A bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 imes 10^{5}t$ 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 in pulse imparated to the bullet?

Question

A bullet is fired from a gun. The force on the bullet is given by $F = 600 - 2 	imes 10^{5}t$ 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 in pulse imparated to the bullet?

Dr. Abubacker · Accepted Answer

(c) $Given\ ,\ F = 600 - 2 	imes 10^{5}\ t = 0$
$\Longrightarrow \ \ t = 3 	imes 10^{- 3}\ s$
$Impulese\ \ I = \int_{0}^{t}{F \bullet dt}$
$= \int_{0}^{3 	imes 10^{- 3}}{\left( 600 - 2 	imes 10^{3}\ t ight)dt}$
$= \lbrack 600\ t - 10^{5}\ t^{2}brack_{0}^{3 	imes 10^{- 3}} = 0.9\ Ns$

Q. 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 in pulse imparated to the bullet?

Explanation

Reviewed by Dr. Abubacker

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