Question

Length of barrel of a gun is l=5.0m and mass of the shell is ${{m}_{0}}=45\text{kg}$. During firing, combustion of the gunpowder takes place at a constant rate $r=2.0\times {{10}^{3}}\,\dfrac{\text{kg}}{\text{s}}$ , the gun-powder is completely transformed into gas of average molar mass $5.0\times {{10}^{-2}}\ \dfrac{\text{kg}}{\text{mol}}$ , temperature T = 1000 K of the gases remains fairly constant and displacement of the shell inside the barrel is proportional to ${{t}^{a}}$ (t is time and a is a constant). Neglecting all forces other than force of the propellant gas during firing, find the muzzle velocity of the shell.

Answer 1

Hint: The internal energy for the monatomic gas is given as:
$\Rightarrow U=\dfrac{3}{2}nRT$
Where, n is the number of the moles of gas, R is the gas constant and T is the temperature.

Complete step-by-step answer:
Given:
Length of the barrel, l = 5.0 m
Mass of the shell, ${{m}_{0}}=45\text{kg}$
Combustion rate, $r=2.0\times {{10}^{3}}\,\dfrac{\text{kg}}{\text{s}}$
Average molar mass, $5.0\times {{10}^{-2}}\ \dfrac{\text{kg}}{\text{mol}}$
Temperature T = 1000 K

Calculation:
The combustion rate is given in the question,
So, the Mass of the powder burnt at time dt is given as,
$\begin{aligned} & \Rightarrow n=\dfrac{r}{M}dt \\\ & \Rightarrow n=\dfrac{2.0\times {{10}^{3}}}{5.0\times {{10}^{-2}}}dt \\\ & \Rightarrow n=0.4\times {{10}^{5}}\times dt\ \text{mole} \\\ \end{aligned}$

The energy of the monatomic gas is given as:
$\begin{aligned} & \Rightarrow U=\dfrac{3}{2}nRT \\\ & \Rightarrow U=\dfrac{3}{2}\times 0.4\times {{10}^{5}}dt\times 1000\times 8.314 \\\ & \Rightarrow U=4.98\times {{10}^{8}}\ dt\ J \\\ \end{aligned}$ … (1)

Now, displacement of the shell inside the barrel is proportional to
$\begin{aligned} & \Rightarrow x\propto {{t}^{a}} \\\ & \Rightarrow x=k{{t}^{a}} \\\ & \Rightarrow \dfrac{x}{dx}=\dfrac{t}{adt} \\\ & \Rightarrow \dfrac{dx}{x}=\dfrac{tdx}{ax} \\\ & \Rightarrow dt=\dfrac{tdx}{ax} \\\ \end{aligned}$

Multiplying both side of equation by we get,
$\Rightarrow 4.98\times {{10}^{8}}\times dt=\dfrac{4.98\times {{10}^{8}}\times tdx}{ax}$ … (2)

Comparing equation (1) and (2):

The total energy transferred is given by,
$\begin{aligned} & \Rightarrow E=\int\limits_{0}^{5}{\dfrac{4.98\times {{10}^{8}}tdx}{ax}} \\\ & \Rightarrow E=\int\limits_{0}^{5}{\dfrac{4.98\times {{10}^{8}}tdx}{5a}} \\\ \end{aligned}$

This energy is balanced by the kinetic energy. So,
$\begin{aligned} & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{4.98\times {{10}^{8}}tdx}{5a} \\\ & \Rightarrow v=\sqrt{\dfrac{2\times 4.98\times {{10}^{8}}tdx}{5am}} \\\ & \Rightarrow v=\sqrt{\dfrac{4.98\times {{10}^{8}}tdx}{45\times 5\times a}} \\\ & \Rightarrow v=0.47\times {{10}^{4}}\sqrt{\dfrac{t}{a}} \\\ \end{aligned}$

Note: The energy of the gas is balanced by the kinetic energy of the shell because the process of the combustion of the gas takes place only when the shell is fired. This combustion of gas converts the gunpowder in the gas.