A racquet ball with mass m = 0.223 kg is moving toward the wall at v = 16.3 m/s and at an angle of θ = 35° 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 = 0.07 s.
Now the racquet ball is moving straight toward the wall at a velocity of vi = 16.3 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.6 m/s. The ball exerts the same average force on the ball as before.
1. What is the magnitude of the change in momentum of the racquet ball?
2. What is the time the ball is in contact with the wall?
3.What is the change in kinetic energy of the racquet ball?
The Attempt at a Solution
The first question I get 6.00kg-m/s by using m(vf-vi) which is 0.223*(-10.6-16.3). Is that correct?
Second question I use ft=Δp which gives me t=Δp/ave force; which i have the values are 6.00/85.1=0.0704s but it seems to me that the time should be lesser cause when travel without any angles it cuts down the time. Is the caluation correct?
The third question is ΔKE=1/2m(vf^2-vi^2) which i get 1/2*0.223*(10.6^2-16.3^2)
which i get -17.1J.
All 3 question just want to check am I doing the correct way,thanks for helping out.