A racquet ball with mass m = 0.252 kg is moving toward the wall at v = 18.3 m/s
ID: 1553722 • Letter: A
Question
A racquet ball with mass m = 0.252 kg is moving toward the wall at v = 18.3 m/s and at an angle of = 26° 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 = 0.06 s.
1) What is the magnitude of the initial momentum of the racquet ball?
kg-m/s
2) What is the magnitude of the change in momentum of the racquet ball?
kg-m/s
3) What is the magnitude of the average force the wall exerts on the racquet ball?
N
4)
Now the racquet ball is moving straight toward the wall at a velocity of vi = 18.3 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -11.3 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?
kg-m/s
5) What is the time the ball is in contact with the wall?
s
6) What is the change in kinetic energy of the racquet ball?
J
Explanation / Answer
1) Pi = m*vi
= 0.252*18.3
= 4.61 kg.m/s
2) change in momentum of the rocket ball = m*(vix - vfx)
= 0.252*(18.3 - (-18.3))*cos(26)
= 8.29 kg.m/s
3) use Impulse momentum theorem,
Impulse = change in momentum
F*dt = change in momentum
F = change in momentum/dt
= 8.29/0.06
= 138 N
4) magnitude of change in momentum = m*(vi - vf)
= 0.252*(18.3 - (-11.3))
= 7.46 kg.m/s
5) use, impulse = change in momentum
F*dt = change in momentum
dt = change in momentum/F
= 7.46/138
= 0.054 s
6) change in kinetic energy of the ball = (1/2)*m*(vf^2 - vi^2)
= (1/2)*0.252*((-11.3)^2 - 18.3^2)
= -26.1 J