Speed of a Body on a Smooth Surface over Time

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SUMMARY

The speed of a body on a smooth horizontal surface, where the mass decreases exponentially with a disintegration constant λ, can be described by the equation v(t) = u(1 - e^(-λt)). This equation is derived from the principles of conservation of momentum, particularly in the context of rocket motion. The discussion emphasizes the importance of understanding logarithmic and exponential functions to solve the problem effectively. The initial velocity of the body is zero, which simplifies the calculations.

PREREQUISITES
  • Understanding of exponential decay and disintegration constants (λ)
  • Familiarity with conservation of momentum principles
  • Knowledge of logarithmic functions
  • Basic physics concepts related to motion and velocity
NEXT STEPS
  • Study the derivation of the rocket equation and its applications
  • Learn about exponential functions and their properties in physics
  • Explore conservation of momentum in different physical systems
  • Investigate real-world applications of mass loss in rocket propulsion
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Physics students, engineers, and anyone interested in understanding motion dynamics and rocket propulsion principles.

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Question:A body of mass is placed on a smooth horizontal surface. The mass of the body is decreased exponentially with disintegration constant λ. Assuming that the mass is ejected backwards with a relative velocity u.If initially the body
was at rest, the speed of the body at time t is:
(a)ue^(-t)
(b)uλt
(c)ue^(-λt)
(d)u{1-e^(-λt)}.
 
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I guess this does not qualify as speedy, but do you know the rocket equation? What you have is a rocket. Figuring out the speed of a rocket is not trivial, but it all depends on conservation of momentum. A decent explanation is given here.

http://ed-thelen.org/rocket-eq.html

The final equation can be written as

[tex]v(t) = v_0 + uln \left[ \frac{M_0}{M(t)} \right][/tex]

In your problem, the initial velocity is zero. If you understand logs and exponentials, you can do the rest.
 

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