A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s
ID: 1283033 • Letter: A
Question
A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s and at an angle of ? = 29 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.
4)
Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.5 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?
Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.5 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? A racquet ball with mass m = 0.255 kg is moving toward the wall at v = 11.5 m/s and at an angle of ? = 29½ 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. 4)Explanation / Answer
Change in momentum = m * (chhange in velocity) = 0.255*(11.5+7.5) = 4.845kg*m/s
The force exerted by wall on ball is same = change in momentum/time of contact in both cases
so in first case change in momo = 5.129kg-m/s over 0.064s
In second case change in mom = 4.845
thus time of contact = 0.0604 s
change in KE = 0.5m*change in square of velocity. = 9.69J