# Impulse and momentum

## (3) Impulse and Linear momentum

• 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 v0 then at any time t velocity of particle is given by the equation
v = v0 + 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 = mv0 + Ft
or
Ft = mv - mv0                   -(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 t1 and t2,then impulse of the force is
I = F(t2-t1)                  -(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 t1 velocity of the particle is v1 and at time t2 velocity of the particle is v2,then
F(t2-t1)=mv2-mv1                  -(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(t2-t1)                  -(5)
Linear momentum=p=mv                  -(6)
where
I=Ixi+Iyj+Izk
F=Fxi+Fyj+Fzk
p=pxi+pyj+pzk
v=vxi+vyj+vzk
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(t2-t1) =mv2-mv1                  -(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 t1 velocity of the particle is v1 and at time t2 velocity of the particle is v2,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 (t2-t1) 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 t2-t1 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