# Kinetic Energy after a Collision

1. May 3, 2013

### student34

1. The problem statement, all variables and given/known data

A 110 kg football player runs at 2.75 m/s directly into a 125 kg player running 2.60 m/s towards him. What is the kinetic energy of the two players if this is a completely inelastic collision.

2. Relevant equations

P = m*v

Ptotal = Pi - Pf

K = 0.5*m*v^2

3. The attempt at a solution

Ptotal = 110kg*2.75m/s - 125kg*2.60m/s = -22.5kgm/s

k = 0.5*(125kg +110kg)*((-22.5kgm/s)/235kg)^2 = 1.07 J

But my textbook's answer is 838 J. I just cannot see where I am wrong.

2. May 3, 2013

### Staff: Mentor

Not sure what you're doing here.

What you need to do is use conservation of momentum to figure out the final speed after they collide. How would you do that?

3. May 3, 2013

### SammyS

Staff Emeritus
Where ever did you get that equation?

4. May 3, 2013

### student34

Oops, but I think that I still have the correct final momentum.

P = PA + PB = mA*VA + mB*VB

Pinitial = Pfinal

5. May 3, 2013

### SammyS

Staff Emeritus
Are you sure you have written the problem exactly as it was given to you ?

It is more common to ask for the amount of Kinetic Energy lost in the collision, --- or the percent loss.

6. May 3, 2013

### student34

I always feel weird putting textbook information word for word, so I'll do it with the reference.

Here it is exactly: "One 110kg football lineman is running to the right at 2.75m/s while another 125kg lineman is running directly toward him at 2.60m/s. What are (a) the magnitude and direction of the net momentum of those two athletes, and (b) their total kinetic energy?"

From Sears and Zemansky's University Physics, Young and Freedman 13th Edition from Pearson

7. May 3, 2013

### student34

Well I added up their kinetic energy before their collision, and it equals the book's answer. So I think they missed something in the question or the wrong answer.

8. May 3, 2013

### Staff: Mentor

Note that there is no mention of a collision, inelastic or otherwise. That's why it often helps to post the question word for word.

9. May 3, 2013

### student34

Ohhh, I see. Ok I'll try it using an elastic collision.

10. May 3, 2013

### student34

It still doesn't make any sense to me because I don't see how I can get their velocities after the collision to find their kinetic energies.

11. May 3, 2013

### Staff: Mentor

Why?

All they ask for is the total momentum and kinetic energy. Unless there are more parts to it, there's no collision involved.

12. May 3, 2013

### student34

Oh god - lol - thank-you.