Map A 82.5-kg linebacker (\"X\") is running at 7.31 m/s directly toward the side

Question

Map A 82.5-kg linebacker ("X") is running at 7.31 m/s directly toward the sideline of a football field. He tackles 95.6-kg running back ("o") moving at 9.27 m/s straight toward the goal line, perpendicular to the original direction of the linebacker. As a result of the collision both players momentarily leave the ground and go out- of-bounds at an angle o relative to the sideline, as shown in the diagrams below. BEFORE IMPACT AFTER IMPACT What is the common speed of the players, immediately after their impact? Number 8.36 m/s What is the angle, , of their motion, relative to the sideline? Number 55.76

Explanation / Answer

Solution) mass m1=82.5 kg

Mass m2=95.6kg

Initial speed u1=7.31m/s

Initial speed u2=9.27m/s

We have m1u1+m2u2=(m1+m2)V

V=m1u1+m2u2/(m1+m2)

V=82.5×7.31+95.6×9.27/(82.5+95.6)

V=8.36m/s

To find angle (phi) we use vector notation

m1u1(i) +m2u2(j)=(m1+m2)V

From here tan(phi)=coefficient of (j)/coefficient of (i)

tan(phi)=m2u2/m1u1

(phi)=tan inverse(m2u2/m1u1)

(phi)=tan inverse(95.6×9.27/82.5×7.31)

(phi)=55.76 degrees

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