How Forces Affect Motion - Chapter 6 Class 9 Exploration

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?
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Since the table moves at
constant velocity
, its acceleration is zero, so the net force is zero. The frictional force must balance the applied force exactly.
Therefore the frictional force =
F
(equal to the applied force, but opposite in direction).
What the motion tells us
What follows
The table moves at a
constant velocity
So its acceleration is
zero
Zero acceleration ⇒
net force = 0
The two horizontal forces must cancel
Applied force =
F
(forward)
Frictional force =
F
(backward)
← Back to: 6.3 The Force of Friction: Often Overlooked but Always Present

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.
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(i)
Remain the same
— with no net force, velocity does not change (first law).
(ii)
Increase
— a force along the motion speeds the ball up.
(iii)
Decrease
— a force opposing the motion slows the ball down.
Each statement checked against Newton's laws
Case
Net force
Velocity of the ball
(i) no net force applied
zero
remains the same
(first law)
(ii) force along the motion
forward
increases
— speeding up
(iii) force opposite the motion
backward
decreases
— slowing down
← Back to: 6.4 Newton's First Law of Motion

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.
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The correct option is
(i) P experiences a net force and Q does not.
On P, the two opposite forces (5 N and 4 N) are unequal, leaving a net force of 1 N. Q moves at constant velocity, so its net force is zero.
Block P
Block Q
5 N right, 4 N left — unequal and opposite
Moving with a
constant velocity
Net force = 5 − 4 =
1 N
to the right
Constant velocity ⇒ net force =
0
P
does
experience a net force
Q does
not
experience a net force
⇒ correct option is
(i)
P experiences a net force and Q does not
← Back to: 6.2 Balanced and Unbalanced Forces

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.)
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Forward force from 95 oarsmen = 95 × 200 = 19000 N.
Backward force from 5 oarsmen = 5 × 200 = 1000 N.
Net force = 19000 − 1000 =
18000 N
, in the forward direction.
Net force on the snake boat
95 oarsmen row backwards
each pushes the water back with 200 N → boat driven
forward
: 95 × 200 =
19000 N
↓
5 oarsmen row the wrong way
5 × 200 =
1000 N
acting
backward
↓
Opposite forces subtract
19000 − 1000
↓
Net force =
18000 N
, in the forward direction
← Back to: 6.2 Balanced and Unbalanced Forces

When a net force acts on an object, we observe that the object accelerates:
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.
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The correct option is
(iv) in the direction of the force, with acceleration proportional to the force acting on the object.
Newton's second law: acceleration is along the net force and proportional to it (and inversely proportional to mass).
Each option checked against F = ma
Option
Claim
Correct?
(i)
opposite to the force,
a
∝
F
No — acceleration is
along
the force
(ii)
opposite to the force,
a
∝
m
No — wrong direction, and
a
∝ 1/
m
(iii)
along the force,
a
∝ 1/
F
No —
a
is
proportional
to
F
(iv)
along the force,
a
∝
F
Yes
— this is F = ma
← Back to: 6.5 Newton's Second Law of Motion

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
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The correct option is
(iii) Object C.
A (straight sloping line) and D (straight line) show constant velocity, and B (flat line) is at rest — all have zero net force. Only C has a curved position-time graph, meaning its velocity is changing, so a net force acts on it.
Reading the four position-time graphs
Object
Shape of the graph
Motion
Net force?
A
straight sloping line
constant velocity
No
B
flat, parallel to time axis
at rest
No
C
curved
line
velocity is
changing
— accelerating
Yes
D
straight line sloping down
constant velocity, backwards
No

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.
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Yes, the boat moves
backwards
(away from the shore). As the sailor pushes the boat backward to jump forward, by Newton's third law the boat pushes the sailor forward and the sailor pushes the boat back, so the boat moves in the opposite direction to the jump.
The sailor and the boat — third law
The sailor pushes on the boat
to jump forward, the feet push the boat
backwards
↓
The boat pushes back on the sailor
an equal and opposite force sends the sailor
forward
↓
The two forces act on different objects
so they do not cancel
↓
Yes
— the boat moves
backwards
, away from the shore
← Back to: 6.6 Newton's Third Law of Motion

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.
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The soft mat or sand increases the time over which the athlete's body stops on landing. A longer stopping time means a smaller acceleration, and so a smaller force acts on the athlete (F = ma), reducing the risk of injury.
Why the landing mat is used
The athlete lands on a soft mat
the mat squashes, so the fall is stopped
slowly
↓
A longer stopping time
the same change in velocity is spread over more time
↓
Smaller acceleration
a =
v-u
t
— a bigger
t
means a smaller
a
↓
F = ma ⇒ a
smaller force
acts on the athlete, so there is less injury
← Back to: 6.5 Newton's Second Law of Motion

A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
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The correct option is
(iv) both exert an equal magnitude of force on each other.
By Newton's third law, the two carts exert equal and opposite forces on each other, regardless of their masses.
Loaded cart
Empty cart
Heavier — more mass
Lighter — less mass
Force on the empty cart =
F
Force on the loaded cart =
F
Third law: the pair is
always equal
, whatever the load
Only the
accelerations
differ, because the masses differ
⇒ correct option is
(iv)
both exert an equal magnitude of force on each other
← Back to: 6.6 Newton's Third Law of Motion

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.
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Since F = ma and the same force acts, the force is constant for every mass. Checking the graph, a × m is the same at each point (for example 1 kg × 10 m s
−2
= 10 N, 2 kg × 5 m s
−2
= 10 N), so F = 10 N throughout.
The force-mass graph is therefore a
straight horizontal line at F = 10 N
, parallel to the mass axis.
What the acceleration-mass graph shows
What the force-mass graph must look like
As
m
rises,
a
falls —
a
∝ 1/
m
Because
F = ma
is the same at every point
1 kg × 10 m s⁻² = 10 N
2 kg × 5 m s⁻² = 10 N
The product
m
×
a
is
constant
So
F = 10 N
for every mass
⇒ the force-mass graph is a
flat horizontal line
at F = 10 N, parallel to the mass axis
← Back to: 6.5 Newton's Second Law of Motion

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.
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From the graph, velocity rises from 10 m s
−1
to 30 m s
−1
in 8 s (a straight line), so the acceleration is
a =
30 - 10
8
=
20
8
= 2.5
m s
-2
.
Force
F = ma = 10 × 2.5 =
25 N
, in the direction of motion.
Force from the velocity-time graph
Read the slope of the line
velocity goes from 10 to 30 m s⁻¹ in 8 s
↓
Slope = acceleration
a =
30-10
8
= 2.5
m s
-2
↓
Apply F = ma
F = 10
kg
× 2.5
m s
-2
↓
F = 25 N
, along the direction of motion
← Back to: 6.5 Newton's Second Law of Motion

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).
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m = 50 g = 0.05 kg, u = 100 m s
−1
, v = 0, s = 50 cm = 0.5 m.
Using
v
2
= u
2
+ 2as
:
0 = (100)
2
+ 2a(0.5)
, so
a =
-10000
1
= -10000
m s
-2
.
Force
F = ma = 0.05 × (-10000) =
-500 N
. The stopping force is 500 N, opposite to the bullet's motion.
Stopping force on the bullet
Convert the units
m = 50 g =
0.05 kg
· s = 50 cm =
0.5 m
· u = 100 m s⁻¹ · v = 0
↓
Use
v
2
= u
2
+ 2as
0 = 100
2
+ 2a(0.5)
→
a = -10000
m s
-2
↓
Apply F = ma
F = 0.05 × (-10000)
↓
F = −500 N
— a stopping force of
500 N
, against the motion
← Back to: 6.5 Newton's Second Law of Motion

An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻¹. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
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Speed = 108 km h
−1
=
108 ×
5
18
= 30
m s
-1
. The ball starts from rest, so u = 0, v = 30 m s
−1
.
Acceleration
a =
F
m
=
800
0.4
= 2000
m s
-2
.
Using v = u + at:
30 = 0 + 2000 t
, so
t =
30
2000
=
0.015 s
.
Time of contact with the football
Convert the speed
108 km h⁻¹ = 108 × 5/18 =
30 m s⁻¹
↓
Find the acceleration
a =
F
m
=
800
0.4
= 2000
m s
-2
↓
Use
v = u + at
with u = 0
30 = 2000 × t
↓
t = 0.015 s
— the boot touches the ball for only 15 milliseconds
← Back to: 6.5 Newton's Second Law of Motion

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?
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Total opposing force = 7 + 3 = 10 N. Acceleration
a =
F
m
=
-10
2
= -5
m s
-2
.
Using
v
2
= u
2
+ 2as
with v = 0, u = 10:
0 = 100 + 2(-5)s
, so
s =
100
10
=
10 m
.
Distance travelled in the rough patch
Add the opposing forces
friction 7 N + extra force 3 N =
10 N
, both against the motion
↓
Find the acceleration
a =
-10
2
= -5
m s
-2
↓
Use
v
2
= u
2
+ 2as
0 = 10
2
+ 2(-5)s
↓
s = 10 m
— the object travels 10 m before coming to rest
← Back to: 6.5 Newton's Second Law of Motion

A tractor pulls a harrow (a ploughing tool) of mass
m
1
with a net force
F
resulting in an acceleration of
a
1
. The same tractor pulls a trolley of mass
m
2
with a force
F
producing an acceleration of
a
2
. 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
a
1
and
a
2
. Ignore friction.
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From F = ma:
m
1
=
F
a
1
and
m
2
=
F
a
2
.
Together the mass is
m
1
+ m
2
=
F
a
1
+
F
a
2
. The new acceleration is
a =
F
m
1
+ m
2
=
F
F
a
1
+
F
a
2
=
1
1
a
1
+
1
a
2
=
a
1
a
2
a
1
+ a
2
.
Tractor, harrow and trolley
Step
Working
Mass of the harrow
F = m
1
a
1
⇒ m
1
=
F
a
1
Mass of the trolley
F = m
2
a
2
⇒ m
2
=
F
a
2
Both pulled together
a =
F
m
1
+ m
2
=
F
F
a
1
+
F
a
2
Resulting acceleration
a =
a
1
a
2
a
1
+ a
2
← Back to: 6.7 Forces Acting on a System of Objects

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.
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Though the forces are equal, the compass needle has a very small mass, so the same force gives it a large acceleration (a = F/m) and it turns easily. The bar magnet is much heavier, so the same force gives it a tiny acceleration that is not noticed.
The compass needle
The bar magnet
Feels a magnetic force of magnitude
F
Feels an equal, opposite force of magnitude
F
Very
small
mass, freely pivoted
Much
larger
mass, resting on the table
a = F/m
is
large
a = F/m
is far too
small
to notice
So the needle
swings round
So the magnet appears
not to move
← Back to: 6.6 Newton's Third Law of Motion

Project 1
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 on an object about to move larger than the friction after motion begins? Is friction on a rolling object less than on a sliding object? Find answers and create an infographic. Such observations help explain why the invention of the wheel was a major milestone.
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Friction increases when surfaces press harder. The friction just before an object starts to move (static) is usually larger than the friction once it is moving (kinetic). Rolling friction is much smaller than sliding friction — which is exactly why wheels make moving loads so much easier, making the wheel a major milestone. Present these findings as a labelled infographic with simple experiments.

Project 2
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 with like poles touching. Release them and record the time taken (using two stopwatches) and distance travelled by each before stopping. Repeat after adding equal masses to both cars. Did the cars travel equal distances in opposite directions? Plot distance versus mass. Analyse and discuss.
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The like poles repel, pushing the cars apart with equal and opposite forces (Newton's third law). Since the cars have equal mass, they get equal accelerations and travel roughly equal distances in opposite directions. Adding equal masses to both reduces the distances equally. Plotting distance against mass shows distance decreasing as mass increases, since heavier cars accelerate less.

Project 3
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 (Fig. 6.43). Add one more turn and repeat. Each extra turn increases the grip, increasing friction and reducing the force needed. The reduction is much larger than expected, showing friction does not increase in a simple linear way — small changes in contact lead to large changes in force. This is also how a rope and post hold large ships at a pier.
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With one turn it is hard to hold the bucket; with each extra turn it becomes dramatically easier, even though you only added one loop. This shows friction from a wrapped rope grows very rapidly (not linearly) with the number of turns. The same effect lets a few turns of rope around a bollard hold a massive ship at a pier with little human effort.

Project 4
It is often instructive to examine how scientific ideas develop over time. If you are interested, explore how Newton formulated the laws of motion by reading excerpts from his original work, the Principia. Both the original text and commentaries are available online.
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Read short, translated excerpts of Newton's Principia along with a modern commentary. Notice how he built on Galileo's thought experiments and stated the laws of motion formally. Write a brief note on how the original wording compares with the way the laws are taught today, and how scientific ideas are refined over time.
📚 Ready to Go Beyond
Newton's laws describe motion across an enormous range of scales, from everyday objects to planets and stars.
They need changing only very close to massive objects, at extremely high speeds near the speed of light, and at very small (atomic) scales.
💭 The Quest Continues …
In the real world, friction is always present whenever surfaces are in contact. How can scientists reduce it?
They use lubricants, smooth coatings or surface texturing, streamline the shapes of objects, and use magnetic levitation to keep surfaces from touching at all.