Question

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

Answer 1

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.