Question

A 0.27 kg softball has a velocity of 14 m/s at an angle of 37° 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)17 m/s, vertically downward, and (b)17 m/s, horizontally back toward the pitcher?

Answer 1

given mass m=0.27 kg

initial velocity V = 14 ( cos 37 i + sin 37 j )

V = Vxi + Vyj = 11.18 i + 8.42 j

when Velcoity V = -17 j m/s

change in momentu dP = mv- mu

dP = m (u-v)=0.27* ( 17 j - (11.18 i + 8.42 j) )

dP = 3.018 i + 2.316 j

so net change in momentum = P^2 = Px^2 + Py^2

P^2 = 3.018^2 + 2.316^2

P = 3.804 Kgm/s

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b)v=-18i

dP = m(u-v)

dP =0.27 * (-18i - (11.18i + 8.42 j)

dP = -7.87 i -8.42 j

net Momentum change P^2 = 7.87^2 + 8.42^2

P = 11.52 kgm/s