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?

Answer 1

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.

Answer 2

it will be 0

Answer 3

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
zackded

Answer 4

The speed is 50m/s