A racquet ball with mass m = 0.248 kg is moving toward the wall at v = 16.5 m/s
ID: 1553621 • Letter: A
Question
A racquet ball with mass m = 0.248 kg is moving toward the wall at v = 16.5 m/s and at an angle of theta = 31 degree with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.73 s. "What is the magnitude of the initial momentum of the racquet ball? What is the magnitude of the change in momentum of the racquet ball? What is the magnitude of the average force the wall exerts on the racquet ball? Now the racquet ball is moving straight toward the wall at a velocity of nu_1 = 16.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at nu_f = -10.9 m/s. The ball exerts the same average force on the ball as before. What is the magnitude of the change in momentum of the racquet ball? What is the time the ball is in contact with the wall? What is the change in kinetic energy of the racquet ball?Explanation / Answer
horizontal velocity Vh = 16.5m/s * cos31º = 14.14 m/s
Energy is conserved because collision is elastic
Incoming velocity = outgoing velocity
F = mv/t = 0.248kg * (-14.14m/s - 14.14m/s) / 0.073s = -96.09N
Change in momentum p
p = mv = 0.248kg * (-10.9m/s - 16.5m/s) = -6.79 kg·m/s
contact time t
t = mv/F = 0.248kg * (-10.9m/s - 16.5m/s) / -96.09N = 0.07 s
Change in kinetic energy
Ek = ½m( (Vf)² - (Vi)²) = ½ * 0.248kg * ( 10.9² - 16.5²)m²/s² = -19.08 J