Is this 2D collision elastic or inelastic? Calculate to find out.

In summary, a rocket of mass M = 30kg traveling at velocity V = 1500m/s horizontally explodes into two chunks of mass m1 = 20kg and m2 = 10kg. The mass of m2 is observed to be moving at an angle of 20 degrees above the horizontal with a speed of 1000m/s just after the explosion. Using the formula for momentum conservation, the magnitude and direction of m2 can be calculated to be 2709 m/s and -14.63 degrees, respectively. In order to determine whether the explosion was elastic or inelastic, the kinetic energy before and after the explosion can be compared. If the values are equal, the collision is elastic, while if they are
  • #1
Touchme
41
0
A rocket of mass M = 30kg is traveling with velocity V = 1500m/s horizontally. When it explodes into two chunks of mass m1 = 20kg and m2 = 10kg. The mass observed to be moving at an angle of (theta1) 20 above the horizontal with a speed of 1000m/s just after the explosion. Find the magnitude and direction of m2.

Do a calculation that allows you to determine whether this explosion was elastic or inelastic.


I know how to find the magnitude and direction of m2. The problem is the second question. I am not sure how to show that the explosion is inelastic. The magnitude and direction of m2 is 2709 m/s and the direction is -14.63.
 
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  • #2
Touchme said:
A rocket of mass M = 30kg is traveling with velocity V = 1500m/s horizontally. When it explodes into two chunks of mass m1 = 20kg and m2 = 10kg. The mass observed to be moving at an angle of (theta1) 20 above the horizontal with a speed of 1000m/s just after the explosion. Find the magnitude and direction of m2.

Do a calculation that allows you to determine whether this explosion was elastic or inelastic.


I know how to find the magnitude and direction of m2. The problem is the second question. I am not sure how to show that the explosion is inelastic. The magnitude and direction of m2 is 2709 m/s and the direction is -14.63.
Compare the kinetic energy right after the explosion to the kinetic energy just before the explosion.
 
  • #3
When I compare the kinetic energy, what values of velocity do I use?
The way I interpret your advice is to find the initial KE(before) which is the rocket 30/2 x 1500^2 = 33750000 J. Now what am I suppose to do? KE(after) = (1/2)(20)(1000)^2 + (1/2)(10)(2709)^2 ?


I also have a few quick questions. If the KE(before) = KE(after), this means elastic, right? and if those are not equal, it is inelastic?
Will the initial momentum of the entire system equal the final momentum of the system or only the components are equal (conserved)?
 
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  • #4
Didn't you already calculate the velocity and direction of the second mass by using Pfinal=Pinitial. Momentum is conserved only in elastic collisions, not inelastic. Therefore, this particular collision is elastic. I want to say that all explosions are elastic...hmmmm...hope this helps:-p
 
  • #5
Sorry but I have to disagree with you. Most explosions are inelastic... Explosions tend to increase KE (chemical potential energy). I am unsure which velocity to use in determining the KE (after). =[
 
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  • #6
Touchme said:
If the KE(before) = KE(after), this means elastic, right?
See above post. Momentum is conserved in elastic collisions, and so is kinetic energy if it is a *perfect* elastic collision. Not all elastic collisions conserve kinetic energy though.
 
  • #7
Yes I agree with those statements; however my question is still unanswered. Which velocity do I use to solve for KE (after)? Without this, I cannot possibly tell if it is elastic or inelastic...
 
  • #8
Okay, you said up there that KE is "chemical potential energy", it is Kinetic Energy actually. For your other question, to calculate the KE of the system after the explosion you have to add together the KE of each mass, therefore, you will use two velocities respectively.
 
  • #9
ok, but which two velocities. The components or the resultant valocities?
 
  • #10
Okay, I am not sure what your saying but the velocities you will use are 2709m/s for M1, and 1000m/s for M2 sorry I am tired right now lol
 
  • #11
matthew baird said:
Didn't you already calculate the velocity and direction of the second mass by using Pfinal=Pinitial. Momentum is conserved only in elastic collisions, not inelastic. Therefore, this particular collision is elastic. I want to say that all explosions are elastic...hmmmm...hope this helps:-p
This is not true. Momentum is conserved in all cases when no external force is present. In this problem, the forces causing the separation of the two objects are internal forces. By Newton's third law, the forces acting on the two objects are equal and opposite at all times. The total change in momentum of the system is zero.
 
  • #12
Touchme said:
When I compare the kinetic energy, what values of velocity do I use?
The way I interpret your advice is to find the initial KE(before) which is the rocket 30/2 x 1500^2 = 33750000 J. Now what am I suppose to do? KE(after) = (1/2)(20)(1000)^2 + (1/2)(10)(2709)^2 ?


I also have a few quick questions. If the KE(before) = KE(after), this means elastic, right? and if those are not equal, it is inelastic?
Will the initial momentum of the entire system equal the final momentum of the system or only the components are equal (conserved)?
You have it right. After the collision the total kinetic energy is the sum of the individual kinetic energies of the two objects. If your first result is correct, then your energy numbers are correct.

Yes, elastic means conservation of kinetic energy. If the before and after kinetic energies are different, the collision is inelastic.
Touchme said:
ok, but which two velocities. The components or the resultant valocities?
For each object, the sum of the squares of the component velocities is the square of the total velocity. You can use the total velocity, or the component velocities (if you use both components). Energy is a scalar (directionless) quantity. It makes no sense to equate "components" of energy before and after a collision. Energy has no components.
 
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Related to Is this 2D collision elastic or inelastic? Calculate to find out.

What is a 2D Collision?

A 2D collision is a type of physical interaction between two objects that occurs on a two-dimensional plane. This can include objects colliding with each other, bouncing off of each other, or transferring energy to each other.

How is 2D Collision different from 3D Collision?

The main difference between 2D and 3D collision is the number of dimensions in which the interaction occurs. In 2D, the objects are moving on a flat plane, while in 3D, the objects are moving in a three-dimensional space. This can affect the types of forces and calculations involved in the collision.

What factors affect the outcome of a 2D Collision?

The outcome of a 2D collision can be affected by several factors, including the mass, velocity, and angle of the objects involved. The type of material the objects are made of can also play a role, as well as any external forces acting on the objects during the collision.

How is 2D Collision studied and analyzed?

Scientists use mathematical equations and computer simulations to study and analyze 2D collisions. These tools allow them to predict the outcome of a collision based on the initial conditions and to understand the forces and energy involved in the interaction.

What are some real-world applications of 2D Collision?

2D collision has many real-world applications, such as in sports, car accidents, and billiards. It is also important in fields such as engineering, where understanding the impact of collisions is crucial in designing structures and vehicles that can withstand forces and protect human life.

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