A 90kg man is standing still on frictionless ice. His friend tosses him a 10kg ball, which has a horizontal velocity of 20m/s. After catching the ball, what is the man's velocity?

The kinetic energy of the ball is 1/2 x m x v^2 = (0.5) (10kg) (20m/s) ^2 = 2000 J. After the man catches the ball, the kinetic energy of the man and the ball is 2000 J, assuming all the ball's kinetic energy has been transferred to the man holding the ball. Hence, 2000 J = 1/2 x (10kg + 90kg) x v^2. Rearranging, v = sqrt (2000J / (0.5) (100 kg)) = 6.32 m/s. So the man's velocity is 6.32 m/s after catching the ball.