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 = 18.0° above the horizontal. If the railway gun was at rest before firing and moved to the right at a speed v = 4.53 m/s immediately after firing, what was the speed of the projectile as it left the barrel (muzzle velocity)?
it is not clear to me the figure, but conservation of momentum in the horizontal direction will work this out.
Use the equation M1V1=M2V2 where M1V1 is the momentum of the rail gun and M2V2 is the mass of the bullet and the horizontal component of its speed.
To answer this question, we can use the principle of conservation of momentum. In this case, we need to consider the total momentum before and after the firing.
Let's start by finding the initial momentum of the gun, mount, and train car system. The initial momentum is given by the product of the total mass and the initial velocity, which is zero since the railway gun was at rest before firing.
Initial momentum = 1.22 × 10^6 kg × 0 m/s = 0 kg·m/s
Now, let's find the final momentum of the gun, mount, and train car system. The final momentum is given by the product of the total mass and the final velocity, which is the velocity of the projectile immediately after firing.
Final momentum = 1.22 × 10^6 kg × 4.53 m/s = 5.5266 × 10^6 kg·m/s
According to the principle of conservation of momentum, the total momentum before and after the firing must be equal. Therefore,
Initial momentum = Final momentum
0 kg·m/s = 5.5266 × 10^6 kg·m/s
Now, let's consider the momentum of the projectile. The momentum of the projectile is given by the product of its mass and velocity.
Momentum of the projectile = mass of the projectile × velocity of the projectile
We need to find the velocity of the projectile, which is the muzzle velocity. But first, we need to find the mass of the projectile.
Mass of the projectile = volume of the projectile × density of the projectile
The volume of the projectile can be calculated using its diameter.
Volume of the projectile = (4/3) × π × (diameter/2)^3
Volume of the projectile = (4/3) × π × (80.0 cm/2)^3 = 268084.777 cm^3
Now, let's convert the volume and density from cm^3 to kg using the appropriate conversion factors.
Volume of the projectile = 268084.777 cm^3 × (1 m/100 cm)^3 = 0.268084777 m^3
Density of the projectile = 7502 kg / (0.268084777 m^3) = 27973.7897 kg/m^3
Mass of the projectile = 0.268084777 m^3 × 27973.7897 kg/m^3 = 7494.459 kg
Finally, we can now calculate the muzzle velocity of the projectile.
0 kg·m/s = 7494.459 kg × muzzle velocity
muzzle velocity = 0 kg·m/s / 7494.459 kg
muzzle velocity = 0 m/s
Therefore, the muzzle velocity of the projectile as it left the barrel is 0 m/s.