To find the average force exerted by the bat on the baseball, we can use the impulse-momentum theorem. The theorem states that the impulse experienced by an object is equal to the change in momentum of that object.
First, we calculate the initial and final momenta of the baseball.
- Calculate initial momentum (\( p_i \)): \[ p_i = m \cdot v_i \] where:
- \( m = 0.14 , \text{kg} \) (mass of the baseball)
- \( v_i = 38 , \text{m/s} \) (initial velocity, thrown horizontally)
\[ p_i = 0.14 , \text{kg} \cdot 38 , \text{m/s} = 5.32 , \text{kg m/s} \]
- Calculate final momentum (\( p_f \)): \[ p_f = m \cdot v_f \] where \( v_f = -38 , \text{m/s} \) (final velocity after being struck by the bat).
\[ p_f = 0.14 , \text{kg} \cdot (-38 , \text{m/s}) = -5.32 , \text{kg m/s} \]
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Calculate change in momentum (\( \Delta p \)): \[ \Delta p = p_f - p_i \] \[ \Delta p = -5.32 , \text{kg m/s} - 5.32 , \text{kg m/s} = -10.64 , \text{kg m/s} \]
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Calculate average force (\( F \)) using the impulse-momentum equation: \[ F \cdot \Delta t = \Delta p \] where \( \Delta t = 0.0006 , \text{s} \) (contact time).
Rearranging for \( F \): \[ F = \frac{\Delta p}{\Delta t} = \frac{-10.64 , \text{kg m/s}}{0.0006 , \text{s}} \]
Calculating \( F \): \[ F = -17600 , \text{N} \]
The negative sign indicates that the force exerted by the bat is in the opposite direction to the initial motion of the baseball. Therefore, the average force the bat exerted on the ball is \( \approx 17600 , \text{N} \) in the direction opposite to the ball's initial motion.