Calculating Impulse from a Graph: How to Solve the Problem

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SUMMARY

The discussion focuses on calculating impulse from a graph by breaking it into two triangles, one representing positive impulse and the other negative. The impulse is defined as the change in momentum, expressed mathematically as Δp=Δ(mv)=mΔv. The relationship between force and impulse is established through the equation F = dp/dt, leading to the integral form ∫dp = Δp = ∫F dt, which represents the signed area under the force curve. Participants confirm that calculating the areas of both triangles separately and combining them is the correct approach, with negative areas contributing to negative impulse.

PREREQUISITES
  • Understanding of basic calculus, specifically integrals
  • Familiarity with the concepts of momentum and impulse
  • Knowledge of force as defined by Newton's second law
  • Ability to interpret graphical representations of physical quantities
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  • Study the fundamentals of integrals in calculus
  • Explore the relationship between force and impulse in physics
  • Learn how to calculate areas under curves for various functions
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Students in physics, particularly those studying mechanics, as well as educators and anyone looking to deepen their understanding of impulse calculations from graphical data.

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Homework Statement


[The problem is in the attached image]


Homework Equations





The Attempt at a Solution


My best guess is to break up the graph into two triangles one positive, one negative and then calculate and add the impulse of both of these triangles... Is that what you're supposed to do?
 

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Impulse is defined as the change in momentum, or Δp=Δ(mv)=mΔv. This quantity is related to the force F by the following logic.

F = dp/dt (you should memorize this equation; it's very useful.)

= d(mv)/dt
= m(dv)/dt
= m(dv/dt)
= ma, which is the definition of force.

Given this simple relationship,

∫dp = Δp = ∫F dt

What is an integral? An integral is the signed area between the x-axis (F=0) and the curve. So yes, calculate them separately, and then glue them together. Don't forget that everything below F=0 counts as negative impulse.
 
Thank you :)
 

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