Finding force without acceleration

AI Thread Summary
The discussion revolves around calculating the average force exerted on a door by bullets fired from a machine gun. For the first scenario, where bullets embed in a wooden door, the average force calculated is 229 N, emphasizing the importance of considering impulse and momentum change. In the second scenario, where bullets rebound elastically from a steel door, the average force remains the same as the first scenario since the mass and final velocity of the bullets do not change. The initial misconception that the door experiences no force due to lack of horizontal acceleration is corrected by understanding the role of impulse. The conversation highlights the significance of applying physics principles correctly to solve real-world problems.
RedDanger
Messages
11
Reaction score
0

Homework Statement


A machine gun fires 22 g bullets horizontally at 260 m/s at a constant rate of
40 bullets/s. (a) If the bullets embed themselves in a thick wooden door, what is the average force
exerted on the door? (b) If the bullets hit a steel door and rebound elastically, what is the average
force exerted on the door?


Homework Equations


F = ma
Vf = Vi + at


The Attempt at a Solution



The door shouldn't experience any force at all, because the bullets are not accelerating horizontally. Is this correct?
 
Physics news on Phys.org
RedDanger said:
The door shouldn't experience any force at all, because the bullets are not accelerating horizontally. Is this correct?
Not at all correct. The bullets are going at some velocity, and then they are stopped by the door right? Thus they have some acceleration.

You should think about this in terms of `impulse' (I)
<br /> I \equiv \Delta mv = F \Delta t<br />
The impulse, which is defined as a change in momentum, is the force applied times the time over which it is applied.
 
Okay, so for the first part I solved for F and got 229N. For the second part, the amount of force exerted on the door should be the same because neither the mass of the bullets nor their final velocity changes due to the type of collision. Is this correct?
 
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