Asked by tom
Two basketball players collide head on. play 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. what is the change in momentum of player a?
b. if the collision lasted 0.1s what is the average force play b must have exerted on player a during the collision?
c. what is the average force that player a must have exerted on player b during the collision?
d. what is the change in momentum of player b?
e. what is the final velocity of player b?
a. what is the change in momentum of player a?
b. if the collision lasted 0.1s what is the average force play b must have exerted on player a during the collision?
c. what is the average force that player a must have exerted on player b during the collision?
d. what is the change in momentum of player b?
e. what is the final velocity of player b?
Answers
Answered by
Damon
original a momentum = 80*2.5
original b momentum = -68*1.2
total momentum before and after
= 80*.5 - 68*1.2
after
= 80*1 + 68 Vb
which is
final momentum a = 80
final momentum b = total - 80
F on a = (final momentum a - initial momentum a) /0.1
original b momentum = -68*1.2
total momentum before and after
= 80*.5 - 68*1.2
after
= 80*1 + 68 Vb
which is
final momentum a = 80
final momentum b = total - 80
F on a = (final momentum a - initial momentum a) /0.1
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