Question

A gun applies a force F on a bullet which is given by$F=\left( 100-0.5\times {{10}^{5}}t \right)N$. The bullet emerges out with speed 400m/s. Then find out the impulses exerted till force on the bullet becomes zero.
$\begin{aligned} & A.\text{ 0}\text{.2 N}\text{.s} \\\ & B.\text{ 0}\text{.3 N}\text{.s} \\\ & C.\text{ 0}\text{.1 N}\text{.s} \\\ & \text{D}\text{. 0}\text{.4 N}\text{.s} \\\ \end{aligned}$

Answer 1

We know that, when a bullet is fired from the gun, both moves in opposite direction because of the force act on bullet by gun which is given by, $F=\left( 100-0.5\times {{10}^{5}} \right)t$. Now to calculate the impulse, use Newton's second law which will give the relation of the force and time with impulse. First calculate period, using F = 0 conditions then using integration, over a period, calculate impulse.

Formula used: $I=\int{Fdt}$

Complete step by step answer:
We have been provided with a gun which is applying force on the bullet. And the force is given by,
$F=\left( 100-0.5\times {{10}^{5}}t \right)N........\left( 1 \right)$
Now we need to find out the impulse exerted on the bullet. We have been provided with the condition that impulse is exerting till force becomes zero that is force (F) =0.
Put F =0 in equation (1), we get
$\begin{aligned} & 0=100-0.5\times {{10}^{5}}t \\\ & 100=0.5\times {{10}^{5}}t \\\ & t=\dfrac{100}{0.5\times {{10}^{5}}} \\\ & t=2\times {{10}^{-3}}\sec .......\left( 2 \right) \\\ \end{aligned}$
We know that impulse is one of the types of force which measures or occurs for very small duration or time. So, according to newton’s second law,
$I=\int{Fdt......\left( 3 \right)}$
Where, I = impulse of bullet,
F = force exerted on bullet
t = time.
Now, we know initially, bullet time is zero and its period is 0 to$2\times {{10}^{-3}}\sec$.
This is the period over which impulse acts on the bullet. Put value in equation (3) from equation (1) and equation (2).
$\begin{aligned} & I=\int\limits_{0}^{2\times {{10}^{-3}}}{\left( 100-0.5\times {{10}^{5}}t \right)}dt \\\ & I=\int\limits_{0}^{2\times {{10}^{-3}}}{100dt-\int\limits_{0}^{2\times {{10}^{-3}}}{0.5\times {{10}^{5}}t\text{ dt}}} \\\ & I={{\left[ 100t \right]}_{0}}^{2\times {{10}^{-3}}}-{{\left[ \dfrac{{{10}^{5}}}{2}\times \dfrac{{{t}^{2}}}{2} \right]}_{0}}^{2\times {{10}^{-3}}} \\\ \end{aligned}$
Put limits, we get
$\begin{aligned} & I=\left[ 100\times 2\times {{10}^{-3}}-\dfrac{{{10}^{5}}}{4}\times 4\times {{10}^{-6}} \right] \\\ & I=2\times {{10}^{-1}}-{{10}^{-1}} \\\ & I={{10}^{-1}}Ns=0.1Ns \\\ \end{aligned}$
Hence, the impulses exerted till force on the bullet becomes zero is ${{10}^{-1}}Ns$ .

So, the correct answer is “Option C”.

Additional Information: In case of a ball hit by a bat or driving a nail into a wooden block with a hammer, the force acts for a short time internally and you might have experienced it. This force varies with time, hence cannot be measured so easily. In such cases, we measure the total effect of the force which is called the impulse of force.
Impulse of the force or impulse is denoted by J. The S.I unit of the impulse is the same as that of the linear momentum i.e., kgm/s or Ns. Dimensions of impulse are$\left[ {{L}^{1}}{{M}^{1}}{{T}^{-1}} \right]$.

Note: Another formula or expression of impulse is equal to change in linear momentum of the body which is given by $ft=mv-mu$. Impulse is a vector quantity. Direction of impulse is always the same as that of the linear momentum. When a bullet is fired from the gun, the gun shows re-collation. As we know guns and bullets move in exactly opposite directions and in this case linear momentum of gun and bullet.