Conservation of Momentum of railway gun

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SUMMARY

The discussion centers on the conservation of momentum as applied to the largest railway gun, Gustav, used in World War II. The gun, with a total mass of 1.22·106 kg, fires a projectile weighing 7502 kg. The initial calculations for the projectile's speed upon leaving the barrel were performed using the formula m1*v1=m2*v2, yielding a muzzle speed of 847.27 m/s. Further analysis involved breaking down the velocity into components, leading to a total velocity of 7.74 m/s, which was later identified as incorrect.

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xdctassonx
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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
I then used vector addition,

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!
 
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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 ✗[/size][/color]

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

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