A 93 kg man holding a 0.653 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.

The final velocity of the man would be 0.157 m/s in the opposite direction of the motion.

Explanation:

Step 1: Data given

Mass of the man (M) = 93 kg

Mass of the ball (m) = 0.653 kg

Speed of the ball (v) = 11.3 m/s

Step 2: The conservation of the momentum

MV + mv = 0

93*V + 0.653*11.3 = 0

93V = - 7.3789

V = - 0.079 m/s

The sign shows that the direction of the man is opposite to the direction of the ball

After rebounded the direction of the ball would change

MV + m * (-v) = (m+M) * vf

- (93*0.079) - (0.653*11.3) = (93+0.653) * vf

-7.347 - 7.3789 = 93.653vf

-14.7259 = 93.653vf

vf = - 0.157 m/s

The final velocity of the man would be 0.157 m/s in the opposite direction of the motion.