What Forces Result from a Football Tackle Involving Inelastic Collision?

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SUMMARY

The discussion focuses on calculating the forces resulting from an inelastic collision between a 110 kg linebacker and a 70 kg halfback during a football tackle. The linebacker moves at 2 m/s while the halfback runs at 4 m/s. The calculations reveal that both players experience a force of -513.4 N, indicating that the direction of the force is opposite to the halfback's initial motion. This negative value is confirmed as valid since force is a vector quantity, and the direction must be considered.

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Homework Statement


A 110 kg football linebacker moving at 2 m/s tackles head-on a 70 kg halfback running 4 m/s. If they have an inelastic collision, and neglecting the effects due to digging in of cleats:
a)If the collision takes .5 sec, what is the force imparted on the halfback?
b) what is the force imparted on the linebacker?

Homework Equations


Ft=[tex]\Delta[/tex]p

The Attempt at a Solution



110(2) + 70(-4) = 110v + 70v
v= 1/3 m/s in halfback's direction

when i solve, I get a negative force. Is this possible?
For both a and b, i get -513.34 N, is this correct?

a)F=(70)(.333 - 4)/.5
F=-513.4 N

b)F=(110)(-.333 - 2)/.5
F=-513.4 N
 
Last edited:
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I'm not going to calculate the numbers, but using variables:

[tex]m_1 v_1 + m_2 v_2 = (m_1 + m_2)v[/tex]

Solve for [tex]v[/tex]:

[tex]\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}=v[/tex]

Using v, you can get change in momentum:

[tex]\Delta p_1=m_1(v_1 - v)[/tex]
[tex]\Delta p_2=m_2(v_2 - v)[/tex]

Then you can calulate the force in each case.

[tex]F_1=\frac{m_1(v_1 - v)}{t}[/tex]
[tex]F_2=\frac{m_2(v_2 - v)}{t}[/tex]

Yes, you can get a negative value for force. Force is a vector, as is velocity.

So initially, assume the halfback is running in the positive x direction, his velocity is positive. The linebacker will be running in the negative x direction with a negative velocity.
 

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