Force and Motion – Class 9 Science Notes (SEE Nepal)

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1. Force

Force is a physical cause that changes or tends to change the state of an object. It can change the speed, direction, shape, or size of an object.

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SI Unit: Newton (N)

Formula:

Force = Mass × Acceleration

F = ma

Types of Force

  • Contact Force (e.g., frictional force, tension)
  • Non-contact Force (e.g., gravitational force, magnetic force)

2. Balanced and Unbalanced Force

Balanced Force: When two or more forces acting on a body cancel each other, and there is no change in motion.

Example: A book lying on a table.

Unbalanced Force: When forces acting on a body are not equal, resulting in a change in motion or direction.

Example: Kicking a football.

3. Motion and Velocity

Motion: A body is said to be in motion if it changes its position with respect to a reference point.

Velocity: The rate of change of displacement. It includes direction.

Velocity = Displacement / Time

  • Uniform Velocity: Equal displacement in equal time intervals.
  • Non-uniform Velocity: Unequal displacement in equal time intervals.

4. Acceleration and Retardation

Acceleration (a): Rate of change of velocity.

a = (v - u) / t

where,

u = initial velocity,

v = final velocity,

t = time

Retardation: Negative acceleration (velocity decreases with time).

5. Graphical Representation of Linear Motion

a. Displacement-Time Graph

  • Straight Line with Positive Slope: Uniform motion
  • Curve: Non-uniform motion
  • Horizontal Line: Body is at rest

b. Velocity-Time Graph

  • Horizontal Line: Uniform velocity
  • Sloped Line: Accelerated motion
  • Area under graph: Displacement
  • Slope of graph: Acceleration

1. Displacement-Time (s-t) Graphs

(a) Uniform Velocity 

  • Graph: Straight line with a constant slope (positive or negative).
  • Interpretation: Slope = Velocity (v=Δs/Δt). Steeper slope = Higher velocity.

Example: A car moves 20 m in 5 s at constant speed. Velocity (v) = 20 m/5 s = 4 m/s.

(b) Non-Uniform Velocity (Accelerated Motion)

  • Graph: Curved line (parabola for constant acceleration).
  • Interpretation: Changing slope = Changing velocity (acceleration).

Example: A car starts from rest and accelerates, covering: 0 m at 0 s, 5 m at 2 s, 20 m at 4 s

2. Velocity-Time (v-t) Graphs

(a) Uniform Acceleration

  • Graph: Straight line with positive slope.
  • Interpretation: Slope = Acceleration (a=Δv/Δt). Area under graph = Displacement.

Example: A car accelerates from 10 m/s to 30 m/s in 5 s. Acceleration (a) = (30−10)/5 = 4 m/s².

(b) Uniform Retardation (Deceleration)

  • Graph: Straight line with negative slope.
  • Interpretation: Slope = Negative acceleration (slowing down).

Example: A car slows from 20 m/s to 0 m/s in 4 s. Retardation = (0−20)/4 = −5 m/s².

(c) Constant Velocity (Zero Acceleration)

  • Graph: Horizontal line (slope = 0).
  • Interpretation: Velocity remains unchanged (no acceleration).

Example: A car moves at 15 m/s for 6 seconds.

Key Takeaways:

Displacement-Time Graph:

  • Slope = Velocity.
  • Straight line = Uniform velocity.
  • Curved line = Acceleration.

Velocity-Time Graph:

  • Slope = Acceleration.
  • Area under graph = Displacement.

Practice: Sketch a v-t graph for a car:

  1. Accelerates from 0 to 10 m/s in 2 s.
  2. Moves at 10 m/s for 3 s.
  3. Decelerates to 0 m/s in 1 s.

6. Equations of Motion

v = u + at

s = {(u + v) × t} / 2

s = ut + ½ at²

v² = u² + 2as

Where:

u = initial velocity,

v = final velocity,

s = displacement,

a = acceleration,

t = time

Derive v = u + at

Given:

  • v = final velocity (in m/s)
  • u = initial velocity (in m/s)
  • a = constant acceleration (in m/s²)
  • t = time (in seconds)

Acceleration (a) is the rate of change of velocity with time:

a = Change in velocity / Time taken

or, a = (v − u) / t

or, at = v – u

or, v = u + at Derived

Derive s = {(u + v) × t} / 2

Given

  • s = displacement (in meters)
  • u = initial velocity (in m/s)
  • v = final velocity (in m/s)
  • t = time (in seconds)

Concept of Average Velocity

For motion with constant acceleration, the average velocity is the arithmetic mean of initial and final velocities:

Average Velocity = (u + v) / 2

Relationship Between Displacement, Average Velocity, and Time

Displacement (s) is given by:

Displacement = Average Velocity × Time

Substitute the average velocity:

s = (u + v) / 2 × t

Derived

Derive s = ut + ½ at²

Given:

  • v = final velocity (in m/s)
  • u = initial velocity (in m/s)
  • a = constant acceleration (in m/s²)
  • t = time (in seconds)
  • s = displacement (in meters)

Using Average Velocity

Average velocity for constant acceleration:

v = (u + v) / 2

or, v = {u + (u + at)} / 2

or, v = (2u + at) / 2

Displacement = Average velocity × time:

s = (2u + at) / 2 × t

or, s = 2ut/2 + at²/2

or, s = ut + ½ at²

Derived

7. Inertia

Inertia is the property of a body to resist changes in its state of rest or uniform motion.

Types of Inertia

Inertia of Rest: A body at rest remains at rest unless acted upon.

Examples:

Sudden Braking of a Vehicle

Scenario: When a moving car stops abruptly.

Observation: Passengers lurch forward.

Why? Their bodies tend to stay in the original state of motion (inertia of rest relative to the car's sudden stop).

Pulling a Tablecloth Trick

Scenario: A tablecloth is yanked quickly from under dishes.

Observation: Dishes stay in place (may wobble but don't move with the cloth).

Why? The dishes resist change due to inertia of rest.

Inertia of Motion: A moving object continues in motion unless stopped.

Examples

Passengers Jerking Forward When a Car Stops

Scenario: A moving car brakes suddenly.

Observation: Passengers lurch forward.

Why? Their bodies resist the change (keep moving at the car's original speed due to inertia of motion).

Athletes Running Before a Long Jump

Scenario: A sprinter takes a run-up before jumping.

Observation: The jumper covers more distance.

Why? The forward motion continues even after the leap (inertia helps maintain horizontal velocity).

8. Newton's Laws of Motion

First Law (Law of Inertia)

A body remains at rest or in uniform motion in the straight line unless acted upon by an external force.

Second Law (Law of Acceleration)

The force acting on a body is equal to the product of its mass and acceleration.

F = ma

Let a body of mass (m) has moved from a point to another point. Let its acceleration be 'a'.

Acceleration of the body is:

  • Directly proportional to force applied.
  • Inversely proportional to mass of the body.

From above equations, we get

a ∝ f/m

or, f ∝ ma

or, f = kma (where k is proportionality constant)

If 1 N force is applied to the body of mass 1 kg accelerating 1 m/s², at that time value of k is also 1. Putting the value of k in equation 3, we get

f = ma

Third Law (Action and Reaction)

For every action, there is an equal and opposite reaction.

Examples:

Walking or Running

Action: Your foot pushes backward on the ground.

Reaction: The ground pushes forward on your foot, making you move.

Why? Without this reaction force, you couldn't walk!

Rocket Launch

Action: Rocket engines expel hot gas downward.

Reaction: The gas pushes the rocket upward (thrust force).

9. Elasticity and Plasticity

Elasticity: The ability of an object to return to its original shape after the removal of deforming force.

Example: Rubber band

Plasticity: The inability of an object to return to its original shape after deformation.

Example: Clay

Feature Elasticity Plasticity
Reversibility It is reversible It is irreversible
Energy It stores energy It dissipates energy
Material Types Rubber, steel (below yield point) Clay, ductile metals
Deformation It is temporary It is permanent

10. Numerical Problems (Examples)

Q1. A car accelerates from 5 m/s to 20 m/s in 3 seconds. Find the acceleration.

Solution:

Given

Initial velocity (u) = 5 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 3 seconds

Acceleration (a) = ?

We have,

a = (v - u) / t

or, a = (20 - 5) / 3

or, a = 5 m/s².

Hence, the acceleration of the car is 5 m/s².

Q2. A bus moving at 30 m/s stops in 10 seconds. What is the retardation?

Solution:

Initial velocity (u) = 30 m/s

Final velocity (v) = 0 m/s

Time taken (t) = 10 seconds

retardaion (-a) = ?

We have,

a = (v - u) / t

or, a = (0 - 30) / 10

or, a = -3 m/s².

Hence, the retardation of the car is 3 m/s².

Q3. A body is accelerated from rest at 2 m/s² for 5 seconds. Find the distance covered.

Solution:

Initial velocity (u) = 0 m/s

Time taken (t) = 5 seconds

Acceleration (a) = 2 m/s²

distance (s) = ?

We have,

s = ut +1/2at2

or, s = 0 x 5 + ½ 2 x 25

or, s = 25 m.

Hence, the distance covered by body is 25 m.

Q4. A car moving at 15 m/s brakes and decelerates at 3m/s². How long does it take to stop? How far does it travel while stopping?

Solution:

Initial velocity (u) = 15 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = -3 m/s²

time taken (t) = ?

distance (s) = ?

We have,

v = u + at

or, t = (v-u)/a

or, t = (0-15)/-3

or, t = 5 seconds

Again

s = ut + ½ at2

or, s = 15 x 5 + ½ (-3) x 25

or, s = 75 - 37.5

or, s = 37.5m

Hence, the distance covered by the car is 37.5m and time taken is 5 seconds.

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