Question

A baseball catcher is catching a fastball that is thrown at 43 m/s (96 mi/h) by the pitcher. If the mass of the ball is 0.15 kg and if the catcher moves his mitt backward toward his body by 8.0 cm as the ball lands in the glove, what is the magnitude of the average force acting on the catcher's mitt? Estimate the time interval required for the catcher to move his hands.

Answer 1

change in KE = 0.5*m*v^2

work done = F*X

from work energy theorem

W = dKE

F*X = 0.5*m*v^2

F*0.08 = 0.5*0.15*43^2

F = 1733.44 N <<<--------answer

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force= rate of change in momentim

F = mv/t

t = mv/F = 0.15*43/1733.44

t = 3.72 *10^-3 s <<-------answer