Question

2) A man (m - 75.4 kg) is initially standing at rest. A large, heave ball (m 18.3 kg) is thrown at him with an initial speed of 4.3 m/s. The ball then bounces off the man goes in the opposite direction at 2.1 m/s. How fast, and in what direction will the man be moving after the collision? (Assume the man and the ball make up an isolated system.) Draw before and after pictures: Write a conservation of momentum equation: Do your algebra Write an answer as a complete sentence: What is the total kinetic energy of the system before the collision? What is the total kinetic energy of the system after the collision? Was this collision: super elastic, elastic, inelastic or totally inelastic?

Answer 1

The problem can be solved from conservation of momentum

mass of man = 75.4 kg

initial velocity of man = 0 m/s

mass of ball = 18.3 kg

initial speed of ball = 4.3 m/s

final speed of ball = 2.1 m/s (in opposite direction)

let velocity of man after collision be v

75.4*0 + 18.3*4.3 = 18.3*-2.1 + 75.4*v

v=1.55 m/s

kinetic energy of system before collision = 0.5*75.4*0²+0.5*18.3*4.3² = 169.2 J

kinetic energy of system after collision = 0.5*75.4*1.55²+0.5*18.3*2.1² = 130.93 J

Since the kinetic energy of system was lost after collision, the collision is inelastic