Respuesta :
Answer:
a) 5 kg m/s
b) 10 kg m/s
Explanation:
GIVEN DATA:
mass of softball 0.30 kg
\theta = 35 degree
velocity = 15 m/s
Along the x axis we have
[tex]P_{xf} - P_{xi} = 0 - 0.3(15)cos 35 = - 3.68\hat i kg m/s[/tex]
along the y axis we have
[tex]P_{yf} - P_{yi} = -(0.3)20 - 0.3(15)sin 35 = - 3.41\hat j kg m/s[/tex]
The magnitude of momentum is
[tex]\Delta P = \sqrt{ (-3.41)^2 + (-3.68)^2} = 5 kg m/s[/tex]
b) Along the x axis we have
[tex]P_{xf} - P_{xi} = -0.3(20) - 0.3(15)cos 35 = - 9.68\hat i kg m/s[/tex]
along the y axis we have
[tex]P_{yf} - P_{yi} = -0 - 0.3(15)sin 35 = -25.58\hat j kg m/s[/tex]
The magnitude of momentum is
[tex]\Delta P = \sqrt{ (-9.68)^2 + (-2.58)^2} = 10 kg m/s[/tex]
The magnitude of the change in momentum of the ball will be 5kgm/s.
How to calculate the momentum?
The magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of 20 m/s, vertically downward will be calculated thus:
x axis = -0.3(15) cos 35 = -3.68
y axis = -0.3(15) sin 35 = -3.41
Magnitude of momentum will be:
= ✓(-3.41)² + ✓(-3.68)²
= 5kgm/s.
The magnitude when the velocity is 20 m/s horizontally back toward the pitcher will be:
= ✓(-9.68)² + ✓(-2.58)²
= 10kg m/s.
Learn more about momentum on:
https://brainly.com/question/1042017
