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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 newton and $t$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?
$9\ N - s$
$58.8\ kg / s$
$0.9\ kg / s$
$178.4\ kg / s$
Solution
When force on the bullet is zero, then $0=600-2 \times 10^{5} t$
or $\quad t=\frac{600}{2 \times 10^{5}}=3 \times 10^{-3} \mathrm{s}$
Now, $\quad I=\int_{0}^{t} F d t=\int_{0}^{3 \times 10^{-3}}\left(600-2 \times 10^{5} t\right) d t$
$=\left[600 t-\frac{2 \times 10^{5} \times t^{2}}{2}\right]_{0}^{3 \times 10^{-3}}$
$=\left[600 \times 3 \times 10^{-3}-10^{5} \times 9 \times 10^{-6}\right]$
$=\left\lceil 18 \times 10^{-1}-0.9\right\rceil=\lceil 1.8-0.9\rceil= 0.9 N-s$