Question

#21 A 0.30 kg softball has a velocity of 15 m/s at an angel of 35o below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while in contact with the bat if the ball leaves with a velocity of (a) 20 m/s, vertically downward, and (b) 20 m/s, horizontally back toward the pitcher?

Answer 1

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

$P_{xf} - P_{xi} = 0 - 0.3(15)cos 35 = - 3.68\hat i kg m/s$

along the y axis we have

$P_{yf} - P_{yi} = -(0.3)20 - 0.3(15)sin 35 = - 3.41\hat j kg m/s$

The magnitude of momentum is

$\Delta P = \sqrt{ (-3.41)^2 + (-3.68)^2} = 5 kg m/s$

b) Along the x axis we have

$P_{xf} - P_{xi} = -0.3(20) - 0.3(15)cos 35 = - 9.68\hat i kg m/s$

along the y axis we have

$P_{yf} - P_{yi} = -0 - 0.3(15)sin 35 = -25.58\hat j kg m/s$

The magnitude of momentum is

$\Delta P = \sqrt{ (-9.68)^2 + (-2.58)^2} = 10 kg m/s$