A bullet of mass 30 g loses 80% of its velocity when it comes out from a sand bag within 0.3 seconds. If its initial velocity of penetration is 25 m/s, the nagnitude of resistive force acting on it is 2N. why and HOW???

Asked by  | 5th Feb, 2012, 02:14: PM

Expert Answer:

The initial velocity= 25 m/s
 
the final velocity= 20% of 25 m/s= 5 m/s
 
so change in velocity= 20 m/s
 
time taken = 0.3 s
 
from 
 
v=u+at
 
5=25+ a *0.3
 
a= - 200/3 m/s2
 
So F= m*a
 
F=0.030*200/3=2 N

Answered by  | 6th Feb, 2012, 04:35: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.