suppose a major league pitcher can throw a baseball at about 45 m/s (about 100mph). IF a baseball has a mass of .55kg. Suppose that the batter hits the ball straight back at the pitcher at 55 m/s. If the bat is in contact with the ball for only .055 second, what size force does the bat exert on the ball?
Force = (momentum change)/(contact time)
= (0.55)(45 + 55)/0.055 Newtons
= 900 N
(about 200 pounds force)
To determine the 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 change in time (Δt).
The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass of the object, and v is the velocity of the object.
In this case, the change in momentum can be calculated by finding the difference between the initial momentum of the ball thrown by the pitcher and the final momentum of the ball when it is hit by the bat.
The initial momentum (p_initial) of the ball thrown by the pitcher can be calculated by multiplying the mass of the ball (m = 0.55 kg) by the initial velocity of the ball (v_initial = 45 m/s). Therefore, p_initial = 0.55 kg * 45 m/s.
The final momentum (p_final) of the ball when it is hit by the bat can be calculated by multiplying the mass of the ball (m = 0.55 kg) by the final velocity of the ball (v_final = -55 m/s, considering the direction of the ball being hit back towards the pitcher). Therefore, p_final = 0.55 kg * (-55 m/s).
The change in momentum (Δp) is then given by the difference between the final and initial momenta: Δp = p_final - p_initial.
Next, we need to calculate the force by dividing the change in momentum by the contact time between the bat and the ball (Δt = 0.055 s). So, the force exerted by the bat on the ball (F) can be calculated using the formula:
F = Δp / Δt
Substituting the values we have determined, the force exerted by the bat on the ball can be calculated.