Introduction

Momentum is defined as the mass of an object times its velocity.  Since mass is a scalar and velocity is a vector the product is a vector in the same direction as the velocity.  The concept of momentum is used in two general types of problems, impulse‑momentum solutions of Newton's 2nd law type problems and conservation of momentum problems.  The principle of the conservation of momentum is one of the most basic and important principles in physics, ranking in importance right alongside Newton's laws of motion.  In theory it is possible to derive the principle of conservation of momentum from Newton's laws of motion or to derive Newton's laws from the principle.

Impulse - Momentum

Impulse is defined as the product of a force and the time that it acts.  Since force is a vector and time is a scalar, impulse is also a vector.  The following discussion illustrates the application of the concept of impulse. Newton's second laws says that:

The definition of acceleration is:

Thus we can write Newton's second law as:

If the mass is constant mΔV =Δ(mV), therefore we have:

Rearrange terms to obtain:

We now define impulse as FΔt and momentum  as mV.  Therefore we have

         impulse = change in momentum

Thus we see that we can use the concept of impulse to write Newton's second law in a slightly different form.  We derived this using Newton's second law in the form F =ma.  One step in the derivation required the mass to be constant. However, what happens if the mass of an object is not constant?  If the mass is not constant then mΔV is not the same as Δ(mV).  Which, if either, of the two forms of Newton's second law (F = ma or FΔt = Δ(mV)) is correct?  It turns out that the second form, FΔt = Δ(mV), is the proper form to use when the mass is not constant.  However, application of the equation to situations when the mass is not constant requires higher mathematics so will not be treated here.

The concepts of impulse and momentum simplify some problems where a force on an object is given as well as the time during which the force acts.  A nice feature of impulse is that successive impulses will add.  Thus if we have a resultant force of 100N acting for 5 seconds and then later a force of 50N acts for 10 seconds, the total impulse is 100X5 + 50X10 =1000N‑sec.  Therefore the momentum changes by 1000kg‑m/sec.  If the mass and initial velocity is known then it is a simple matter to find the final velocity.  Examples of applications of the impulse‑momentum equations are given in the solved problems section.

We can also use the impulse‑momentum equation to explain in a qualitative way many phenomena involving forces on moving objects.  A given change in momentum can be produced by a large force acting over a small time or a smaller force acting over a longer time.  For example, a passenger riding in a car always wears his seatbelt because he knows that in case of collision the force to stop his forward motion will act over a longer time so the required force will be less. With the seat belt on, the force begins to act almost at the instant of collision and continues to act until the passenger and car have stopped.  Without the seatbelt the passenger keeps moving as the car stops until he makes contact with part of the car, probably between his head and the nearly stationary windshield.  In this case he will stop in a much smaller time so a much larger force is required.  Furthermore this force may act on a much less durable part of the passenger's anatomy than the seat belt does.

Sample Problem #1

A child's wagon has a mass of 25kg with a small child riding in it.  It is rolling across a horizontal surface with a speed of 1.5 .  The friction force is constant at 5N.  The child's older brother applies a steady force of 20 N for 5 s, finds that he can't keep up and reduces the force to 10 N for 10 more seconds.  How fast is the wagon moving after the 15sec have elapsed?

Sample Solution #1

We will solve this problem by using the equation

The force used must be the resultant force so before finding the impulse 5N must be subtracted from each applied force to find the resultant force.  The two impulses may now be calculated and added and the total set equal to the change in momentum as follows:

Since ΔV = V₂ - V₁ we have

You might try to do this problem without using the concepts of impulse and momentum.  Simply find the acceleration for each period using Newton's second law.  Then use the kinematics equations for each interval to calculate the final velocity for that interval.  You will see that the method above is somewhat shorter.

Sample Problem #2

A baseball having a mass of 150 g and a speed of 30 strikes a bat.  The batter hits the ball just hard enough so that it rebounds with a speed of 30 in the opposite direction.  If the ball remains in contact with the bat for 0.1 seconds, calculate the average force acting on the ball.

Sample Solution #2

Before making your table of values please notice one thing.  The speed of the ball is the same before and after being struck by the bat but the velocity and therefore the momentum are not.  To denote the fact that the ball is moving in the opposite direction after it has been struck use a negative sign for the final velocity.

This problem will use the equation impulse = change in momentum as follows:

Units

Before starting problems a word regarding units is in order.  Since momentum is mass times velocity the units will have to be a mass multiplied by a velocity.  Typical units might be (preferred) or .  The units for impulse will be a force unit multiplied by a time.  These might be N·s (preferred) or dyne·s.  The student should be able to demonstrate that a N·s is equal to a .  In impulse = momentum type problems be sure to use N·s if the momentum is in .  In conservation of momentum type problems (in the next section), units may be mixed to a certain extent.  For example, you may  measure mass in kg and velocity in as long as the same units are used on both sides of the equation.

Kinetic energy (explained in the section on elastic collisions) clearly must have units of mass multiplied by velocity squared.  Usually the unit used is a .  You will learn in a later section that this unit is called a Joule (J).

The use of English units may present some problem.  Forces are measured in pounds (lbs), and speeds in . Impulse then may be measured in lb·sec.  However, at this time you do not know an English unit of mass.  This problem can be solved by recalling that the weight of an object is equal to its mass multiplied by the acceleration of gravity:

Thus whenever you want the mass of an object in English units you should divide its weight by 32.2 (on the earth) and express the units as or .  This unit actually has a name, the “slug”.

Normally the use of English units is avoided in physics but in this case there are many examples which are going to have much more meaning to you if they are expressed in units with which you are familiar.

The negative sign denotes direction, the force acts in the direction opposite to the original velocity.

1.

A 1500kg car moving at a speed of 20m/sec runs into a rock wall and stops in 0.04sec. What was the initial momentum? What was the final momentum? What average force acts on the car during the collision?

2.

A 1kg object initially at rest is acted upon by a force of 50.0N and acquires a speed of 0.10m/sec. During what time did the force act? What was the impulse?

3.

A 100g bullet is fired horizontally at a 1.8kg block resting on a horizontal smooth surface. The bullet is embedded in the block. If the initial speed of the bullet is 400m/sec, what is the speed of the block and bullet after impact? What fraction, if any, of the initial kinetic energy is lost in impact?

4.

A 40kg object is initially at rest. It is given a blow of impulse 250Ns. What speed does the object acquire?

5.

A 100kg man jumps into a swimming pool from a height of 5m. It takes 0.4 sec for the water to reduce his velocity to zero. What average force did the water exert on the man?

6.

Consider the changes in motion produced by the following forces: an object (5kg) moving on the x‑axis is acted on for 2sec by a constant force of 10N toward the right, then for 2 sec by a constant force of 20N to the right and then for 10 seconds by a constant force of 10N to the left. If the object was originally moving to the right at a speed of 2 m/sec, what will be its final speed?