A 0.11-g bullet leaves a pistol at 323 m/s, while a similar bullet leaves a rifle at 396 m/s....

Question:

A 0.11-g bullet leaves a pistol at 323 m/s, while a similar bullet leaves a rifle at 396 m/s.

Explain the difference in exit speeds of the two bullets, assuming that the forces exerted on the bullets by the expanding gases have the same magnitude.

Work-Energy.

Suppose we fire a caliber 38 rifle and a caliber 38 pistol. It is clear that the speed of the projectile fired by the rifle is certainly greater than the speed of the same projectile fired by the pistol. The key is the barrel.

Answer and Explanation:

{eq}\text{Known data:}\\ v_i = 0\\ v_p = 323\,m/s\\ v_r = 396\,m/s\\ {/eq}

The work performed by the expanding gases on the bullet as it moves through the barrel of a firearm is defined by the following expression:

{eq}W = \int_0^l {{F_{\left( l \right)}}dl} {/eq}

The transfer of energy to the bullet by virtue of the force of the gases is a function of the length of the barrel (l).

On the other hand, the work-energy theorem establishes that the net work performed by the gases is equal to the kinetic energy change of the bullet.

{eq}W = \frac{1}{2}m\left( {{v^2} - {v_i}^2} \right)\,\,\,\,\,\,\,\,{v_i} = 0 \\ W = \frac{1}{2}m\left( {{v^2} - {0^2}} \right)\, \\ v = \sqrt {\dfrac{{2W}}{m}} {/eq}

The speed of the bullet increases with the work done by the gases, and the work increases with the length of the barrel of the firearm.

Therefore, if a bullet is fired from a rifle (long barrel), it will come out at a higher speed than if it is fired from a pistol (short barrel).


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