Bullet pendulum problem w/ Calculus - AP Physics C

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SUMMARY

The discussion centers on solving the bullet pendulum problem using calculus, specifically focusing on the equations of motion under damping forces. The participant successfully derived the velocity function v(t) = v_0 * e^(-bt/m) and the position function as position = m*v_0/b * (1 - e^(-bt/m)). The confusion arose regarding the bounds of integration when transitioning from velocity to position, which was clarified through the use of indefinite integrals and the application of limits as time approaches infinity.

PREREQUISITES
  • Understanding of Newton's second law of motion (F_net = m * a)
  • Familiarity with calculus concepts, particularly integration and limits
  • Knowledge of exponential decay functions in physics
  • Basic principles of energy conservation, including kinetic and potential energy equations
NEXT STEPS
  • Study the derivation of exponential decay in damped harmonic motion
  • Learn about the application of definite vs. indefinite integrals in physics problems
  • Explore the concept of limits in calculus, particularly in the context of asymptotic behavior
  • Investigate other applications of Newton's laws in systems with damping forces
USEFUL FOR

Students and educators in AP Physics C, particularly those focusing on mechanics and calculus applications in physics. This discussion is also beneficial for anyone looking to deepen their understanding of motion under damping forces.

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Homework Statement



See problem here:
http://i184.photobucket.com/albums/x153/spl10246/problem.png

Solution:
http://i184.photobucket.com/albums/x153/spl10246/solution1.png
http://i184.photobucket.com/albums/x153/spl10246/solution2.png


Homework Equations



p = m * v
K = .5 * m * v^2
U = m * g * h
a = dv/dt
v = dx/dt

The Attempt at a Solution



I've gotten A and C fine, I'm still working to get B. D is the trouble part.

F_net = m * a
Only force acting is -bv

m * a = -bv
m * dv/dt = -bv

dv/dt = -bv / m

dv = -bv/m * dt

dv/v = -b/m * dt



At this point the published solution makes no sense to me at all. I am used to just putting an indefinite integral on both sides at this point. My confusion is about the bounds. Why is the left side v_0 to v?

I understand the steps after setting up the integrals (including evaluating the integrals and using log properties and so on).
 
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Okay: I finally arrived at the right answer by just using indefinite integrals and including a constant of integration that I solved for. That got me v(t) = v_0 * e ^ (-bt/m)

Then I did the same thing to get position from velocity, and used indefinite integrals again...
I got position = m*v_0 / b * (1 - e^(-bt/m))

Taking the limit as t-> infinity (b/c the block never fully stops, v(t) has no zeros but does a horizontal asymptote) yields m*v_0/b, which is the answer.
 

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