p0328 - 3.28. CHAPTER 3, PROBLEM 28 237 3.28 Chapter 3,...

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Unformatted text preview: 3.28. CHAPTER 3, PROBLEM 28 237 3.28 Chapter 3, Problem 28 Problem: A CSI researcher fires a bullet horizontally into a bullet recovery tank with initial speed vo . The magnitude of the drag force on the bullet is D = mkv 2 , where m is the bullet’s mass, v is instantaneous speed and k is a constant of dimensions length−1 . Assume the motion is so rapid that any vertical deflection due to gravity is negligible. (a) Determine the speed and position of the bullet in terms vo , k and t. (b) The length of a typical bullet recovery tank is L = 101 in and k = 0.092 in−1 for a light 45-caliber bullet. Determine the bullet’s speed when it reaches the end of the tank if its initial speed is vo = 1100 ft/sec. 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............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. x So ut on: (a) New on’s Second Law e s us ha ma = −D Us ng he fac ha a = dv/dt we have dv = −kv 2 dt Thus we f nd v vo =⇒ a = −kv 2 =⇒ dv = −kdt v2 1 v v=v v=vo dv = −k v2 t dt 0 =⇒ − = −kt Eva ua ng he ef hand s de y e ds 1 1 − = −kt vo v Therefore rearrang ng erms he speed s v= The bu e ’s pos on fo ows from vo dx =v= dt 1 + vo kt Therefore eva ua ng he n egra s we f nd x= 1 n (1 + vo kt) k x t vo 1 + vo kt =⇒ dx = 0 0 vo dt 1 + vo kt 238 CHAPTER 3. FORCE AND ACCELERATION (b) Note first from Part (a) that 1 + vo kt = vo /v so that the bullet’s position is given by x= So, when the bullet reaches the end of the tank, v = vo e−kL For the given values of vo = 1100 ft/sec, k = 0.092 in−1 and L = 101 in, we have v= 1100 ft sec e−(0.092 in−1 )(101 in) 1 vo n k v = 0.10 ft sec ...
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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