Inelastic Collision of a football tackle

[llauren84]

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.

Homework Equations

m₁v₁=m₁+m₂*v₂
v₂= m₁/(m₁+m₂)*v₁

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= $$\sqrt{5.00^2+3.00^2}$$m/s= 5.83m/s

v₂= m₁/(m₁+m₂)*v₁
=(90kg/185kg)*5.83 m/s =2.84 m/s
v_2x= 2.84*sin40degrees=1.82i
v_2y= 2.84*cos40degrees=2.16j
So, the vector for the velocity after impact is (1.82i+2.16j)m/s

(c)w=$$\Delta$$K=(1/2)m*$$\Delta$$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]

(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]

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.

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.

 

[llauren84]

So part a and c are correct?

 

[LowlyPion]

So part a and c are correct?

A is ok. Perhaps you should correct b to determine whether c is?

 

[llauren84]

Right.

So, I googled the Inelastic Collision to get the formulas.

 

[llauren84]

Part B:

m₁v_1i+m₂v_2i=(m₁+m₂)v_f
vf=[m₁v_1i+m₂v_2i/(m₁+m₂)
I did the x direction and then the y direction:
v_fx=1.56m/s
v_fy=1.86m/s
Used pythagorean theroum.
v_f=2.43m/s
tan^-1(v_fy/v_fx)=50.0degrees
So, it is 50 degrees from east.

For part C:

w=1/2 m₁+m₂ (v_f²-v_i²)
w=(1/2)(185kg)(2.43²-5.83²)m²/s²
w=-2.60kJ

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

 

[LowlyPion]

Part B:

m₁v_1i+m₂v_2i=(m₁+m₂)v_f
vf=[m₁v_1i+m₂v_2i/(m₁+m₂)
I did the x direction and then the y direction:
v_fx=1.56m/s
v_fy=1.86m/s
Used pythagorean theroum.
v_f=2.43m/s
tan^-1(v_fy/v_fx)=50.0degrees
So, it is 50 degrees from east.

For part C:

w=1/2 m₁+m₂ (v_f²-v_i²)
w=(1/2)(185kg)(2.43²-5.83²)m²/s²
w=-2.60kJ

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

You may be confused, but it looks OK.

 

[llauren84]

Thanks so much.