Asked by Sami
A 60kg man standing on a stationary 40kg boat throws a 0.2kg ball with a velocity of m/s. Assuming there is no friction between the man and the boat, what is the speed of the boat after the man throws the ball?
Answers
Answered by
Chanz
Momentum before = 0, so it must be zero after:
(60+40)v + (.2)....oops, you didn't put the speed of the ball after. No solution for you.
(60+40)v + (.2)....oops, you didn't put the speed of the ball after. No solution for you.
Answered by
bon
it will be 0
Answered by
Zack
So the Pa=-Pb
At first the intial was zero so
0=(60+40)(v)+(0.2)(1)
100v=0.2
100v/100=0.2/100
☝️ Cancel each other
V=0.2/100
At least this is my thoughts if you any other ideas share me
[email protected]
At first the intial was zero so
0=(60+40)(v)+(0.2)(1)
100v=0.2
100v/100=0.2/100
☝️ Cancel each other
V=0.2/100
At least this is my thoughts if you any other ideas share me
[email protected]
Answered by
Elizabeth
The speed is 50m/s
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