Change in total kinetic energy in a conservation of momentum problem?

In summary, the conversation discusses a collision between two pucks on a frictionless, horizontal air table. The speed of puck A before the collision is calculated using the equation (m_2*v_2)/(m_1) - (v_1). The change in total kinetic energy of the system is determined by subtracting the kinetic energy of puck A after the collision and the kinetic energy of puck B after the collision from the initial kinetic energy of puck A. The equations for kinetic energy and momentum are discussed, and the correct method for calculating the change in kinetic energy is explained.
  • #1
erik-the-red
89
1
Question:

On a frictionless, horizontal air table, puck A (with mass [tex]m_1[/tex]) is moving toward puck B (with mass [tex]m_2[/tex]), that is initially at rest. After the collision, puck A has a velocity of [tex]v_1[/tex] to the left, and puck B has velocity [tex]v_2[/tex] to the right.

1. What was the speed of puck A before the collision?

My answer to this part of the question was correct. It is [tex](m_2*v_2)/(m_1) - (v_1)[/tex].

2. Calculate the change in the total kinetic energy of the system that occurs during the collision.

I think this depends on the first part of the question.

So, I'm thinking:

[tex]\Delta K= K_f - K_i[/tex]
[tex](1/2)(m)(v_f^2 - v_i^2)[/tex]

I should sum up the velocities, right?

[tex](1/2)(m_1+m_2)((v_2 - v_1)^2 - ((m_2*v_2)/(m_1) - (v_1))^2)[/tex]

But, that is not correct.
 
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  • #2
I'm not very clear with what was meant by your last equation. In anycase, you basically do:

Change in kinetic energy = Kinetic energy of puckA before collision - Kinetic energy of puckA after collision - kinetic energy of puckB after collision.

KE is a function of only mass and velocity. Since all are known, carefully plugging in the numbers should do the trick.
 
  • #3
erik-the-red,

When calculating kinetic energy, you first square the velocity of each particle, multiply by the particles mass divided by two and then add. You have added the final velocities first and then squared and this is wrong.
 
  • #4
I inputed [tex](1/2)(m_1)((m_2*v_2)/(m_1) - (v_1))^2 - (1/2)(m_1)(v_1)^2 - (1/2)(m_2)(v_2)^2[/tex] and it was incorrect.

My understanding of change in total kinetic energy, [tex]\Delta K[/tex], is the difference between the final and initial kinetic energies.

Physics Monkey, you told me to add. Why is this so?
 
  • #5
[itex]\Delta K = K_f - K_i[/itex], not [itex]K_i - K_f[/itex].
 
  • #6
Try applying conservation of momentum and kinetic energy.

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html#4

Consider the momentum of the masses to left and right. Remember momentum is proportional to velocity, so it is a vector quantity. Energy is a scalar.
 
  • #7
From the looks of it, the problem now isn't with the physics, but rather the math. Try cleaning it up, do things step by step and plug in the numbers. Like what Physics Monkey said, you are doing some incorrect operations by trying to rush the calculation. One by one,

[tex]KE_1 = \frac{1}{2}m_1v_1^2[/tex]
 
  • #8
It looks like I definitely did [tex]K_i - K_f[/tex], which obviously is not correct.

So, I decided to analyze [tex]K_f[/tex] first.

I got [tex]K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2[/tex].

Then, I decided to analzye [tex]K_i[/tex].

I got [tex]K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)[/tex]
This simplifies to [tex](1/2)(m_2*v_2 - m_1*v_1)[/tex].

I took the difference [tex]K_f - K_i = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2 - (1/2)(m_2*v_2 - m_1*v_1)[/tex] and it was not correct.

I really think I did everything right this time. ?
 
  • #9
erik-the-red said:
It looks like I definitely did [tex]K_i - K_f[/tex], which obviously is not correct.
Other than that (which gives the wrong sign) I don't see anything wrong with what you posted in #4.

So, I decided to analyze [tex]K_f[/tex] first.

I got [tex]K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2[/tex].
OK.

Then, I decided to analzye [tex]K_i[/tex].

I got [tex]K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)[/tex]
This simplifies to [tex](1/2)(m_2*v_2 - m_1*v_1)[/tex].
You forgot to square the speed.
 
Last edited:
  • #10
Thanks so much, Doc Al. I wanted to simplify my answer, but I forgot that it's [tex]v^2[/tex].
 

1. What is the relationship between conservation of momentum and change in total kinetic energy in a problem?

The conservation of momentum principle states that in a closed system, the total momentum of all objects remains constant. This means that while individual objects may experience changes in momentum, the sum of their momenta will always be the same. In a problem involving conservation of momentum, the change in total kinetic energy is typically used to determine the final velocities of the objects involved.

2. How do you calculate the change in total kinetic energy in a conservation of momentum problem?

To calculate the change in total kinetic energy, you first need to determine the initial kinetic energy of all objects involved in the problem. This can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Next, you need to determine the final kinetic energy of all objects, using the same formula. Finally, the change in total kinetic energy is calculated by subtracting the initial kinetic energy from the final kinetic energy.

3. Can the total kinetic energy change in a conservation of momentum problem?

In most cases, the total kinetic energy will remain constant in a conservation of momentum problem. This is because the conservation of momentum principle states that the total momentum is conserved, and kinetic energy is directly related to momentum. However, there are some cases where external forces may act on the objects, causing a change in their kinetic energy and subsequently a change in the total kinetic energy.

4. What factors can affect the change in total kinetic energy in a conservation of momentum problem?

The change in total kinetic energy in a conservation of momentum problem can be affected by the masses and velocities of the objects involved. The larger the mass and/or velocity of an object, the greater its kinetic energy will be. Additionally, external forces or collisions can also impact the change in total kinetic energy.

5. How does the law of conservation of energy relate to the change in total kinetic energy in a conservation of momentum problem?

The law of conservation of energy states that energy cannot be created or destroyed, it can only be transferred or transformed from one form to another. This applies to the change in total kinetic energy in a conservation of momentum problem, as the total kinetic energy will remain constant unless external forces or collisions act on the objects, causing a transfer or transformation of energy. In other words, the total kinetic energy may change, but the total energy in the system will always remain the same.

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