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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 —2 x 105/

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@ gopal

 

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avinash.dongre

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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

Posted by

Safeer PP

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