# Physics 11th Newton's Laws of Motion

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## NEWTON'S LAWS OF MOTION

**(1) Introduction**

- While studying kinematics,we have already studied about the position ,displacement and acceleration of a moving particle
- Here in this chapter,we would take our understanding one step further to learn abouts origins of accleration or force
- Here we will specifically consider the cause behind the moving objects i.e what causes the objects to move
- Thus we will learn the theory of motion based on the ideas of mass and force and the laws connecting these physical concepts to the kinematics quantities
- So we will begin by stating the Newton's law's of motion which are of critical importance in classical mechanics
- Laws of motion as stated by Newtons in his principle are

(i) Every body continues in its state of rest or uniform motion in straight line ,unlesss compelled to change that state by force imposed upon it

(ii) Change of motion of an objects is proportional to the force acting on it and is made in the direction of the straight line along the direction of force

(iii) To every action there is always an equal and opposite reaction

**(2) Force**

- Concept of force is central to all of physics whether it is classical physics,nuclear physics,quantum physics or any other form of physics
- So what is force? when we push or pull anybody we are said to exert force on the body
- Push or pull applied on a body does not exactly define the force in general.We can define force as an influence causing a body at rest or moving with constant velocity to undergo an accleration
- There are many ways in which one body can exert force on another body

Few examples are given below

(a)Stretched springs exerts force on the bodies attached to its ends

(b)Compressed air in a container exerts force on the walls of the container

(c) Force can be used to deform a flexible object

There are lots of examples you could find looking around yourself - Force of gravitational attraction exerted by earth is a kind of force that acts on every physical body on the earth and is called the weight of the body
- Mechanical and gravitation forces are not the only forces present infact all the forces in Universe are based on four fundamental forces

(i) Strong and weak forces: These are forces at very short distance (10^{-05}m) and are responsible for interaction between neutrons and proton in atomic nucleus

(ii) Electromagnetic forces: EM force acts between electric charges

(iii) Gravitational force -it acts between the masses - In mechanics we will only study about the mechanical and gravitational forces
- Force is a vector quatity and it needs both the magnitude as well as direction for its complete description
- SI unit of force is Newton (N) and CGS unit is dyne where

1 dyne= 10^{05}N

**(3) Newton's First law of Motion**

- We have already stated Newton's First law of motion which says that a body would continue to be in state of rest or continue to move with constant velocity unless acted upon by a net external force
- Here the net external force on the body is the vector sum of all the extenal forces acting on the body
- When the body at rest or in a state of motion with uniform velocity then in both the cases acceleration is zero.This implies that

a=0 for F=0 - When net forces i.e vector sum of all the forces acting on the body is zero.the body is said to be in equilbrium .When rotational motion is involved <,net torque on body should also be zero i.e their is no change in either translational or rotational motion
- Since forces can be combined according to the rules of vector addition.Thus for a body to be in equilibrium
**R**=Σ**F**=0

or in component form

ΣF_{x}=0

ΣF_{y}=0

These are the condition for the body in translational equilbrium - We will discuss about rotational equilibrium while studying torque and rotational motion
- Thus Newton's First law of motion quantitatively defines the concept of force as a influence that changes the state of motion of the body
- It does not say anything about what has to be done to keep object moving that is once the body gains motion by the application of force would it always
- According to first law if we completely eliminates frictional forces,no forward force at all would be required to keep an object ( say a block on table)

**(4) Inertia and Mass**

- From First law of motion an object at rest would not move unless it is acted upon by a force
- This inherent property of objects to remain at rest unless acted upon by a force is called intertia rest
- Now consider the case of an object moving with uniform velocity along the straight line .Again from Newton's law it would continue to move with uniform
- This inherent property by virtue of which a body in state of uniform motion tend to maintain its uniform motion is called inertia of motion
- Combining these two statements 'The property of an object to remain in state of rest or uniform rectilinear motion unless acted upon by a force is called inertia
- Mass of any body is the measure of inertia .For example if we apply equal amount of force on two objects of different mass (say m
_{1}and

m_{2}such that m_{1}> m_{2}) Tnen acceleration of both the object would be different (i.e , a_{1}< a_{2}) - Acceleration of object having larger mass would be lesser then the acceleration of object having smaller mass
- Thus larger the mass of the body ,smaller would be the acceleration and larger would be the inertia
- Newton's first law of motion revealing this fundamental property of matter i.e inertia is also known as law of inertia

**(5) Newton's second law of motion**

- Newton's first law of motion qualitatively defines the concept of force and the principle of inertia
- For an body at rest,application of force causes a changed in its existing state and application of force on a body moving with uniform velocity would give the body under consideration as acceleration
- Newton's second law of motion is a relation between force and acceleration
- Newton's second law of motion says that

" The net force on a body is equal to the product of mass and acceleration of the body"

Mathematically**F**=m_{net}**a**(1)

Where F_{net}is the vector sum of all the forces acting on the body - Above equation -(1) can be resolved along x,y and z components .Thus in component form

F_{netx}=ma_{x}

F_{nety}=ma_{y}

F_{netz}=ma_{z} - Component of accleration along a given axis is caused only by the net component of force along that axis only not by the components of force along other
- Newton's second law of motion is completely consistent with newton first law of motion as from equation (1) F=0 implies that a=0
- For a body moving under the influence of force,acceleration at any instant is determined by the force at that instant not by the prev ious motion of the
- Newton's second law of motion is strictly applicable to a single particle .In case of rigid bodies or system of particles,it refers to total external forces acting on the system excluding the internal forces in the system.

**(6)Newton's third law of motion**

- Statement of newton's third law of motion is
**" To every action there is always an equal and opposite reaction"**. - Thus,whenever a body exerts force on another then another object exert an equal force on previous body but in opposite direction
- Force example motion of rocket depends on the third law of motion i.e, action and reaction .Rocket exerts action force on gas jet in backward direction
- Force of action and reaction acts on different objects i.e,

Force object 1 exerts on object 2= Force object 2 exerts on object 1

i.e,

F_{12}=-F_{21}

Action=-(Reaction) - According to newtonian mechanics force is always a mutual interaction between the bodies and force always occurs in pairs
- Equal and opposite mutual forces between two bodies is the basic idae between Newton's third law of motion
- While considering a system of particles ,internal force always cancel away in pairs i.e consider two particles in a body if F
_{12}and

F_{21}are internal forces between partcile system 1 and 2 then they add up to give a null internal force.Same way internal forces for the particles

**(7) Applying Newton's law of motion**

- Newton's law of motion ,we studied in earliar topics are the foundation of mechanics and now we look forward to solve problems in mechanics
- In general,we deal with mechanical systems consisting of different objects exerting force on each other
- While solving a problem choose any part of the assembly and apply the laws of motion to that part including all the forces on the choosen part of the assembly due to remaining parts of the assembly
- Following steps can be followed while solving the problems in mechanics

1)Read the problem carefully

2) Draw a schematic diagrom showing parts of the assembly for example it may be a single particle or two blocks connected to string going over pulley etc

3) Identify the object of prime interest and make a list of all the forces acting on the concerned object due to all other objects of the assembly and exclude the force applied by the object of prime interest on the other parts of the assembly

4) Indicate the forces acting on the concerned object with arrow and Lable each force for example tension on the object under consideration can be labelled by letter T

5) Draw a free body diagram of the object of interest based on the labelled picture.Free body diagram for the object under consideration shows all the forces exerted on this object by the other bodies.Do not forget to consider weight W=mg of the body while labelling the forces acting on the body

6) If additional objects are involved draw seperate free body diagrma for them also

7)Resolve the rectangular components of all the forces acting on the body

8) Write Newton second law of equation for the body and solve them to find out the unknowm quantities

9) Do not forget to employ Newton's third law of motion for action reaction pair which results in null resultant force - Following solved example would clearly illustrate how to apply Newton's laws of motion follwowing the above given procedure

**Solved Example :**

Question:

A horizontal forces of magnitude 500N pulls two blocks of masses m

m

Given that force is applied on the block m

Let T be the tension in the string and a be the accleration of each mass .Now we will draw free body diagrams for each masses

Weights of the blocks m

m

m

Dividing 1 by 2

we get

T=m

Substituting the given values

T=166.7 N

Using value of T in equation 1 ,we find

a=16.67 m/s

Above sample problem shows how to solve a typical mechanics problem.Similarly by adopting given procedure we can solve other such problems

A horizontal forces of magnitude 500N pulls two blocks of masses m

_{1}=10 kg and m_{2}=20 kg which are connected by the light inextensible string and lying on the horizontal frictionless surface.Find the tension in the strings and acceleration of each mass when forces is applied on massm

_{2}?**Solution:**Given that force is applied on the block m

_{2}as shown in the figure belowLet T be the tension in the string and a be the accleration of each mass .Now we will draw free body diagrams for each masses

Weights of the blocks m

_{1}g and m_{2}g are balanced by their normal reaction R_{1}and R_{2}respectivley.The equations of motion of the two massed are found using Newton's second law of motionm

_{1}a=T ...............................(1)m

_{2}a=F-T ............................(2)Dividing 1 by 2

we get

T=m

_{1}F/(m_{1}+m_{2})Substituting the given values

T=166.7 N

Using value of T in equation 1 ,we find

a=16.67 m/s

^{2}Above sample problem shows how to solve a typical mechanics problem.Similarly by adopting given procedure we can solve other such problems

**(8) Inertial frame of Reference**

- Inertial frame of refrences are those frames of reference in which newton's first and second law of motion is always hold true
- A frame of reference in which Newton's law are not valid is called non-inertial frame of refrence
- In an inertial frame if a body is not acted by external force ,it continues to be in state of rest or uniform translatory motion.Thus in an inertial frame if the body is not acted upon by an external force then accleration would be zero
**a**=(d^{2}**r**/dt^{2})=0 - If a frame is inertial frame ,then all those frames which are moving with constant velocity relative to the previous frame are also inertial frames
- Inertial frame of refrence are necessarly unaccelerated frames because if the frame is acclerated the particle moving with uniform velocity will appear

**(9) Fictitious ( or Pseudo ) Forces**

- We already know about Non-inertial frame of reference .All the accelrated and rotatig frame of reference are non-inertial frame of refrence
- Consider an interial frame of reference S and let S
^{'}be any other frame moving with accleration w.r.t to frame S as shown in the below figure

- Now if no external forces are acting on particle P .Then its acceleration would be zero in Frame S but in frame S
^{'},an observer will find an acceleration -**a**acting on the particle._{0} - The observer force on particle P of mass m in Frame S
^{'}is -m**a**_{0} - But in reality no such force is acting on the particle and particle appears to be accelerated in this non-inertial frame of reference.Such one force is known as Pseudo or Fictitious Force. Hence Pseudo Force on particle is
**F**=-m_{P}**a**_{0} - Now if we apply F
_{i}on the particle and a_{i}is the observed accleration of particle in S frame(Inertial frame) The according to Newton's law**F**=m_{i}**a**_{i} - For calculating net force in accelerated frame consider both the frames S and S
^{'}coincid at time t=0 .After time t let**r**and_{i}**r**be the position vector of the particle in frame S and S_{n}^{'}respectively

The relation between**r**and_{i}**r**is_{n}**r**=_{i}**r**+(1/2)_{n}**a**t_{0}^{2}Where**a**is the acceleration of frame S_{0}^{'}wrt frame S

Differentiating the equation w.r.t time twice**a**=_{i}**a**+_{n}**a**_{0}

or m**a**-m_{i}**a**=m_{0}**a**_{n}

=>**F**+_{0}**F**=_{P}**F**_{N}

This equation gives obesrved force in acclerated frame of reference

**Solved Examples**

**Question 1**Force exerted on a body changes it's

(a) Direction of motion

(b) Momentum

(c) Kinetic energy

(d) All the above

**Solution**: A body acted upon by a certain force produces acceleration i.e. it undergoes change in it's velocity. hence choice (d) is correct

**Question 2**there are two statements

A Rate of change of momentum correspondes to force

B Rate of change of momentum corresponds to Kinetic Energy

Which of the following is correct

(a) A only

(b) B only

(c) Both A and B are correct

(d) Both A and B are wrong

**Solution**

F=dp/dt so a is correct

**Question 3**A truck and car are moving on a plane road with same kinetic Energy. They are brought to rest by application of brakes which provide equal retarding forces. Which of the following statements is true.

(a) Distance travelled by truck is shorter then car before comming to rest

(b) Distance travelled by car is shorter then truck before comming to rest

(c) Distance travelled depends on individual velocity of both the vehicles

(d) Both will travel same distance before comming to rest

**Solution 3**

By work energy therorm we know

KE

_{f}-KE

_{i}=Workdone

So in this case

Intial KE=retarting force * distance travelled

Since kinectic energy is same and retarding force is equal

Distance travelled will be same

**Question 4**A block A of mass m

_{1}is released from top of smooth inclined plane and it slides down the plane. Another block of mass m

_{2}such that m

_{2}> m

_{1}is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?

(a) Both blocks will reach ground at same time

(b) Both blocks will reach ground with the same speed

(c) speed of both the blocks when they reach ground will depend on their masses

(d) Block A reaches ground brfore block B

**Solution 4**

since velocity of block when it reaches the ground is given by v=(2gh)

^{1/2}and it is independent of the mass the correct choice will be (b).

**Question 5**Which of the following observer is/are non interial

a.A child revolving in the merry ground

b. A driver in a train moving with constant velocity v

c. A passenger in a bus which is slowing down to a stop

d None of these

**Solution 5**

Since centripetal acceleration is present in merry round it is non interial frame of refrence

Since bus is slowing down to a stop,it means it is experiencing a retardation ,so it is non interial frame of refrence

**Question 6**there are two statements

A Newtons first law in valid from the pilot in an aircraft which is taking off

B Newtons first law in valid from the observer in a train moving with constant velocity

Which of the following is correct

(a) A only

(b) B only

(c) Both A and B are correct

(d) Both A and B are wrong

**Solution 6**

We know that Newtons law is invalid in Non interial frame of refrence, now since acceleration is present in Ist case..So answer is B

**Question 7**A IITJEE text book of mass M rests flat on a horizontal table of mass m placed on the ground.Let R

_{X->Y}be the constant force exerted by the body x on body Y.According to Newton third law,which of the following is an action-reaction pair of forces?

a. (M+m)g and R

_{table->book}

b. R

_{ground->table}and mg+R

_{book->table}

c. R

_{ground->table}and R

_{table->ground}

d Mg and R

_{table->book}

**Solution 7**

Action reaction pair acts on diffrent body and always are in oppsite direction.We will have to draw free body diagram for each part in this case.

**Question 8**Choose the correct alternative

a. The acceleration of a moving particle is always in the direction of its velocity

b. Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 360 between them

c.Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 180 between them

d. If the acceleration vector is always perpendicular to the velocity vector of a moving particle,the velocity vector does not change

**Solution 8**

In circular motion,acceleration vector is perpendicular to velocity and velocity is also changing.

Now acceleration and velocity can have any angle between

**Question 9**there are two statements

A Newtons first law in valid from the pilot in an aircraft which is taking off

B Newtons first law in valid from the observer in a train moving with constant velocity

Which of the following is correct

(a) A only

(b) B only

(c) Both A and B are correct

(d) Both A and B are wrong

**Solution 9**Newtons law is valid from Intertial frame of refrence

when the plane is taking off,it has acceleration so it can not be intertial frame of refrence

When the observer is train moving with constant velocity,so no acceleration is invloved.So it is Intertial frame.Hence Newton law is valid

**Question 10**.A body of mass 5 kg starts from the origin with an intial velocity u=30i+40j m/sec.If the constant force acts on the body F=-(i+5j) N.The time in which y component of the velocity becomes zero is

a. 5 sec

b. 80 sec

c. 20 sec

d. 40 sec

**Solution 10**.

Force=-(i+5j) N

Mass =5 Kg

intial velocity u=30i+40j m/sec

Acceleration:-

a=F/m=-(i+5j)/5=-.2i-j

Y component of intial velocity is 40 m/s

Acceleration in that direction -1 m/s

^{2}

So

v=u+at

0=40-1t

or t=40 sec

Hence d is correct

**Question 11**.A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.The motion of the particle takes place in a plane.it follows that

a. velocity is constant

b. accelerattion is constant

c. KE is constant

d. Particle moves in a circular path

**Solution 11**

Since force is always perpendicular to the body,workdone by it will be zero

So kinetic energy is constant

It is the case of uniform circular motion

Velocity magnitude is constant but direction is constantly changing

Acceleration is constany but direction is constantly changing

Hence c and d are constant

**Question 12**A small mass m is suspended from one end of a vertical string. and then whirled in a horizontal circle at a constant speed v.

Which of the followings is true

a.The strings stays vertical

b.The string becomes inclined to the vertical.

c.There is no force on mass m except its weight

d.The angle of inclination of the string does not depend on the v

e.The centripetal force on m is mg

**Solution 12**

Since the body is rotating on a horizontal circle,String will be inclined to the horizontal

It is shown in figure

The forces acting on the body are weight and tension in the string

Vertical component of the tension will balance out the weight of the body

Horizontal component will provide the required centripetal force for the circular motion

if T is the tension and W be the weight and θ be the angle of inclination

then

Tcosθ =Wg

And Tsinθ =mv

^{2}/R

So angle of inclination depends on the speed of the motion

So only correct answer is b

**Question 13**.The tension in cable supporting an elevator is equal to th weight of the elevetor.The elevator may be

a. going up with incresing speed

b. going down with increasing speed

c. going up with uniform speed

d. going down with uniform speed.

**Solution 13**

Since the tension is the string equals weight of the elevator,It means resultant force is zero

So acceleration must be zero

Hence c and d are correct answer

**Question 14**A refrence frame attached to the earth

a.is an inertial frame by defination

b.cannot be inertial frame as earth is revolving around the sun

c.is an intertial frame because Newton's law are applicable in this frame

d.cannot be intertial frame because earth is rotating about its own axis

**Solution 14**

Earth cannot be inertial frame as it is revolving around the sun and also rotating about its axis

**Question 15**By means of rope ,a body of weight W is moved vertically upward with constant acceleration a .Find the tensile force in the rope

a. W(1+a/g)

b W(1-a/g)

c W

d. none of the above

**Solution 15**

Forces acting on the body

Tensile force vertically upward=S

Weight of the body vertically downward=W

Body has acceleration a vertically upwards

Net force on the body must provide the acceleration

so S-W=(W/g)a

or S=W(1+a/g)