Question

A 0.33 kg softball has a velocity of 15 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)21 m/s, vertically downward, and (b)21 m/s, horizontally back toward the pitcher?

Answer 1

here,

mass , m = 0.33 kg

the initial velocity , u = 15 m/s * ( cos(46) i - sin(46) j)

u = 10.42 m/s i - 10.8 m/s j

a)

when final velocity , v = - 21 m/s j

the change in momentum , dP = Pf - Pi

dP = m * ( v - u)

dP = 0.33 * ( - 21 j - ( 10.42 i - 10.8 j))

dP = (- 3.44 i - 3.67 j ) kg.m/s

the magnitude of change in momentum , |dP| = sqrt(3.44^2 + 3.67^2) = 4.8 kg.m/s

b)

when final velocity , v = - 21 m/s i

the change in momentum , dP = Pf - Pi

dP = m * ( v - u)

dP = 0.33 * ( - 21 i - ( 10.42 i - 10.8 j))

dP = (- 10.4 i + 3.564 j ) kg.m/s

the magnitude of change in momentum , |dP| = sqrt(10.4^2 + 3.564^2) = 11 kg.m/s