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Helppp!!!!! Please!!
A 0.144 kg baseball approaches a batter with a speed of 20m/s . The batter lines the ball directly back to the pitcher with a speed of 30m/s . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat

Sagot :

onyebuchinnaji

Answer:

  • The impulse exerted by the ball is 7.2 kg.m/s
  • The average force exerted on the ball by the bat is 600 N.

Explanation:

Given;

mass of the baseball, m = 0.144 kg

velocity of the baseball, v₁ = 20 m/s

velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)

The impulse exerted by the ball is calculated as follows;

J = ΔP = mv₁ - mv₂

ΔP = m(v₁ - v₂)

ΔP = 0.144 [20 - (-30)]

ΔP = 0.144 ( 20 + 30 )

ΔP = 0.144 (50)

ΔP = 7.2 kg.m/s

The average force exerted on the ball by the bat is calculated as;

[tex]F = \frac{\Delta P }{t} \\\\where; t \ is \ time \ of \ contact= 0.012 \ s\\\\F = \frac{7.2}{0.012} \\\\F = 600 \ N[/tex]

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