Finding time of a collision using impulse and conservation of momentum

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SUMMARY

The discussion focuses on calculating the duration of a collision for a 1400 kg car that stops uniformly from a speed of 25.0 m/s over a distance of 1.20 m. The correct time of collision is determined to be 0.096 seconds using kinematics, while an attempt to use impulse resulted in an incorrect time of 0.048 seconds. The discrepancy arises because impulse calculations require knowledge of the force acting on the object, which is not provided in this scenario. The conversation highlights the importance of using appropriate methods for solving physics problems based on the information available.

PREREQUISITES
  • Understanding of kinematics equations, specifically vf = vi + at and vf² = vi² + 2ax
  • Knowledge of impulse and momentum concepts, including I = mΔv and I = Δp
  • Familiarity with the concept of uniform acceleration and its implications in collision scenarios
  • Basic understanding of average vs. instantaneous velocity in physics
NEXT STEPS
  • Study the derivation and applications of kinematics equations in collision problems
  • Learn about impulse-momentum theorem and its practical applications in physics
  • Explore the relationship between force, mass, and acceleration in collision scenarios
  • Investigate the differences between average and instantaneous velocity and their relevance in physics calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision analysis, as well as educators looking for examples of impulse and momentum applications in real-world scenarios.

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Homework Statement



The front of a 1400 kg car is designed to absorb the shock of a collision by having a "crumple zone" in which the front 1.20 m of the car collapses in absorbing the shock of a collision.

(a) If a car traveling 25.0 m/s stops uniformly in 1.20 m, how long does the collision last?
answer in s

NOTE: I know this is not part of the template, but I needed to put it in somewhere. I realize this problem can be easily solved using kinematics equations and constant acceleration, but I'm trying to figure out why using the impulse doesn't work. See attempt section for more.

Homework Equations



vf = vi + at
vf2 = vi2 + 2ax
I= m\Deltav
I= m(vf) - m(vi)

The Attempt at a Solution



CORRECT ANSWER:
0= 25 + at
0= 252 + 2ax
a = -252/2ax = -260.4 m/s2
t = -25/a = 0.096 s

TRIAL USING IMPULSE:
I = 0 - m(25) = -35 000
I = m\Deltax/\Deltat *I realize this may be where the problem lies
1.2m/-35 000 = t
t = 0.048 s *I also notice that this is exactly 1/2 of the right answer
 
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The thing is that there isn't much point in trying to use impulse for this problem. If you knew the force acting on the object as well as its starting velocity, impulse would be the way to go. Without knowing the force though, you would have to find it with kinematics, at which point you might as well also just use kinematics to correctly solve the problem.

Also, yes, the problem you had in your trial was doing
I = m∆x/∆t
Impulse is defined as
I= ∆p = m∆v = m(v2-v1)
where v2 and v1 are instantaneous velocities at particular times. You used an average velocity with ∆x/∆t.
 
Thanks, I think I understand. I somewhat knew that I was working in circles, but not knowing why I couldn't get the answer was bugging me to no end.
 

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