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EN4 : Dynamics and Vibrations

4 Impulse and Momentum

In the previous chapter, we investigated what happens when we integrate a force through a distance and introduced the concept of work. We were then able to show that the work done by the net force was equal to the change in the kinetic energy of the particle. In addition, we were able to show that the work done by external forces was equal to the change the total energy of the particle. We then inferred that in an isolated conservative mechanical system, the total mechanical energy is conserved.

In this chapter we will take a different strategy, in that we will integrate forces with respect to time. This will give us a new concept called ``impulse''. Just as we had defined kinetic energy, we will define a new quantity called momentum which will depend on the mass and velocity of the body. We will then see that impulse on a particle will be the change in momentum, just as work was equal to change in kinetic energy. We will also arrive at a very important result that if there are no forces acting on a particle, its momentum will be conserved.

4.1 Impulse

Let tex2html_wrap_inline508 be the net force that is acting on a particle. Note that the force is assumed to be a function of time. The impulse of a force in a time interval tex2html_wrap_inline510 to tex2html_wrap_inline512 is defined as the time integral of the force


Note that impulse, unlike work, is a vector quantity. The SI units for impulse is Newton times seconds or Ns.

4.2 Momentum

The linear momentum tex2html_wrap_inline518 of a particle of mass m with velocity is tex2html_wrap_inline522 is defined as


Note that this is a vector quantity, and has SI units of kilograms times meters per sec, or kgm/s.

4.3 Impulse-Momentum Relationship

As usual, the first question that comes to mind is ``so what?''. Well, we did not define these quantities in futility. These quantities are related via a relation just as work and kinetic energy were related. To see this, lets take a closer look at Newton's Law. We know that the force applied produces an acceleration, or,


In other words we can restate Newton's Law as ``The rate of change of momentum of a particle is the applied force''. This is a more general statement of Newton's Law and is more useful than the old tex2html_wrap_inline536 . To see this think of a system in which the mass is changing, such as a rocket. We shall, however, not deal with systems such as this but remain focussed on systems where the mass is constant.

Having restated Newton's Law in this manner, we are now in a position to derive the relationship between impulse and momentum. Let us integrate the new form of Newton's Law from time tex2html_wrap_inline510 to tex2html_wrap_inline512 . We get


Thus we get a rather interesting result (that looks very similar to the work energy theorem): impulse is equal to the change in momentum.

Note also that when there is no force acting on the particle, its momentum is conserved.

Let us look at some examples that will help gel our ideas about impulse and momentum.

4.3.1 Examples

Example 1: A force acts on a particle of mass m initially at rest for a time interval T, during which it is has a functional form tex2html_wrap_inline556 . Compute the velocity of the ball at t = T. The impulse tex2html_wrap_inline560 . Thus tex2html_wrap_inline562 or, the velocity at time T is tex2html_wrap_inline566 .


Example 2: A ball is dropped from a height H. After it bounces off the floor, the ball rises back to height h. Compute the impulse exerted by the floor on the ball. Let tex2html_wrap_inline510 be the instant at the the ball first makes contact with the ground. At this instant the velocity vector of the ball is given by tex2html_wrap_inline574 which implies that the momentum at this instant is tex2html_wrap_inline576 . Now, let tex2html_wrap_inline512 be the time instant at which the ball just leaves the floor. The velocity vector of the ball at this instant is tex2html_wrap_inline580 (How?). The momentum at the instant tex2html_wrap_inline512 is therefore tex2html_wrap_inline584 . Now by impulse-momentum relationship, we have that


Thus the impulse in this case is


4.4 Impacts

So far we have considered only one-particle problems in this course. In the remainder of this lecture we will look at two particles moving in a straight line - what happens when they collide, etc.

Consider two masses tex2html_wrap_inline594 and tex2html_wrap_inline596 moving along a straight line such that they have velocities tex2html_wrap_inline598 and tex2html_wrap_inline600 respectively. Assuming that both tex2html_wrap_inline602 and tex2html_wrap_inline604 are positive (i.e.,) they are both moving from left to right, and that tex2html_wrap_inline606 , we see that the mass tex2html_wrap_inline594 will eventually collide with mass tex2html_wrap_inline596 . When they collide they apply some forces on each other and separate after some time. Let us say they have final velocities tex2html_wrap_inline612 and tex2html_wrap_inline614 respectively. Our task will be to determine these two quantities. Let us see how we can apply all that we have learned so far in this course to determine the answer to this problem.


Let us assume that the bodies come in contact at time tex2html_wrap_inline510 and remain in contact until some later time tex2html_wrap_inline512 . During this time interval the mass tex2html_wrap_inline594 applies a force tex2html_wrap_inline508 on mass tex2html_wrap_inline596 . By Newton's third law, the mass tex2html_wrap_inline596 must exert an equal but opposite force on the mass tex2html_wrap_inline594 . We know from the impulse-momentum relation for mass tex2html_wrap_inline596 that


Impulse momentum relation for mass tex2html_wrap_inline594 gives


From these two relations, we get that


This give us a very powerful statement: The total momentum of the system of two particles is conserved in collisions!. Taking the required components along the tex2html_wrap_inline654 -direction, we get


But alas, this gives us only one equation for the two unknowns tex2html_wrap_inline656 and tex2html_wrap_inline658 . Thus application of the fundamental principles that we have learned so far gives us insufficient information to solve this problem.

This is the first time in we are encountering such a problem. So far when we talked, say about dropping a ball of mass m we did not have to say if the ball was made of steel or wood or plastic. In the case of impacts this is not the case! We cannot solve the problem without the knowledge of some material property. This material property is called the coefficient of restitution. This is defined as


or said in words, it is the ratio of the magnitude of relative velocity after impact (relative velocity of separation) to the value before impact (relative velocity of approach). This quantity depends on the materials that the balls are made of. It is always between 0 and 1. Typical numbers are 0.8 for glass/glass, 0.6 for steel/steel, very close to zero for clay on clay. What is the physical origin of the coefficient of restitution? When this collide they deform. In the deformation process some of the kinetic energy of the balls is converted to other forms of energy such as heat and elastic energy.

Let us look at the limiting cases. When e is unity, we say that the collision is perfectly elastic. In such cases there is no loss of kinetic energy. When e is zero we say that the collision is perfectly plastic. In this case the bodies stick together after collision.

Let us look at some examples:

Example 1: A ball of mass m traveling with a velocity tex2html_wrap_inline668 collides with an identical ball which is at rest. If the coefficient of restitution is e, compute the final velocities of the two balls.

From the principle of conservation of momentum we know that


We also know that the coefficient of restitution is e, and thus


Solving for tex2html_wrap_inline656 and tex2html_wrap_inline658 , we get that


In the case where e = 1, the mass tex2html_wrap_inline594 comes to rest and the mass tex2html_wrap_inline596 move with the velocity tex2html_wrap_inline668 . When e=0, the mass tex2html_wrap_inline594 and tex2html_wrap_inline596 stick together and move with the velocity tex2html_wrap_inline690 .

Example 2: A ball of mass m is dropped on a cement floor from a height H. If the ball bounces back to a height h, compute the coefficient of restitution e between the ball and the cement floor.

We know from Example 2 of Section 4.3 that the velocity of the ball (take this as `` tex2html_wrap_inline594 '') just before impact is tex2html_wrap_inline702 and just after impact is tex2html_wrap_inline704 . Noting that the velocity of earth (or the cement floor), which is `` tex2html_wrap_inline596 '' in this problem, before and after impact is zero, we get that