Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s. If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

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