two basketball players collide head on. play a weighs 80 kg and is travelling 2.
ID: 2115456 • Letter: T
Question
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?
PLEASE SHOW WORK
THANK YOU
Explanation / Answer
a) Change in momentum = mv2 - mv1 = 80*(1.0 - 2.5) = -120 kgm/s
-ve sign because change in mometum is in left direction.
b) Average force = Change in Momentum / Time = -120/0.1 = -1200N
c) 1200 N
d) Change in momentum = 1200 * 0.1 = 120 kgm/s
e) mv2 - mv1 = 1200 => 68(v2 - (-1.2)) = 1200
=> v2 + 1.2 = 17.647
=> v2 = 16.447 m/s
final velocity is 16.447 m/s to the right.
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