To explain the terms impulse and momentum consider a particle of mass m is moving along x-axis under the action of constant force F as shown below in the figure
If at time t=0 ,velocity of the particle is v_{0} then at any time t velocity of particle is given by the equation v = v_{0} + at
where a = F/m
can be determined from the newton's second law of motion .Putting value of acceleration in above equation\
we get
mv = mv_{0} + Ft
or Ft = mv - mv_{0} -(1)
right side of the equation Ft, is the product of force and the time during which the force acts and is known as the impulse
Thus
Impulse= Ft
If a constant force acts on a body during a time from t_{1} and t_{2},then impulse of the force is I = F(t_{2}-t_{1}) -(2)
Thus impulse received during an impact is defined as the product of the force and time interval during which it acts
Again consider left hand side of the equation (1) which is the difference of the product of mass and velocity of the particle at two different times t=0 and t=t
This product of mass and velocity is known as linear momentum and is represented by the symbol p. Mathematically p = mv --(3)
physically equation (1) states that the impulse of force from time t=0 to t=t is equal to the change in linear momentum during
If at time t_{1} velocity of the particle is v_{1} and at time t_{2} velocity of the particle is v_{2},then
F(t_{2}-t_{1})=mv_{2}-mv_{1} -(4)
so far we have considered the case of the particle moving in a straight line i.e along x-axis and quantities involved F,v, and a were all scalars
If we call these quantities as components of the vectors F,v and a along x-axis and generalize the definitions of momentum and impulse so that the motion now is not constrained along one -direction ,Thus we got
Impulse=I=F(t_{2}-t_{1}) -(5)
Linear momentum=p=mv -(6)
where I=I_{x}i+I_{y}j+I_{z}k F=F_{x}i+F_{y}j+F_{z}k p=p_{x}i+p_{y}j+p_{z}k v=v_{x}i+v_{y}j+v_{z}k
are expressed in terms of their components along x,y and z axis and also in terms of unit vectors
On generalizing equation (4) using respective vectors quantities we get the equation F(t_{2}-t_{1}) =mv_{2}-mv_{1} -(7)
So far while discussing Impulse and momentum we have considered force acting on particle is constant in direction and magnitude
In general ,the magnitude of the force may vary with time or both the direction and magnitude may vary with time
Consider a particle of mass m moving in a three-dimensional space and is acted upon by the varying resultant force F. Now from newtons second law of motion we know that F=m(dv/dt)
or Fdt=mdv
If at time t_{1} velocity of the particle is v_{1} and at time t_{2} velocity of the particle is v_{2},then from above equation we have
Integral on the left hand side of the equation (8) is the impulse of the force F in the time interval (t_{2}-t_{1}) and is a vector quantity,Thus
Above integral can be calculated easily if the Force F is some known function of time t i.e.,
F=F(t)
Integral on the right side is when evaluated gives the product of the mass of the particle and change in the velocity of the particle
using equation (9) and (10) to rewrite the equation (8) we get
Equivalent equations of equation (11) for particle moving in space are
Thus we conclude that impulse of force F during the time interval t_{2}-t_{1} is equal to the change in the linear momentum of the body on which its acts
SI units of impulse is Ns or Kgms^{-1}
Example 1
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 5 m/s
If the mass of the ball is 0.20 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball) Solution
Impulse =change in momentum
=.20 * 5 - (-.20* 5)
=2 N s
Watch this tutorial for more information on How to solve momentum/Impulse problems