A BULLET OF MASS 5g TRAVELS WITH A SPEED OF 500ms-1. IF IT PENETRATES A FIXED TRAGET WHICH OFFERS RESISTIVE FORCE OF 1000 N TO THE MOTION OF BULLET FIND THE SPEED WITH WHICH THE BULLET EMERGES OUT OF TARGET IF TARGET IS OF THICKNESS 0.5m

Asked by mpandey358246.mp | 5th May, 2020, 02:12: PM

Expert Answer:

mass of bullet, m = 5 g = 0.005 kg =5 × 10-3 kg 
v = 500 m/s 
 
Initially the kinetic energy of bullet = 1/2 (mv2)
 
 
K i n e t i c space e n e r g y space o f space b u l l e t space b e f o r e space h i t t i n g space t h e space t a r g e t space equals space 1 half cross times open parentheses 5 cross times 10 to the power of negative 3 end exponent close parentheses space cross times space open parentheses 500 close parentheses squared space equals space 625 space J space

E n e r g y space s p e n t space w h i l e space m o v i n g space a g a i n s t space t h e space r e s i s t i v e space f o r c e space o f space 1000 space N space o f f e r e d space w h i l e space p e n e t r a t i n g space t h r o u g h space t a r g e t
equals space R e s i s t i v e space f o r c e space cross times space t h i c k n e s s space o f space t a r g e t space equals space 1000 space cross times space 0.5 space equals space 500 space J space

therefore space T h u s space k i n e t i c space e n e r g y space o f space b u l l e t space a f t e r space e m e r g i n g space o u t space o f space t h e space t a r g e t space equals space 625 space minus space 500 space J space equals space 125 space J space

T h u s comma space
i f space t h e space s p e e d space o f space b u l l e t space i s space v subscript 1 comma space t h e n space k i n e t i c space e n e r g y space equals space 1 half m v subscript 1 superscript 2 space
125 space equals space 1 half cross times open parentheses 5 cross times 10 to the power of negative 3 end exponent close parentheses space cross times space open parentheses v subscript 1 superscript 2 space close parentheses space

v subscript 1 space equals space square root of fraction numerator 2 cross times 125 over denominator 5 cross times 10 to the power of negative 3 end exponent end fraction end root space equals space 223.60 space m divided by s space

Answered by Shiwani Sawant | 5th May, 2020, 05:25: PM