7.1. Momentum http://www.ck12.org
FIGURE 7.6
(A) Ball caught by catcher. (B) Batter
bunting ball.
The change in momentum of the ball,∆p, is:pf−pi= 0 −(+ 6. 81 ) =− 6. 81 kgs·m
Case 2:The ball hits the bat with a horizontal speed of 46.95 m/s and it leaves the bat with a horizontal speed of
46.95 m/s.
What would you guess is the magnitude of the horizontal momentum change of the ball?
a. 0
b. 6. 81 kgs·m
c. 13. 6 kgs·m
It is not uncommon to choose answer A but that is incorrect. Had the problem been stated using velocity instead of
speed and direction, the correct answer may have been more obvious. That is:
The ball hits the bat with a horizontalvelocity of 46.95 m/sand it leaves the bat with a horizontal velocity of-46.95
m/s.
Remember that speed is only the magnitude of velocity, but momentum is a vector quantity. Therefore, we must
always consider the velocity of objects when working with momentum.
Let’s work out Case 2 withFigure7.7 in mind.
FIGURE 7.7
If we arbitrarily assume the ball’s velocity to the batter is positive, as inFigure7.7, then
pi=mvi= ( 0 .145 kg)( 46 .9 m/s) = 6. 81
kg·m
s
The final momentum of the ball is negative since it moves in the opposite direction.