Calculating Recoil Velocity: Female Character in Diamonds Are Forever

  • Thread starter Thread starter pdubb
  • Start date Start date
  • Tags Tags
    Velocity
AI Thread Summary
The discussion focuses on calculating the recoil velocity of a female character in Diamonds Are Forever after firing a gun. For the first scenario, using the bullet's mass of 0.013 kg and its velocity of +702 m/s, the recoil velocity is calculated as -0.19 m/s. In the second scenario, where a blank cartridge ejects a mass of 5.0 x 10^-4 kg at the same velocity, the recoil velocity can be determined using the same formula. The key takeaway is that the method for calculating recoil remains consistent, with only the mass of the bullet changing. Understanding this principle is essential for solving both parts of the problem accurately.
pdubb
Messages
6
Reaction score
0
The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires the gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is 0.013 kg, and its velocity is +702 m/s. Her mass (including the gun) is 48 kg.

(a) What recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place?

(b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of 5.0 10-4 kg at a velocity of +702 m/s?

i know that you take (mass of the bulllet) x (velocity of the bullet) / (mass of the girl) to find the girl's velocity, and it equals -.19 for the first one but i don't know how to do the second part, or if it should also be a negative number.

any help would be appreciated!
 
Physics news on Phys.org
The only thing that changes is the mass of the bullet. So figure her velocity in exactly the same way for both parts.
 
that works! thanks!
 
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