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 system, there are no exceptions to this law. This law becomes a powerful tool 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 affect 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 post-collision velocity (v) can be determined using a momentum table as shown below.

	Before Collision	After Collision
Person	0	(60 kg) • v
Medicine ball	(15 kg) • (20 km/hr) = 300 kg • km/hr	(15 kg) • v
Total	300 kg • km/hr	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 expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball. To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system 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.

A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.

Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity (v).

The collision between the ball and the catcher's mitt occurs in an isolated system, total system momentum is conserved. Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt). The table below depicts this principle of momentum conservation.

	Before Collision	After Collision
Ball	0.15 kg • 45 m/s = 6.75 kg•m/s	(0.15 kg) • v
Catcher's Mitt	0	(0.25 kg) • v
Total	6.75 kg•m/s	6.75 kg•m/s

Observe in the table above that the known information about the mass and velocity of baseball and the catcher's mitt 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 0.15 • v and 0.25 • v are used for the after-collision momentum of the baseball and catcher's mitt. To determine v (the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results:

0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s

0.40 kg • v = 6.75 kg•m/s

v = 16.9 m/s

Using algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.

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 ofenergy 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.

Now we will consider the analysis of a collision in which the two objects do not stick together. In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together.

A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).

The collision can be analyzed using a momentum table similar to the above situations.

	Before Collision	After Collision
Truck	3000 • 10 = 30 000	3000 • v
Car	0	1000 • 15 = 15 000
Total	30 000	30 000

Observe in the table above that the known information about the mass and velocity of the truck and car 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. The after-collision velocity of the car is used (in conjunction with its mass) to determine its momentum after the collision. Finally, the expression 3000•v was used for the after-collision momentum of the truck (v is the velocity of the truck after the collision). To determine v (the velocity of the truck), the sum of the individual after-collision momentum of the two objects is set equal to the total momentum. The following equation results:

3000*v + 15 000 = 30 000

3000*v = 15 000

v = 5.0 m/s

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 lost momentum (slowed down) 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 that are worth the practice. However, be certain that you don't come to believe that physics is merely an applied mathematics course that 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 that provides the basis of their solution.

The next section of this lesson involves examples of problems that 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 that follow.

The Cart and The Brick

Part B

 

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum lost by one object equals the momentum gained by the other object.

The animation below portrays the collision between a 1.0-kg cart and a 2-kg dropped brick. It will be assumed that there are no net external forces acting upon the two objects involved in the collision. The only net force acting upon the two objects (the cart and the dropped brick) are internal forces - the force of friction between the cart and the dropped brick. The before- and after-collision velocities and momentum are shown in the data tables.

In the collision between the cart and the dropped brick, total system momentum is conserved. Before the collision, the momentum of the cart is 60 kg*cm/s and the momentum of the dropped brick is 0 kg*cm/s; the total system momentum is 60 kg*cm/s. After the collision, the momentum of the cart is 20.0 kg*cm/s and the momentum of the dropped brick is 40.0 kg*cm/s; the total system momentum is 60 kg*cm/s. The momentum of the cart-dropped brick system is conserved. The momentum lost by the cart (40 kg*cm/s) is gained by the dropped brick.

Momentum Conservation Principle

One of the most powerful laws in physics is the law ofmomentum conservation. The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) isconserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof.

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that theimpulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

A useful analogy for understanding momentum conservation involves a money transaction between two people. Let's refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money that each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack's pocket to Jill's pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack's money plus Jill's money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.

A useful means of depicting the transfer and the conservation of money between Jack and Jill is by means of a table.

The table shows the amount of money possessed by the two individuals before and after the interaction. It also shows the total amount of money before and after the interaction. Note that the total amount of money ($200) is the same before and after the interaction - it is conserved. Finally, the table shows the change in the amount of money possessed by the two individuals. Note that the change in Jack's money account (-$50) is equal to and opposite of the change in Jill's money account (+$50).

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. A common physics lab involves the dropping of a brick upon a cart in motion.

The dropped brick is at rest and begins with zero momentum. The loaded cart (a cart with a brick on it) is in motion with considerable momentum. The actual momentum of the loaded cart can be determined using the velocity (often determined by a ticker tape analysis) and the mass. The total amount of momentum is the sum of the dropped brick's momentum (0 units) and the loaded cart's momentum. After the collision, the momenta of the two separate objects (dropped brick and loaded cart) can be determined from their measured mass and their velocity (often found from a ticker tape analysis). If momentum is conserved during the collision, then the sum of the dropped brick's and loaded cart's momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick. Momentum data for the interaction between the dropped brick and the loaded cart could be depicted in a table similar to the money table above.

	Before Collision Momentum	After Collision Momentum	Change in Momentum
Dropped Brick	0 units	14 units	+14 units
Loaded Cart	45 units	31 units	-14 units
Total	45 units	45 units

Note that the loaded cart lost 14 units of momentum and the dropped brick gained 14 units of momentum. Note also that the total momentum of the system (45 units) was the same before the collision as it was after the collision.

Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand. The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West (review the section on adding vectors if necessary). Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.

Now suppose that a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The momentum of the medicine ball is 80 kg*m/s before the collision. The momentum of the clown is 0 m/s before the collision. The total momentum of the system before the collision is 80 kg*m/s. Therefore, the total momentum of the system after the collision must also be 80 kg*m/s. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg*m/s. Momentum is conserved in the collision.

Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram. Later in Lesson 2, we will use the momentum conservation principle to solve problems in which the after-collision velocity of objects is predicted.

Two Cars in 2-Dimensional Collision

Inelastic Collision

 

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

The animation below portrays the inelastic collision between two 1000-kg cars. The before- and after-collision velocities and momentum are shown in the data tables.

In the collision between the two cars, total system momentum is conserved. Yet this might not be apparent without an understanding of the vector nature of momentum. Momentum, like all vector quantities, has both a magnitude (size) and a direction. When considering the total momentum of the system before the collision, the individual momentum of the two cars must be added as vectors. That is, 20 000 kg*m/s, East must be added to 10 000 kg*m/s, North. The sum of these two vectors is not 30 000 kg*m/s; this would only be the case if the two momentum vectors had the same direction. Instead, the sum of 20 000 kg*m/s, East and 10 000 kg*m/s, North is 22 361 kg*m/s at an angle of 26.6 North of East. Since the two momentum vectors are at right angles, their sum can be found using the Pythagorean theorem; the direction can be found using SOH CAH TOA (specifically, the tangent function). The value 22 361 kg*m/s is the total momentum of the system before the collision; and since momentum is conserved, it is also the total momentum of the system after the collision. Since the cars have equal mass, the total system momentum is shared equally by each individual car. In order to determine the momentum of either individual car, this total system momentum must be divided by two (approx. 11 200 kg*m/s). Once the momentum of the individual cars are known, the after-collision velocity is determined by simply dividing momentum by mass (v=p/m).

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 250 000 Joules (200 000 J for the eastbound car plus 50 000 J for the northbound car). After the collision, the total system kinetic energy is 125 000 Joules (62 500 J for each car). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.