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

Question:

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

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:

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

I should sum up the velocities, right?

$$(1/2)(m_1+m_2)((v_2 - v_1)^2 - ((m_2*v_2)/(m_1) - (v_1))^2)$$

But, that is not correct.
 Recognitions: Homework Help 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.
 Recognitions: Homework Help Science Advisor 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.

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

I inputed $$(1/2)(m_1)((m_2*v_2)/(m_1) - (v_1))^2 - (1/2)(m_1)(v_1)^2 - (1/2)(m_2)(v_2)^2$$ and it was incorrect.

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

Physics Monkey, you told me to add. Why is this so?
 Mentor Blog Entries: 1 $\Delta K = K_f - K_i$, not $K_i - K_f$.

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.../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.
 Recognitions: Homework Help 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, $$KE_1 = \frac{1}{2}m_1v_1^2$$
 It looks like I definitely did $$K_i - K_f$$, which obviously is not correct. So, I decided to analyze $$K_f$$ first. I got $$K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2$$. Then, I decided to analzye $$K_i$$. I got $$K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)$$ This simplifies to $$(1/2)(m_2*v_2 - m_1*v_1)$$. I took the difference $$K_f - K_i = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2 - (1/2)(m_2*v_2 - m_1*v_1)$$ and it was not correct. I really think I did everything right this time. ?

Mentor
Blog Entries: 1
 Quote by erik-the-red It looks like I definitely did $$K_i - K_f$$, 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 $$K_f$$ first. I got $$K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2$$.
OK.

 Then, I decided to analzye $$K_i$$. I got $$K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)$$ This simplifies to $$(1/2)(m_2*v_2 - m_1*v_1)$$.
You forgot to square the speed.
 Thanks so much, Doc Al. I wanted to simplify my answer, but I forgot that it's $$v^2$$.