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1 week, 4 days ago

# A bullet is fired from a gun The force is given by

A bullet is fired from a gun. The force is given by $F=600-2\times10^{5}t$. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

F 600 â€”2 x 105/
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S safeer
Answered 1 week, 4 days ago

We have given,                 $F=600-2*10^5tJ$

At the bullet leaves the barrel, the force on the bullet becomes zero.                 So,

$F=600-2*10^5t=0$

$t=600/2*10^5=3*10^{-3}s$

Then, average impulse imparted to the bullet

$I=\int_{0}^{t}Fdt\Rightarrow \int_{0}^{3*10^{-3}} (600-2*10^5t)dt$

$=\int_{0}^{3*10^{-3}}[600t-(2*10^5t^2/2)] \Rightarrow I=600*3*10^{-3}-10^5*(3*10^{-3})^2 = 1.8 - 0.9 = 0.9 Ns$

A Avinash Dongre
Answered 1 week, 4 days ago

@ gopal

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