# Momentum and impulse of a bullet

In summary, the conversation discusses finding the momentum of a bullet after being fired from a gun and the value of the force required to bring the gun to rest in 1.2 seconds. The equation for impulse is used, but there is confusion about the use of the equation and the concept of uniform rising and falling forces. The equation v=u+at is also used, but there is uncertainty about its applicability in this situation. The concept of uniform rising and falling forces is discussed, but there is no clear conclusion.

## 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.

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

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.

## What is momentum and impulse?

Momentum is a measure of an object's motion, calculated by multiplying its mass by its velocity. Impulse is the change in an object's momentum over time.

## How is momentum and impulse related to a bullet?

When a bullet is fired from a gun, it has a certain amount of momentum. The force of the gunpowder explosion creates an impulse that accelerates the bullet to a high velocity, giving it a large amount of momentum.

## What factors affect the momentum of a bullet?

The mass and velocity of the bullet are the main factors that affect its momentum. A heavier bullet or a faster bullet will have a greater momentum. The direction of the bullet's motion also affects its momentum.

## What is the conservation of momentum and how does it apply to a bullet?

The law of conservation of momentum states that in a closed system, the total momentum remains constant. This means that the momentum of a bullet before it is fired is equal to the momentum of the bullet and gun after it is fired. This also applies to a bullet hitting a target, as the momentum of the bullet is transferred to the target.

## How does the momentum and impulse of a bullet affect its penetration power?

The momentum and impulse of a bullet play a significant role in its penetration power. A bullet with a higher momentum and impulse will have a greater force upon impact, allowing it to penetrate deeper into a target. This is why larger, heavier bullets are often used for hunting and self-defense purposes.

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