Conservation of Momentum of railway gun

Click For Summary
The discussion centers on the conservation of momentum related to the railway gun Gustav, which had a total mass of 1.22 million kg and fired a projectile weighing 7502 kg. The gun moved at 5.21 m/s after firing, leading to calculations for the projectile's muzzle speed. Initial calculations using momentum conservation yielded a speed of 847.27 m/s for the projectile, but the user encountered issues when attempting to break the velocity into components. Further vector addition calculations suggested a speed of 7.74 m/s, which was also incorrect. The thread highlights the complexities of applying conservation of momentum in this scenario.
xdctassonx
Messages
6
Reaction score
0
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!
 
Physics news on Phys.org
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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
9
Views
5K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 19 ·
Replies
19
Views
3K