Physics 325 Spring 2011 Problem Session 2 Solutions

# Physics 325 Spring 2011 Problem Session 2 Solutions - t = 3...

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Physics 325 Spring 2011 Discussion 2 January 31, 2011 At t = 0 , an object of mass m is at x = 0 , moving with velocity v 0 . The only force acting on m is a friction force - bv 1 / 2 , where v is the object’s velocity. (a) Calculate v ( x ) and x ( t ) . (b) What is the maximum distance travelled by the object? (c) Suppose now there is another force K (constant) acting on the object. What is v as t → ∞ ? Solutions : (a) dv dt = dv dx dx dt = v dv dx = - b m v - Z v v 0 vdv = b m Z t 0 dt 2 3 ( v 3 / 2 0 - v 3 / 2 ) = b m x v = v 0 ± 1 - 3 b 2 mv 3 / 2 0 x ² 2 / 3 v = v 0 (1 - αx ) 2 / 3 , where α = 3 b 2 mv 3 / 2 0 . Now, v = dx/dt , so dx dt = v 0 (1 - αx ) 2 / 3 Z x 0 dx (1 - αx ) 2 / 3 = v 0 Z t 0 dt - 3 α ³ (1 - αx ) 2 / 3 - 1 ´ = v 0 t x ( t ) = 1 α ( 1 - (1 - 1 3 αv 0 t ) 3 ) 1

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According to the expression of v ( x ) , the object stops moving when it reaches x = 1 , which occurs at
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Unformatted text preview: t = 3 / ( αv ) . For any time afterwards, v = 0 , so x stays constant at 1 /α . Thus, x ( t ) = ( 1 α ± 1-(1-1 3 αv t ) 3 ) , for ≤ t ≤ 3 αv 1 α , for t > 3 αv (b) From previous part, x max = 1 α . (c) As t → ∞ , the velocity will reach the terminal velocity, regardless of v . When v = v term = terminal velocity, the acceleration is zero, so we have m dv dt = 0 = K-b √ v term , or v term = K 2 b 2 . 2...
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## This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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Physics 325 Spring 2011 Problem Session 2 Solutions - t = 3...

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