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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. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

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

when it leaves the barrel at time t=t_{0}, force becomes 0

so, 0 = 600 -2\times 10^{5}t

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

Impulse \ imparted =\int_{0}^{t_{0}}F.dt=\int_{0}^{t_{0}}(600-2\times 10^{5}t)dt

Impulse \ imparted = 600t_{0}-\frac{2\times 10^{5}t_{0}^{2}}{2}=0.9\ Ns

Posted by

Vishwash Tetarwal

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