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Inelastic Collision of a football tackle

  • Thread starter llauren84
  • Start date
1. Homework Statement
In an American football game, a 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. Account for the missing energy.

2. Homework Equations
m1v1=m1+m2*v2
v2= m1/(m1+m2)*v1

3. The Attempt at a Solution
(a) Because they are stuck together, it is a perfectly inelastic collision.

(b)Theta= 40degrees = tan-1(3/5)
m1=90.0kg
m2=95.0kg
v1=(5.00i+3.00j)m/s= [tex]\sqrt{5.00^2+3.00^2}[/tex]m/s= 5.83m/s

v2= m1/(m1+m2)*v1
=(90kg/185kg)*5.83 m/s =2.84 m/s
v2x= 2.84*sin40degrees=1.82i
v2y= 2.84*cos40degrees=2.16j
So, the vector for the velocity after impact is (1.82i+2.16j)m/s

(c)w=[tex]\Delta[/tex]K=(1/2)m*[tex]\Delta[/tex]v2
w=(1/2)(m1+m2)(v22-v12)
w=-2.405kJ
This is the heat generated from the collision.

I really worked out the problem, but I don't know if it's correct. Please check it for me. Thanks.
 

rl.bhat

Homework Helper
4,433
5
(b)Theta= 40degrees = tan-1(3/5)
m1=90.0kg
m2=95.0kg
v1=(5.00i+3.00j)m/s= LaTeX Code: \\sqrt{5.00^2+3.00^2} m/s= 5.83m/s

This is not correct.
Find the initial momentum of each player and find the resultant momentum of the tackled players.
 

LowlyPion

Homework Helper
3,079
4
In an American football game, a 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. Account for the missing energy.

2. Homework Equations
m1v1=m1+m2*v2
v2= m1/(m1+m2)*v1
Your equations don't accurately reflect the collision. You are dealing with 2 dimensions here. Perhaps if you separate out the components and deal with them independently it will become clearer?

For instance can treat them in x as though one was stationary. And again in y as though the other is stationary, yielding your 2 velocities, and then you can use Pythagoras to determine the resultant from the x,y as you were attempting.
 
So part a and c are correct?
 

LowlyPion

Homework Helper
3,079
4
So part a and c are correct?
A is ok. Perhaps you should correct b to determine whether c is?
 
Right.

So, I googled the Inelastic Collision to get the formulas.
 
Last edited:
Part B:

m1v1i+m2v2i=(m1+m2)vf
vf=[m1v1i+m2v2i/(m1+m2)
I did the x direction and then the y direction:
vfx=1.56m/s
vfy=1.86m/s
Used pythagorean theroum.
vf=2.43m/s
tan-1(vfy/vfx)=50.0degrees
So, it is 50 degrees from east.

For part C:

w=1/2 m1+m2 (vf2-vi2)
w=(1/2)(185kg)(2.432-5.832)m2/s2
w=-2.60kJ

Is this correct? Do I need to break the work into componants too? I am so confused.
 

LowlyPion

Homework Helper
3,079
4
Part B:

m1v1i+m2v2i=(m1+m2)vf
vf=[m1v1i+m2v2i/(m1+m2)
I did the x direction and then the y direction:
vfx=1.56m/s
vfy=1.86m/s
Used pythagorean theroum.
vf=2.43m/s
tan-1(vfy/vfx)=50.0degrees
So, it is 50 degrees from east.

For part C:

w=1/2 m1+m2 (vf2-vi2)
w=(1/2)(185kg)(2.432-5.832)m2/s2
w=-2.60kJ

Is this correct? Do I need to break the work into componants too? I am so confused.
You may be confused, but it looks OK.
 
Thanks so much.
 

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