Two basketball players collide head on. Player A weighs 80 kg and is travelling
ID: 2145093 • Letter: T
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
PS. Second time asking question. Some dude just copied and pasted the answers someone else put down to a different version of this question. Don't be that guy.
Explanation / Answer
momentum conservation
m1u1+m2u2 = m1v1+m2v2 , u1(initial vel of A) = 2.5, u2((initial vel of B) = -1.2, v1(final vel of A) = 1 , v2(final vel of A) put these in eqn
80*2.5 -68*1.2 = 80*1 +68 *v2
200 - 81.6 - 80 = 68 v2
solving we get , v2 = 0.5647 m/sec to the right
so D)final velocity of B is v2 = 0.5647 m/sec to the right
A) average force exerted on player A = change in momentum of A / time = (m1v1 -m1u1)/0.1 = -120/0.1 = -1200 N , so 1200 N towrds left
B)average force exerted on player B = change in momentum of B / time = (m2v2 - m2u2)/t = 120/.1 = 1200 N , so 1200 N towards right
C) change in momentum for B is 120 Kgm/sec