*♦ 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,

P^{→ }= mv^{→}

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

For a body ,

dP^{→}/dt ∝ F^{→}

d(mv^{→})/dt ∝ F^{→}

md(v^{→})/dt ∝ F^{→}

F^{→} = k.m (d(v^{→})/dt)

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

Thus, F^{→} = m a ^{→}

**Illustration 1 :** 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)√(2x10x20) kg m/sec

ΔP = 20 kg m/sec.

The average force experienced by the body = F_{av} = ΔP/Δt 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,

s = ut + (1/2) at^{2}

⇒ h = (1/2)gt^{2}

(Since , u = 0)

t = √(2h/g) and ΔP = m√(2gh)

F_{av} = ΔP/Δt

Putting the general values of ΔP & t we obtain

F_{av} =m√(2gh)/√(2h/g) = 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 1 :**

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 ?