A baseball pitcher throws a 145 g baseball parallel to the ground with a speed of 27.31 m/s in the +x direction. The batter imparts an impulse of –10.43 to the ball so that the ball returns along the same path in the –x direction. What is the velocity of the ball after leaving the bat?
Answer : v = 44.62 m/s in -x direction
Mass of ball, m = 145g = 0.145 kg
Initial velocity of ball , u = 27.31 m/s
Impulse , I = -10.43 kg*m/s
Final velocity of ball, v = ?
Initial momentum = mu = 0.145 * 27.31 = 3.96 kg*m/s
final momentum = mv = - 0.145v kg*m/s (-ve sign because final momentum in -x direction)
Impulse = change in momentum
-10.43 = -0.145v - 3.96
v = (-10.43 + 3.96) / -0.145
v = 6.47 / 0.145
v = 44.62 m/s in -x direction
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