A60Kg 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? Where are come the mass of velocity is 100
We can use the principle of conservation of momentum to solve this problem. The total momentum of the system (man, boat, and ball) before the throw is zero, since everything is stationary. After the man throws the ball, the momentum of the ball is 0.2 x 100 = 20 kg m/s (using the mass and velocity given in the question). To conserve momentum, the man and boat together must have an equal and opposite momentum of -20 kg m/s.
Let's call the velocity of the boat after the throw "v". We can set up an equation for conservation of momentum:
(mass of man + mass of boat) x 0 = mass of man x 0 + mass of boat x 0 + 0.2 x 100 + (mass of man + mass of boat + 0.2) x v
Simplifying and solving for v, we get:
v = -20 / (60 + 40.2) = -0.186 m/s
So the boat moves backwards at a speed of 0.186 m/s after the man throws the ball. Note that the negative sign indicates that the boat is moving opposite to the direction of the ball.