This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapt. 9 Momentum 93 kinetic energy should obey 1 2 m 1 v 2 1 F arms m 1 , where is v 1 the final speed of the arms in the swing just before the collision with the ball. One can see that v 2 1 , and therefore v 1 , is independent of mass given our assumptions. Thus all batters swing with about the same speed given our assumptions. We take v 1 to be a constant. Note the model may not be adequate, but we can see where it takes us. a) Given the horizontal range formula for projectile motion x = v 2 g sin(2 ) , how does one change the launch velocity v to increase travel distance? What does your answer imply for hit baseballs? b) Assume the batterball collision is a 1dimensional elastic collisionwhich is probably not so bad. The batter is object 1 and the ball is object 2. The final direction of the ball is positive direction which means the balls initial velocity is negative. The final velocity of object 2 is given by the elastic collision formula v 2 = ( m 2 m 1...
View
Full
Document
 Fall '06
 Buchler
 Physics, Energy, Kinetic Energy, Mass, Momentum

Click to edit the document details