Momentum and impulse of a bullet

Click For Summary

Homework Help Overview

The problem involves a bullet fired from a gun, focusing on the concepts of momentum and impulse. The bullet has a specified mass and velocity, and the scenario includes the gun being brought to rest by a force that varies over time.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of momentum and the relationship between impulse and force. Questions arise regarding the interpretation of the force that rises and falls uniformly, and whether the average force can be used in the context of non-constant acceleration.

Discussion Status

There is an ongoing exploration of the problem, with participants offering different interpretations and calculations. Some guidance has been provided regarding the average force, but there is no explicit consensus on the approach or the correctness of the reasoning presented.

Contextual Notes

Participants question the implications of the problem's phrasing, particularly regarding the uniform rise and fall of force, and whether the average deceleration can be applied in this scenario. There is uncertainty about the assumptions made in the calculations.

mattgad
Messages
15
Reaction score
0

Homework Statement



A bullet of mass 0.03kg is fired from a gun with a horizontal velocity of 400 ms^-1.
Find the momentum of the bullet after it is fired. If the gun is then brought to rest in 1.2s by a horizontal force which rises uniformly from zero to B N and then falls uniformly to zero, find the value of B.

Homework Equations



Impulse = force * time.

The Attempt at a Solution



Momentum of the bullet = 0.03 * 400 = 12 kg ms^-1, taking the direction of the bullet to be positive. So momentum of the gun is also 12 kg ms^-1.

Now, the problem, my friends have all jumped in and said as I = ft, then 12 = F*1.2, so F = 10 N. Why is this? How have we gone from the information given about rising uniformly to B and then back again, to just using that equation? Also, why is the impulse just the same as the momentum?

Also, I was thinking, don't we need to double this? Is 10 N not the force to rise up, and the same is needed to come back down?

Hope this post makes sense, thanks for help.
 
Physics news on Phys.org
I'm not sure if I am correct or not, but I'm giving my input.:smile:

The deceleration of the gun is not constant. It increases to a certain max value and then decreases back to zero.
If that max value is 'B' Newtons, I guess the average force over the time interval is \frac{B}{2} Newton.

avg. deceleration of gun = \frac{B}{2m_{gun}}

v=u+at
0=\frac{12}{m_{gun}}-\frac{B}{2m_{gun}}t

which gives B=20 N.

But I doubt if I can use the average deceleration of gun like I did ? :rolleyes:
__________________________
I may be wrong
 
Thanks for your input, although I thought you could only use v = u + at when acceleration is constant.

Anyone else?
 
What does "rises uniformly" and "falls uniformly" mean? I'm guessing it means the rate of change of this horizontal force F is constant for some interval of time. In other words, for [0, t'], dF/dt = c and for (t', 1.2], dF/dt = c' where c and c' are constants. Is c = c'? Is t' = 0.6?
 
I would assume that too, but I'm not too sure. I'm not sure how to proceed with that.
 

Similar threads

Replies
7
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 28 ·
Replies
28
Views
9K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
Replies
7
Views
2K