A 79.5-kg linebacker (\"X\") is running at 6.59 m/s directly toward the sideline

Question

A 79.5-kg linebacker ("X") is running at 6.59 m/s directly toward the sideline of a football feild. He tackles a 87.8-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 relative to the sideline.

What is the common speed of the players, immediatly after their impact? in m/s

What is the angle of their motion, relative to the sideline? in degrees

Explanation / Answer

Conservation of Momentum in X-direction;

79.5*6.59 = (79.5+87.8)*Vx

Vx = 3.13 m/sec

conservation of momentum along Y-Direction

87.8*9.05 = (79.5+87.8)*Vy

Vy = 4.75 m/sec

so,there common velocity;

=sqrt(3.13^2 + 4.75^2)

=5.688 m/s

Angle relative to the sideline;

=arctan(Vx/Vy)

=33.382 degree

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