An official major league baseball has a mass of 0.14 kg . A pitcher throws a 40 \( \mathrm{~m} / \mathrm{s} \) fastball which is hit by the batter straight back up the middle at a speed of \( 46 \mathrm{~m} / \mathrm{s} \). What is the impulse of the ball during the collision with the bat? Woint Numeric 1 point If this collision occurs during a time of 0.012 seconds, what is the av force exerted by the bat on the ball? Type your answer.
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To calculate the impulse of the ball during the collision with the bat, we can use the impulse-momentum theorem, which states that the impulse experienced by an object is equal to the change in momentum of that object. 1. **Calculate the initial momentum:** - Mass \( m = 0.14 \, \text{kg} \) - Initial velocity of the ball \( v_i = 40 \, \text{m/s} \) (toward the batter, considered positive) - Final velocity of the ball \( v_f = -46 \, \text{m/s} \) (after being hit, in the opposite direction) Change in momentum (impulse) \( J = m(v_f - v_i) \) \[ J = 0.14 \, \text{kg} \times (-46 \, \text{m/s} - 40 \, \text{m/s}) = 0.14 \, \text{kg} \times (-86 \, \text{m/s}) = -12.04 \, \text{kg·m/s} \] The impulse of the ball during the collision is approximately **-12.04 kg·m/s**. 2. **Calculate the average force exerted by the bat on the ball:** \[ F = \frac{J}{\Delta t} \] where Δt is the time duration of the collision (0.012 seconds). \[ F = \frac{-12.04 \, \text{kg·m/s}}{0.012 \, \text{s}} \approx -1003.33 \, \text{N} \] The average force exerted by the bat on the ball is approximately **-1003.33 N** (the negative sign indicates the direction opposite to the ball's initial motion).