A 0.35-kilogram volleyball is hit with a horizontal velocity of 3.0 m/s. One of

Question

A 0.35-kilogram volleyball is hit with a horizontal velocity of 3.0 m/s. One of the players on the front line jumps up and hits it back with a horizontal velocity of 7.0 m/s. If the contact time is 0.05 s, what was the average force applied by the player? If a 0.6-kilogram ball is dropped from height of 9 m, what is the momentum of the ball just before it hits the ground? Two bodies of masses 8 kg and 5 kg move along the x-axis in opposite directions with velocities of 23 m/s and -6 m/s, respectively. They collide and stick together. Find the velocity after collision.

Explanation / Answer

1) Average force = rate of change in momentum

                             =   0.35 * (3 + 7)/0.05

                             = 70 N

2) momentum of ball just before it hits ground = 0.6 * sqrt(2 * 9.8 * 9)

                                                                           = 7.969 kg-m/sec

3) velocity after collision = (8 * 23 - 5 * 6)/13

                                        = 11.846 m/sec

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