Kinetic Energy after a Collision

In summary, the question asks for the total momentum and kinetic energy of two football players running towards each other without mentioning any collision. Therefore, there is no need to calculate for a collision. The total momentum is the sum of their individual momentums and the total kinetic energy can be calculated using the mass and velocities of both players.
  • #1
student34
639
21

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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  • #2
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?
 
  • #3
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.
 
  • #4
Oops, but I think that I still have the correct final momentum.

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

Pinitial = Pfinal
 
  • #5
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
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
 
  • #7
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
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.
 
  • #9
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.
 

1. What is kinetic energy after a collision?

Kinetic energy after a collision refers to the energy possessed by an object in motion after it has collided with another object.

2. How is kinetic energy after a collision calculated?

Kinetic energy after a collision 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.

3. Is kinetic energy conserved after a collision?

According to the law of conservation of energy, total energy is always conserved in a closed system. Therefore, in an isolated collision, the total kinetic energy before and after the collision will be the same.

4. What factors affect the kinetic energy after a collision?

The kinetic energy after a collision is affected by the mass and velocity of the objects involved in the collision. The angle and type of collision (elastic or inelastic) can also impact the kinetic energy.

5. How does kinetic energy after a collision relate to real-life scenarios?

In real-life scenarios, kinetic energy after a collision can be seen in car accidents, sports collisions, and any other instance where objects collide. Understanding the concept can help engineers and scientists design safer and more efficient systems and equipment.

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