Momentum and impulse of a bullet

AI Thread Summary
The momentum of the bullet after being fired is calculated as 12 kg m/s, derived from its mass of 0.03 kg and velocity of 400 m/s. The discussion revolves around determining the force B needed to bring the gun to rest in 1.2 seconds, with some participants suggesting that the average force is simply F = 10 N, derived from impulse equations. However, the deceleration of the gun is not constant, leading to the conclusion that the average force over the time interval is actually B/2, resulting in B being calculated as 20 N. There is confusion regarding the implications of "rises uniformly" and "falls uniformly," with participants questioning whether the average deceleration can be applied in this scenario. The conversation highlights the complexities of applying kinematic equations when acceleration is not constant.
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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.
 
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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.
 
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