A 60 kg man standing on a stationary 40 kg 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? The velocity 50m/s is where are you come from?
We can use the law of conservation of momentum to solve this problem. The total momentum before the ball is thrown is zero (since everything is stationary), and the total momentum after the ball is thrown must also be zero (since there are no external forces acting on the system).
The momentum of the man and boat before the ball is thrown is:
p1 = (60 kg + 40 kg) * 0 m/s = 0 kg m/s
The momentum of the ball is:
p2 = 0.2 kg * 50 m/s = 10 kg m/s
Therefore, the momentum of the man, boat, and ball after the ball is thrown must also be zero:
p3 = (60 kg + 40 kg + 0.2 kg) * v = 0 kg m/s
Solving for v, we get:
v = 0 m/s
This means that the boat does not move at all after the man throws the ball. The velocity of 50m/s mentioned in the question is not relevant to this problem.