Question

A bullet is fired from a gun. The force on the bullet is given by $F=600-2\times {10}^{5}t$, where F is in N and t in sec. The force on bullet becomes zero as soon as it leaves barrel. What is average impulse imparted to bullet?

Answer 1

$F=600-2\times {10}^{5}t$

$0=600-2\times {10}^{5}t$

$t=600/(2\times {10}^5)$

$t=3\times {10}^{-3}s$

$t=3milliseconds$.

Time is taken by bullet leave the barrel is 3 milliseconds

$F.idt=(6t\times {10}^{5}t).dt$

$J=[600t-(2\times {10}^5\frac{t²}{2}\left)\right]$

$J=[600\times (3\times {10}^{-3})-(2\times {10}^5\times \frac{(3\times {10}^{-3}{)}^2}{2}\left)\right]$

$J=0.9Ns$