Kinetic Energy after a Collision

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SUMMARY

The discussion centers on calculating the kinetic energy of two football players involved in a completely inelastic collision. The players have masses of 110 kg and 125 kg, with velocities of 2.75 m/s and 2.60 m/s, respectively. The initial momentum was incorrectly calculated as -22.5 kg·m/s, leading to a kinetic energy result of 1.07 J, which contradicts the textbook's answer of 838 J. The correct approach involves using conservation of momentum to determine the final velocity after the collision, which was not initially considered.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with kinetic energy formulas (K = 0.5 * m * v^2)
  • Knowledge of inelastic collision concepts
  • Ability to perform vector calculations for momentum
NEXT STEPS
  • Study the conservation of momentum in inelastic collisions
  • Learn how to calculate final velocities after collisions
  • Explore kinetic energy loss in collisions
  • Review examples of elastic vs. inelastic collisions in physics
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Physics students, educators, and anyone interested in understanding collision dynamics and energy conservation principles in mechanics.

student34
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Homework Statement



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.

Homework Equations



P = m*v

Ptotal = Pi - Pf

K = 0.5*m*v^2

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.
 
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student34 said:
Ptotal = 110kg*2.75m/s - 125kg*2.60m/s = -22.5kgm/s

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?
 
student34 said:

Homework Statement



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.

Homework Equations



P = m*v

Ptotal = Pi - Pf

Where ever did you get that equation?
K = 0.5*m*v^2

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

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

Pinitial = Pfinal
 
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.
 
SammyS said:
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.

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
 
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.
 
student34 said:
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?"
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.
 
Doc Al said:
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.

Ohhh, I see. Ok I'll try it using an elastic collision.
 
  • #10
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
student34 said:
Ok I'll try it using an elastic collision.
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
Doc Al said:
Why?

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

Oh god - lol - thank-you.
 

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