[Physics] Detecting Collisions

[+chorpeac MVC]

Ok I have two objects, each with mass, velocity, and acceleration.

How can you determine a collision using the normal vectors, assuming you are in 3D space? Do you use matrices?

[vincent]

nevermind :blush:

[+chorpeac MVC]

what's up? No response ripgut? :)

[vincent]

Let's see if this helps, keep in mind i don't know much about physics my courses won't start till later next year:

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.

Granny (m=80 kg) whizzes around the rink with a velocity of 6 m/s. She suddenly collides with Ambrose (m=40 kg) who is at rest directly in her path. Rather than knock him over, she picks him up and continues in motion without "braking." Determine the velocity of Granny and Ambrose. Assume that no external forces act on the system so that it is an isolated system.

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.

the collision between Granny and Ambrose occurs in an isolated system, total system momentum is conserved. The total momentum before the collision (possessed solely by Granny) equals the total momentum after the collision (shared between Granny and Ambrose). The table below depicts this principle of momentum conservation.

                   Before Collision       After Collision
         Granny         80 * 6 = 480           80 * v
------------------------------------------------------
         Ambrose       0                           40 * v
------------------------------------------------------
         Total            480                        480
------------------------------------------------------

Observe in the table above that the known information about the mass and velocity of Granny and Ambrose 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 "80*v" and "40*v" was used for the after-collision momentum of the Granny and Ambrose. To solve the problem for "v" (the velocity of both persons after the collision), set the sum of the individual momentum of the two objects equal to the total momentum. The following equation results:

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.

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 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 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).

   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 solve the problem for "v" (the velocity of the truck), set the sum of the individual after-collision momentum of the two objects 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 slowed down (lost momentum) and the car has gained momentum.

Animation

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.

Edited November 15, 2005 by ripgut