HW12-Linear Momentum Item 5 ? : 50126 Part A A major league catcher catches a fa

Question

HW12-Linear Momentum Item 5 ? : 50126 Part A A major league catcher catches a fastball moving at 95.0 mi/h and his hand and glove recoil 10.0 cm in bringing the ball to rest. If it took 0.00410 s to bring the ball (with a mass of 270 g) to rest in the glove, what is the of the change in momentum of the ball? kg .m/s Submit Request Answer Part B What is the direction of the change in momentum of the ball? the same as the initial velocity of the ball opposite to the initial velocity of the ball Previous Answers Correct Part C Find the magnitude of the average force the ball exerts on the hand and glove. Submit Request Answer

Explanation / Answer

1 mi/h = 0.44704 m/s

95 mi/h = 95*0.44704 = 42.46 m/s

a) change in momentum of the ball is dp = (m*v) = (270*10^-3*42.46) = 11.46 kg m/s

b) opposite to the initial velocity of the ball

C) Favg *dt = dp

Favg*0.00410 = 11.46

Favg = 11.46/0.00410 = 2795.12 N

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