Question

To a close approximation, the pressure behind a rifle bullet varies inversely with the position x of the bullet along the barrel. Thus the acceleration of the bullet may be written as a=k/x where k is a constant. If the bullet starts from rest at x=7.5 mm and if the muzzle velocity of the bullet is 600 m/s at the end of the 750mm barrel, compute the acceleration of the bullet as it pass the midpoint of the barrel at x=375mm

Answer 1

We use the definition for the acceleration to solve the equation and find the value for k:

$a=\cfrac{dv}{dt}=\cfrac{dv}{dx}\cfrac{dx}{dt}=v\cfrac{dv}{dx}=\cfrac{k}{x} \\ \cfrac{k}{x}=v\cfrac{dv}{dx} \implies k \int{\cfrac{dx}{x}}= \int{v}{dv}$

We have to use the limits for the integration to calculate k:

$k \int_{x_1=7.5\,mm}^{x_2=750\,mm}{\frac{dx}{x}}= \int_{v_1=0\,m/s}^{v_2=600\,m/s} {{v}{dv}} \\ \implies k \Big[{\ln{x}}\Big] _{x_1=7.5\,mm}^{x_2=750\,mm}= \frac{1}{2}\Big[v^2 \Big]_{v_1=0\,m/s}^{v_2=600\,m/s} \\ \text{} \\ \therefore k= \dfrac{ (600\,m/s)^2-(0\,m/s)^2}{2\ln(\frac{750\,mm}{7.5\,mm})} \\ \text{ } \\ k=39086.50337\, \frac{m^2}{s^2}$

Now we substitute the value for k and x = 375 mm = 0.375 m to find the acceleration of the bullet at that point:

$a=\cfrac{k}{x}=\cfrac{39086.50337\, \frac{m^2}{s^2}}{0.375\,m} \\ \therefore a_{x=375\,mm}=104230.676 \cfrac{m}{s^2}$

In conclusion, the acceleration of the bullet when it passesthe midpoint of the barrel is a = 1.04231 X 10⁵ m/s².

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.