A bullet is fired from a gun. The force on the bullet is given by F= 600 - 2 x 10^5 t
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first we shall find the time taken by the bullet to come out of te barrel.
this will help us calculate the impulse on the bullet
the required time is 3*10-3 s.
now we can calculate the impulse by two ways.
the first way is integration which is the most common method.
J = ∫F(t)dt = ∫{600-(2*105t)}dt
the limits shall be from t = 0 to t = 3*10-3 s.
the other way is very simple. it is the method of average.
when a quantity has a linear variation we can take the average of the initial and final quantiites to evaluate such an integral
that is if F is a linear function of t then
∫0TF(t)dt = <F(t)>T = {F(t = 0) + F(t = T)}/2.
both these ways give the same answer as 300 N-s.
this will help us calculate the impulse on the bullet
the required time is 3*10-3 s.
now we can calculate the impulse by two ways.
the first way is integration which is the most common method.
J = ∫F(t)dt = ∫{600-(2*105t)}dt
the limits shall be from t = 0 to t = 3*10-3 s.
the other way is very simple. it is the method of average.
when a quantity has a linear variation we can take the average of the initial and final quantiites to evaluate such an integral
that is if F is a linear function of t then
∫0TF(t)dt = <F(t)>T = {F(t = 0) + F(t = T)}/2.
both these ways give the same answer as 300 N-s.
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