{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p0328

# p0328 -

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. L vo D v ....................................................................................................................... ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ...................................................................................................................... ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................ ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. . ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................. 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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online