A 77.9-kg linebacker (\"X\") is running at 7.43 m/s directly toward the sideline

Question

A 77.9-kg linebacker ("X") is running at 7.43 m/s directly toward the sideline of a football field. He tackles a 86.5-kg running back ("O") moving at 9.71 m/s straight the goa 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, as shown in the diagrams below AFTER IMPACT BEFORE IMPACT What is the common speed of the players, immediately after their impact? Number What is the angle, p, of their motion, relative to the sideline? Number

Explanation / Answer

By conservation of momentum

For the linebacker, M = 77.9*7.43 = 578.797

For the running back, M = 86.5*9.71 = 839.915

Since they have a common speed after the collision, they must be attached to each other after collision

So,

total mass = 77.9+86.5 = 164.4kg

by using there mass we find the final velocity

164.4*v1 = 578.797

v1= 578.797/164.4 = 3.52m/s

164.4*v2= 839.915

v2 = 839.915/164.4 = 5.10m/s

magnitude of their final velocity

vf = (v1^2+v2^2) = (3.52^2+5.10^2) = 6.2m/s

tanc = v1/v2 = 3.52/5.10 = 0.69

where, c is angle of their motion

c = 34.6deg

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