiment again, but this time hold the force constant and change the mass of the cart.

  1. Repeat Activity 6.3 with one change: keep the mass of the cup and the objects in it constant, so the pulling force stays the same.
  2. Double the mass of the cart by adding objects inside it, and measure that mass on a weighing scale.
  3. Record the time over the same distance s, exactly as before, from the slow-motion video.
  4. Use a2 ÷ a1 = T12 ÷ T22 to compare the two accelerations.

Activity 6.4: Let us experiment (Demonstration activity) — Page 1036.5 Newton’s Second Law of Motion

Activity
Q1Using the values of time measured, find the ratio of acceleration for these two cases. Do you find that for the same force, when you increased the mass of the cart, the acceleration decreased?
Answer

Yes. The heavier cart takes longer over the same distance, which means its acceleration is smaller.

Both runs start from rest and cover the same distance s
s = ½ a1T12 = ½ a2T22
a2 ÷ a1 = T12 ÷ T22

Doubling the cart’s mass makes it slower, so T2 > T1
→ T12 ÷ T22 < 1 → a2 < a1

Threads of Curiosity — Page 1046.5 Newton’s Second Law of Motion

Threads of Curiosity
Q1How much does a force of 1 N feel?
Answer

About the weight of a 100 g mass resting in your palm — a small apple, or a standard 100 g packet.

m = 100 g = 0.1 kg, g = 9.8 m s–2
Weight, F = mg
F = 0.1 kg × 9.8 m s–2
F = 0.98 N ≈ 1 N

The mass presses down on your palm with about 1 N, and your palm pushes it back up with about 1 N. The two forces balance, which is why the mass stays at rest.

Pause and Ponder — Page 1066.5 Newton’s Second Law of Motion

Pause and Ponder
Q6A 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?
Answer

The net force is zero.

m = 100 g = 0.1 kg, velocity is constant at 0.5 m s–1
Change in velocity = 0 → a = 0 m s–2

F = ma
F = 0.1 kg × 0 m s–2
F = 0 N
Q7Two 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.
Answer

You need the larger force for the heavier child — the one with the greater mass.

F = ma
Acceleration a is to be the same for both children
So F is directly proportional to m

Lighter child, m1: F1 = m1a
Heavier child, m2 (> m1): F2 = m2a > F1
Q8How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Answer

The bubble wrap or hay stretches out the stopping time of every jolt, and a longer stopping time means a smaller force on the glass.

During a bump, the glass has to lose its velocity: from u to 0
a = (v – u) ÷ t = (0 – u) ÷ t
F = ma = –mu ÷ t

The mass m and the velocity change u are fixed by the jolt
So larger t → smaller |a| → smaller |F|

Activity 6.5: Let us explore — Page 1076.6 Newton’s Third Law of Motion

Activity
Q1Now, using both your hands, push the table away from you, i.e., apply a force on the table in the forward direction as shown in Fig. 6.23a. What happens to you? Does the chair you are sitting upon move in the opposite direction?
Force applied by the girl on the table Force applied by the table on the girl Table Chair with wheels
Fig. 6.23a, page 107 — pushing the table in the forward direction, and the pair of equal and opposite forces that results (redrawn sketch).
Answer

You (with the chair) move backwards — yes, the chair moves in the direction opposite to your push.

You push the table forwards with force F
The table pushes you backwards with the same force F
Your feet are off the floor, so no friction holds you
Net force on you + chair = F backwards → you accelerate backwards
Q2Now, try to pull the table towards you, i.e., apply a force on the table in the direction opposite to that in step 2 (Fig. 6.23b). In which direction does your chair move now?
Force applied by the girl on the table Force applied by the table on the girl Table Chair with wheels
Fig. 6.23b, page 107 — pulling the table towards you; both arrows of the pair flip over (redrawn sketch).
Answer

The chair moves forwards, towards the table.

You pull the table towards you (backwards, from the table’s point of view)
The table pulls you towards it with an equal force
Net force on you + chair points towards the table → you roll forwards
Q3What conclusion can you draw from this activity?
Answer

Whenever you exert a force on the table, the table exerts an equal force on you in the opposite direction — at the same instant. This is 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 object

In-text Questions — Page 1076.6 Newton’s Third Law of Motion

Q1If you are sitting on a bicycle and you want to move forward without pedalling, what do you do (Fig. 6.24)?
Answer

You push the ground backwards with your feet. By Newton’s third law, the ground pushes your feet forwards with an equal and opposite force, and this force moves you forward along with the bicycle.

Activity 6.6: Let us verify — Page 1086.6 Newton’s Third Law of Motion

Activity
Q1Imagine that you are pulling the free end of the other spring balance with your other hand. Predict what will be the readings of their scales if the spring balances are stationary.
Answer

Prediction: both scales will read exactly the same number of newtons.

Balance A pulls Balance B with force FAB
Balance B pulls Balance A with force FBA
Newton’s third law: |FAB| = |FBA|
Each spring balance shows the magnitude of the force pulling it
→ Reading A = Reading B
Q2Now, carry out step 3. Repeat it multiple times by varying the magnitude of the force applied by you. Is your observation same as your prediction?
Answer

Yes. Whatever force you pull with — small, medium or large — the two scales read the same value every single time.

TrialReading of balance AReading of balance BConclusion
Gentle pulle.g. 2 N2 NEqual
Medium pulle.g. 5 N5 NEqual
Hard pulle.g. 9 N9 NEqual

Activity 6.7: Let us understand — Page 1096.6 Newton’s Third Law of Motion

Activity
Q1Remove the thread tied to the neck of the balloon and observe in which direction the straw and the balloon move.
Answer

The balloon and straw shoot along the thread in the direction opposite to the direction in which the air rushes out of the neck.

Stretched balloon pushes air molecules out through the neck, with force F
Escaping air pushes the balloon material the other way, with force F
The two forces act on different objects — air and balloon — so they do not cancel
Net force on balloon + straw is forwards → it accelerates along the thread

In-text Questions — Page 1096.6 Newton’s Third Law of Motion

Q1What will happen if the engine of a rocket moving in the space, fires in the direction of its motion?
Answer

The rocket slows down.

Exhaust gas is thrown out forwards (along the direction of motion)
By Newton’s third law, the gas pushes the rocket backwards
Net force on the rocket is opposite to its velocity
→ acceleration is negative (retardation) → speed decreases

Pause and Ponder — Page 1106.6 Newton’s Third Law of Motion

Pause and Ponder
Q9Why does a fireperson sometimes struggle when holding the pipe issuing water?
Answer

Because the water pushes the hose backwards with a force equal to the force the hose uses to push the water forwards — and that backward force can be very large.

Hose pushes water forwards at high speed, with force F
Newton’s third law → water pushes hose backwards with force F
The fireperson must supply a forward force of F just to keep the hose still
Several kilograms of water are pushed out at high speed every second → large F
Q10Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.
Answer

By firing its engine and expelling gas — the only way to get a force when there is nothing outside to push against.

Engine expels exhaust gas in some direction with force F
By Newton’s third law, the gas pushes the spacecraft with force F in the opposite direction
That is the net force, since gravity is negligible
a = F ÷ m → the velocity changes along the direction opposite to the exhaust
  • To speed up: fire the exhaust backwards, opposite to the direction of motion.
  • To slow down: fire the exhaust forwards, along the direction of motion.
  • To turn: fire small side thrusters, so the reaction force acts sideways and changes the direction of the velocity.

In-text Questions — Page 1106.6 Newton’s Third Law of Motion

Q1You have learnt about contact and non-contact forces earlier. Is Newton’s third law applicable only for contact forces?
Answer

No. Newton’s third law applies to all types of forces, contact or non-contact, that we come across in everyday mechanical situations. For example:

  • Magnetic force: two bar magnets exert equal and opposite forces on each other (Fig. 6.31).
  • Electrostatic force: two similarly charged balloons push each other apart with equal and opposite forces (Fig. 6.32).
  • Gravitational force: the Earth pulls a fruit down, and the fruit pulls the Earth up with an equal force (Fig. 6.33).

In-text Questions — Page 1116.7 Forces Acting on a System of Objects

Q1But can we apply these laws to two or more objects connected together?
Answer

Yes. Two or more connected objects can be treated as a single system and Newton’s laws applied to that system as a whole.

Forces inside the system (internal forces) — e.g. the tension T in the connecting string — need not be considered
Only forces from outside the system (external forces) matter — e.g. the applied force F

a = F ÷ (mass of the system) = F ÷ (m1 + m2) … Eq. 6.4
Q2How can we find the acceleration of each box?
Answer

Treat the two boxes and the string as one system. Both boxes are tied together, so both have the same acceleration:

External force on the system = F (tension T is internal)
Mass of the system = m1 + m2

a = F ÷ (m1 + m2) … Eq. 6.4
This single value of a is the acceleration of both Box 1 and Box 2

Revise, Reflect, Refine — Pages 112 – 114End-of-chapter exercise

Revise, Reflect, Refine
Q1Using 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?
Answer

The frictional force is F — equal in magnitude to the applied force, and opposite in direction.

Velocity is constant → a = 0 m s–2
Net horizontal force = ma = m × 0 = 0 N

Horizontal forces: applied force F (forwards), friction Ff (backwards)
F – Ff = 0
Ff = F, directed opposite to the motion
Q2For 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 on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
Answer
PartCorrect optionReason
(i) No net forceremain the samea = F/m = 0 ÷ m = 0 m s–2; velocity cannot change (Newton’s first law)
(ii) Net force along the motionincreaseAcceleration is along the velocity, so the ball speeds up
(iii) Net force opposite to the motiondecreaseAcceleration is opposite to the velocity — a retardation, so the ball slows down
Q3Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. 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 statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.
P 5 N 4 N (a) Q (b)
Fig. 6.36, page 112 — (a) block P with 5 N and 4 N acting in opposite directions, (b) block Q moving with a constant velocity.
Answer

The correct statement is (i) — P experiences a net force and Q does not experience a net force.

Block P: 5 N to the right, 4 N to the left (Fig. 6.36a)
The forces are opposite, so subtract them
Net force = 5 N – 4 N = 1 N to the right → non-zero

Block Q: moving with constant velocity (Fig. 6.36b)
a = 0 m s–2 → Net force = ma = 0 N
Q4While 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.)
Answer

The net force is 18 000 N in the forward direction.

Force by each oarsman = 200 N

Forward group: 95 oarsmen
Fforward = 95 × 200 N = 19 000 N

Opposing group: 5 oarsmen
Fbackward = 5 × 200 N = 1 000 N

The two groups push in opposite directions, so subtract
Net force = 19 000 N – 1 000 N = 18 000 N, forwards
Q5When a net force acts on an object, we observe that the object accelerates:
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.
Answer

The correct option is (iv) — in the direction of force, with acceleration proportional to the force acting on the object.

Newton’s second law: a = F ÷ m, i.e. F = ma
Direction of a = direction of the net force F
For fixed m: a ∝ F (double the force, double the acceleration)
For fixed F: a ∝ 1 ÷ m (double the mass, halve the acceleration)
Q6The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D
Position 0 Time Object A Position 0 Time Object B Position 0 Time Object C Position 0 Time Object D
Fig. 6.37, page 113 — position–time graphs of the four objects A, B, C and D.
Answer

The correct option is (iii) — Object C.

ObjectShape of the position–time graphWhat the slope tells youNet force
AStraight line, sloping upwardsConstant positive velocityZero
BHorizontal straight linePosition not changing — the object is at restZero
CCurve that gets steeper and steeperVelocity increasing — the object is acceleratingNon-zero
DStraight line, sloping downwardsConstant negative velocity (moving back at a steady rate)Zero
Q7A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.
Shore Sailor Boat
Fig. 6.38, page 113 — a sailor jumping forward from a small boat on to the shore (redrawn sketch).
Answer

Yes — the boat moves backwards, away from the shore.

Sailor pushes the boat backwards with force F (in order to jump forward)
By Newton’s third law, the boat pushes the sailor forwards with force F
Water offers very little friction, so the backward force on the boat is nearly unopposed
aboat = F ÷ mboat, directed away from the shore
Q8During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.
Landing mat Bar
Fig. 6.39, page 113 — an athlete clearing the bar in a high jump, with the landing mat below (redrawn sketch).
Answer

The mat increases the time over which the athlete’s velocity falls to zero, which reduces the acceleration and therefore the force on the athlete.

Athlete lands with downward velocity u and is brought to v = 0
a = (v – u) ÷ t = –u ÷ t
F = ma = –mu ÷ t

m and u are the same whether the surface is hard or soft
Soft mat → much larger t → much smaller |a| → much smaller |F|
Q9A 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

The correct option is (iv) — both carts exert forces of equal magnitude on each other.

Newton’s third law: whenever one object exerts a force on a second,
the second simultaneously exerts an equal and opposite force on the first

F (loaded on empty) = F (empty on loaded) in magnitude
The two forces are opposite in direction and act on different carts

But the accelerations are not equal:
aempty = F ÷ mempty, aloaded = F ÷ mloaded
mloaded > mempty → aempty > aloaded
Q10The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
2.5 5.0 7.5 10.0 1 2 3 4 5 0 Mass (kg) Acceleration (m s⁻²)
Fig. 6.40, page 114 — the acceleration–mass graph, redrawn from the printed values (1 kg, 10.0 m s⁻²), (2 kg, 5.0 m s⁻²), (4 kg, 2.5 m s⁻²) and (5 kg, 2.0 m s⁻²).
Answer

The force–mass graph is a horizontal straight line at F = 10 N, because the same force was applied to every object.

Read pairs of values off Fig. 6.40, then use F = ma:

m = 1 kg, a = 10.0 m s–2 → F = 1 kg × 10.0 m s–2 = 10 N
m = 2 kg, a = 5.0 m s–2 → F = 2 kg × 5.0 m s–2 = 10 N
m = 4 kg, a = 2.5 m s–2 → F = 4 kg × 2.5 m s–2 = 10 N
m = 5 kg, a = 2.0 m s–2 → F = 5 kg × 2.0 m s–2 = 10 N

F = 10 N for every mass
10 5 15 1 2 3 4 5 Mass (kg) Force (N) F = 10 N, constant
Force–mass graph for Fig. 6.40: a straight line parallel to the mass axis at F = 10 N.
Q11The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
10 20 30 4 8 0 Time (s) Velocity (m s⁻¹)
Fig. 6.41, page 114 — the velocity–time graph, redrawn from the printed values: a straight line through (0 s, 10 m s⁻¹), (4 s, 20 m s⁻¹) and (8 s, 30 m s⁻¹).
Answer

The force acting on the object is 25 N, in the direction of motion.

From Fig. 6.41, the graph is a straight line through the points
(t = 0 s, v = 10 m s–1), (t = 4 s, v = 20 m s–1), (t = 8 s, v = 30 m s–1)

Acceleration = slope of the velocity–time graph = (v – u) ÷ t
a = (30 m s–1 – 10 m s–1) ÷ (8 s – 0 s)
a = 20 m s–1 ÷ 8 s = 2.5 m s–2

Newton’s second law: F = ma
F = 10 kg × 2.5 m s–2
F = 25 kg m s–2 = 25 N
Q12A 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).
Answer

The stopping force is 500 N, acting opposite to the bullet’s motion.

Convert to SI units first
m = 50 g = 0.05 kg, u = 100 m s–1, v = 0 m s–1, s = 50 cm = 0.5 m

Step 1 — find the acceleration using v2 = u2 + 2as
(0 m s–1)2 = (100 m s–1)2 + 2 × a × 0.5 m
0 = 10 000 m2 s–2 + (1.0 m) a
a = –10 000 m2 s–2 ÷ 1.0 m = –10 000 m s–2

Step 2 — find the force using F = ma
F = 0.05 kg × (–10 000 m s–2)
F = –500 N
The minus sign means the force opposes the motion; its magnitude is 500 N
Q13An 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.
Answer

The time of contact is 0.015 s, that is 15 milliseconds.

Step 1 — convert the speed to SI units
v = 108 km h–1 = 108 × 1000 m ÷ 3600 s = 30 m s–1
The ball starts from rest, so u = 0 m s–1

Step 2 — find the acceleration from F = ma
a = F ÷ m = 800 N ÷ 0.4 kg
a = 2000 kg m s–2 ÷ kg = 2000 m s–2

Step 3 — find the contact time from v = u + at
30 m s–1 = 0 m s–1 + (2000 m s–2) t
t = 30 m s–1 ÷ 2000 m s–2
t = 0.015 s = 15 ms
Q14An 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?
Answer

The object travels 10 m in the rough patch before stopping.

Step 1 — find the net force. Both forces oppose the motion, so they add.
Fnet = 7 N + 3 N = 10 N, directed opposite to the motion

Step 2 — find the acceleration using F = ma
a = F ÷ m = –10 N ÷ 2 kg
a = –5 m s–2 (negative because it retards the motion)

Step 3 — find the distance using v2 = u2 + 2as, with u = 10 m s–1, v = 0
(0 m s–1)2 = (10 m s–1)2 + 2 × (–5 m s–2) × s
0 = 100 m2 s–2 – (10 m s–2) s
s = 100 m2 s–2 ÷ 10 m s–2 = 10 m
Q15A 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.
Answer

The resulting acceleration is a = a1a2 ÷ (a1 + a2).

Step 1 — write each mass in terms of F and its acceleration using F = ma
Harrow alone: F = m1a1 → m1 = F ÷ a1
Trolley alone: F = m2a2 → m2 = F ÷ a2

Step 2 — treat harrow + trolley as one system (Eq. 6.4)
a = F ÷ (m1 + m2)

Step 3 — substitute the masses
a = F ÷ (F/a1 + F/a2)
a = F ÷ [ F (1/a1 + 1/a2) ]
a = 1 ÷ (1/a1 + 1/a2)
a = a1a2 ÷ (a1 + a2)
Q16When 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 of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
N S Bar magnet N S E W Compass needle Magnetic compass
Fig. 6.42, page 114 — the pole of a bar magnet held near a magnetic compass; the needle swings round to face the magnet (redrawn sketch).
Answer

Because equal forces do not produce equal accelerations — the compass needle has a far smaller mass and almost no friction opposing it, so the same force moves it easily.

Newton’s third law: |F on needle| = |F on bar magnet| = F

Newton’s second law: a = F ÷ m
Needle: very small mass mneedle → large acceleration
Bar magnet: much larger mass mmagnet → very small acceleration

The Journey Beyond — Page 115Project work

The Journey Beyond
Q1You know that the force of friction depends on the nature of the surfaces in contact. Does it also depend on how hard the surfaces press each other? Is the friction acting on an object that is about to move larger than the friction after motion begins? Is the friction which acts on a rolling object less than that on a sliding object? Find answers to these questions and create an infographic. Such observations help explain why the invention of the wheel was a major milestone in human history.
Answer

The answers are: yes; yes; and yes. Here is how to establish each one experimentally, and what you should find.

QuestionHow to test itWhat you find
Does friction depend on how hard the surfaces press?Pull the same wooden block with a spring balance, then stack a second identical block on top and pull again on the same surfaceYes — the reading roughly doubles when the pressing force doubles
Is friction just before motion larger than after motion starts?Watch the spring balance carefully: note the peak reading at the instant of slipping, then the steady reading while slidingYes — the reading falls slightly once sliding begins
Is rolling friction less than sliding friction?Drag the block, then put it on four marbles or toy-car wheels and pull again with the same spring balanceYes — the rolling reading is much smaller

What a good infographic must contain:

  • A labelled diagram of the spring balance test, with the reading marked in N.
  • A bar chart comparing your readings: one block, two blocks; just-about-to-move, sliding; sliding, rolling.
  • One clear sentence of conclusion under each comparison.
  • A short “so what” panel: ball bearings in cycle hubs, wheels on carts and suitcases, why brakes and tyre treads need more friction.
Q2Take two toy cars of equal mass and stick a bar magnet on top of each (Fig. 6.32). Fix a metre scale on a smooth surface. Place the cars near the midpoint of the metre scale with the like poles touching. Release the cars and record the time taken (using two stopwatches), and distance travelled by each before coming to a rest. Repeat the experiment after adding equal masses to both cars. Did the cars travel equal distances in opposite directions? Plot a graph of distance travelled versus mass. Analyse and discuss your findings.
Bar magnet Toy car Toy car
Fig. 6.31, page 110 — two toy cars, each carrying a bar magnet, exerting equal and opposite magnetic forces on each other (redrawn sketch).
Answer

Yes — the two cars travel very nearly equal distances in opposite directions, and both distances shrink as you add mass.

Like poles repel, so each car pushes the other away
Newton’s third law: |F on car A| = |F on car B| = F
Equal masses → a = F ÷ m is the same for both → equal speeds, opposite directions

Adding equal masses to both: F is unchanged, m is larger
→ a is smaller → each car leaves with a smaller speed → shorter distance

What the graph should look like. Plot distance travelled (y-axis) against total mass of a car (x-axis). You get a falling curve — greater mass, shorter run — not a straight line.

TrialMass of each carDistance moved by car ADistance moved by car B
1mlargestalmost the same as A
22msmalleralmost the same as A
33msmallestalmost the same as A
Q3Wrap a rope once around a rough tree branch or post. Attach a heavy bucket to one end and try to hold it by the other end (Fig. 6.43). Now, add one more turn of the rope and repeat. You will find that each extra turn increases the ‘grip’ between the rope and the branch, increasing the friction and reducing the force required, making it much easier to hold the same load. The reduction in effort is much larger than you might expect from just adding one turn. This shows that friction does not increase in a simple linear way, small changes in contact can lead to large changes in force. In the same way, friction between a rope and a post allows large ships to be held safely at a pier.
Rope wrapped once round the branch Heavy bucket Branch
Fig. 6.43, page 115 — a rope taken once round a rough tree branch, a heavy bucket hanging from one end and the other end held by hand (redrawn sketch).
Answer

What you will observe: with no turn you must hold nearly the full weight of the bucket. With one turn it becomes noticeably easier; with two turns you can hold the same bucket with a small fraction of the effort; with three or four turns you can hold it with two fingers.

Number of turnsEffort needed to hold the same bucketWhat is happening
0Almost the full weightNo friction from the post; your hand alone supports the load
1Noticeably lessFriction along the wrapped length carries part of the load
2A small fractionFriction from the second turn acts on an already-reduced tension
3 or moreVery littleThe reduction compounds turn after turn
Q4It is often instructive to examine how scientific ideas develop over time. If you are interested, explore how Newton formulated the laws of motion by reading excerpts from his original work, the Principia. Both the original text and commentaries are available online.
Answer

Method for this project: read a short excerpt in translation, compare Newton’s own words with the modern statement in this chapter, and write about what changed and what did not.

Sample answer (an outline you can build on):

IdeaBefore NewtonNewton’s contribution
Motion needs a causeIt was mistakenly believed a force was needed to keep an object moving at constant velocityOnly a change of velocity needs a force; steady motion needs none
InertiaGalileo argued by thought experiment that a body would keep moving if all impediments were removedNewton named the property inertia and made it the first law
Force and accelerationNo quantitative ruleThe second law, stated as the rate of change of momentum being proportional to the net force
InteractionForces were thought of as acting on one bodyThe third law: forces always come in equal and opposite pairs on two different bodies

Chapter 6 of 12 – Science

Also for This Chapter

NCERT Solutions Class 9 Science Chapter 6 How Forces Affect Motion – Overview

ClassClass 9
SubjectScience
Chapter6 – How Forces Affect Motion
Textbook pages94 – 115
Questions solved62
Session2027
FormatFull answers on this page + PDF

NCERT Solutions Class 9 Science Chapter 6 – FAQs

Where can I download NCERT Solutions Class 9 Science Chapter 6 How Forces Affect Motion PDF?

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Is this for the 2027 session?

Yes, it follows the latest NCERT Class 9 Science book for 2027.

How many questions are solved in NCERT Class 9 Science Chapter 6?

All 62 questions of the chapter (textbook pages 94 – 115) – the in-text questions, activities and the end-of-chapter exercises.

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Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.

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