How Much Impulse Is Needed to Stop a Baseball with Unrealistic Momentum?

  • Thread starter Thread starter slimyspacewit
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
To stop a baseball with a momentum of 900 kg m/s, an impulse equal to that momentum is required, meaning the outfielder must apply an impulse of 900 Ns. Given a maximum force of 100 N, the time needed to exert this force can be calculated using the impulse-momentum theorem. However, the momentum value presented seems unrealistic, as it implies an extremely high velocity for a baseball, estimated between 3000 to 4500 m/s. This suggests a potential error in the initial momentum figure, as typical baseballs have a mass of about 0.2 to 0.3 kg. Clarifying the momentum value is essential for accurate calculations.
slimyspacewit
Messages
1
Reaction score
0
If a baseball were flying towards an outfielder with a momentum of 900 kg m/s. What impulse would the outfielder need to apply to bring the ball to a stop? AND If the maximum force the dude can exert is 100N for what length of time will he have to exert this force?
 
Physics news on Phys.org
This sounds like a homework question.
Please post it in the homework forums and they can help you there.
 
I think there is something wrong with your number. I don't know the mass of a baseball but I would guess about 0.2 to 0.3 kg.
To have a momentum of 900 kg m/s means a velocity of 3000 to 4500 m/s and that does not sound reasonable
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top