#### Study Material and Notes ofÂ Ch 9 Force and Laws of Motion Class 9th Science

**Topics in the Chapter**

â€¢ Introduction

â€¢ Effect of Force

â€¢ Balanced Force

â€¢ Unbalanced Force

â€¢ Laws of Motion

â€¢ Newton's Laws of motion

â†’ First law of Motion

â€¢ Mass and Inertia

â€¢ Momentum

â†’ Momentum and Mass

â†’ Momentum of an object which is in the state of rest

â†’ Unit of momentum

â†’ Numericals based on momentum

â€¢ Second Law of Motion

â†’ Proof of Newton's First law of motion from Second Law

â€¢ Third law of Motion

â†’ Law of conservation of Momentum

â†’ Force: It is the force that enables us to do any work. To do anything, either we pull or push the object. Therefore, pull or push is called force.

â€¢ Effect of Force

â€¢ Balanced Force

â€¢ Unbalanced Force

â€¢ Laws of Motion

â€¢ Newton's Laws of motion

â†’ First law of Motion

â€¢ Mass and Inertia

â€¢ Momentum

â†’ Momentum and Mass

â†’ Momentum of an object which is in the state of rest

â†’ Unit of momentum

â†’ Numericals based on momentum

â€¢ Second Law of Motion

â†’ Proof of Newton's First law of motion from Second Law

â€¢ Third law of Motion

â†’ Law of conservation of Momentum

**Introduction**â†’ Force: It is the force that enables us to do any work. To do anything, either we pull or push the object. Therefore, pull or push is called force.

Example, to open a door, either we push or pull it. A drawer is pulled to open andÂ pushed to close.

**Effect of Force**

â†’ Force can make a stationary body in object.

For example: a football can be set to move by kicking it i.e. by applying a force.

For example: a football can be set to move by kicking it i.e. by applying a force.

â†’ Force can stop a moving body. For example, by applying brakes, a running cycle or a running vehicle can be stopped.

â†’ Force can change the direction of a moving object.

For example: by applying force i.e. by moving handle, the direction of a running bicycle can be changed. Similarly by moving steering, the direction of a running vehicle is changed.

For example: by applying force i.e. by moving handle, the direction of a running bicycle can be changed. Similarly by moving steering, the direction of a running vehicle is changed.

â†’ Force can change the speed of a moving body. By accelerating, the speed of a running vehicle can be increased or by applying brakes the speed of a running vehicle can be decreased.

â†’ Force can change the shape and size of an object.

For example: by hammering, a block of metal can be turned into a thin sheet. By hammering, a stone can be broken into pieces.

For example: by hammering, a block of metal can be turned into a thin sheet. By hammering, a stone can be broken into pieces.

â€¢ Forces are mainly of two types:

(i) Balanced forces

(ii) Unbalanced forces

**Balanced Forces**

â†’ If the resultant of applied forces is equal to zero, it is called balanced forces.

Example: In the tug of war if both the team apply similar magnitude of forces in opposite directions, rope does not move in either side.

This happens because of balanced forces in which resultant of applied forces become zero.

â†’ Balanced forces do not cause any change of state of an object. Balanced forces are equal in magnitude and opposite in direction.

â†’ Balanced forces can change the shape and size of an object.

For example: When forces are applied from both sides over a balloon, the size and shape of balloon is changed.

For example: When forces are applied from both sides over a balloon, the size and shape of balloon is changed.

**Unbalanced Forces**

â†’ If the resultant of applied forces are greater than zero, the forces are called unbalanced forces.

â†’ An object in rest can be moved because of applying balanced forces.

â†’ An object in rest can be moved because of applying balanced forces.

â€¢ Unbalanced forces can do the following:

â†’ Move a stationary object

â†’ Increase the speed of a moving object

â†’ Decrease the speed of a moving object

â†’ Stop a moving object

â†’ Change the shape and size of an object

**Laws of Motion**

**â€¢ Galileo Galilei:**Galileo first of all said that object move with a constant speed when no foces act on them.

â†’ This means if an object is moving on a frictionless path and no other force is acting upon it, the object would be moving forever. That is, there is no unbalanced force working on the object.

â†’ But practically it is not possible for any object. Because to attain the condition of zero, unbalanced force is impossible.

â†’ Force of friction, force of air and many other forces are always acting upon an object.

â†’ Force of friction, force of air and many other forces are always acting upon an object.

**Newtonâ€™s Laws of Motion**

â†’ Newton studied the ideas of Galileo and gave the three laws of motion. These laws are known as Newtonâ€™s laws of motion.

**Newtonâ€™s First Law of Motion (Law of Inertia)**

â†’ Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force.

Explanation: If any object is in the state of rest, then it will remain in rest until a external force is applied to change its state. Similarly, an object will remain in motion until any external force is applied over it to change its state.

This means all objects resist to in changing their state. The state of any object can be changed by applying external forces only.

This means all objects resist to in changing their state. The state of any object can be changed by applying external forces only.

__â€¢ Newtonâ€™s First Law of Motion in Everyday Life__

(i) A person standing in a bus falls backward when bus starts moving suddenly.

â†’ This happens because the person and bus both are in rest while bus is not moving, but as the bus starts moving, the legs of the person start moving along with bus but rest portion of his body has the tendency to remain in rest. Because of this, the person falls backward; if he is not alert.

â†’ This happens because the person and bus both are in rest while bus is not moving, but as the bus starts moving, the legs of the person start moving along with bus but rest portion of his body has the tendency to remain in rest. Because of this, the person falls backward; if he is not alert.

(ii) A person standing in a moving bus falls forward if driver applies brakes suddenly.

â†’ This happens because when bus is moving, the person standing in it is also in motion along with bus. But when driver applies brakes the speed of bus decreases suddenly or bus comes in the state of rest suddenly, in this condition the legs of the person which are in contact with the bus come in rest while the rest part of his body have the tendency to remain in motion. Because of this person falls forward if he is not alert.

(iii) Before hanging the wet clothes over laundry line, usually many jerks are given to the clothes to get them dried quickly. Because of jerks, droplets of water from the pores of the cloth falls on the ground and reduced amount of water in clothes dries them quickly.

â†’ This happens because when suddenly clothes are made in motion by giving jerks, the water droplets in it have the tendency to remain in rest and they are separated from clothes and fall on the ground.

â†’ This happens because when suddenly clothes are made in motion by giving jerks, the water droplets in it have the tendency to remain in rest and they are separated from clothes and fall on the ground.

(iv) When the pile of coin on the carom-board is hit by a striker, coin only at the bottom moves away leaving rest of the pile of coin at same place.

â†’ This happens because when the pile is struck with a striker, the coin at the bottom comes in motion while rest of the coin in the pile has the tendency to remain in the rest and they vertically falls the carom-board and remain at same place.

â†’ This happens because when the pile is struck with a striker, the coin at the bottom comes in motion while rest of the coin in the pile has the tendency to remain in the rest and they vertically falls the carom-board and remain at same place.

**Mass and Inertia**

â†’ The property of an object because of which it resists to get disturb its state is called inertia.

â†’ Inertia of an object is measured by its mass. Inertia is directly proportional to the mass. This means inertia increases with increase in mass and decreases with decrease in mass.

â†’ A heavy object will have more inertia than the lighter one. In other words, the natural tendency of an object that resists the change in state of motion or rest of the object is called inertia.

â†’ Inertia of an object is measured by its mass. Inertia is directly proportional to the mass. This means inertia increases with increase in mass and decreases with decrease in mass.

â†’ A heavy object will have more inertia than the lighter one. In other words, the natural tendency of an object that resists the change in state of motion or rest of the object is called inertia.

â†’ Since a heavy object has more inertia, thus it is difficult to push or pull a heavy box over the ground than the lighter one.

**Momentum**

â†’ Momentum is the power of motion of an object.

â†’ The product of velocity and mass is called the momentum. Momentum is denoted by â€˜pâ€™.

â†’ Therefore, Â Momentum of the object = Mass Ã— Velocity (p = m Ã— v)

where, p = momentum, m = mass of the object and v = velocity of the object.

__â€¢ Some explanations to understand the momentum:__

â†’ A person get injured in the case of hitting by a moving object, such as stone, pebbles or anything because of momentum of the object.

â†’ Even a small bullet is able to kill a person when it is fired from a gun because of its momentum due to great velocity.

â†’ A person get injured severely when hit by a moving vehicle because of momentum of vehicle due to mass and velocity.

**Momentum and Mass**

â†’ Since momentum is the product of mass and velocity (p = m Ã— v) of an object. This means momentum is directly proportional to mass and velocity. Momentum increases with increase of either mass or velocity of an object.

â†’ This means if a lighter and a heavier object is moving with same velocity, then heavier object will have more momentum than the lighter one.

â†’ If a small object is moving with great velocity, it has tremendous momentum. And because of momentum, it can harm an object more severely.

Example: a small bullet having a little mass even kills a person when it is fired from a gun.

Example: a small bullet having a little mass even kills a person when it is fired from a gun.

â†’ Usually, road accidents prove more fatal because of high speed than in slower speed. This happens because vehicles running with high speed have greater momentum compared to a vehicle running with slower speed.

**Momentum of an object which is in the state of rest**

Let an object with mass â€˜mâ€™ is in the rest.

Since, object is in rest, therefore, its velocity, v = 0

Now we know that,

Momentum = mass Ã— velocity

â‡’ p = m Ã— 0 = 0

Thus, the momentum of an object in the rest i.e. non-moving, is equal to zero.

**Unit of momentum**

â†’ SI unit of mass = kg

â†’ SI unit of velocity = m/s

We know that,

Momentum (p) = m Ã— v

Momentum (p) = m Ã— v

âˆ´ p = kg Ã— m/s â‡’ p = kgm/s

**Numerical Problems Based on Momentum**

**Type I: Calculation of Momentum**

**Example 1:**What will be the momentum of a stone having mass of 10 kg when it is thrown with a velocity of 2 m/s?

**Solution**

Mass (m) = 10 kg

Velocity (v) = 2 m/s

We know that,

Momentum (p) = Mass (m) Ã— Velocity (v)

Momentum (p) = Mass (m) Ã— Velocity (v)

âˆ´ p = 10 kg Ã— 2 m/s = 20 kg m/s

Thus, the momentum of the stone = 20 kg m/s.

**Example 2:**Calculate the momentum of a bullet of 25 g when it is fired from a gun with a velocity of 100 m/s.

**Solution**

Given,

Velocity of the bullet (v) = 100 m/s

Mass of the bullet (m) = 25 g = 25/1000 kg = 0.025 kg

Since, p = m Ã— v

âˆ´ p = 0.025 Ã— 100 = 2.5 kg m/s

Momentum of the bullet Â = 2.5 kg m/s.

**Example 3:**Calculate the momentum of a bullet having mass of 25 g is thrown using hand with a velocity of 0.1 m/s.

**Solution**

Given,

Velocity of the bullet (v) = 0.1 m/s

Mass of the bullet (m) = 25 g = 25/1000 kg = 0.025 kg

Momentum (p) = Mass (m) Ã— Velocity (v)

âˆ´ p = 0.025 kg Ã— 0.1 m/s

â‡’ p = 0.0025 kg m/s

Momentum of the bullet = 0.0025 kg m/s.

**Example 4:**The mass of a goods lorry is 4000 kg and the mass of goods loaded on it is 20000 kg. If the lorry is moving with a velocity of 2 m/s, what will be its momentum?**Solution**

Given,

Velocity (v) = 2 m/s

Mass of lorry = 4000 kg

Mass of goods on the lorry = 20000 kg

Mass of goods on the lorry = 20000 kg

âˆ´ Total mass (m) on the lorry = 4000 kg + 20000 kg = 24000 kg

Momentum (p) = Mass (m) Ã— Velocity (v)

âˆ´ p = 24000 kg Ã— 2 m/s

â‡’ p = 48000 kg m/s

Momentum of the lorry = 48000 kg m/s.

**Example 5:**A car having mass of 1000 kg is moving with a velocity of 0.5 m/s. What will be its momentum?

**Solution**

Given,

Velocity of the car (v) = 0.5 m/s

Mass of the car (m) = 1000 kg

Momentum (p) = Mass (m) Ã— Velocity (v)

âˆ´ p = 1000 kg Ã— 0.5 m/s = 500 kg m/s

Momentum of the car = 500 kg m/s.

**Statement of Second Law of Motion**

â†’ Rate of change of momentum of an object is proportional to applied unbalanced force in the direction of force.

**Mathematical expression**

Suppose,

Mass of an object = m kg

Initial velocity of an object = u m/s

Final velocity of an object = v m/s

âˆ´ Initial momentum, p

Final momentum, p

_{1}Â = muFinal momentum, p

_{2}Â = mv
âˆ´ Change in momentum = Final momentum â€“ Initial momentum

= mv â€“ mu

= m(v â€“ u)

âˆ´ Rate of change of momentum =Â Change in momentum/Time taken

= m(v-u)/t

â€¢ Â According to 2nd law, this rate of change is momentum is directly proportional to force.

We know that, a = (v-u)/tÂ (From 1st equation of motion)

âˆ´ F = kma

where, k is a constant. Its value can be assumed as 1.

âˆ´ F = 1 Ã— m Ã— a = ma

â€¢ SI unit = kg m/s 2 or Newton

**1 Newton:**When an acceleration of 1 m/s

^{2}Â is seen in a body of mass 1 kg, then the

force applied on the body is said to be 1 Newton.

**Proof of Newtonâ€™s First Law of Motion from Second Law**

â†’ First law states that if external force F = 0, then a moving body keeps moving with the same velocity, or a body at rest continues to be at rest.

âˆ´ F = 0

We know, F = m(v-u)/t

__(i) A body is moving with initial velocity u then,__

m(v-u)/t = 0 â‡’ v â€“ u = 0

âˆ´ v = u

Thus, final velocity is also same.

__(ii) A body is at rest i.e., u = 0__

Therefore, from above u = v = 0

So, the body will continue to be at rest.

**Third Law of Motion**

â†’ For every action there is an equal an opposite reaction.

**Applications**

(i) Walking is enabled by 3rd law.

(ii) A boat moves back when we deboard it.

(iii) A gun recoils.

(iv) Rowing of a boat.

**Law of Conservation of Momentum**

â†’ When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting.

â€¢ Initial momentum = Final momentum

Suppose, two objects A and B each of mass m

_{1}Â and mass m_{2}Â are moving initially with velocities u_{1}Â and u_{2}, strike each other after time t and start moving with velocities v_{1}and v_{2}Â respectively.
Now,

Initial momentum of object A = m

_{1}u_{1}
Initial momentum of object B = m

_{2}u_{2}
Final momentum of object A = m

_{1}v_{1}
Final momentum of object B = m

_{2}v_{2}
So,

Rate of change of momentum in A,

F1 = (m

_{1}v_{1}Â - m_{1}u_{1})t = m_{1}(v_{1}Â - u_{1})/t Â ....(i)
Rate of change of momentum in B,

F2 = (m

_{2}v_{2}Â - m_{2}u_{2})t =Â Â m_{2}(v_{2}Â - u_{2})/t Â ....(ii)
We know from 3rd law of motion,

F

_{1}Â = âˆ’F_{2}
So, m

_{1}(v_{1}Â - u_{1})/t = -m_{2}(v_{2}Â - u_{2})/t
â‡’ m

_{1}v_{1}Â â€“ m_{2}u_{2}Â = âˆ’m_{2}v_{2}Â + m_{1}u_{1}
â‡’ m

_{1}u_{1}Â + m_{2}u_{2}Â = m_{1}v_{1}Â + m_{2}v_{2}
Thus, Initial momentum = Final momentum

**Example 1:**A bullet of mass 20 g is fired horizontally with a velocity of 150 m/s from a pistol of mass 2 kg. Find the recoil velocity of the pistol.

**Solution**

Given,

Mass (m

_{1}) of bullet = 20 g = 0.02 kg
Mass (m

_{2}) of pistol = 2 kg
Initially bullet is inside the gun and it is not moving.

âˆµ Mass = m

_{1}+m_{2}Â = (0.02 + 2) kg = 2.02 kg
and u

_{1}Â = 0
âˆ´ Initial momentum = 2.02 Ã— 0 = 0 Â ....(i)

Let the velocity of pistol be v

_{2}Â and v_{1}Â for bullet = 150
âˆ´ Final momentum = m

_{1}v_{1}Â + m_{2}v_{2Â }= 0.02Ã—150 + 2v_{2}Â ...(ii)
We know that,

Initial momentum = Final momentum

âˆ´ (0.02Ã—150)/100Â + 2v

_{2}Â = 0 Â [From equations (i) and (ii)]
â‡’ 3 + 2v

_{2}Â = 0
â‡’ 2v

_{2}Â = âˆ’3
â‡’ v

_{2}Â = âˆ’1.5 m/s
(âˆ’)ve sign indicates that gun recoils in direction opposite to that of the bullet.

**Example 2:**Two hockey players viz A of mass 50 kg is moving with a velocity of 4 m/s and another one B belonging to opposite team with mass 60 kg is moving with 3 m/s, get entangled while chasing and fall down. Find the velocity with which they fall down and in which direction?

**Solution**

Given,

m

_{A}Â = 50 kg, u_{A}Â = 4 m/s
m

_{B}Â = 60 kg, u_{B}Â = 3 m/s
Initial momentum A = m

_{A}u_{A}= 50 Ã— 4 = 200 kgm/s
Initial momentum B = m

_{B}u_{B}= 60 Ã— 3 = 180 kgm/s
âˆ´ Total initial momentum = 200 + 180 = 380 kgm/s ....(i)

Final momentum = (m

_{A}Â + m_{B})v = (50 + 60)v = 110v Â ....(ii)
According to the law of conservation of momentum,

380 Â = 110v

â‡’ v = 380/110 =Â 3.454 m/sÂ

**NCERT Solutions of Force and Laws of Motion**