A pitcher pitches a baseball of mass 0.20 kg. The baseball arrives at home plate

Question

A pitcher pitches a baseball of mass 0.20 kg. The baseball arrives at home plate with a speed of 40 m/s and is batted directly back to the pitcher (horizontally) with a return speed of 50 m/s. a.What is the magnitude of the change of momentum of the baseball? b.If the bat remains in contact with the ball for 0.12 s, what is the magnitude of the average force of the bat on the ball? c.If instead of going directly back to the pitcher (horizontally), the ball comes off the bat at an angle of 30o above the horizontal with a speed of 50 m/s, what is the magnitude of the average force of the bat on the ball?

Explanation / Answer

the momentum is P=mv the change would be delta P = 0.2*(60-40)=0.2*20=4 the magnitude of the force is given by deltaP/delta t = F as a vector so F=0.15*(6.5-3.5)/0.025=0.15*3/0.025 Ek = kinetic energy Ep = potential energy delta Ek = - delta Ep Ek - 0 = - mg (0-2) Ek=7*9.8*2

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