posted by Anonymous .
A ping pong ball (m= 6g) bounces off a table. Before striking the table the ball has a velocity of 4m/s in a direction pheta below the x-axis. It leaves the table with the same speed moving in a direction pheta above the x axis. What impulse does the table exert on the tennis ball. If the ball is in contact with the table for .o1 seconds what average force does the table exert on the ball?
I know I=p1-p2, but i don't understand how to deal with finding the angles
The impulse is dealing with the changed momentum, it only changes in the vertical. So, figure the vertical components>
v=4sinTheta in each case. One velocity is downward, one is upward. So subtracting them changes the sign, so they add.
Impulse = deltamomentum