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.
| Chapter | Download | |
|---|---|---|
| Solutions PDF (English medium) | How Forces Affect Motion | View PDF |
| Solutions PDF (Hindi medium) | गति पर बलों का प्रभाव | View PDF |
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
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.
Water pushes paddle forwards with the same force F
The two forces act on different objects — water and paddle — so they do not cancel
The empty canoe moves faster.
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
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
The two forces on the barbell are:
- The gravitational force (weight) of the barbell, exerted by the Earth, acting vertically downwards.
- The upward force applied by the weightlifter’s hands on the bar.
Yes — while she holds it steady, the two forces are balanced.
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
Not necessarily. The object moves only if your push is larger than the force of friction.
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
Yes — but only because friction is present. The force is needed to cancel friction, not to keep the motion going.
→ 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
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
Yes. The stack slows down steadily and stops after covering some distance from C.
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
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.
Yes. On polished marble or tiles the distance is the largest of the three and the slowing is the gentlest.
Ranking of deceleration (a = Ffriction ÷ m): wooden > laminated > marble
Ranking of distance travelled: marble > laminated > wooden
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.
| Surface | Nature | Force of friction | Distance travelled from C |
|---|---|---|---|
| Wooden table top | Roughest of the three | Largest | Smallest |
| Laminated table top | Smoother | Smaller | Larger |
| Polished marble / tile | Smoothest | Smallest | Largest |
In-text Questions — Page 996.3 The Force of Friction: Often Overlooked but Always Present
Measure it directly with a spring balance, as Activity 6.2 does.
- Lay the spring balance horizontally on the surface and check that it reads zero.
- Hook a wooden block to it and pull with a slowly increasing force.
- 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.
- 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
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).
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
The net force on the block is zero.
→ acceleration a = 0 m s–2
F = ma = m × 0 = 0 N
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.
Yes to all three. The readings differ from surface to surface, and they line up exactly with Activity 6.1.
| Surface | Distance in Activity 6.1 | Spring balance reading in Activity 6.2 | Force of friction |
|---|---|---|---|
| Polished marble / tile | Largest | Smallest | Smallest |
| Laminated table top | Medium | Medium | Medium |
| Wooden table top | Smallest | Largest | Largest |
Think as a Scientist — Page 1006.3 The Force of Friction: Often Overlooked but Always Present
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.
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
No. The net force on it is zero.
→ acceleration a = 0 m s–2
Net F = ma = m × 0 m s–2 = 0 N
(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.
(i) and (ii) are possible. (iii) is not possible.
| Situation | Possible? | Reason |
|---|---|---|
| (i) Remains at rest if at rest | Yes | Zero net force means zero acceleration; a body at rest has no reason to start moving |
| (ii) Keeps moving with constant velocity | Yes | Zero acceleration means the velocity cannot change in magnitude or direction |
| (iii) Moving with constant acceleration | No | Any acceleration a ≠ 0 needs a net force F = ma ≠ 0 — a contradiction |
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.
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
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):
- Make a cart with ball-bearing wheels and tie a thread to it.
- 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.
- 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.
- Time the cart over a fixed distance s using a slow-motion video — call it T1.
- 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
Analysis: both runs start from rest and cover the same distance s.
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
Run the same experiment again, but this time hold the force constant and change the mass of the cart.
- 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.
- Double the mass of the cart by adding objects inside it, and measure that mass on a weighing scale.
- Record the time over the same distance s, exactly as before, from the slow-motion video.
- 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
Yes. The heavier cart takes longer over the same distance, which means its acceleration is smaller.
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
About the weight of a 100 g mass resting in your palm — a small apple, or a standard 100 g packet.
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
The net force is zero.
Change in velocity = 0 → a = 0 m s–2
F = ma
F = 0.1 kg × 0 m s–2
F = 0 N
You need the larger force for the heavier child — the one with the greater mass.
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
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.
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
You (with the chair) move backwards — yes, the chair moves in the direction opposite to your push.
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
The chair moves forwards, towards the table.
The table pulls you towards it with an equal force
Net force on you + chair points towards the table → you roll forwards
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.
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
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
Prediction: both scales will read exactly the same number of newtons.
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
Yes. Whatever force you pull with — small, medium or large — the two scales read the same value every single time.
| Trial | Reading of balance A | Reading of balance B | Conclusion |
|---|---|---|---|
| Gentle pull | e.g. 2 N | 2 N | Equal |
| Medium pull | e.g. 5 N | 5 N | Equal |
| Hard pull | e.g. 9 N | 9 N | Equal |
Activity 6.7: Let us understand — Page 1096.6 Newton’s Third Law of Motion
The balloon and straw shoot along the thread in the direction opposite to the direction in which the air rushes out of the neck.
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
The rocket slows down.
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
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.
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
By firing its engine and expelling gas — the only way to get a force when there is nothing outside to push against.
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
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
Yes. Two or more connected objects can be treated as a single system and Newton’s laws applied to that system as a whole.
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
Treat the two boxes and the string as one system. Both boxes are tied together, so both have the same acceleration:
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
The frictional force is F — equal in magnitude to the applied force, and opposite in direction.
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
(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.
| Part | Correct option | Reason |
|---|---|---|
| (i) No net force | remain the same | a = F/m = 0 ÷ m = 0 m s–2; velocity cannot change (Newton’s first law) |
| (ii) Net force along the motion | increase | Acceleration is along the velocity, so the ball speeds up |
| (iii) Net force opposite to the motion | decrease | Acceleration is opposite to the velocity — a retardation, so the ball slows down |
(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.
The correct statement is (i) — P experiences a net force and Q does not experience a net force.
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
The net force is 18 000 N in the forward direction.
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
(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.
The correct option is (iv) — in the direction of force, with acceleration proportional to the force acting on the object.
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)
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D
The correct option is (iii) — Object C.
| Object | Shape of the position–time graph | What the slope tells you | Net force |
|---|---|---|---|
| A | Straight line, sloping upwards | Constant positive velocity | Zero |
| B | Horizontal straight line | Position not changing — the object is at rest | Zero |
| C | Curve that gets steeper and steeper | Velocity increasing — the object is accelerating | Non-zero |
| D | Straight line, sloping downwards | Constant negative velocity (moving back at a steady rate) | Zero |
Yes — the boat moves backwards, away from the shore.
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
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.
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|
(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.
The correct option is (iv) — both carts exert forces of equal magnitude on each other.
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
The force–mass graph is a horizontal straight line at F = 10 N, because the same force was applied to every object.
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
The force acting on the object is 25 N, in the direction of motion.
(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
The stopping force is 500 N, acting opposite to the bullet’s motion.
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
The time of contact is 0.015 s, that is 15 milliseconds.
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
The object travels 10 m in the rough patch before stopping.
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
The resulting acceleration is a = a1a2 ÷ (a1 + a2).
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)
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 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 answers are: yes; yes; and yes. Here is how to establish each one experimentally, and what you should find.
| Question | How to test it | What 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 surface | Yes — 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 sliding | Yes — 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 balance | Yes — 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.
Yes — the two cars travel very nearly equal distances in opposite directions, and both distances shrink as you add mass.
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.
| Trial | Mass of each car | Distance moved by car A | Distance moved by car B |
|---|---|---|---|
| 1 | m | largest | almost the same as A |
| 2 | 2m | smaller | almost the same as A |
| 3 | 3m | smallest | almost the same as A |
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 turns | Effort needed to hold the same bucket | What is happening |
|---|---|---|
| 0 | Almost the full weight | No friction from the post; your hand alone supports the load |
| 1 | Noticeably less | Friction along the wrapped length carries part of the load |
| 2 | A small fraction | Friction from the second turn acts on an already-reduced tension |
| 3 or more | Very little | The reduction compounds turn after turn |
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):
| Idea | Before Newton | Newton’s contribution |
|---|---|---|
| Motion needs a cause | It was mistakenly believed a force was needed to keep an object moving at constant velocity | Only a change of velocity needs a force; steady motion needs none |
| Inertia | Galileo argued by thought experiment that a body would keep moving if all impediments were removed | Newton named the property inertia and made it the first law |
| Force and acceleration | No quantitative rule | The second law, stated as the rate of change of momentum being proportional to the net force |
| Interaction | Forces were thought of as acting on one body | The third law: forces always come in equal and opposite pairs on two different bodies |
Chapter 6 of 12 – Science
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- Next: Chapter 7 Work Energy and Simple Machines
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Also for This Chapter
NCERT Solutions Class 9 Science Chapter 6 How Forces Affect Motion – Overview
| Class | Class 9 |
| Subject | Science |
| Chapter | 6 – How Forces Affect Motion |
| Textbook pages | 94 – 115 |
| Questions solved | 62 |
| Session | 2027 |
| Format | Full 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?
Read every answer on this page, or download the free PDF from the table at the top.
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.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.