A 0.140-kg baseball is pitched horizontally at 36.7 m/s. When a player hits the ball, it moves at the
same speed, but in the opposite direction. If the bat and the ball are in contact for 0.450 ms, calculate
the average force the bat exerts on the ball.
knickerbocker over here
To calculate the average force exerted by the bat on the ball, we can use Newton's second law of motion, which states that force (F) is equal to the change in momentum (Δp) divided by the time interval (Δt) over which the force is exerted.
First, we need to calculate the initial momentum (p₁) of the baseball. Since the baseball is moving horizontally, the initial momentum is given by the equation:
p₁ = m₁ * v₁
where m₁ is the mass of the baseball and v₁ is the velocity at which it is pitched.
p₁ = 0.140 kg * 36.7 m/s = 5.138 kg·m/s
Next, we need to determine the final momentum (p₂) of the baseball after it is hit. Since the ball moves at the same speed but in the opposite direction, the final momentum is given by the equation:
p₂ = -m₁ * v₁
where -m₁ represents the opposite direction.
p₂ = -0.140 kg * 36.7 m/s = -5.138 kg·m/s
Now, we can calculate the change in momentum (Δp) of the baseball:
Δp = p₂ - p₁ = (-5.138 kg·m/s) - (5.138 kg·m/s) = -10.276 kg·m/s
Given that the time interval (Δt) over which the force is exerted is 0.450 milliseconds, we need to convert it to seconds:
Δt = 0.450 ms = 0.450 × 10^(-3) s = 4.5 × 10^(-4) s
Finally, we can calculate the average force (F) using the equation:
F = Δp / Δt
F = (-10.276 kg·m/s) / (4.5 × 10^(-4) s) ≈ -22,835 N
Therefore, the average force exerted by the bat on the ball is approximately 22,835 N. Note that the negative sign indicates that the force is in the opposite direction of the initial velocity.
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