On a very muddy football field, a 100 kg linebacker tackles an 80 kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.6 m/s north and the halfback is sliding with a velocity of 7.4 m/s east.
your question is incomplete but similar question is given in a text book so i think you want to ask following
What is the magnitude of the velocity at which the two players
move together immediately after the collision?
ans-
p1 = m1v1 = 100*8.6 = 860
p2 = m2v2 = 80*7.4 = 592
applying conservation of momentum and suppose ' i ' toward north direction and ' j ' toward east direction.
(860)i+(592)j= (100+80)V
V= ( 860/180)i+(592/180)j and resultant velocity is square root of
( 860/180)^2+(592/180)^2=5.79m/s and direction is arctan
(3.28/4.77)=34.51 degree from east.
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