How Far Will Bullet Push Me Back at 70kg, 5g, 340m/s?

[DardanIsufi]

a bullet travels at 340m/s , my weight is 70kg , the weight of the bullet is 5g , let's assume that the bullet hits me and neater me or the bullet take damage , the speed loss of the bullet is zero , and here is my question : how many meters will the bullet push me back in this case ?
sorry for my English ;)

 

[berkeman]

a bullet travels at 340m/s , my weight is 70kg , the weight of the bullet is 5g , let's assume that the bullet hits me and neater me or the bullet take damage , the speed loss of the bullet is zero , and here is my question : how many meters will the bullet push me back in this case ?
sorry for my English ;)

Welcome to the PF.

Is this question for schoolwork? If so, it needs to be posted in our Introductory Physics Homework forum, and you will fill out the HH Template that you are provided there. :-)

 

[DardanIsufi]

Tank You ;)
It's not for school , I'm just curious to know how it's sloved .

 

[berkeman]

Tank You ;)
It's not for school , I'm just curious to know how it's sloved .

You use Conservation of Momentum for an Inelastic collision. Search down this wikipedia page for Inelastic Collision:

http://en.wikipedia.org/wiki/Conservation_of_momentum#Conservation

:-)

 

[DardanIsufi]

thank you but this is not really what I would like to know , with this formula(m₁v₁=-m₂v₂) I can calculate the speed but not the distance that the bullet would push me back

 

[ellipsis]

It depends on the surface you are on. If you're in space, or on ice (where friction is zero), you would be pushed back indefinitely.

Calculating how long it takes to come to a stop dependent on an initial velocity is a problem in friction, not momentum.

 

[FactChecker]

That is not possible to calculate. The momentum of a bullet on the target is less than the kick of the gun on the shooter. You know that the shooter resists and usually doesn't really get moved backward by the kick of the gun. The same is true of the target, only he is not pushed as much. If you assume that the target doesn't resist, then he gets some velocity (that can be calculated using conservation of momentum) and he would keep moving forever till something stopped him. So that distance can not be calculated unless you assume things about what will stop him.

 

[jbriggs444]

If the change in momentum of the bullet is zero, the change in momentum of the target is also zero. There is no push-back from the impact of the bullet on the target because there is no impact at all.

 

[FactChecker]

If the change in momentum of the bullet is zero, the change in momentum of the target is also zero. There is no push-back from the impact of the bullet on the target because there is no impact at all.

Oh, I missed that: "the speed loss of the bullet is zero". Is that what the OP really meant? I erroneously assumed that the bullet's final speed is zero.

 

[ellipsis]

It is a malformed question.

 

[Loren]

About as far as the person that shot you gets thrown back. For every action there is an equal...

People only get thrown back after being shot in Hollywood, where the laws of physics simply do not work. Other examples would be those loud explosions in space.

However, your problem is weird in that you imply that both the bullet and you are traveling at the same velocity, but it still hits you... My guess is that you must be a Hollywood actor in some film. So, you will end up flying out the bar room door onto the dirt in some Western movie. Am I right?

 

[Vanadium 50]

the speed loss of the bullet is zero

Then you don't move back at all.

 

[Po2]

BMAPhysics 6n .22 caliber bullet being shot into a block of wood. A ballistic pendulum.

or

or
Bullet and Block Inelastic Collision HD -- College, AP, MCAT Physics

 

[DardanIsufi]

thank you for answering , but I think nobody really understands my question ,forget about that "bullet speed loss thing" . I know that if the bullet drills thru my body I wouldn't be pushed back at all , that's why I assumed that neither me or the bullet takes damage in the collision

 

[jbriggs444]

If you already knew that you would not be pushed back at all, why did you ask:

how many meters will the bullet push me back in this case ?

 

[FactChecker]

thank you for answering , but I think nobody really understands my question ,forget about that "bullet speed loss thing" . I know that if the bullet drills thru my body I wouldn't be pushed back at all , that's why I assumed that neither me or the bullet takes damage in the collision

Ok. If the bullet stops in the body, the only thing you can easily calculate is the immediately resulting velocity of the combined body and bullet. Conservation of momentum says that is 340*5/(70,000+5) = 0.0243 m/sec. How far they go and what stops them is a completely separate problem.

 

[Loren]

thank you for answering , but I think nobody really understands my question ,forget about that "bullet speed loss thing" . I know that if the bullet drills thru my body I wouldn't be pushed back at all , that's why I assumed that neither me or the bullet takes damage in the collision

The bullet will deform (they are designed to). Both you and your bullet will be in a higher state of entropy when it is all over.

 

[DardanIsufi]

is my question really so hard to understand or to hard to solve ;) , its very simple we have the bullet and a human , both of them are made by a prefect material that can not be deformed , bullet hits the human with a certain speed while the human is standing on the ground, and we know the weights of the human and the bullet , I just would like to know the distance that the bullet would push the human back if it hits him , ok let's say that there is no air to make it easier , that means that the only force that would stop the human after a certain distance is the friction force between ground an human

 

[Loren]

is my question really so hard to understand

I would say yes. If all these people can't understand your question, then maybe it's how you phrased the question.

or to hard to solve ;) , its very simple we have the bullet and a human , both of them are made by a prefect material that can not be deformed , bullet hits the human with a certain speed while the human is standing on the ground, and we know the weights of the human and the bullet , I just would like to know the distance that the bullet would push the human back if it hits him , ok let's say that there is no air to make it easier , that means that the only force that would stop the human after a certain distance is the friction force between ground an human

is my question really so hard to understand or to hard to solve ;) , its very simple we have the bullet and a human , both of them are made by a prefect material that can not be deformed , bullet hits the human with a certain speed while the human is standing on the ground, and we know the weights of the human and the bullet , I just would like to know the distance that the bullet would push the human back if it hits him , ok let's say that there is no air to make it easier , that means that the only force that would stop the human after a certain distance is the friction force between ground an human

The problem is a simple physics problem. You have not specified whether it is a perfectly elastic collision or not. If it is elastic, then the problem is simply the same physics as two pool balls colliding (one much larger in mass than the other), which follows the conservation of momentum. The final velocities of bullet and person are different in both direction and speed, but the difference in velocity between the body and the bullet is exactly equal to the initial velocity of the bullet.

v'1 = [(m1 - m2) / (m1 + m2) ] • v1 ; v'2 = [(2m1) / (m1 + m2) ] • v1

If the collision is perfectly in-elastic, then the conservation of momentum still holds, but the final velocity of both the person and bullet are equal and in the same direction the bullet was initially traveling, but obviously much, much slower due to the mass of the larger body.

Vf = [m1 / (m1 + m2)] • V_bullet

How simple is that?

Well, not that simple because both examples assume that the person is on a frictionless surface, which is not the way it normally goes.

However, let's apply common sense to the problem before working any numbers. Since the conservation of momentum still applies, the the force of the bullet at its terminal muzzle velocity must equal the same force that the rifle or pistol imparts back onto the shooter. Does the shooter fly backward when they fire off the round? No! There is a slight kick in the pistol or rifle, but the shooter doesn't fly backwards. I was a national competitive high power rifle shooter and I never flew backwards after pulling the trigger.

So, there is no way the target will fly backwards, you will drop if you are stunned or killed or you might fall backwards (you may fall forward, too), you may remain standing, but the force of the bullet is too small to push you backwards (except in Hollywood).

Since you are standing on a surface with friction, the problem is more complex, because your body bends, sways, etc. You are not a solid plank of wood, but again, the net force of the bullet is too weak.

The bottom line is the question is simply not a good question and I hope you can now understand why.

 

[DardanIsufi]

ok , I understand now , thank you

 

[Vanadium 50]

is my question really so hard to understand

Considering that this thread has gone on for 20 posts, the answer is evidently "yes".

 

[jbriggs444]

The problem with making the human and bullet

both of them are made by a prefect material that can not be deformed

is that it turns the problem into the equivalent of an irresistible force meeting an immovable object. It is a contradiction in terms. It turns out that how hard something is (how much force it takes to deform it by a given amount) is not equivalent to how elastic something is (how much of the energy of impact is returned upon rebound). What is needed for this problem, as a number of posters have pointed out, is knowing how elastic the collision is.