How Much Force Exerts on a Bullet in a Rifle Barrel?

Thread starter SailorMoon14
Start date Sep 20, 2006
Tags

Bullet Force

AI Thread Summary

To calculate the force exerted on a bullet in a rifle barrel, the mass of the bullet (4.5 g) and its exit speed (318 m/s) are essential. The force can be determined using the formula F = ma, where 'm' is the mass and 'a' is the acceleration, which can be derived from the change in velocity over the distance of the barrel (0.75 m). The work done on the bullet, represented as force times distance, also plays a crucial role in understanding the bullet's motion. Ultimately, the discussion emphasizes the relationship between force, distance, and the bullet's acceleration within the rifle barrel. Understanding these dynamics is key to analyzing bullet behavior in firearms.

Sep 20, 2006

SailorMoon14

A 4.5 g bullet leaves the muzzle of a rifle with a speed of 318 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.75 m long barrel of the rifle? The answer is in Newtons.

Thanks for any input :)

Last edited: Sep 20, 2006

Physics news on Phys.org

Sep 21, 2006

Galileo

Science Advisor

Homework Helper

What do you know about the bullet after a constant has force as acted upon it over a given distance? (force times distance = ?)

Thread 'Errors and significant figures'

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...

View full post »

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...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »