in a baseball game a 0.2 kg ball moving at 12 m/s is hit by a bat. after the impact the ball moves in the opposite direction with a velocity of 18 m/s. if the force exerted on the ball of the batter is 670 N. how long is the time of contact

Answer 1

Force x (contact time) = Momentum change
= 0.2 kg*(12 + 18)m/s = 6.0 kg*m/s

Contact time = 6.0/670 = 9.0*10^-3 s

Answer 2

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