To find the average force exerted by the bat on the baseball, we can use the impulse-momentum theorem, which states that the impulse (change in momentum) is equal to the average force multiplied by the time the force is applied.
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Calculate the initial and final momentum of the baseball:
The initial velocity \( v_i \) of the baseball is \( 38 , \text{m/s} \) (thrown horizontally), and after being struck, the final velocity \( v_f \) is \( -38 , \text{m/s} \) (indicating it is now moving in the opposite direction).
The mass \( m \) of the baseball is \( 0.14 , \text{kg} \).
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Initial momentum \( p_i \): \[ p_i = m \times v_i = 0.14 , \text{kg} \times 38 , \text{m/s} = 5.32 , \text{kg⋅m/s} \]
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Final momentum \( p_f \): \[ p_f = m \times v_f = 0.14 , \text{kg} \times (-38 , \text{m/s}) = -5.32 , \text{kg⋅m/s} \]
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Calculate the change in momentum \( \Delta p \): \[ \Delta p = p_f - p_i = -5.32 , \text{kg⋅m/s} - 5.32 , \text{kg⋅m/s} = -10.64 , \text{kg⋅m/s} \]
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Calculate the average force \( F \): The impulse can also be expressed as: \[ F \Delta t = \Delta p \] where \( \Delta t = 0.0006 , \text{s} \).
From this, we can solve for \( F \): \[ F = \frac{\Delta p}{\Delta t} = \frac{-10.64 , \text{kg⋅m/s}}{0.0006 , \text{s}} \approx -17,733.33 , \text{N} \]
Therefore, the average force the bat exerted on the ball is approximately \(-17,733 , \text{N}\).
Final answer: \(-17,733 , \text{N}\)