Calculating Average Force in Nonelastic Collisions with No External Forces

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To calculate the average force during a nonelastic collision between a baseball and a bat, the relevant equation involves the change in momentum over the contact time. The baseball, weighing 0.145 kg, is pitched at 35.0 m/s and hit back at 57.0 m/s, leading to a change in velocity of 22 m/s. The correct calculation for average force is derived from the formula m*(v(f) - v(i))/Δt, where Δt is the contact time of 2.50×10^-3 s. Clarification was needed regarding the direction of velocity changes, emphasizing that the change in velocity can indeed be negative depending on the frame of reference. The final average force calculation should accurately reflect these considerations.
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A 0.145-kg baseball pitched at 35.0m/s is hit on a horizontal line drive straight back toward the pitcher at 57.0 m/s.
If the contact time between bat and ball is 2.50×10−3s , calculate the average force between the ball and bat during contact.

The equation has something to do with nonelastic collisions but it has no external forces so it should have something like m*(v(f)-v(i)/change in t

I tried to plug and chug but it wasn't correct. ((57-22)*.145)/(2.5*(10^-3) The answer I came out with was 1276
 
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nhmockus said:
((57-22)*.145)/(2.5*(10^-3)

Where did the 22 come from?
 
Haha that's what i found. Sorry, it should be 57-35 to = 22. I did it correctly in my calculations
 
nhmockus said:
Haha that's what i found. Sorry, it should be 57-35 to = 22. I did it correctly in my calculations

Think about which direction the velocity is going in, are you sure the change in velocity is 57-35=22?
 
Why wouldn't it be? Or are you saying that it could be negative?
 
If I am traveling east at 5m/s, then I turn around and start traveling west at 3m/s, what is the change in my velocity?
 

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