A 76.5-kg linebacker (\"X\") is running at 6.59 m/s directly toward the sideline

Question

A 76.5-kg linebacker ("X") is running at 6.59 m/s directly toward the sideline of a football field. He tackles a 86.5-kg running back ("O") moving at 9.05 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 pi 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, pi, of their motion, relative to the sideline?

Explanation / Answer

Mass of X is m = 76.5 kg

Mass of O is M = 86.5 kg

Initial velocity of X is u = (6.59 m/s) j

Initial velocity of O is U = (9.05 m/s ) i

Where i , j are the unit vectors along right and upward directions respectively.

Let the combined velocity of the both players after collision be v.

From law of conservationn of momentum ,

    mu + MU = (m+M) v

(76.5x 6.59 j ) +(86.5 x9.05 i ) = (76.5 + 86.5) v

504.135 j +782.825 i = 163 v

           v = 3.092 j + 4.802 i

              = 4.802 i + 3.092 j

Common speed of the players after collision v = (4.802 2 +3.092 2) 1/2

                                                                   = 5.711 m/s

phi = tan -1 (3.092 / 4.802 )

        = tan -1 ( 0.6438)

        = 32.77 o

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