Collisions: Difference between revisions

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vfinal = 2.5 / 1.2 = 2.08333
vfinal = 2.5 / 1.2 = 2.08333


Thus, the final velocity is 2.08 m/s.
Thus, the final velocity is 2.08 m/s. Since the two balls stick together, this is an example of a [[Maximally Inelastic Collision]]


===Middling===
===Middling===

Revision as of 16:10, 17 April 2016

SECOND EDIT BY SUNGJAE HYUN IN 2016-04-16 ADDITIONS MADE BY (currently in progress, adding computational model and difficult problem) Sam Webster 4-17-16

This topic covers Collisions, a comprehensive way to combine the Momentum and Energy Principles.

The Main Idea

Collisions are special types of contact interactions between objects. From a physics standpoint, collisions are a way to combine the Momentum and Energy Principles. In the case of collisions, if we choose a system of the two objects interacting, the change in momentum of the system and the change in energy of the system are both zero. With this in mind, calculations with collisions become very simple. We are able to choose a system of only the two objects (excluding external forces such as the Earth's gravity) because the collision takes place during such a short time that the external forces have a negligible effect. There are two types of collisions, Elastic Collisions and Inelastic Collisions.

A Mathematical Model

[math]\displaystyle{ {E}_{f} = {E}_{i} }[/math] where [math]\displaystyle{ {E}_{f} }[/math] is the total final energy of the system and [math]\displaystyle{ {E}_{i} }[/math] is the total initial energy of the system.

[math]\displaystyle{ {p}_{f} = {p}_{i} }[/math] where [math]\displaystyle{ {p}_{f} }[/math] is the total final momentum of the system and [math]\displaystyle{ {p}_{i} }[/math] is the total initial momentum of the system.

A Computational Model

The following situation was used for this model created using python:

Two objects are initially moving towards a third stationary object. Object 1 (in red) at position (-10,0,-2)m has a mass of 2kg and velocity (10,0,2)m/s. Object 2 (in blue) at position (0,6,3)m has a mass of 4kg and velocity (0,-6,3)m/s. Object 3 (in orange) at the origin has a mass of 5kg. The objects collide at the origin and after the collision it is observed that object 1 has embedded in object 3 (hence it is no longer visible in the model) and the combined object moves with velocity (5,-6,0)m/s (This is an example of a Maximally Inelastic Collision). The final velocity of object 2 is calculated and used in the rest of the simulation. (NOTE: the white sphere visible after the collision represents the origin to add dimension to the model and the delay during the collision is there to emphasis when and where the collision happens. Also use the link below the model for a more detailed explanation of the calculations and concepts.).

Check out the code used for the simulation at [1]. The code is commented pretty heavily (but it does assume some knowledge of python and python animations) in hopes that future students can modify some of the variables to experiment with creating different results.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Example 1) A 0.5 kg soccer ball is moving with a speed of 5 m/s directly toward to 0.7 kg basket ball which is at rest. When two balls collide and stick together what will their final velocity be?

___

m1 = 0.5 kg v1,init = 5 m/s m2 = 0.7 kg v2,init = 0 m/s

So m1 * v1 + m2 * + v2 = (m1 + m2) * vfinal

LHS = 0.5 * 5 + 0.7 * 0 = 2.5 kg*m/s RHS = (0.5 + 0.7) * vfinal = 1.2 * vfinal

LHS = RHS in inelastic collision

2.5 = 1.2 * vfinal

vfinal = 2.5 / 1.2 = 2.08333

Thus, the final velocity is 2.08 m/s. Since the two balls stick together, this is an example of a Maximally Inelastic Collision

Middling

Example 2) A bullet of 50 caliber machine gun is 42 grams. It strikes a wooden target block of mass 10 kg stationed on a friction-less surface. The wooden block gains velocity of 1.8 m/s after being embedded with the bullet. What was the velocity of the bullet before it collided with the target?



m1 = 42/1000 = 0.042 kg v1,init = ? m2 = 10 kg v2,init = 0 m1,2 = 0.042kg + 10kg vfinal = 1.8 m/s

m1 * v1,init + m2 * v2,init = (m1 + m2) * vfinal

0.042 kg * v1,init = (0.042 + 10) * 1.8 m/s v1,init = 18.0756 / 0.042 = 430.37 m/s

Final Answer: The bullet's speed is 430.37 m/s.

Difficult

Example 3) Two rocks in space are approaching a third stationary rock. Rock 1 has mass 30kg moves with velocity (13,-10,-2)m/s. Rock 2 has mass 60kg and moves with velocity (-3,23,3)m/s. Rock 3 has mass 35kg and is stationary. The rocks collide at rock 3's location. After the collision it is observed that Rock 1 has embedded into Rock 3, but a chunk of mass 15kg has broken off of the combined rock, Rock 3'. It is also noticed that Rock 2 is now moving with velocity (-2,12,2)m/s. (NOTE: due to the difficulty and use of three dimensions it is suggested to use a computer to solve this problem. Therefore the sample solution will involve code, but conceptually everything remains the same.) First determine the momentum of Rock 3' plus the momentum of the 15kg chunk.


Later Rock 3' is observed moving with velocity (x,y,z), what is the velocity of the 15kg chunk?


With the information you know, determine the combined change in internal energy for all of the rocks.

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