A 0.25 kg softball has a velocity of 13 m/s at an angle of 46° below the horizontal

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

B.)

Explanation: The object's motion (Kinetic energy) let's the molecules bounce.