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