Two basketball players collide head on. Player A weighs 80 kg and is travelling

Question

Two basketball players collide head on. Player A weighs 80 kg and is travelling 2.5m/s to the right while player B weighs 68 kg and is travelling 1.2 m/s to the left. after collision player A is travelling at 1.0 m/s to the right.

A.  If the collision lasted 0.1s what is the average force player B must have exerted on player A during the collision?

B. What is the average force that player A must have exerted on player B during the collision?

C. What is the change in momentum of player B?

D. What is the final velocity of player B?

PLEASE SHOW EQUATION USED, PLUG IN NUMERS, THEN SOLVE
THANK YOU

Explanation / Answer

change in momentum of player a = 80*(2.5+1) = 280 kgm/s

force on a = change in momentum / time = 280/0.1 = 2800 N

force on b must be equal and opposite ti forceon a,

Force on b = 2800 N

change in momentum of b = change in momentum of a =280 kgm/s

for final velocity of b,

change in velocit = 280/68= 4.1176 M/.s

so ,

final velocity of b = 4.1176-1.2 = 2.917 m/s towards right

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