A tennis ball of mass 0.0610 kg is served. It strikes the ground with a velocity of 52.5 m/s (117 mi/h) at an angle of 21.6° below the horizontal. Just after the bounce it is moving at 50.5 m/s at an angle of 17.9° above the horizontal. If the interaction with the ground lasts 0.0693 s, what average force did the ground exert on the ball?
2007-03-19 05:15:28 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
dP/dt = M dv/dt = Ma = force of impact; dv/dt = 2.5 m/sec//.0693 sec = change in velocity over time (acceleration), and M = .061 kg = ball mass. (You can do the math.)
Lessons: The angle of incidence/rebound has no bearing on this problem because the velocities of interest are those of the path of the ball before and after hitting the court. Both of these were given.
Force is equal to the time derivative of the momentum (P). Thus, F = dP/dt = d(Mv)/dt = dM/dt*v + Mdv/dt; where if M = constant, as in the case of the ball, we have dM/dt = 0; so that F = Mdv/dt.
QUESTION: Can you think of a case where dM/dt <> 0? [Hint: think Cape Canaveral]
2007-03-19 07:44:22 · answer #1 · answered by oldprof 7 · 0⤊ 0⤋
calculate the momentum of ball just before and after the collosion.
the difference( note that the momentum mass*velocity is a vector so do the vector calculations) divided by time is the average force the ground exerts on ball:
force*time=momentum=mass*velocity
2007-03-19 12:33:18 · answer #2 · answered by corr man 1 · 0⤊ 0⤋
f average =(change in momentum)/time
time=0.0693s
change in momentum=mass( 52.5sin(21.6)+50.5sin(17.9) )
mass=0.061kg
2007-03-19 12:21:09 · answer #3 · answered by tarundeep300 3 · 0⤊ 0⤋