Question

A bullet is fired from a gun. The force on the bullet is given by $600-2*10^5$,where$F$isinnewtonand$t$ is in second. The force on bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is

• A. $6Ns$
• B. $0.9Ns$
• C. $18Ns$
• D. $1.8Ns$

Answer 1

The correct option is A $6Ns$

$F=6w-2\times {10}^{5}t$

$\Rightarrow 0=6w-2\times {10}^{5}t$

$\Rightarrow t=\frac{600}{2\times {10}^5}=3\times {10}^{-3}s$

$\Rightarrow t=3$ milliseconds.

Time taken by the bullet to leave barrel is $3$ milliseconds.

$\Rightarrow F.dt=(600-2\times {10}^{5}t)dt$

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

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

$=0.9Ns.$

Hence, the answer is $0.9Ns.$