A 83.1-kg linebacker (\"X\") is running at 7.91 m/s directly toward the sideline

Question

A 83.1-kg linebacker ("X") is running at 7.91 m/s directly toward the sideline of a football field. He tackles 89.1-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 phi relative to the sideline, as shown in the diagrams below. What is the common speed of the players, immediately after their impact? What is the angle, phi, of their motion, rotative to the sideline?

Explanation / Answer

taking to the right along i and upward along j vector,

m1 = 83.1 kg

v1i = 7.91m/s j

m2 = 89.1 kg

v2i = 9.27 m/s i

Applying momentum conservation for the collision,

m1 v1i + m2 v2i = (m1 + m2) v

(83.1 x 7.91 j) + (89.1 x 9.27 j) = (83.1 + 89.1) v

v = 3.82j + 4.80i

speed = sqrt(3.82^2 + 4.80^2) = 6.13 m/s ..........Ans

theta = tan^-1(3.82 / 4.80)

= 38.5 deg ..............Ans

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