# Laws of Motion

**Newton’s Law of Motion**

There are three laws of motion which was given by Newton.

**First Law**

It states that, “Everybody in the universe continues it’s state of rest or uniform motion in a straight line unless an external force acts on it to change it’s state.”**Second law**

It states that, “The rate of change of momentum of a body is directly proportional to the resultant force that takes place in the direction of force.”**Third Law**

It states that, “To every action , there is equal and opposite direction.”

**Definition of force from the first law**

According to Newton’s First Law of Motion, “An inert object cannot change it’s state of rest or uniform motion . To change the state an external agency is needed which is known as force.”

Thus, Force is defined as the external agency that changes or tends to change the state of a body either at rest or uniform motion in a straight line.

**Measurement of force from Second Law**

Let us consider a body of mass 'm' is moving initially with velocity 'u' is acted upon by constant resultant force 'F' for the time 't' so that the final velocity becomes 'v' as shown in the above figure. If 'a' is the acceleration produced on the body then,

\(a= \frac{v-u}{t}............(i)\)

Also,

Initial momentum = mu

Final momentum = mv

Then,

Change in momentum= mv-mu

Rate of change in momentum= \(\frac{mv-mu}{t}\)

From second law of motion,

\(F \propto \frac{mv-mu}{t}\)

\(or, F \propto \frac{m(v-u)}{t}\)

\(or, F \propto m.a\)

\(or, F =k m.a\)

where k is the proportionality constant.

When k=1

F= m.a

Hence, the second law of motion gives the measurement of force.

**Inertia**

Inertia is the inability of a body to change it's state of rest or uniform motion. Inertia is of two types:

**Inertia of Rest**

The inability of a body to change it's state of rest is defined as inertia of rest. For example: Passenger standing inside jerks backward when bus suddenly starts, falling of fruits when branches are shaken etc.**Inertia of Motion**

The inability of a body to change it's state of uniform motion is defined as inertia of motion. For example: Passenger standing inside jerks forward when bus suddenly stops etc.

**Momentum**

Momentum of a body can be defines as the product of it's mass and velocity. It's unit is kg m/s.

If a body of mass 'm' is travelling with velocity 'v' then,

\(Momentum (p) = mass(m) \times velocity(v)\)

**Impulse**

The impulse of a body is defined as the product of force and time for which the force is applied. It's unit is Ns.

If the force 'F' acts on a body for time 't' then ,

\(Impulse(I) = Force(F) \times time(t)\)

**Relation between Impulse and Momentum**

Let us consider a body of a mass 'm' is moving with initial velocity 'v' due to constant resultant force 'F' for time 't' so that the body gains a final velocity of 'v'. If 'a' is the acceleration produced on a body,

\(a=\frac{v-u}{t}\)

Also, \(F= m.a\)

\(or, F = m (\frac{v-u}{t})\)

\(or, F.t = mv - mu \)

\(or, I(Impulse) = \triangle p\)

Hence, impulse of a body is equal to change in momentum of that body.

**Conservation of Linear Momentum**

It states that "The total linear momentum of a body or system remains constant when no external force acts on the system."

Let us consider a body A with mass 'm_{a}' moving with initial velocity 'u_{a}' collides with another body B of mass 'm_{b}' moving with velocity 'u_{b}' as shown in the figure. Also, let us consider 'v_{a}' and 'v_{b}' be their velocities after collision . If 'F_{b}' force is exerted by A on B for time 't'. According to Newton's Third Law of Motion, the body B also exert force 'F_{a}' on body A for same time,

F_{b} = - F_{a}

or, F_{b} t = - F_{a} t

or, Impulse of body B = -Impulse of body A

or, Change in momentum of B = -Change in momentum of A

or, m_{b} v_{b}- m_{b} u_{b} = - (m_{a} v_{a} - m_{a} u_{a})

or, m_{b} v_{b} + m_{a} v_{a} = m_{a} u_{a} + m_{b} u_{b}

or, Total momentum after collision = Total momentum before collision

Hence this proves that the total linear momentum of a system is overall constant when no external force is acted on the system.

**Second Law as Complete Law**

Both first law and third law are defined by second law. Hence, second law of motion is also called as the complete law.

**For First Law**

We know that,

F = m.a

For no any external force, F= 0

m.a = 0

Since, the mass of a body cannot be zero

a= 0

Hence, the body continues its state of rest or motion when no external force is applied to it.**For Third Law**

Let us consider a body A collides with body B for time 'T'.

Change in momentum of A ,

\(\triangle P_{a} = F_{a} t \)

Change in momentum of B,

\(\triangle P_{b} = F_{b} t \)

Total change in momentum,

\(\triangle P_{a} + \triangle P_{b} = \triangle P \)

For no any external force, \(\triangle P = 0\)

\(\triangle P_{a} + \triangle P_{b} = 0 \)

\(or, F_{a} t + F_{b} t = 0\)

\(or, (F_{a} + F_{b}) t = 0\)

Since, time cannot be zero.

\(F_{a} + F_{b} = 0\)

\(F_{a} = -F_{b}\)

Hence , the action and reaction are equal but opposite in direction.

**Mass- Pulley System**

Let us consider a smaller mass m and heavier mass M are suspended at two ends of string which passes over a pulley as shown in the figure. The smaller mass moves upward and the heavier mass moves downward with the same acceleration 'a'. If 'T' be the tension of string,

For smaller mass,

T= mg + ma (Force and weight in opposite direction)

or, T - mg = ma............(i)

For heavier mass,

T= Mg - Ma (Force and weight in same direction)

or, Mg - T = Ma............(ii)

Adding equation (i) and (ii) we get,

Mg - mg = Ma + ma

or, (M-m)g = (M+m) a

or, a = \((\frac{M-m}{M+m}) g\)

Substuting the value of a ,

\(T = mg + ma\)

\(= mg + m (\frac{M-m}{M+m}) g\)

\(= (1 + \frac{M-m}{M+m}) mg\)

\(= (\frac{M+m+M-m}{M+m})mg\)

\(= (\frac{2Mm}{M+m})g\)

**Apparent weight of person in lift**

Let us consider a person of mass 'm' is standing on the surface of lift. The weight of a person acts vertically downward and the reaction R acts vertically upward.

- Lift at stationary

If the lift is at stationary then we can write,

R = mg - Lift moving upward with acceleration

If the lift is moving up with acceleration a then we have,

R= mg + ma - Lift moving downward with acceleration

If the lift is moving down with acceleration a then we have,

R= mg - ma - Lift moving up or down with uniform velocity

If the lift is moving up or down with uniform velocity(a=0),

R = mg - Lift is free falling

If the lift is at freefall then a=g

R = mg - ma

= mg - mg

=0

Hence person feels weightlessness. - Lift moving upward with a=g

R= mg + ma

=mg + mg

=2mg