**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 :__

__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 **

**The acceleration of the ball is **

**The term 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**

** are antiparallel**

**⇒ F.Δt = m(v _{1} + v_{2}) ….(1)**

**Since, the ball falls through a height h , **

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

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

**= 1.05 N-sec.**

**Impulsive force**

**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:**

**Where , **

**Where , **

**On putting the values**

**⇒ F = 105 N.**