# Class 9 Force and Laws of Motion Notes

Notes Important Questions NCERT Solutions MCQ Quiz – 1 MCQ Quiz – 2**Force and Laws of Motion Notes**

## Force:-

A push or pull on an object is called force.

## A force can change-

- the state of motion of an object i.e, motion to rest or, rest to motion.
- the speed of the object
- the direction of motion
- the shape of the object

## Contact Force:-

The force exerted by one object on another object due to the physical contact between them is called contact force.

**Ex:-** muscular force, frictional force etc.

## Non-contact Force:-

The force exerted by one object on another object without any physical contact between them is called non-contact force.

**Ex:-** Magnetic force, electrostatic force, gravitational force etc.

## Balanced Forces:-

If the net force acting on an object is zero, then the forces acting on the object are said to be balanced.

## Unbalanced Forces:-

If the net force acting on an object is not zero, then the forces acting on the object are said to be unbalanced.

## Newton’s First Law of Motion (Law of Inertia):-

An object remains in a state of rest or of uniform motion in a straight line unless or until an unbalanced force is applied.

** Or,** All objects resist a change in their state of motion.

### Inertia:-

The natural tendency of an object to resist a change in its state of motion or of rest is called inertia.

*Inertia is classified as:-*

**Inertia of rest:**Ex – A person seating in a car leans backward when the car starts all of a sudden**Inertia of motion:**Ex – A person seating in a moving car leans forward when the car stops all of a sudden**Inertia of direction:**Ex – Whenever a bus takes a sharp turn, the passengers experience a force acting away from the centre of the curve

- Mass of an object is a measure of its inertia.
- Inertia ∝ Mass

### Momentum:-

The momentum of an object is defined as the product of its mass and velocity.

**p = mv**- S.I. unit:- kg m/s

## Newton’s Second Law of Motion:-

The force applied on an object is directly proportional to the rate of change of momentum in the direction of force.

### Mathematical Formulation of Second Law of Motion:-

Consider an object of mass ‘m’ moving in a straight line with initial velocity ‘u’. It is uniformly accelerated to final velocity ‘v’ in time ‘t’.

So, initial momentum = p_{1} = mu and final momentum = p_{2} =mv

A/c to second law of motion,

Force applied ∝ Change in momentum/time

or, F ∝ (p_{2} – p_{1})/t

or, F ∝ (mv – mu)/t

or, F ∝ m(v-u)/t

or, F ∝ ma **[**since, **a = (v-u)/t]**

or, F = k m a where, k is constant of proportionality

Here, k = 1 So, **F = ma**

- S.I. unit of force:-
**kg m/s**or^{2}**Newton(N)** **1 Newton =**1**kg m/s**=^{2}**10**^{5}**Dyne****1 Newton of force**:- The amount of force is said to be 1 newton that produces an acceleration of 1 m/s^{2}in an object of 1 kg mass.

### Proof of First Law of Motion Using Second Law of Motion:-

From second law of motion,

F = ma

or, F = m(v-u)/t

or, Ft/m = v-u

when, **F = 0**

then, v-u = 0

or, **v = u**

Here, we see that when external applied force is 0, then there is no change in its state of motion or of rest which is the first law of motion.

## Newton’s Third Law of Motion:-

It states that every action has equal and opposite reaction.

Ex:- A forward force on the bullet and recoil of the gun.

## Law of Conservation of Momentum:-

The sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them.

### Mathematical Proof of Law of Conservation of Momentum:-

Consider two objects having masses m_{A} and m_{B} moving with initial velocities u_{A} and u_{B} respectively in the same direction such that u_{A} > u_{B}. After collision, let their final velocities be v_{A} and v_{B} respectively.

.

From 2nd law of motion,

Force exerted by A on B = F_{AB} = m_{A}(v_{A}-u_{A})/t

Force exerted by B on A = F_{BA} = m_{B}(v_{B}-u_{B})/t

From 3rd law of motion,

F_{AB} = -F_{BA}

or, m_{A}(v_{A}-u_{A})/t = -m_{B}(v_{B}-u_{B})/t

or, m_{A}v_{A} – m_{A}u_{A} = -m_{B}v_{B} + m_{B}u_{B}

or, **m _{A}u_{A} + m_{B}u_{B} = m_{A}v_{A} + m_{B}v_{B}**

## Comments