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A 220 lb running back is running along the sideline of the football field toward the goal line. The running back is 1.0 m from the sideline running at 8.0 m/s toward the goal. A 180 lb defender runs across the field toward the running back 2.0 m from the goal line. Just before the collision, both players leap into the air. The running back's horizontal speed remains 8.0 m/s. What is the minimum possible horizontal velocity for the defender in order to ensure the players cross out of bounds before crossing the goal line?

System: The running back and the defender, treated as point particles. External influences are negligible compared to the internal forces of the collision, and so are neglected.

Model: [Momentum and Impulse].

Approach: We are interested only in the horizontal components of the velocity. If we treat the players as point masses, then the defender's goal is for the horizontal velocity after the collision to be directed at a minimum of 30 degrees from the running back's original direction of motion, as shown in the figure.

It is important to understand that the 1m and 2m legs of the triangle drawn in black are not the components of the final velocity. They are distances, not velocities. The vector triangle formed by the final velocity (shown in purple) and its x- and y- components must be similar (in a technical math sense) to the distance triangle. Thus, the angle in the distance triangle (30° in this case) must be the same as the angle of the velocity vector.

PICTURE

Since we are assuming external forces are negligible during the collision, we will have constant horizontal momentum:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ p_

Unknown macro: {x,i}

^

Unknown macro: {rm system}

=p_

^

Unknown macro: {RB}

+p_

Unknown macro: {x,i}

^

Unknown macro: {DF}

= p_

Unknown macro: {x,f}

^

Unknown macro: {rm system}

= p_

^

+p_

Unknown macro: {x,f}

^

Unknown macro: {DF}

] [p_

Unknown macro: {x,i}

^

Unknown macro: {rm system}

=p_

^

Unknown macro: {RB}

+p_

Unknown macro: {x,i}

^

= p_

^

Unknown macro: {rm system}

= p_

Unknown macro: {x,f}

^

Unknown macro: {RB}

+p_

^

Unknown macro: {DF}

] \end

where RB denotes the running back and DF the defender. For the case at hand, these equations can be simplified to read:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ m^

Unknown macro: {RB}

v_

Unknown macro: {x,i}

^

= (m^

Unknown macro: {RB}

+m^

Unknown macro: {DF}

)v_

Unknown macro: {x,f}

]
[ m^

v_

Unknown macro: {y,i}

^

Unknown macro: {DF}

= (m^

+m^

Unknown macro: {DF}

) v_

Unknown macro: {y,f}

] \end

We can immediately solve the x direction equation:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ v_

Unknown macro: {x,f}

= \frac{m^

Unknown macro: {RB}

v_

Unknown macro: {x,i}

{RB}}{m

+m^{DF}} ]\end

but the y equation contains two unknowns. We can resolve this problem by using the constraints (discussed above) that in order to end up out of bounds:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ v_

Unknown macro: {y,f}

= v_

Unknown macro: {x,f}

\tan\theta = \frac{m^

Unknown macro: {RB}

v_

Unknown macro: {x,i}

{RB}}{m

+m^{DF}} \tan\theta]\end

With this information, we can solve to find:

Unknown macro: {latex}

\begin

Unknown macro: {large}

[ v_

Unknown macro: {y,i}

^

Unknown macro: {DF}

= \frac{m^

Unknown macro: {RB}

+m^{DF}}{m^{DF}}v_

Unknown macro: {y,f}

= \frac{m^

v_

Unknown macro: {x,i}

{RB}}{m{DF}} \tan\theta = ....] \end

so that the horizontal component of the defender's velocity must be at least ....

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