A 74.1-kg linebacker (\"X\") is running at 7.19 m/s directly toward the sideline

Question

A 74.1-kg linebacker ("X") is running at 7.19 m/s directly toward the sideline of a football field. He tackles a 85.2 kg running back ("O") moving at 9.49 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, relative to the sideline?

Explanation / Answer

Let Vf be the common speed

momentum along X is

Px = 85.2 x 9.49 = 808.55 kg-m/s

along Y

Py = 74.1 x 7.19 = 532.78 kg-m/s

P = sqrt [808.55^2 + 532.78^2] = 968.3 kg-m/s

From conservation of momentum

Pi = Pf

Pi = (m1 + m2) vf

968.3 = (74.1 + 85.2) vf

vf = 968.3/159.3 = 6.08 m/s

Hence, vf = 6.08 m/s

the angle will be

tan(phi) = vf/v2

phi = tan-1 (6.08/9.49) = 32.45 deg

Hence, phi = 32.45 deg

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