Impulse with a triangular force profile on a gun

[mattgad]

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.

 

[Google_Spider]

I'm not sure if I am correct or not, but I'm giving my input.

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 ?
__________________________
I may be wrong

 

[mattgad]

Thanks for your input, although I thought you could only use v = u + at when acceleration is constant.

Anyone else?

 

[e(ho0n3]

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?

 

[mattgad]

I would assume that too, but I'm not too sure. I'm not sure how to proceed with that.