Find Ratio of vb/vc for Person Homework Statement

  • Thread starter Thread starter lpcampbe
  • Start date Start date
AI Thread Summary
To find the ratio vb/vc, momentum conservation principles must be applied, as kinetic energy is not relevant in this scenario. When the bullet is fired, the momentum of the system (person and bullet) remains constant, leading to the equation m_b * v_b = m_p * v_c, where m_b is the mass of the bullet, v_b is its velocity, m_p is the mass of the person, and v_c is the recoil speed. Given the masses and velocities of both the bullet and the blank cartridge, the ratio of the recoil speeds can be derived from their respective momenta. The mass of the person can be assumed constant, simplifying the calculations. Ultimately, the solution hinges on understanding momentum conservation rather than kinetic energy.
lpcampbe
Messages
15
Reaction score
0

Homework Statement


A person is standing on a sheet of ice so slippery that friction may be ignored. This individual fires a gun parallel to the ground. When a standard cartridge is used , a 17-g bullet is shot forward with a speed of 280 m/s, and the person recoils with a speed of vc. When a blank cartridge is used , a mass of 0.13g is shot forward with a speed of 53 m/s , and the recoil speed is vb. Find the ratio vb/vc.

mass of bullet = 0.017 kg
mass of blank = 1.3e-4 kg

Homework Equations


KE = 1/2mv^2 ?


The Attempt at a Solution


I tried using that equation and solved for KE of the bullet (666.4 J) and the KE of the blank (0.1825 J) but I don't know if that's useful or not. I'm not given the mass of the person, and I don't just want to assume 60 kg. I know I'm supposed to attempt to do the problem before asking for help, but I don't even know where to start.
 
Physics news on Phys.org
Here, remember that momentum is always conserved. Because of this, you can say that the total momentum change is 0.

Also keep in mind that the mass of the man never changes in either situation. Think about this.
 
The kinetic energy (KE) would not be helpful here, becase chemical energy is being converted into kinetic energy when the gun is fired. Momentum, however, is helpful.

Think of the system (man plus pistol plus bullet) splitting in two. The two parts (man plus pistol) and (bullet) will have equal but opposite momenta. Their speeds will therefore be in the inverse ratio of their masses. In practice, the bullet would be so small, that the man and his paraphernalia may be assumed to have the same mass throughout the experiment.
 
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...
Back
Top