**Concept :**

We observe a lot of collisions in our daily life. One of the characteristic of a collision is that there is sudden change in velocity of the object.

**For example,** when an α-particle passes by the nucleus of a hydrogen atom in Rutherford’s experiment, it gets deflected in a very short time.

Deflection means a change in direction of motion, a change in velocity.

Therefore the α-particle is said to be accelerating. This acceleration is caused by strong electrostatic force of repulsion between α-particle and the nucleus in a very short time. In this process, the particles do not touch each other.

Let us take **another example.** When a rubber ball strikes a floor, it remains in contact with the floor for very short time in which it changes its velocity.

This is an example of collision where physical contact takes place between the colliding bodies.

*Collision is a process in which two bodies mutually interact and exchange their energy & momentum.*

** Impulse is the change in momentum produced by an impulsive force in a short interval of time. **

Note that impulsive force is a large force that acts on a body for a short interval of time.

__Impulse :__

Take the example of a ball striking with the ground with certain velocity, say v_{1}^{→} . During a very short time δt , it changes its velocity to v_{2}^{→} , say.

The change in velocity of the ball is

Δv^{→} = v_{2}^{→} – v_{1}^{→}

=> The acceleration of the ball is

a^{→} = Δv^{→}/Δt

⇒ a^{→} = dv/dt

Since a^{→} = F_{net }/m according to Newton’s 2nd Law of motion

**The term ∫F.dt is known as impulse.**

⇒ The impulse of a force F in a given time interval Δt is equal to the product of the force and the time interval.

This is numerically equal to the change of momentum of the body due to the action of the force F during that time internal.

**Illustration :** A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its total mechanical energy. If it remains in contact with the ground for δt = 0.01 sec., find the impulse of the impact force.

**Solution:** The change in momentum ΔP of the ball during Δt = 0.01 sec. is known as the impulse (F Δt) of the force of impact

⇒ F Δt = ΔP

ΔP^{→} | = |mv_{2}^{→} – mv_{1}^{→}|

= m(v_{1} + v_{2}) are antiparallel)

⇒ Fδt = m(v_{1} + v_{2}) ….(1)

Since, the ball falls through a height h , √2gh .

As the ball losses 75 percent of its total mechanical energy that is kinetic energy.

⇒ 1/2 mv_{2}^{2} = 1/2 mv_{1}^{2} (1 – 75/100)

⇒ v_{2} = v_{1} /2 = √2gh/2 = √(gh/2)

Substitution of the values of v_{1} and v_{2} in (1) yields

Fδt = 90 x 10^{-3} x 3√(gh/2)

= 90 x 10^{-3} x 3√(9.8×10/2)

= 1.05 N-sec.

### Impulsive force

Following the previous example, we find that the average force exerted by the ground on the ball during time δt is given as F_{net} = m(v_{2} – v_{1})/δt

since (v_{2} – v_{1}) is finite and the time internal δt is infinitesimal, their ratio becomes very large.

Therefore F_{net} is a very large force acting on the ball for very short time.

This impact force is known as **impulsive force.** The impact force or impulsive force F is much greater than the external forces like weight (mg) and frictional force, in general.

**Illustration :** A ball of mass = 100 gm is released from a height h_{1} = 2.5 m with the ground level and then rebounds to a height h_{2} = 0.625 m. The time of contact of the ball and the ground is δt = 0.01 sec. Find the impulsive (impact) force offered by the ball on the ground.

**Solution:**

F = m(v_{2}^{→} – v_{1}^{→})/δt

where | v_{2}^{→} – v_{1}^{→}|

= v_{2} + v_{1} ,

where v_{1} = √2gh_{1} and v_{2} = √2gh_{2}

⇒ F =

⇒ F =

⇒ F = 105 N.