# Impulse - Momentum Theorem

## Impluse And Momentum of Class 11

In the previous chapter, we have learnt that work done by a force brings about change in kinetic energy of a particle. Let us see what physical quantity changes due to impulse of a force.

According to Newton's second law, the net force acting on a particle is equal to the product of mass and acceleration.

Since a = , therefore

= m

Substituting the value of net force in equation (9.3), we get

=

or=

Notice when we change the variable of integration from t to v, we must also change the time limits of the integral to the corresponding limits of v.

For constant mass,

=

or= (9.4)

The quantity

(9.5)

is a vector and is called the momentum of a particle of mass m moving with velocity .

Thus=

or= Δ (9.6)

**Equation (9.6)** shows that the net impulse of forces acting on a particle is equal to the change in momentum of the particle. This is called the Impulse-Momentum Theorem.

**Example 9.3**

A 2 kg is moving at a speed of 6 m/s. How large a force F is needed to stop the block in a time of 0.5 ms?

**Solution**

Impulse on block = Change in momentum of block

Ft = mvf - mvi

F(5 × 10-4) = 2(0) - (2)(6)

orF = -2.4 × 104 N

The negative sign indicates that the force opposes the motion.

**Example 9.4**

A ball falling with velocity m/s subjected to a net impulse

= (0.6 + 0.18 ) Ns. If the ball has a mass of 275 g, calculate its velocity immediately following the impulse.

**Solution**

Using Impulse - Momentum Theorem

m - m=

or

Thus, = -0.65 - 0.35 + 0.6 + 0.18 /0.275

or = (-0.65 - 0.35 ) + (2.18 + 0.655)

or = (1.53 + 0.305 ) m/s