A(n) 0.42 kg softball is pitched at a speed of 12 m/s. The batter hits it back directly at the pitcher at a speed of 27 m/s. The bat acts on the ball for 0.016 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N · s. 011 (part 2 of 2) 10.0 points What is the magnitude of the average force exerted by the bat on the ball? Answer in units of N.

Question

Henry · Accepted Answer

Correction: Impulse = M*(V-Vo) = 0.42*(-27-12) = -16.4 kg*m/s

Henry · Answer

V = Vo + a*t = -27 m/s. 12 + a*0.016 = -27 0.016a = -27 - 12 = -39 a = -2438 m/s^2 Impulse = M*V = 0.42 * (-27) = -11.34 F = M*a = 0.42 * (-2438) = 1024 N.