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 impulse imparted to a bullet?
(A) 9N-s
(B) 0
(C) 0.9 N-s
(D) 1.8 N-s

Hint: Impulse is the product of force acts and it is equal to the total change in the momentum.
i.e. \[\vec I = \vec F \bullet t\]
Impulse is a vector quantity, its direction being the same as that of the given force. Units are dyne-second (dyn s) in c.g.s. system and newton-second in SI.
Its dimensional formula is \[[ML{T^{ - 1}}]\].
If the force acts for time t during which momentum of the body changes from \[{\vec p_1}{\text{ to }}{\vec p_2}\] then impulse I is denoted by:
\[I = \int\limits_0^t {\vec Fdt} \]

Complete step by step answer:
Force acting on the bullet is given as
\[\Rightarrow F = 600 - 2 \times {10^5}t\]
\[ \Rightarrow t = \dfrac{{600}}{{2 \times {{10}^5}}}\]
\[\Rightarrow t = 3 \times {10^{ - 3}}s\]
Now impulse imparted by the bullet is \[I = \int\limits_0^t {\vec Fdt} \]
So, we get
\[ \Rightarrow I = |600t - \dfrac{{2 \times {{10}^5}{t^2}}}{2}|_0^{3 \times {{10}^{ - 3}}}\]
\[\therefore I = 0.9Ns\]

Hence, Option (C) is the correct answer.

Note: Alternative approach
Impulse is measured by the total change in momentum that the force produces in a given time. According to Newton’s second law of motion
\[\vec F = \dfrac{{d\vec p}}{{dt}}\]
\[ \Rightarrow d\vec p = \vec Fdt\]
If the force acts for time t during which momentum of the body charges from \[{\vec p_1}{\text{ to }}{\vec p_2}\], then
\[\int\limits_0^t {\vec Fdt = \int\limits_{{p_1}}^{{p_2}} {d\vec p} } = {\vec p_2} - {\vec p_1}\]
\[ \Rightarrow \vec I = \vec F\int\limits_0^t {dt} = {\vec p_2} - {\vec p_1}\]
\[\therefore \vec I = \vec Ft = {\vec p_2} - {\vec p_1}\]
Hence the above equation represents Impulse-Momentum Theorem, states that a given change in momentum can be produced either by applying small force for large time or large force for the small time.