A .145kg baseball pitched at 39m/s is hit on a horizontal line drive straight back toward the pitched at 52m/s. If the average force the bat exerts on the ball is 13,200N, how long did the collision last?
You are asking the same basic question over and over slightly different ways.
average force = change in momentum / change in time
Just do it.
To find the duration of the collision, we can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the net impulse applied to it. In this case, the impulse experienced by the baseball is equal to the force multiplied by the time of collision.
The impulse is given by the change in momentum, which can be calculated using the initial and final velocities of the baseball:
Change in momentum = mass * (final velocity - initial velocity)
Given:
Mass of the baseball (m) = 0.145 kg
Initial velocity of the baseball (u) = 39 m/s
Final velocity of the baseball (v) = -52 m/s (since it is moving back towards the pitcher)
Change in momentum = 0.145 * (-52 - 39)
To find the time of collision, we can rearrange the equation:
Impulse = Force * time
Since impulse = change in momentum, we can substitute the value of impulse:
Force * time = 0.145 * (-52 - 39)
Now we can solve for time:
time = (0.145 * (-52 - 39)) / 13,200
Therefore, the duration of the collision is the calculated value of time. Let's calculate it.
time = (0.145 * (-91)) / 13,200
time = -0.00100275
Since time can't be negative, we take the absolute value:
time = 0.00100275 seconds
Therefore, the collision lasted approximately 0.00100275 seconds.