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Solution
We are given that the combustion of gunpowder takes place at a constant rate $r$, such that the mass of the gas $m_g$ at time $t$ is $m_g = rt$. The temperature $T$ is constant.
Using the Ideal Gas Law $PV = nRT$, where $n = \frac{rt}{M}$ and $V = Ax$ ($A$ is the cross-sectional area):
$$ P(Ax) = \left( \frac{rt}{M} \right) RT \implies P = \frac{rRT \cdot t}{MAx} $$
The force acting on the shell is $F = PA$. By Newton’s Second Law ($F = m_0 a$):
$$ m_0 \frac{d^2x}{dt^2} = \left( \frac{rRT \cdot t}{MAx} \right) A = \frac{rRT}{M} \frac{t}{x} $$
We are given that displacement is proportional to $t^\alpha$. Let $x = C t^\alpha$.
Velocity $v = \frac{dx}{dt} = C \alpha t^{\alpha-1}$
Acceleration $a = \frac{d^2x}{dt^2} = C \alpha (\alpha-1) t^{\alpha-2}$
Substituting $x$ and $a$ into the equation of motion:
$$ m_0 [C \alpha (\alpha-1) t^{\alpha-2}] = \frac{rRT}{M} \frac{t}{C t^\alpha} $$
$$ m_0 C \alpha (\alpha-1) t^{\alpha-2} = \frac{rRT}{MC} t^{1-\alpha} $$
Comparing powers of $t$: $\alpha – 2 = 1 – \alpha \implies 2\alpha = 3 \implies \alpha = \frac{3}{2}$.
Comparing coefficients with $\alpha = 3/2$:
$$ m_0 C \left(\frac{3}{2}\right) \left(\frac{1}{2}\right) = \frac{rRT}{MC} $$
$$ C^2 = \frac{4 rRT}{3 m_0 M} \implies C = \left( \frac{4 rRT}{3 m_0 M} \right)^{1/2} $$
Now, we find the velocity at the end of the barrel ($x = l$).
$$ v = \frac{dx}{dt} = \frac{3}{2} C t^{1/2} $$
From $x = C t^{3/2} = l$, we have $t^{1/2} = (l/C)^{1/3}$. Substituting this into the velocity expression:
$$ v = \frac{3}{2} C \left( \frac{l}{C} \right)^{1/3} = \frac{3}{2} C^{2/3} l^{1/3} $$
$$ v = \frac{3}{2} \left( \frac{4 rRT}{3 m_0 M} \right)^{1/3} l^{1/3} = \frac{3}{2} \left( \frac{4 rRT l}{3 m_0 M} \right)^{1/3} $$
Numerical Substitution:
Values: $l = 5.0\, \text{m}, m_0 = 45\, \text{kg}, r = 2.0 \times 10^3\, \text{kg/s}, M = 5.0 \times 10^{-2}\, \text{kg/mol}, T = 1000\, \text{K}, R \approx 8.314\, \text{J/mol K}$.
$$ v = \frac{3}{2} \left( \frac{4 (2000)(8.314)(1000)(5)}{3(45)(0.05)} \right)^{1/3} $$
$$ v = 1.5 \left( \frac{332,560,000}{6.75} \right)^{1/3} \approx 1.5 (49,268,148)^{1/3} $$
$$ v \approx 1.5 \times 366.6 \approx 550\, \text{m/s} $$