A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.3 m/s
ID: 1511093 • Letter: A
Question
A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.3 m/s and at an angle of theta = 28" 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 = 0.064 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 v_i = 11.3 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at v_f = -7.3 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
1) Pi = m*vi
= 0.255*11.3
= 2.8815 kg.m/s
2) change in momentum of the ball = m*delta_vx
= 2*m*vi*cos(theta)
= 2*0.255*11.3*cos(28)
= 5.09 kg.m/s
3) Apply, Favg*delta_t = m*delta_vx
Favg = m*delta_vx/dt
= 5.09/0.064
= 79.5 N
4) |delta_P| = |m*(Vf - Vi)|
= |0.255*(-7.3 - 11.3)|
= 4.743 kg.m/s
5) Favg*dt = delta_P
dt = delta_P/Favg
= 4.743/79.5
= 0.06 s
6) delta_KE = 0.5*m*(vf^2 - vi^2)
= 0.5*0.255*(7.3^2 - 11.3^2)
= -9.486 J