Reggie Jackson stepped up to the plate and hit a .15 kg fast ball traveling at 36m/s, the impact caused the ball to leave his bat with a velocity of 45 m/s in the opposite direction. If the impact lasted for .002 seconds, with what force did Jackson exert on the baseball?
F=MA=M*(v2-v1)/T where M=.15kg, v1=-36m/s, v2=45m/s, and T=.002s
F=(.15Kg)*(45m/s-(-36m/s))/(.002s)
F=(.15Kg)*(81m/s)/(.002s)
F=(12.15Kgm/s)/(.002s)
F=(6075Kgm/s^2)
F=6075N
wiener
To calculate the force exerted by Reggie Jackson on the baseball, we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (mv) of an object and is given by the equation F = Δp/Δt.
First, let's calculate the initial momentum of the baseball before the impact. We can use the formula p = mv, where p is momentum, m is mass, and v is velocity.
Given:
Mass of the baseball (m) = 0.15 kg
Initial velocity of the baseball (v₁) = 36 m/s
Initial momentum (p₁) = m * v₁
= 0.15 kg * 36 m/s
= 5.4 kg·m/s
Next, let's calculate the final momentum of the baseball after the impact. We know that the baseball's final velocity (v₂) is 45 m/s, but we need to consider that it is in the opposite direction. Therefore, the final velocity will be negative.
Final momentum (p₂) = m * v₂
= 0.15 kg * (-45 m/s)
= -6.75 kg·m/s
Now, we can calculate the change in momentum (Δp), which is simply the final momentum minus the initial momentum.
Change in momentum (Δp) = p₂ - p₁
= -6.75 kg·m/s - 5.4 kg·m/s
= -12.15 kg·m/s
Finally, to find the force exerted by Reggie Jackson on the baseball (F), we divide the change in momentum by the duration of the impact (Δt).
Given:
Duration of the impact (Δt) = 0.002 seconds
Force (F) = Δp/Δt
= (-12.15 kg·m/s)/(0.002 seconds)
= -6075 N
The negative sign indicates that the force is in the opposite direction of the initial velocity.
So, Reggie Jackson exerted a force of -6075 Newtons on the baseball during the impact.