The largest railway gun ever built was called Gustav and was used briefly in World War II. The gun, mount, and train car had a total mass of 1.22 106 kg. The gun fired a projectile that was 80.0 cm in diameter and weighed 7502 kg. In the firing illustrated in the figure, the gun has been elevated at è = 18.2° above the horizontal. If the railway gun was at rest before firing and moved to the right at a speed v = 4.27 m/s immediately after firing, what was the speed of the projectile as it left the barrel (muzzle velocity)?
To find the muzzle velocity of the projectile, we can use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before firing is equal to the total momentum after firing.
Before firing, the gun and the projectile are at rest, so the total momentum is zero. After firing, the gun moves to the right with a speed of 4.27 m/s.
Let's assume the muzzle velocity of the projectile is v_p. The mass of the gun, mount, and train car is 1.22 × 10^6 kg, and the mass of the projectile is 7502 kg.
Using the conservation of momentum equation:
(0) = (1.22 × 10^6 kg) × (4.27 m/s) + (7502 kg) × (v_p)
Simplifying the equation:
0 = 5.2134 × 10^6 kg·m/s + 7502 kg·m/s × v_p
To solve for v_p, we isolate it on one side of the equation:
-5.2134 × 10^6 kg·m/s = 7502 kg·m/s × v_p
Dividing both sides by 7502 kg·m/s:
v_p = -5.2134 × 10^6 kg·m/s / 7502 kg·m/s
v_p = -695.963 m/s
The negative sign indicates that the projectile was fired in the opposite direction of the gun's motion (since the gun moves to the right, the projectile moves to the left).
Therefore, the speed of the projectile as it left the barrel (muzzle velocity) is approximately 695.963 m/s in the opposite direction of the gun's motion.