Question

A 60kg man standing on a stationary 4okg boat throws a 0.2kg ball with a velocity  of 50 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

5/32 or 0.16 m/s

Step-by-step explanation:

m1u1 + m2u2 = m1v1 + m2v2

(64x0) + (0.2 x 0) = (64 x v1) + (0.2 x 50)

v = (0.2 x 50) / 64

v = 5/32 or 0.16 m/s

(ignore negative sign as it means that the boat is moving the opposite direction of ball, that's why the answer was coming negative)

I hope my answer helps you.