A racquet ball with mass m = 0.222 kg is moving toward the wall at v = 17.5 m/s and at an angle of θ = 32° wit?
A racquet ball with mass m = 0.222 kg is moving toward the wall at v = 17.5 m/s and at an angle of θ = 32° 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 = 0.072 s.
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 vi = 17.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -11.2 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
please help as soon as possible
Answers & Comments
Verified answer
horizontal velocity Vh = 17.5m/s * cos32º = 14.8 m/s
"perfectly elastic" means that energy is conserved, so incoming velocity = outgoing velocity
F = mΔv/Δt = 0.222kg * (-14.8m/s - 14.8m/s) / 0.072s = -91.5 N ← avg force
Δp = mΔv = 0.222kg * (-11.2m/s - 17.5m/s) = -6.37 kg·m/s ← change in momentum
Δt = mΔv/F = 0.222kg * (-11.2m/s - 17.5m/s) / -91.5N = 0.07 s ← contact time
ΔEk = ½m( (Vf)² - (Vi)²) = ½ * 0.222kg * ( 11.2² - 17.5²)m²/s² = -20 J ← ΔEk