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(A) The force on the bullet is given by $F = 600 - 2 	imes 10^5 t$. When the bullet leaves the barrel,the force becomes zero. Setting $F = 0$,we get

Vedclass · Accepted Answer

$0.9$

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(A) The force on the bullet is given by $F = 600 - 2 	imes 10^5 t$. When the bullet leaves the barrel,the force becomes zero. Setting $F = 0$,we get $600 - 2 	imes 10^5 t = 0$. Solving for $t$,$t = \frac{600}{2 	imes 10^5} = 3 	imes 10^{-3} \ s$. Impulse $I$ is defined as the integral of force over time: $I = \int_{0}^{t} F \ dt$. Substituting the values: $I = \int_{0}^{3 	imes 10^{-3}} (600 - 2 	imes 10^5 t) \ dt$. Integrating,we get $I = [600t - 10^5 t^2]_0^{3 	imes 10^{-3}}$. Calculating the value: $I = 600(3 	imes 10^{-3}) - 10^5(3 	imes 10^{-3})^2$. $I = 1.8 - 10^5(9 	imes 10^{-6}) = 1.8 - 0.9 = 0.9 \ N-s$.

$A$ 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 $t$ is in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the impulse imparted to the bullet (in $N-s$)?

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