What Forces Result from a Football Tackle Involving Inelastic Collision?

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In an inelastic collision between a 110 kg linebacker and a 70 kg halfback, the final velocity of both players moving together is calculated to be 1/3 m/s in the halfback's direction. The force imparted on the halfback is calculated to be -513.4 N, while the linebacker experiences the same magnitude of force in the opposite direction. Negative force values are acceptable as force is a vector quantity, indicating direction. The calculations involve changes in momentum over time, using the equation Ft = Δp. Understanding the directionality of forces and velocities is crucial in analyzing the collision dynamics.
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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=\Deltap

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
 
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I'm not going to calculate the numbers, but using variables:

m_1 v_1 + m_2 v_2 = (m_1 + m_2)v

Solve for v:

\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}=v

Using v, you can get change in momentum:

\Delta p_1=m_1(v_1 - v)
\Delta p_2=m_2(v_2 - v)

Then you can calulate the force in each case.

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

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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