NCERT Solutions Class 9 Science Chapter 6 How Forces Affect Motion

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 9 · SCIENCE

NCERT Solutions

Chapter 6: How Forces Affect
Motion

NCERT Textbook — Exploration

BOOK PAGES SECTIONS QUESTIONS MEDIUM

94 – 115 24 60 English

Solutions, notes, sample papers & more at 63 pages

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

CLASS 9 · SCIENCE · EXPLORATION

NCERT Solutions — Chapter 6: How Forces Affect
Motion
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.

TEXTBOOK BOOK PAGES

Exploration (Class 9) 94 – 115

SECTIONS QUESTIONS

24 60

MEDIUM

English

Think It Over — Page 94
Chapter opener

THINK IT OVER

Q1 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?

Because the water pushes back. This is Newton's third law of motion at work — the paddle and
the water form an action–reaction pair.

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

The forward force on the paddle is passed on through the canoeist's arms to the canoe, and the
canoe accelerates forward.

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why pushing harder makes it faster: a harder push means a larger F on the water,
so by the third law the water pushes the paddle forward with a larger F too. By the
second law, a = F/m — with the mass of the canoe fixed, a larger forward force gives
a larger acceleration, so the canoe gains speed more quickly and reaches a higher
velocity in each stroke.

canoe

paddle pushes water back
water pushes paddle
forward

The action–reaction pair that drives a canoe. The two arrows are equal in size but act on different
bodies.

Q2 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 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

Page 2 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why it happens: acceleration is inversely proportional to mass. Adding a passenger
increases the total mass of the canoe-plus-load, so the same force produces a
smaller acceleration. In the same stroke time the empty canoe therefore gains more
velocity.

Check it yourself: if the empty canoe plus canoeist is 100 kg and the passenger
adds 50 kg, a 300 N paddle force gives a = 300 N ÷ 100 kg = 3 m s–2 for the empty
canoe and a = 300 N ÷ 150 kg = 2 m s–2 for the loaded one — two-thirds as much.

In-text Questions — Page 95
6.2 Balanced and Unbalanced Forces

Q1 In such cases, what is the effect of forces when more than one force is acting on an
object at rest or in motion?

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.

Same direction: net F = F1 + F2

Opposite directions: net F = |F1 – F2|, along the larger force

Tip: when you push a box across a floor, the applied force and friction both act. The
box moves only when your push is larger than friction, because only then is the net
force non-zero and forward.

Pause and Ponder — Page 97

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Class 9 Science Chapter 6 How Forces Affect Motion
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6.2 Balanced and Unbalanced Forces
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PAUSE AND PONDER

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why it happens: "steady" means the barbell is at rest, so its velocity is not changing
and its acceleration is zero. By Newton's second law, a = F/m, zero acceleration for a
non-zero mass requires the net force to be zero. Hence the upward push of her
hands must exactly equal the downward weight.

force by hands (up)

weight mg (down)

net force = 0
Forces on a barbell held steady: equal in magnitude, opposite in direction, so they balance.

Did you know? Example 6.4 works out the number: a 30 kg barbell needs F = mg =
30 kg × 9.8 m s–2 = 294 N upwards.

Page 5 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q2 Two 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?

R S

Table

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).

No, the forces are not balanced at that instant. Player R exerted the larger force.

Why it happens: the joined hands only stay still while the two pushes are equal and
opposite — balanced. The moment the hands actually move, the velocity has
changed from zero to something, so there is an acceleration and therefore a non-
zero net force. The net force, and hence the motion, is along the direction of the
larger force. In Fig. 6.9, R is on the left of the picture and the arms tilt out of the page
towards you — that is R's pushing direction, so R is pushing harder.

Tip: "the arms have not moved yet" is the balanced case; "the arms are moving" is
always the unbalanced case. Look at the direction of the movement — it tells you
whose force is bigger.

Page 6 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

In-text Questions — Page 97
6.3 The Force of Friction: Often Overlooked but Always Present

Q1 Will 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?

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

Net force = Fapplied – Ffriction

If Fapplied ≤ Ffriction → net force is zero → the object stays at rest

If Fapplied > Ffriction → net force is forwards → the object accelerates forwards

Why it happens: friction arises between the bottom surface of the object and the
floor and always acts opposite to the direction in which the object tends to move. Up
to a limit, it simply matches whatever you apply, so the two forces stay balanced and
nothing happens. Only once you exceed that limit does a net force appear, and the
object starts to move.

Q2 Does it mean that you need to continuously apply a force to keep it moving?

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

Page 7 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why it happens: this is exactly what Newton's first law says. Motion itself needs no
force to continue; only a change of velocity needs one. On a real floor friction keeps
trying to change the velocity, so you must keep supplying an equal and opposite
force to hold the net force at zero. On a perfectly frictionless floor no force at all
would be needed once the object was moving.

Did you know? For ages people believed a force was needed to keep an object
moving at all. In the 17th century Galileo Galilei argued, from thought experiments,
that if all impediments to motion are removed a body would continue moving
indefinitely.

What if … — Page 98
6.3 The Force of Friction: Often Overlooked but Always Present

WHAT IF …

Q1 the force of friction disappears in the world? How will the motion of objects be
impacted?

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.
Nothing would stay where you put it. Knots, nails, screws, a glass on a slightly tilted table,
a rope round a post — all rely on friction.

Why it happens: friction is the only horizontal contact force in most everyday
situations. Remove it and the horizontal net force on a moving body becomes zero,
so its velocity can never change by itself; and we lose the very force we use to push
against the ground.

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

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Tip: friction is often called a nuisance because it wastes energy, but this question

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Activity 6.1: Let us investigate — Page 98
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6.3 The Force of Friction: Often Overlooked but Always Present

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q2 Does 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?

stack of 4 coins
C
C

A B A B
rubber band

(b) marks A, B and C (c) band pulled back to C

Fig. 6.12 (b) and (c), page 98 — redrawn sketch of the Activity 6.1 set-up.

Yes to both. On the laminated top the stack goes further, and its velocity falls more gradually.

Why it happens: the rubber band is stretched to the same mark C every time, so the
coins start with the same velocity in both trials. What changes is the surface. A
laminated top is smoother than a wooden one, so the force of friction on the coins is
smaller. Since a = F/m, a smaller retarding force gives a smaller deceleration, so the
velocity drops more slowly and the coins travel a longer distance before reaching
rest.

Tip: keeping A, B and C at exactly the same separations is what makes the
comparison fair — it guarantees the same starting velocity in every trial.

Q3 Does the stack of coins travel an even larger distance and its velocity decrease even
more slowly?

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

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Why it happens: a polished marble or tiled floor is smoother still, so friction on the
coins is smallest. The smaller the retarding force, the smaller the deceleration, and
the longer the stack keeps moving before its velocity reaches zero.

Q4 What conclusion do you draw from your observations?

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 DISTANCE TRAVELLED
FRICTION FROM C

Wooden table top Roughest of the Largest Smallest
three

Laminated table top Smoother Smaller Larger

Polished marble / Smoothest Smallest Largest
tile

Why it happens: the coins always start with the same velocity, so the only thing that
can change the stopping distance is the retarding force. Different surfaces give
different frictional forces, hence different decelerations and different distances.
Extending the trend, a surface with zero friction would let the coins move on for ever
— which is precisely the thought experiment on page 100 and the content of
Newton's first law.

In-text Questions — Page 99

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

6.3 The Force of Friction: Often Overlooked but Always Present

Q1 But how can you check if the force of friction is indeed different for different
surfaces?

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.

Why this measures friction: just as the block begins to move it is still travelling with
essentially no change in velocity, so its acceleration is zero and the net force on it is
zero. The only two horizontal forces are the pull of the spring and friction, so they
must be equal in magnitude. Whatever the spring reads, friction equals.

Check it yourself: the surface on which the stack of coins travelled the largest
distance in Activity 6.1 should give the smallest spring balance reading here. Two
different activities, one consistent answer.

Activity 6.2: Let us measure — Page 99
6.3 The Force of Friction: Often Overlooked but Always Present

ACTIVITY

Q1 Pull 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?

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).

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Tip: pull steadily and horizontally. A jerky or slanted pull adds a vertical component
and spoils the reading.

Q2 If the velocity of the block is neither increasing nor decreasing, what can you say
about the net force acting on the block?

wooden block
spring balance
pull

string

Fig. 6.14, page 99 — redrawn sketch: a block being pulled by a spring balance.

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

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why the two activities agree: Activity 6.1 measures friction indirectly — a bigger
friction means a bigger deceleration (a = Ffriction/m) and hence a shorter stopping
distance. Activity 6.2 measures the same friction directly in newtons. Both must, and
do, rank the surfaces in the same order. That agreement is what makes the
conclusion trustworthy.

Think as a Scientist — Page 100
6.3 The Force of Friction: Often Overlooked but Always Present

THINK AS A SCIENTIST

Q1 Now, 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?

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

Why it happens: in the real activity it is friction, and only friction, that removes the
object's velocity. Take friction away and nothing is left to change the velocity.
Newton's first law then applies exactly: an object in motion continues to move with
constant velocity unless a net force acts upon it.

Page 15 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Did you know? This is the very argument Galileo made in the 17th century, before
Newton stated the law: if a body moves along a horizontal plane and all
impediments to its motion are removed, it will continue to move indefinitely. A
thought experiment let him reach a correct conclusion about conditions he could
never create in his laboratory.

Pause and Ponder — Page 101
6.4 Newton's First Law of Motion

PAUSE AND PONDER

Q3 An object is moving with a constant velocity. Is there a net force acting upon it?

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

Why it happens: a net force is what changes velocity, not what maintains it.
Individual forces may well be acting — a car moving at a steady 60 km h–1 has
engine thrust, friction, weight and the normal force on it — but they cancel one
another exactly, leaving a zero net force.

Tip: "constant velocity" also covers being at rest (v = 0 held constant). Both cases
mean zero acceleration and hence zero net force.

Page 16 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q4 Suppose, 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.

(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 Yes Zero acceleration means the velocity cannot change in
constant velocity magnitude or direction

(iii) Moving with constant No Any acceleration a ≠ 0 needs a net force F = ma ≠ 0 — a
acceleration contradiction

Why (iii) fails: acceleration and net force are locked together by F = ma. If a is
constant and non-zero, then F = ma is constant and non-zero too. You cannot have
acceleration without a net force to cause it.

Q5 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.

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

Example — pushing a heavy box at a steady speed.

Page 17 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Forwards: your applied push, say 40 N

Backwards: force of friction, 40 N

Vertically: weight mg down, normal force N up, N = mg
Net force = 0 N in both directions → the box slides with constant velocity

Why it happens: Newton's first law is about the net force, not about individual
forces. Four separate forces act on the pushed box, yet the object behaves exactly as
if none acted, because they cancel in two opposite pairs. Other everyday examples: a
tug of war where both teams pull equally hard, and a ball floating on water where
the buoyant force balances the weight.

Think as a Scientist — Page 102
6.5 Newton's Second Law of Motion

THINK AS A SCIENTIST

Q1 Now, 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?

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

Weight of the hanging load, F = mg

Double the mass in the cup → double its weight → double the pulling force on the cart

Keep the cart, the surface and the distance travelled the same, so only the force changes

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.

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Class 9 Science Chapter 6 How Forces Affect Motion
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4. Time the cart over a fixed distance s using a slow-motion video — call it T1.
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5. Double the mass in the cup (so the force doubles) and time it again — T2.
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Why this is a fair test: the mass of the cart, the surface, the starting velocity (u = 0)
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Activity 6.3: Let us experiment (Demonstration activity) — Page 102
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6.5 Newton's Second Law of Motion
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ACTIVITY

Q1 Using the values of the time measured, let us do some analysis.
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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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.

Worked illustration: suppose T1 = 1.4 s and T2 = 1.0 s

a2 ÷ a1 = (1.4 s)2 ÷ (1.0 s)2 = 1.96 s2 ÷ 1.00 s2 ≈ 2

Doubling the force roughly doubles the acceleration

Conclusion: for an object of fixed mass, the acceleration increases as the net force
applied on it increases — the acceleration is proportional to the net force. Notice
that you never had to measure the acceleration itself; the fixed distance and u = 0 let
the timing do all the work.

Tip: in practice the increase comes out a little less than exactly two. Apart from
measurement error, friction between the cart's wheels and the table takes away part
of the applied force.

Think as a Scientist — Page 103

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

6.5 Newton's Second Law of Motion

THINK AS A SCIENTIST

Q1 Apart 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?

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.

Why this design works: a hypothesis about mass can only be tested if mass is the
only thing you change. The force (weight of the cup and its contents), the surface,
the starting velocity u = 0 and the distance s are all kept fixed, so any change in the
timing must come from the change in mass. This is the same "change one variable at
a time" logic that made Activity 6.3 a fair test of force.

Activity 6.4: Let us experiment (Demonstration activity) — Page 103
6.5 Newton's Second Law of Motion

ACTIVITY

Q1 Using 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?

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

Page 21 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Worked illustration: T1 = 1.0 s, T2 = 1.4 s

a2 ÷ a1 = (1.0 s)2 ÷ (1.4 s)2 = 1.00 ÷ 1.96 ≈ 0.51 ≈ ½

Doubling the mass roughly halves the acceleration

What it means: for a given magnitude of force, the acceleration produced is
inversely related to the mass of the object. Put together with Activity 6.3 (a ∝ F for
fixed m), you get Newton's second law: a = F ÷ m, or F = ma.

Tip: the measured ratio is usually a little off exactly ½ — friction at the wheels and
timing error from the video both contribute.

Threads of Curiosity — Page 104
6.5 Newton's Second Law of Motion

THREADS OF CURIOSITY

Q1 How much does a force of 1 N feel?

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

Page 22 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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.

Why this is worth knowing: newtons only become real when you can feel one.
Once "1 N ≈ the pull of a 100 g mass" is fixed in your mind, the numbers in problems
stop being abstract — a 3000 N force on a sports car is the weight of about 300 kg,
and a 500 N stopping force on a bullet is enormous for something of mass 50 g.

Pause and Ponder — Page 106
6.5 Newton's Second Law of Motion

PAUSE AND PONDER

Q6 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?

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=0N

Page 23 of 63

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

co m
m.
Why the mass and the speed do not matter: Newton's second law connects force

m l a se
to acceleration, not to velocity. However fast the car is moving, if that velocity is not
o
c is no acceleration, so there can be no net force. Forces
changing.there a g such as
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Tip: the numbers 100 g and 0.5 m s–1 are there to tempt you into a calculation. Read
. ag
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the word "constant" first — it settles the
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Two children of different masses are sitting on identical swings. To impart identical
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You need the larger force for the heavier child — the one with the greater mass.
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Acceleration a is to be the same for both children

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So F is directly proportional to m
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ag Heavier child, m2 (> m1): F2 = m2a > F1
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Why it happens: mass measures how strongly an object resists a change in its
motion — its inertia. To produce the same change of velocity in the same time, you
must overcome more inertia in the heavier child, and that needs a bigger push.

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Check it yourself: for a = 2 m s–2, a 20 kg child needs F = 20 kg × 2 m s–2 = 40 N,
a
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–2

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q8 How are glass items packed for transportation using a bubble wrap or hay
protected from damage?

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|

Why it happens: a hard crate would stop the glass in a few milliseconds; the
compressible bubbles or springy hay let it come to rest over a much longer time as
they squash. The same change of velocity spread over a longer time means a much
smaller acceleration, and by F = ma a much smaller force on the glass — small
enough to stay below the force that would crack it. The padding also spreads the
force over a larger area of the glass.

Did you know? Exactly the same principle protects a cricket fielder who pulls his
hands back while catching, a car passenger saved by an airbag, and a high jumper
landing on a foam mat.

Activity 6.5: Let us explore — Page 107

Page 25 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

6.6 Newton's Third Law of Motion

ACTIVITY

Q1 Now, 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 Force applied by
the girl on the table 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).

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

Page 26 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why it happens: a force never acts alone — at least two objects must interact. When
your hands press on the table, the table presses back on your hands with an equal
and opposite force. The heavy table hardly budges because of its large mass and the
friction under it, but you are on wheels, so the backward force on you is essentially
unopposed and you roll away.

Tip: lifting your legs off the floor matters. With your feet down, friction from the
floor would balance the table's push and you would not move.

Q2 Now, 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 Force applied by
the girl on the table 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).

The chair moves forwards, towards the table.

Page 27 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Why it happens: the direction of the reaction force always flips with the direction of
the action force, because the pair is equal and opposite. Push the table away and it
pushes you away; pull it towards you and it pulls you towards it. Both times the force
on you is opposite to the force you apply on the table.

Q3 What conclusion can you draw from this activity?

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

The key point students miss: the two forces of the pair act on two different objects
— one on the table, one on you. That is why they do not cancel each other and why
you actually move. Two equal and opposite forces balance only when they act on the
same object.

Try This: the same thing explains why you can start a bicycle moving without
pedalling by pushing the ground backwards with your feet, and why you walk at all
— your foot pushes the ground back, and the ground (through friction) pushes you
forward.

Activity 6.6: Let us verify — Page 108

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

6.6 Newton's Third Law of Motion
co m
e m.
m as
ACTIVITY

.co a g l
a s em that you are pulling the free end of the other spring balance with your
gl
Imagine
a
Q1
other hand. Predict what will be the readings of their scales if the spring balances
are stationary.

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

co m
Balance A pulls Balance B with force FAB
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em third law: |F | = |F |
Balance B pulls Balance A with force FBA

a s
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Each spring balance shows the magnitude of the force pulling it
m a
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force one experiences comes entirely from the other. Newton's third law says such a

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READING OF BALANCE A READING OF BALANCE B CONCLUSION
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Medium pull e.g. 5 N 5N

Hard pull e.g. 9 N
a g l 9N Equal

com
m .
m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

What this proves: the forces the two balances apply on each other, in opposite
directions, are equal in magnitude — and they stay equal no matter how the
magnitude is changed. Repeating with different forces is what turns a single lucky
reading into evidence for a general law: Newton's third law of motion.

Tip: both balances must be horizontal and reading zero to start with, otherwise their
own weights add a systematic error to the comparison.

Activity 6.7: Let us understand — Page 109
6.6 Newton's Third Law of Motion

ACTIVITY

Q1 Remove the thread tied to the neck of the balloon and observe in which direction
the straw and the balloon move.

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

Why it happens: as the stretched rubber shrinks it squeezes the air inside and
expels it backwards. By Newton's third law the escaping air pushes forward on the
balloon with an equal force. The straw threaded on the taut line simply keeps the
motion in a straight line so you can see the effect clearly.

Page 30 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

straw

balloon
air rushes out
balloon moves this way
The balloon rocket: air is expelled one way, and the balloon is pushed the other way by an equal
force.

Did you know? A rocket works in exactly this way. Its engine expels hot gas
downwards; the gas pushes the rocket upwards with an equal force, and because
that upward force is larger than the rocket's weight, the net force is upwards and the
rocket lifts off.

In-text Questions — Page 109
6.6 Newton's Third Law of Motion

Q1 What will happen if the engine of a rocket moving in the space, fires in the direction
of its motion?

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

Page 31 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why it happens: a rocket engine does not push against anything outside — it works
purely by throwing mass in one direction and receiving an equal push in the other.
Point the exhaust forwards and the reaction force points backwards, which is a
retarding force. This is the only way to brake in space, where there is no air and no
ground to push against.

Did you know? The Vikram lander of Chandrayaan-3 used exactly this method to
slow down and reach the velocity needed for a soft landing near the south pole of
the Moon.

Pause and Ponder — Page 110
6.6 Newton's Third Law of Motion

PAUSE AND PONDER

Q9 Why does a fireperson sometimes struggle when holding the pipe issuing water?

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

Fast-moving water = large change of momentum every second = large F

Why the force is so big: a fire hose throws out several kilograms of water every
second at a very high speed. Producing that large change of velocity in the water in
so short a time needs a large force (F = ma), and the reaction on the hose is equally
large. If the fireperson does not brace hard against it, the net force on hose-plus-
person is backwards and they are pushed back — which is why firefighters often
hold the nozzle in pairs and lean into it.

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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Tip: a rocket, a canoe paddle and a fire hose are the same physics: throw mass one
way, get pushed the other way.

Q10 Suppose 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.

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.

Why nothing else would work: with negligible gravity and no air, no external agent
is touching the spacecraft. Newton's first law then guarantees its velocity cannot
change on its own. The craft must therefore carry its own reaction mass — the fuel
— and throw it away to get a force.

In-text Questions — Page 111
6.7 Forces Acting on a System of Objects

Q1 But can we apply these laws to two or more objects connected together?

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

Page 33 of 63

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

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

co m
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a = F ÷ (mass of the system) = F ÷ (m1 + m2) … Eq. 6.4
e m
g l as
a
Why internal forces can be ignored: by Newton's third law they always come in
equal and opposite pairs — the string pulls Box 1 backwards and Box 2 forwards
co m
with the same T. Added over the whole system, each pair cancels, so they cannot
em.
m l as
.co g
change the motion of the system as a whole. The two boxes then accelerate exactly
m a
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like one object of mass m1 + m2.

a g
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Tip: your own body is a system too. While walking, your arms and legs move in a
a s
m .co
complicated way, yet your overall motion can be studied by treating your body as a
agl
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single object. Science often becomes simpler when we stop looking at parts and
start looking at the whole. a g

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How can we find the acceleration of each box?
a g l
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Treat the two boxes and the string as one system. Both boxes are tied together, so both have
se m
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the same acceleration:
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External force on the system = F (tension T is internal)

Mass of the system = m1 + m2
co m
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This single value of a is the acceleration of both Box 1 and Box 2
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Worked example. Let m1 = 3 kg, m2 = 2 kg and F = 10 N on a frictionless surface.
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a

co m
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m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

a = 10 N ÷ (3 kg + 2 kg) = 10 N ÷ 5 kg = 2 m s–2

Tension, from Box 2 alone: T = m2a = 2 kg × 2 m s–2 = 4 N

Check with Box 1: F – T = 10 N – 4 N = 6 N = m1a = 3 kg × 2 m s–2 ✓

a = F ÷ (m₁ + m₂), the same for both boxes

string F
Box 2 Box 1

T (internal pair — cancels)

The two tensions form an equal and opposite internal pair, so only F acts on the system from outside.

Tip: the other external forces — the total weight (m1 + m2)g downwards and the
total normal force (N1 + N2) upwards — balance each other, so they do not enter the
horizontal equation.

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

REVISE, REFLECT, REFINE

Q1 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 frictional force is F — equal in magnitude to the applied force, and opposite in direction.

Page 35 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Why it happens: constant velocity is the signature of zero net force (Newton's first
law). Since the only two horizontal forces on the table are your push and friction,
they must cancel exactly. Note that friction here is not "less than" your push — if it
were, the table would be speeding up.

Tip: if F were larger than friction the table would accelerate; if it were smaller, the
table would slow down. Constant velocity pins them to being equal.

Q2 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 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 REASON
OPTION

(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 increase Acceleration is along the velocity, so the ball speeds up
motion

(iii) Net force opposite to decrease Acceleration is opposite to the velocity — a retardation,
the motion so the ball slows down

Page 36 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why the direction of the force decides everything: Newton's second law says the
acceleration is always along the net force. If that direction matches the velocity, each
second adds to the speed; if it opposes the velocity, each second subtracts from it.
"Frictionless" is stated so that you know no hidden retarding force is acting —
whatever happens is entirely due to the net force named in the question.

Q3 Two 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.

5N P 4N Q

(a) (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.

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

Page 37 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

5N 4N
P Q

constant velocity → net
net 1 N → (a)
force = 0 (b)

P has unbalanced forces (net 1 N to the right); Q moves at constant velocity, so its net force is zero.

Why Q has no net force even though it is moving: motion does not require a force
— only a change of motion does. Q's velocity is constant, so its acceleration is zero,
so by F = ma the net force on it must be zero.

Q4 While 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.)

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

Page 38 of 63

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Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

co m
e m.
Force by each oarsman = 200 N
m l as
m .co a g
l a se
g
Forward group: 95 oarsmen
aF
forward = 95 × 200 N = 19 000 N

com
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Opposing group: 5 oarsmen
g l as
Fbackward = 5 × 200 N = 1 000 N
a

co m
em.
m l as
.co g
The two groups push in opposite directions, so subtract
m a
l a se
Net force = 19 000 N – 1 000 N = 18 000 N, forwards
a g

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Why it happens: forces in the same direction add and forces in opposite directions
subtract, with the net force along the.clarger total. The 5 mistaken oarsmen do not
s m
ecancel
g l a
a
merely fail to help — they actively 5 of the correct oarsmen, so the crew loses
2 × 1 000 N = 2 000 N compared with all 100 rowing together (which would give 20
000 N).
co m
m .
m as e
.co a g l
Check it yourself: the effective number of oarsmen is 95 – 5 = 90, and 90 × 200 N =
s m 000 N — the same answer, reached faster.
e18
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a
se m
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m .c we observe that the object accelerates: (i) a
sofe force, with acceleration proportional to the force acting
Q5 When a net force acts on an object,

l a
ag to the direction of force, with acceleration proportional
opposite to the direction
on the object. (ii) opposite
to the mass of the object. (iii) in the direction of force, with acceleration inversely
m
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proportional to the force acting on the object. (iv) in the direction of force, with
acceleration proportional to the force acting on the object.m
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emANSWER ag
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force acting on the object.
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co m
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m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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)

OPTION VERDICT WHAT IS WRONG

(i) Wrong Direction is wrong; acceleration is along the net force

(ii) Wrong Direction is wrong, and a is inversely — not directly — related to mass

(iii) Wrong a is directly proportional to force, not inversely

(iv) Correct Matches a = F/m exactly

The evidence behind it: Activity 6.3 showed that with the mass fixed, doubling the
force roughly doubled the acceleration. Activity 6.4 showed that with the force fixed,
doubling the mass roughly halved it. Together they give a = F/m.

Q6 The 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

Position

Position

Position

0 Time 0 Time 0 Time 0 Time

Object A Object B Object C Object D

Fig. 6.37, page 113 — position–time graphs of the four objects A, B, C and D.

The correct option is (iii) — Object C.

Page 40 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

OBJECT SHAPE OF THE POSITION– WHAT THE SLOPE TELLS YOU NET
TIME GRAPH FORCE

A Straight line, sloping upwards Constant positive velocity Zero

B Horizontal straight line Position not changing — the object is at Zero
rest

C Curve that gets steeper and Velocity increasing — the object is Non-zero
steeper accelerating

D Straight line, sloping downwards Constant negative velocity (moving Zero
back at a steady rate)

Slope of a position–time graph = velocity

Straight line (A, B, D) → constant slope → constant velocity → a = 0 → net F = ma = 0 N

Curved line (C) → changing slope → changing velocity → a ≠ 0 → net F ≠ 0

Object A Object B Object C Object D
Position on the vertical axis, time on the horizontal axis in each
graph

Only C is curved. A changing slope means a changing velocity, and only a changing velocity needs a
net force.

The trap in this question: D slopes downwards, which looks dramatic, but a straight
line of any slope means a constant velocity — the object is simply moving in the
negative direction at a steady rate. Steepness is speed; curvature is acceleration.

Page 41 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q7 A 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.

Sailor

Shore

Boat

Fig. 6.38, page 113 — a sailor jumping forward from a small boat on to the shore
(redrawn sketch).

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

Why the boat moves so noticeably: the boat is small, so its mass is not very much
larger than the sailor's, and it floats on water where friction is almost negligible. The
same equal-and-opposite force therefore produces an easily visible acceleration of
the boat. If the sailor jumped from a heavy ship tied to a pier, the same force would
give a far smaller acceleration and you would not notice it.

Page 42 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Tip: this is why a sailor is told to moor the boat first. If the boat slides back as you
push off, part of your push is wasted and you may fall into the water.

Q8 During 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.

Bar

Landing mat

Fig. 6.39, page 113 — an athlete clearing the bar in a high jump, with the landing mat
below (redrawn sketch).

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|

Page 43 of 63

as e
Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

Feel the size of the effect. Take m = 60 kg and u = 6 m s–1.
co m
e m.
m l as
.co a g
LANDING SURFACE STOPPING TIME T A=U÷T F = MA

se m
g l a 0.02 s
a
Hard ground 300 m s–2 18 000 N

Foam mat / sand bed 0.50 s 12 m s–2 720 N

co m
em . ag
g l as
Why it happens: the mat cannot change how fast the athlete arrives, only how
a
gently they are brought to rest. Because the foam or sand compresses, the same

m
change of velocity is spread over a far longer time, so the acceleration — and by F =
co
m.
ma the force on bones and joints — drops by a factor of about 25 in this example.

as e
om l
The same reasoning explains a fielder pulling his hands back while catching, an
airbag.cin a car and bubble wrap round glassware. a g
a s em
a gl

om a s
. c agl
Q9 A hand cart loaded with vegetables collides with an identical but empty hand cart.

s e
During the collision: (i) the loadedm cart exerts a force of larger magnitude on the
g a exerts a force of larger magnitude on the loaded
empty cart. (ii) the empty lcart
a
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.
co m
m .
m as e
.co a g l
se m
a
The correct option is (iv) — both carts exert forces of equal magnitude on each other.

ag l
Newton's third law: whenever one object exerts a force on a second,
se m
comand opposite force on the first g l a
. a
em
the second simultaneously exerts an equal
a s
agl
F (loaded on empty) = F (empty on loaded) in magnitude

co m
The two forces are opposite in direction and act on different carts
m .
m as e
.co a g l
se m But the accelerations are not equal:
g l a
a c
aempty = F ÷ mempty, aloaded = F ÷ mloaded
m .
m a s e
. co agl
mloaded > mempty → aempty > aloaded
e m
g l as
a

co m
m .
m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Why the answer feels wrong at first: we watch the empty cart get flung away and
conclude it was hit harder. What we are actually seeing is its larger acceleration, not a
larger force. The forces are equal; the masses are not, and a = F/m does the rest.
Newton's third law makes no reference at all to the masses of the interacting bodies.

Did you know? The same reasoning explains Example 6.7 — the Earth and a falling
fruit pull each other with equal forces, but the Earth's enormous mass makes its
acceleration far too small to notice.

Q10 The 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.

10.0
Acceleration (m s⁻²)

7.5

5.0

2.5

0 1 2 3 4 5
Mass (kg)

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).

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

Page 45 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Force (N)

15
F = 10 N, constant
10

5

1 2 3 4 5
Mass (kg)
Force–mass graph for Fig. 6.40: a straight line parallel to the mass axis at F = 10 N.

Why the two graphs look so different: the acceleration–mass graph in Fig. 6.40 is a
falling curve because a = F/m — with F fixed, a is inversely proportional to m, so the
curve drops steeply and then flattens. Multiplying each point by its own mass
undoes that inverse relation exactly, and the product F comes out the same
everywhere. A horizontal force–mass line is therefore the graphical statement of "the
same force was used on every object".

Page 46 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q11 The 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.

30
Velocity (m s⁻¹)

20

10

0 4 8
Time (s)

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).

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

Page 47 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Check it yourself: use the other pair of points — a = (20 – 10) m s–1 ÷ (4 – 0) s = 2.5
m s–2. Same answer, which confirms the line really is straight and the acceleration
constant.

Why the slope gives the acceleration: acceleration is defined as change of velocity
divided by the time taken, and that is exactly what "rise over run" measures on a
velocity–time graph. Because the line is straight, the slope — and therefore the force
— is the same throughout the motion. A curved velocity–time graph would mean a
changing force.

Q12 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).

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

Page 48 of 63

as e
Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

co m
e m.
Convert to SI units first
m l as
.co
m = 50 g = 0.05 kg, u = 100 m s–1, v = 0 m s–1, s = 50 cm = 0.5 m
m a g
l a se
a g
Step 1 — find the acceleration using v2 = u2 + 2as

co m
(0 m s–1)2 = (100 m s–1)2 + 2 × a × 0.5 m
e m . ag
g l as
a
0 = 10 000 m2 s–2 + (1.0 m) a

a = –10 000 m2 s–2 ÷ 1.0 m = –10 000 m s–2

co m
em.
m l as
.co
Step 2 — find the force using F = ma
m a g
Fs=e 0.05 kg × (–10 000 m s–2)
a
agl
F = –500 N
m a s
.co agl
The minus sign means the force opposes the motion; its magnitude is 500 N

se m
g l a
a
Why the force is so large: the bullet loses all 100 m s–1 of its velocity within just half
a metre of wood. Such an enormous change of velocity over such a short distance
co m
m .
means a huge retardation, and F = ma turns that into a large force even though the
m as e
.co a g l
bullet weighs only 50 g — about 0.49 N. The stopping force is more than a thousand

s e m the bullet's own weight.
times

agla
se m
a
Tip: always convert grams to kilograms and centimetres to metres before
substituting, otherwise the newtonco m not come out. 50 g = 0.05 kg and 50 cm = 0.5 g l
. will
a
a s em
agl
m.

co m
m .
sN.e The mass of the football
Q13 An ace footballer converted a penalty shot by kicking the football with a speed of

o m l a
g foot and the ball.
.c was 0.4 kg. Calculate the time of contact betweenatheir
108 km h–1. The estimated force they imparted was 800

se m
g l a
a c
m .
m a s e
co agl
The time of contact is 0.015 s, that is 15 milliseconds.

m .
as e
a g l

co m
m .
m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Why the time is so short: 800 N acting on a mass of only 0.4 kg gives a colossal
acceleration of 2000 m s–2 — about 200 times g. At that rate the ball reaches 30 m s–
1 in a fraction of a heartbeat. This is why a football leaves the boot almost instantly

and why high-speed cameras are needed to see the ball deform against the foot.

Tip: to convert km h–1 to m s–1, multiply by 5/18. Here 108 × 5/18 = 30 m s–1.

Q14 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?

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

Page 50 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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

Why the two forces add: friction acts opposite to the motion, and the extra 3 N is
stated to oppose the motion too. Forces pointing the same way always add in
magnitude, so the object feels a single 10 N retarding force. Before the rough patch
the object had constant velocity, which tells you its net force was zero there — the
rough patch is what makes the net force non-zero.

Check it yourself: the stopping time is t = (v – u)/a = (0 – 10 m s–1) ÷ (–5 m s–2) = 2 s,
and the average velocity is (10 + 0)/2 = 5 m s–1. Distance = 5 m s–1 × 2 s = 10 m ✓

Q15 A 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.

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

Page 51 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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)

A quick sanity check with numbers. Let a1 = 6 m s–2 and a2 = 3 m s–2.

a = (6 × 3) ÷ (6 + 3) m s–2 = 18 ÷ 9 = 2 m s–2
This is smaller than both a1 and a2 — as it must be, since the same force now pulls a larger

total mass

Why the reciprocals appear: masses add, but accelerations do not. Since a = F/m,
mass is inversely related to acceleration for a fixed force. Adding the masses
therefore means adding the reciprocals of the accelerations — which is why the
answer has the same form as resistors in parallel or lenses in contact.

Page 52 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q16 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 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.

Bar magnet

N
Compass needle

S
N

W E

S

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).

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.

Page 53 of 63

as e
Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

co m
e m.
Newton's third law: |F on needle| = |F on bar magnet| = F
m l as
m .co a g
l a se
g
Newton's second law: a = F ÷ m
aNeedle: very small mass m
needle → large acceleration

co m
. ag
Bar magnet: much larger mass mmagnet → very small acceleration
e m
g l as
a
Mass. A compass needle is a tiny sliver of magnetised metal of a fraction of a gram; the bar
magnet is many times heavier. Since a = F/m, the same F gives the needle a far bigger
acceleration.
. com
m a s emits turning. The
gl easily balances the
Friction. The needle is balanced on a sharp pivot, so almost nothing resists

. co lies on a table, where friction between it and the surface
a
em magnetic pull, keeping its net force at zero.
bar magnet

a s
gl
small

a
s
The general principle: Newton's third law fixes only the forces, never the outcomes.
m a
.co agl
What each body then does depends on its own mass and on the other forces acting

a s em
on it. Example 6.7 makes the same point on a grand scale — the Earth and a falling

a l the Earth's huge mass makes its acceleration
fruit pull each other equally,gbut
unmeasurably small. Example 6.8 makes it with a gun and a bullet: the same 2 N
gives the bullet 20 m s–2 but the gun only 0.4 m s–2.
co m
m .
o m l a se
.cThis: hold the bar magnet loosely on a smooth surface,
a gor float it on a piece of
m
asethermocol in water, and bring the compass near. Now you will see the magnet move
Try

agl too — because you have removed the friction that was holding it.
se m
com g l a
m . a
gl 115
The Journey Beyond —aPage
ase

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


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Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Project work

THE JOURNEY BEYOND

Q1 You 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.

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 Pull the same wooden block with a spring Yes — the reading roughly
how hard the surfaces balance, then stack a second identical block doubles when the pressing
press? on top and pull again on the same surface force doubles

Is friction just before Watch the spring balance carefully: note the Yes — the reading falls
motion larger than after peak reading at the instant of slipping, then slightly once sliding begins
motion starts? the steady reading while sliding (static friction > sliding
friction)

Is rolling friction less than Drag the block, then put it on four marbles Yes — the rolling reading is
sliding friction? or toy-car wheels and pull again with the much smaller
same spring balance

Why the wheel changed history: rolling friction is a small fraction of sliding friction,
because the rolling body does not have to be dragged over the surface irregularities
— it lifts over them, and only a small region is in contact at any moment. Dragging a
loaded sledge over the ground needs a large, continuous force; putting it on wheels
cuts that force enormously, so one bullock or one person can move a load that
would otherwise need many.

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.

Page 55 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

A short "so what" panel: ball bearings in cycle hubs, wheels on carts and suitcases, why
brakes and tyre treads need more friction.

Q2 Take 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). Note: the question prints “Fig.
6.32”, but Fig. 6.32 in the book is the pair of charged balloons; the toy cars with bar
magnets are Fig. 6.31. In this activity the like poles are made to touch, so the two forces
point outwards instead of towards each other.

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

Page 56 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

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.

TRIAL MASS OF EACH DISTANCE MOVED BY CAR DISTANCE MOVED BY CAR
CAR A 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

Analysis: the equal distances confirm Newton's third law — the magnetic push on
each car is the same size. The shrinking distances confirm the second law — for the
same force, a heavier car gets a smaller acceleration, so it starts off slower and
friction brings it to rest sooner. Small differences between the two cars in the same
trial are worth discussing honestly: unequal magnet strengths, slightly different
wheel friction, an uneven surface, or reaction time on the two stopwatches.

Tip: release both cars at exactly the same instant, and start both stopwatches
together. Repeat every trial three times and use the average — a single run of an
experiment like this is not reliable.

Page 57 of 63

Class 9 Science Chapter 6 How Forces Affect Motion AglaSem · NCERT Solutions

Q3 Wrap 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
Branch

Heavy bucket

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).

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.

Page 58 of 63

ase
Class 9 Science Chapter 6 How Forces Affect Motion
a g l AglaSem · NCERT Solutions

co m
m.
NUMBER OF EFFORT NEEDED TO HOLD WHAT IS HAPPENING

as e
com
TURNS THE SAME BUCKET

. a g l
m
ase
0 Almost the full weight No friction from the post; your hand alone

agl
supports the load

1 Noticeably less Friction along the wrapped length carries
m
.co ag
part of the load

a sem
agl
2 A small fraction Friction from the second turn acts on an
already-reduced tension

3 or more Very little
co m
The reduction compounds turn after turn

se m.
o m l a
g acts along the whole
m .c reduction is so much bigger than expected: friction
Why the
a
l a se of rope in contact with the post, and at every point it removes a fraction of the
length
a g tension still remaining. So each new turn does not subtract a fixed amount of force

s
— it multiplies down whatever tension survived the previous turn. Repeated
m a
.co agl
multiplication by a fraction falls away very fast, which is exactly what "friction does

se m
a
not increase in a simple linear way" means here.

a g l

m
Safety and honesty in the report: use a light bucket first, keep your feet clear, and
. co
m
record the effort with a spring balance rather than by feel if you can. Then note

o m l a se
where you see the same idea at work: a ship's mooring line taking a few turns round
g cart's rope is wound
.c at a pier, a rock climber's belay, and the way a bullock
a
m
aseround a peg.
a bollard

agl
se m
com g l a
m . a
ase
It is often instructive to examine how scientific ideas develop over time. If you are

gl
Q4

a
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.
com
m .
m as e
.co
a g l
s e mMethod for this project: read a short excerpt in translation, compare Newton's own words with
agla the modern statement in this chapter, and write about what changed and what did not.
.c
1. Find an English translation of the Philosophiæ Naturalis Principia Mathematica (1687) and read
s e m
. c om
only the "Axioms, or Laws of Motion" at the start — it is barely two pages.
a gla
m law in terms of motion — what we now call momentum,
2. Note that Newton stated hisesecond
s
a
asl F = ma. Section 6.5 of this chapter mentions this fuller form in the
mass × velocity — not g
a
Ready to Go Beyond box.

co m
m .
m as e
.co


a g l Page 59 of 63