A 0.25 kg softball has a velocity of 13 m/s at an angle of 46° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a)20 m/s, vertically downward, and (b)20 m/s, horizontally back toward the pitcher?

a. 2.959kgm/s

b. 7.63kgm/s

Explanation:

We were given Mass= 0.25kg

Velocity = 13m/s

Angle = 46°

a) Along the x axis we have =

0- 0.25(13)cos(46°) = (- 2.26)i kgm/s

Along the y axis we have =

-0.25(20) +0.25(13)sin(46°) = (-1.91)j kgm/s

Magnitude of change in momentum

=√ i²+ j²

= √ (-2.26)²+ ( -1.91)²

= √8.7557

= 2.959 kgm/s

b.

Along the x axis

-0.25(20) - 0.25(13)cos(46°) = (-7.26)i kgm/s

Along the y axis

0 + 0.25(13)sin(46°) = (-2.34)j kgm/s

Magnitude of change in momentum

= √ i²+ j²

= √( -7.26)² + (-2.34)²

= √58.1832

= 7.63kgm/s

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

the answer would be c