# Conservation of Momentum of railway gun

• xdctassonx
In summary: Using the law of conservation of momentum, we can equate this to the momentum of the gun after firing, which is 7502*v2. Therefore, v2 = 6.36*106/7502 = 847.27 m/s.To find the y component of the projectile's momentum, we can use the trigonometric relationship cos(24.7) = 5.21/y. Solving for y, we get y = 5.73 m/s. Therefore, the total muzzle speed of the projectile is √(5.73^2+5.21^2) = 7.74 m/s. In summary, the largest
xdctassonx
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 24.7° above the horizontal.

a) If the railway gun was at rest before firing and moved to the right at a speed of 5.21 m/s immediately after firing, what was the speed of the projectile as it left the barrel (muzzle speed)? Assume that the wheel axles are frictionless.

b) How far will the projectile travel if air resistance is neglected?

m1*v1=m2*v2

I have attempted this with conservation of momentum.

(1.22*10^6kg)(5.21m/s)=7502(v2)
v2=847.27
I then realized i had to break it up into components. I know that the x component of the projectile would be 5.21 m/s so to find the y component I did this:
cos(24.7)=5.21/y
ycos(24.7)=5.21
y=5.73m/s

v=√(5.73^2+5.21^2)
v=7.74 m/s

This answer was incorrect, I also tried entering it as a negative number which was wrong.
Any help would be appreciated!

xdctassonx said:
I have attempted this with conservation of momentum.

(1.22*10^6kg)(5.21m/s)=7502(v2)
v2=847.27
I then realized i had to break it up into components. I know that the x component of
the projectile would be 5.21 m/s

The x component of the projectile's momentum = 5.21 * 1.22*106

## 1. What is the conservation of momentum principle and how does it apply to railway guns?

The conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant, regardless of any external forces acting on it. In the case of railway guns, this means that the total momentum of the gun and its projectile before and after firing must be equal, in order to conserve momentum.

## 2. How does the mass of the railway gun and its projectile affect the conservation of momentum?

The mass of the railway gun and its projectile play a crucial role in the conservation of momentum. According to the principle, the total momentum must remain constant, so if the mass of the gun increases, the velocity of the projectile must decrease in order to maintain momentum. Similarly, a lighter gun would result in a faster-moving projectile.

## 3. What other factors can affect the conservation of momentum in railway guns?

In addition to mass, there are other factors that can affect the conservation of momentum in railway guns. These include the velocity of the gun, the angle of elevation, and external forces such as air resistance. In order to maintain conservation of momentum, these factors must be taken into account when designing and firing a railway gun.

## 4. Can the conservation of momentum be violated in railway guns?

No, the conservation of momentum is a fundamental law of physics and cannot be violated. If the total momentum before and after firing a railway gun is not equal, then it is likely that there are external forces or factors that were not properly accounted for.

## 5. How is the conservation of momentum used in designing and optimizing railway guns?

The conservation of momentum is a crucial principle in designing and optimizing railway guns. By understanding how mass, velocity, and other factors affect momentum, engineers can make informed decisions about the size, weight, elevation angle, and other aspects of the gun and its projectile. This allows for more efficient and accurate firing of the railway gun.

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