*♦ Learn about : Newton’s first law of motion , Newton’s second law of motion & Numerical problems based on it ♦*

### Newton’s first law of motion :

Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an externally impressed force to act otherwise.

The tendency of material bodies to maintain their velocity constant is known as inertia Mass is a measure of inertia. When an external force acts on a body, it changes its velocity.

__Newton’s second law of motion :__

The rate of change of momentum of a body is directly proportional to the applied force and acts in the direction of the force.

Momentum is the **‘quantity of motion’** present in a body. It is a vector quantity and for small enough velocities (v « c, the velocity of light), it is given by,

, where m is the mass of the body and v^{→} its velocity.

For a body ,

where k is a constant. With proper choice of units, k = 1

Thus ,

**Illustration :** A body of mass m = 1 kg falls from a height h = 20 m from the ground level

(a) What is the magnitude of the total change in momentum of the body before it strikes the ground ?

(b) What is the corresponding average force experienced by it?

(g = 10m/sec^{2}).

**Solution:**

(a) Since the body falls from rest (u = 0) through a distance h before striking the ground, the speed v of the body is given by

v^{2} = u^{2} + 2 as ; putting a = g and s = h

we obtain v = √(2gh)

The magnitude of total change in momentum of the body

= ΔP = | mv – 0 | = mv

where v = √(2gh)

⇒ ΔP = m√(2gh)

ΔP = (1)√(2×10×20) kg m/sec

ΔP = 20 kg m/sec.

The average force experienced by the body =

, where Δt = time of motion of the body = t(say)

We know ΔP = 20 kg m/sec.

Therefore we will have to find t using the given data. We know from kinematics that,

, (Since , u = 0)

Putting the general values of ΔP & t we obtain

F_{av} = mg

where mg is the weight (W) of the body and is directed vertically downwards. Therefore the body experiences a constant vertically downward force of magnitude mg.

**Exercise :**

A body of mass m = 2 kg changes its velocity v_{1} = (i + 2j + 3k) m/sec to v_{2 }= (-2i + 2j – k) m/sec during 2 seconds.

(a) Find the magnitude of the change in momentum of the body .

(b) What is the magnitude of the average force experienced by the body ?

* Newton’s third law of motion*

**Every action has an equal and opposite reaction.** This implies that if a force is applied by a body A on a body B, then a reaction force acts on A due to B.

**According to the third law,**

, is they are equal in magnitude but opposite in direction.

**This law is true for all forces** – contact forces (such as friction)and non-contact forces (such as gravitation and electrostatics ) but does not apply to **pseudo forces.**

**Illustration : **Two blocks of mass 2 kg and 4 kg are kept in contact with each other on a smooth horizontal surface. A horizontal force of 12 N is applied on the first block due to which the pair moves with a constant acceleration. Calculate the force acting between the blocks.

**Solution :** Let a be the common acceleration of the blocks,

R = Force between the two blocks

From FBD of Ist block

F – R = m_{1} a . .. . (i)

From FBD of block 2,

R = m_{2} a . . . .(ii)

From (i) & (ii),

F = (m_{1} + m_{2}). a

a = F / (m_{1} + m_{2})

= 12/6 = 2 m/sec^{2}

Force between blocks

= R = m_{2} a

= (4 ×2) N = 8 N.

__Application of Newton’s laws of motion__

**Learn about : Application of laws of motion : Techniques & Approach**

A separate diagram of the body is drawn showing all the different forces exerted by the bodies in the environment. This diagram is known as the **free body diagram.**

Application of Newton’s Laws to any system (consisting of one or more objects) can be done by following a systematic method.

**We Recommend the following steps in the order given below :**

**(i)** Draw the complete free body diagram (FBD), showing all the forces acting on each separate body.

**(ii)** Select proper coordinates for analysing the motion of each body.

Include any pseudo-forces within the FBD, if required.

**(iii)** If there are any constraints, write the proper constraint equations.

**(iv)** Apply Newton’s 2nd law of motion : F^{->} = ma^{->} for each body. This leads to a system of equations.

**(v)** Solve these equations:

**(a)** Identify the known and unknown quantities. Check that the no. of equations equals the number of unknowns.

**(b)** Check the equations using dimensional analysis.

**(c)** After solving, check the final solution using back substitution.

**(vi)** If the velocity (v^{→} ) or, position ( x^{→}) is required, proceed from a knowledge of the acceleration ( a^{->}) as found from equations in step (v) and apply kinematics: dv^{→}/dt = a^{→} (known) and finally integrate.