Cannon Question: Loose Cannon Velocity

  • Thread starter Thread starter JessGold
  • Start date Start date
  • Tags Tags
    Cannon
AI Thread Summary
The discussion focuses on a physics problem involving a cannon and the conservation of momentum and energy. When the cannon is bolted to the Earth, it fires a shell with a specific velocity, and the kinetic energy imparted to the system is considered. Once unbolted, the cannon can recoil freely, affecting the velocity of the fired shell. The key point is that the same kinetic energy is imparted to the system in both scenarios, which allows for the calculation of the shell's velocity when the cannon is loose. Understanding these conservation laws is crucial for solving the problem accurately.
JessGold
Messages
1
Reaction score
0

Homework Statement


A cannon of mass 5.80 x 10^3 kg is rigidly bolted to the Earth so it can recoil only by a negligible amount. The cannon fires an 85.0-kg shell horizontally with an initial velocity of +551 m/s. Suppose the cannon is then unbolted from the Earth and no external force hinders its recoil. What would be the velocity of a shell fired by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Two conservation laws apply here. One of them is already implicitly mentioned in the hint. The other one has to do with the masses and velocities.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top