A bullet is fired gun. The force on bullet is F=600-2xx10^(5)t newton. The force reduces to zero just when bullet leaves barrel. Find he impulse imparted to bullet

A bullet is fired gun. The force on bullet is F=600-2xx10^(5)t newton.

1.	A bullet is fired gun. The force on bullet is F=600-2xx10^(5)t newton. The force reduces to zero just when bullet leaves barrel. Find he impulse imparted to bullet
Answer» Solution :`F=600-2xx10^(5)t` F becomes zero as soon as the bullet leaves the barrel.`0 = 600-2xx10^(5)t, 6=2xx10^(5)t` `t=3XX10^(-3)s` Impusle `= int_(0)^(t)Fdt` `= int_(0)^(t)(600-2xx10^(5)t)dt=[600t-cancel(2)xx10^(5)(t^(2))/(cancel(2))]^(3xx10^(-3))` `= 600xx3xx10^(-3)-10^(5)xx9xx10^(-6)=0.9 NS`