Question

A 82.3 kg linebacker "X" is running at 7.79 m/s directly toward the sideline of a football field. He tackles a 89.1 kg running back "O" moving at 8.83 m/s straight toward the goal line, perpendicular to the orginal direction of the linebacker. As a result of the collision both players momentarily leave the ground and go out-of-bounds at an angle relative to the sideline.

(a) What is the common speed of the players , immediately after their impact? (I know what to do for this one it is part B I am stuck on).

(m₁V_i1)(m₂V_i2)/(m₁+m₂)= Final Velocity

(82.3 kg)(7.79 m/s)+(89.1 kg)(8.83 m/s)/ (82.3 kg)+(89.1kg)=V_f

V_f= 8.83 m/s

(b) What is thier angle, relative to the sideline?

For this do I just take the inverse tangent of part a to get my angle? please help.

Answer 1

a)

Apply law of conservation of momentum along \(\mathrm{x}\) -direction,

$$ \begin{array}{c} m_{1} v_{1}=\left(m_{1}+m_{2}\right) v_{x} \\ (82.3 \mathrm{~kg})(7.79 \mathrm{~m} / \mathrm{s})=(82.3 \mathrm{~kg}+89.1 \mathrm{~kg}) v_{x} \\ v_{x}=3.74 \mathrm{~m} / \mathrm{s} \end{array} $$

And apply law of conservation of momentum along y-direction,

$$ m_{2} v_{2}=\left(m_{1}+m_{2}\right) v_{y} $$

$$ \begin{array}{l} (89.1 \mathrm{~kg})(8.83 \mathrm{~m} / \mathrm{s})=(82.3 \mathrm{~kg}+89.1 \mathrm{~kg}) v_{y} \\ v_{y}=4.59 \mathrm{~m} / \mathrm{s} \end{array} $$

The common speed of the player is

$$ v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(3.74 \mathrm{~m} / \mathrm{s})^{2}+(4.59 \mathrm{~m} / \mathrm{s})^{2}}=5.92 \mathrm{~m} / \mathrm{s} $$

b)

The angle made by the common ve locity with y-axis is

$$ \theta=\tan ^{-1}\left(\frac{v_{x}}{v_{y}}\right)=\tan ^{-1}\left(\frac{3.74 \mathrm{~m} / \mathrm{s}}{4.59 \mathrm{~m} / \mathrm{s}}\right)=39.2^{\circ} $$