Lesson 2: The Law of Momentum Conservation

Using Equations as a "Recipe" for Algebraic Problem-Solving

As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. The momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in an isolated systems, there are no exceptions to this law. This law becomes a powerful law in physics because it allows for predictions of the before- and after-collision velocities (or mass) of an object. In this portion of Lesson 2, the law of momentum conservation will be used to make such predictions. The law of momentum conservation will be combined with the use of a "momentum table" and some algebra skills to solve problems involving collisions occurring in isolated systems.

Consider the following problem:

A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity ("v") across the ice.

If it can be assumed that the effect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system. Momentum should be conserved and the problem can be solved for v by use of a momentum table as shown below.

	Before Collision	After Collision
Person	0	60 * v
Medicine ball	300	15 * v
Total	300	300

Observe in the table above that the known information about the mass and velocity of the two objects was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expression "60*v" and "15*v" was used for the after-collision momentum of the person and the medicine ball. To solve the problem for "v" (the velocity of both the objects after the collision), set the sum of the individual momentum of the two objects equal to the total momentum. The following equation results:

60*v + 15*v = 300
75*v = 300
v = 4 km/hr

Using algebra skills, it can be shown that v=4 km/hr. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision. (NOTE: The unit km/hr is the unit on the answer since the original velocity as stated in the question had units of km/hr.)

Now consider a similar problem involving momentum conservation.

Before the collision, Granny has momentum and Ambrose does not. The collision causes Granny to lose momentum and Ambrose to gain momentum. After the collision, the Granny and Ambrose move with the same velocity ("v") across the rink.

80*v + 40*v = 480
120*v = 480
v = 4 m/s

Using algebra skills, it can be shown that v = 4 m/s. Both Granny and Ambrose move across the ice with a velocity of 4 m/s after the collision. (NOTE: The unit m/s is the unit on the answer since the original velocity as stated in the question had units of m/s.)

The two collisions above are examples of inelastic collisions. Technically, an inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. In an inelastic collision, the kinetic energy of the colliding objects is transformed into other non-mechanical forms of energy such as heat energy and sound energy. The subject of energy will be treated in a later unit of The Physics Classroom. To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision.

Using algebra skills, it can be shown that v = 5.0 m/s. The truck's velocity immediately after the collision is 5.0 m/s. As predicted, the truck has slowed down (lost momentum) and the car has gained momentum.

The three problems above illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass-velocity information. There are additional practice problems (with accompanying solutions) later in this lesson which are worth the practice. However, be certain that you don't come to believe that physics is merely an applied mathematics course which is devoid of concepts. For certain, mathematics is applied in physics; however, physics is about concepts and the variety of means in which they are represented. Mathematical representations are just one of the many representations of physics concepts. Avoid merely treating these collision problems as mere mathematical exercises. Take the time to understand the concept of momentum conservation which provides the basis of their solution.

The next section of this lesson involves examples of problems which provide a real test of your conceptual understanding of momentum conservation in collisions. Before proceeding with the practice problems, be sure to try a few of the more conceptual questions which follow.

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