A 0.15 kg baseball is moving at 49 m/s toward home plate when it is hit with the bat. The ball is in contact with the bat for 0.0022 s, and leaves the bat at 25 m/s directly away from home plate. Determine the average force exerted by the bat on the ball.
a = (Vf-Vo)/t = (-25-49)/0.0022 = -33.64
m/s^2.
F = M*a.
Note: The force will be negative, because it opposes the initial motion.
To determine the average force exerted by the bat on the ball, you can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (p) of an object. The formula for momentum is:
p = m * v
Where:
p is the momentum
m is the mass of the object
v is the velocity of the object
In this case, the baseball's momentum before being hit is given by:
p1 = m * v1
And the baseball's momentum after being hit is given by:
p2 = m * v2
The change in momentum (Δp) is the difference between the final and initial momenta:
Δp = p2 - p1
The average force (F) can be calculated using the formula:
F = Δp / Δt
Where Δt is the time during which the force is applied. In this case, it is the contact time between the bat and the ball.
Now, let's calculate the values:
Given:
Mass of the baseball (m) = 0.15 kg
Initial velocity of the baseball (v1) = 49 m/s
Final velocity of the baseball (v2) = -25 m/s (negative because it is moving away from home plate)
Contact time (Δt) = 0.0022 s
First, calculate the initial momentum (p1):
p1 = m * v1
p1 = 0.15 kg * 49 m/s
p1 = 7.35 kg·m/s
Next, calculate the final momentum (p2):
p2 = m * v2
p2 = 0.15 kg * (-25 m/s)
p2 = -3.75 kg·m/s
Now, find the change in momentum (Δp):
Δp = p2 - p1
Δp = -3.75 kg·m/s - 7.35 kg·m/s
Δp = -11.1 kg·m/s
Finally, calculate the average force:
F = Δp / Δt
F = -11.1 kg·m/s / 0.0022 s
F ≈ -5045 N
Therefore, the average force exerted by the bat on the ball is approximately -5045 Newtons. Note that the negative sign indicates that the force is in the opposite direction (away from home plate) as the initial motion of the ball (towards home plate).