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A 0.144kg baseball is pitched horizontally at 38.0 m/s. After the baseball is hit by the bat, it moves at the same speed, but in the opposite direction. What is the change in momentum of the ball? What is the impulse delivered by the bat? If the ball and the bat were in contact for 0.80 ms, what was the average force the bat exerted on the ball?

Respuesta :

Answer:

Given:

mass of the ball m = 0.144 kg

velocity v = 38 m/s

now,  change in momentum

P = m v- ( - mv)

  = 2 mv

  =2 x (0.144) x (38)

  = 10.944 kg-m/s

Impulse J= F. Δt

change in momentum is equal to impulse

J = 10.944 kg-m/s

we know force is equal to change in momentum per unit time

[tex]F = \dfrac{P}{t}[/tex]

[tex]F = \dfrac{10.944}{0.8\times 10^{-3}}[/tex]

F = 13.68 x 10³ N

F = 13.68 kN

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