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(C) The force applied on the bullet is given by $F(t) = 100 - 0.5 	imes 10^{5} t$. First,we find the time $t$ when the force becomes zero: $100 - 0.5

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$0.1$

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(C) The force applied on the bullet is given by $F(t) = 100 - 0.5 	imes 10^{5} t$. First,we find the time $t$ when the force becomes zero: $100 - 0.5 	imes 10^{5} t = 0$ $0.5 	imes 10^{5} t = 100$ $t = \frac{100}{0.5 	imes 10^{5}} = 2 	imes 10^{-3} \ s$. Impulse $I$ is defined as the integral of force over time: $I = \int_{0}^{t} F dt = \int_{0}^{2 	imes 10^{-3}} (100 - 0.5 	imes 10^{5} t) dt$. Integrating the expression: $I = [100t - \frac{0.5 	imes 10^{5} t^{2}}{2}]_{0}^{2 	imes 10^{-3}}$. Substituting the value of $t$: $I = 100(2 	imes 10^{-3}) - 0.25 	imes 10^{5} (2 	imes 10^{-3})^{2}$ $I = 0.2 - 0.25 	imes 10^{5} (4 	imes 10^{-6})$ $I = 0.2 - 0.25 	imes 0.4$ $I = 0.2 - 0.1 = 0.1 \ N \cdot s$.

$A$ gun applies a force $F$ on a bullet which is given by $F = (100 - 0.5 \times 10^{5} t) \ N$. The bullet emerges out with a speed of $400 \ m/s$. Find the impulse exerted until the force on the bullet becomes zero (in $N \cdot s$).

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