When a bullet is fired from the back of train

[PainterGuy]

TL;DR Summary

What would happen when a bullet is fired from the back of train which is moving as fast as the bullet?

Hi,

I was reading this article, https://science.howstuffworks.com/question456.htm, about firing a bullet from the back of train which is moving as fast as the bullet.

The following is taken from the article mentioned above.

If you shoot the bullet off the back of the train, the bullet will still be moving away from you and the gun at 1,000 mph, but now the speed of the train will subtract from the speed of the bullet. Relative to the ground, the bullet will not be moving at all, and it will drop straight to the ground.

So, ideally, the bullet will fall straight to the ground as it exits the barrel. Would a person firing the bullet feel any recoil in such a scenario? I don't think there'd be any recoil. Could you please confirm this? Thank you!

 

[kuruman]

Yes, the person will feel a recoil because the gun does its thing without knowing that it's on the train.
Yes, the bullet will fall on the ground as you said.

 

[hutchphd]

Yes he would feel recoil. The bullet in the gun must receive impulse to slow it to zero.
This is also why a rocket engine with exhaust velocity of 1000 ft/s can accelerate a rocket to 7 miles/s.

 

[Ibix]

Mythbusters practical demo:

 

[kuruman]

It works in reverse too, sort of.

 

[PeroK]

I don't think there'd be any recoil. Could you please confirm this? Thank you!

Newton's second law tells you the force required to accelerate the bullet (or decelerate in the ground frame):
$$ F = ma$$
Newton's third law tells you there is an equal and opposite recoil force.

 

[A.T.]

I don't think there'd be any recoil.

You should make yourself familiar with Galilean invariance. It's a very important principle that answers such questions immediately:
https://en.wikipedia.org/wiki/Galilean_invariance

 

[vanhees71]

And it's very important to clearly distinguish the frames.

If I understand it right, you consider a train moving with constant velocity $\vec{v}=v \vec{e}_x$ wrt. Earth. Now we consider an inertial frame of reference $(t,x,y,z)$ (restframe of the Earth) and another one $(t',x',y',z')$ as the restframe of the train. Then any motion can be described in both frames, and the transformation law between the coordinates, the Galileo transformation, is
$$t=t', \quad \vec{x}=\vec{x}'+\vec{v} t,$$
where I have assumed that the frames coincide at time $t=0$, and we use the same basis vectors in both frames, $\vec{e}_x'=\vec{e}_x$,...

The equation of motion of your projectile (neglecting somewhat very unrealistic air resistance) in the Earth frame of reference is given by
$$m \ddot{\vec{x}}=-g \vec{e}_z.$$
Using the Galileo transformation you get
$$m \ddot{\vec{x}}'=-g \vec{e}_z.$$
Now consider the case where the projectile, as seen from the train frame, is shot with initial velocity $\vec{u}_0':=\dot{x}'(0)=-\vec{v}$ at $t=0$ from the rear of the train which at this moment is at $\vec{x}=\vec{x}'=0$.

Then the solution of the equations of motion reads
$$\vec{x}'(t)=-\frac{g}{2} t^2 \vec{e}_x-\vec{v} t.$$
To get the solution you can either directly use the Galileo transformation directly
$$\vec{x}(t)=\vec{x}'(t)+\vec{v} t=-\frac{g}{2} t^2 \vec{e}_z$$
or you transform the initial condition to the Earth frame first and then solve the equations of motion.

Of course we have the transformation property for the velocities immediately from the Galileo transformation given above, i.e.,
$$\vec{u}(t)=\dot{\vec{x}}(t)=\frac{\mathrm{d}}{\mathrm{d} t} [\vec{x}'(t)+\vec{v} t]=\vec{u}'(t)+\vec{v}.$$
Thus the initial conditions in the Earth frame are $\vec{x}(0)=0$, \quad $\vec{u}_0=\vec{u}_0'+\vec{v}=0$, and solving the equations of motion we again get
$$\vec{x}(t)=-\frac{g}{2} t^2 \vec{e}_z.$$
That both ways to calculate the motion of the projectile in the Earth frame give the same result is, of course, an example for the Galileo invariance of Newtonian mechanics.

 

[A.T.]

I don't think there'd be any recoil.

You should make yourself familiar with Galilean invariance. It's a very important principle that answers such questions immediately:
https://en.wikipedia.org/wiki/Galilean_invariance

Just to clarify: By immediate answer I don't mean some derivation like the one posted above. I mean a simple argument like:

If the gun has a recoil when at rest relative to the ground, then it has the same recoil when moving at constant velocity relative to the ground.

This is based on the approximation that the local ground can be considered inertial for the spatial and temporal extend of the experiment.

 

[hmmm27]

So, ideally, the bullet will fall straight to the ground as it exits the barrel. Would a person firing the bullet feel any recoil in such a scenario?

<unnecessary pedantry>
"As it exits the barrel" is a little problematic : there's some pressurized gas behind it when the bullet leaves the barrel, which will actually speed it up a little for a few inches (think: the flame you see coming out the barrel when a shot is fired), against the surrounding air before it dissipates.

Since in the scenario, you're firing with the 1,000mph wind, that stream of spent gas will have more effect on the bullet, and - since it's normally also pushing against an ambient air velocity of zero, that particular portion of "recoil" will be lessened.

Not zero total recoil, though ; not anywhere near : powder go boom, bullet goes one way, gun goes the other ; Newton wins, shoulder loses.

Also, the problem of how the hell are you standing up (or laying down) in a 1,000mph wind in the first place.
<\up>

 

[DaveC426913]

"As it exits the barrel" is a little problematic...

Maybe out of context it would be, but the context was provided:

So, ideally, the bullet will fall straight to the ground...

 

[hmmm27]

Maybe out of context it would be, but the context was provided:

Uh-huh ; here, borrow my pedantry brackets But, muzzle blast is usually a noticeable chunk of total recoil, and I imagine the difference in a 1k mph wind would be a noticeable portion of that.

 

[nasu]

Uh-huh ; here, borrow my pedantry brackets But, muzzle blast is usually a noticeable chunk of total recoil, and I imagine the difference in a 1k mph wind would be a noticeable portion of that.

This is interesting but not relevant to the question in the OP. The air moves relative to the gun in both reference frames. The comparison is not between a situation with wind versus one without wind. Whatever the effect of wind on recoil, it will be present in both reference frames.

 

[PeroK]

Uh-huh ; here, borrow my pedantry brackets But, muzzle blast is usually a noticeable chunk of total recoil, and I imagine the difference in a 1k mph wind would be a noticeable portion of that.

The last time I was on a train there wasn't a 200km/h wind blowing through the carriages! Assuming the rifle was fired from inside the train, the bullet would simply fall out into the slipstream of the train.

 

[hmmm27]

Ah, in my mind it took place on a flat-car : seemed more plausible than actually inside a coach, eh.

 

[hutchphd]

But for the muzzle blast momentum is also conserved. I do not understand why the wind will affect the recoil. Suppose you are shooting a blank (wad): would the recoil be wind dependent ? Why?

 

[hmmm27]

Bullet aside, muzzle blast (>1k velocity, overall) comes out into a 1k tailwind (as opposed to still air), so less pushback, so less of that element of recoil.

 

[Ibix]

I hesitate to offer a simplification, but we could always avoid the whole gas propulsion argument and consider a catapult...

 

[jbriggs444]

Ah, in my mind it took place on a flat-car : seemed more plausible than actually inside a coach, eh.

How about on the back platform of the caboose. Imagine some presidential figure waving to the passing crowd as he stands there...

moving at 1000 miles per hour, sheltered from the wind by the rest of the caboose...

while he takes a bead on the one guy who is not cheering but is instead...

going "neener, neener, can't shoot me".

 

[PainterGuy]

Thank you, everyone! I really appreciate your help.

 

[sysprog]

Yes, the person will feel a recoil because the gun does its thing without knowing that it's on the train.
Yes, the bullet will fall on the ground as you said.

If the shooter tosses a pebble off the back of the train at the same time as he fires the bullet, the bullet will wind up far away from the pebble, having traveled much further away from the train than the pebble did.

 

[Quester]

If the shooter tosses a pebble off the back of the train at the same time as he fires the bullet, the bullet will wind up far away from the pebble, having traveled much further away from the train than the pebble did.

Since a person can't toss a pebble at a speed higher than about a hundred miles/hour, the pebble will be traveling about 900 miles/hour in the direction of the train's travel relative to the ground. If it takes the pebble a half a second to fall to the ground, the pebble will hit the ground over 600 feet from the bullet in the direction of the train's travel.

 

[Shane Kennedy]

Summary:: What would happen when a bullet is fired from the back of train which is moving as fast as the bullet?

Hi,

I was reading this article, https://science.howstuffworks.com/question456.htm, about firing a bullet from the back of train which is moving as fast as the bullet.

The following is taken from the article mentioned above.So, ideally, the bullet will fall straight to the ground as it exits the barrel. Would a person firing the bullet feel any recoil in such a scenario? I don't think there'd be any recoil. Could you please confirm this? Thank you!

From your perspective, the bullet would follow it's normal parabolic (?) trajectory, just without air resistance.

 

[Quester]

From your perspective, the bullet would follow it's normal parabolic (?) trajectory, just without air resistance.

As in the video in Post #4, the trajectory for the bullet would be a vertical line (straight down).

 

[Ibix]

As in the video in Post #4, the trajectory for the bullet would be a vertical line (straight down).

From the ground perspective, yes. I believe @Shane Kennedy was thinking of "you" as being on the truck with the gun. There's a gun camera shot from the truck bed in the video too, where you see the ball arcing away on a parabolic (or clearly curved, anyway) trajectory.

 

[RandyD123]

Summary:: What would happen when a bullet is fired from the back of train which is moving as fast as the bullet?

Hi,

I was reading this article, https://science.howstuffworks.com/question456.htm, about firing a bullet from the back of train which is moving as fast as the bullet.

The following is taken from the article mentioned above.So, ideally, the bullet will fall straight to the ground as it exits the barrel. Would a person firing the bullet feel any recoil in such a scenario? I don't think there'd be any recoil. Could you please confirm this? Thank you!

Reminds me of the experiment that if you put a gun in a vice at X height off the ground and you hold a bullet in your hand at the same height, and you fire the gun and at the same time drop the bullet, both bullets hit the ground at the same time.

 

[sysprog]

Reminds me of the experiment that if you put a gun in a vice at X height off the ground and you hold a bullet in your hand at the same height, and you fire the gun and at the same time drop the bullet, both bullets hit the ground at the same time.

. . . provided that the vice holds the gun exactly parallel to the ground, and that the drop is strictly a drop, i.e. orthogonal to the ground . . .

 

[jbriggs444]

. . . provided that the vice holds the gun exactly parallel to the ground, and that the drop is strictly a drop, i.e. orthogonal to the ground . . .

And that we are ignoring air resistance.

If air resistance goes as the square of velocity (as it does for bullet speeds) then the vertical resistance on a bullet that is moving at high speed horizontally is greater than the vertical resistance on a bullet that is falling straight down. Instead of being able to look at the horizontal and vertical components separately, the differential equations of motion in the x and y directions are coupled and stuff gets complicated.

If that paragraph slid right past, consider a simple scenario with a three-four-five triangle. We have quadratic drag. Air resistance goes as the square of velocity: $F=v^2$. Pick your favorite units. The triangle is laid out with the 4 on the horizontal and the 3 on the vertical. The hypotenuse is diagonally downward from our firing point.

We shoot one bullet down the hypotenuse of this three-four-five triangle. It is moving horizontally at a velocity of 4 and vertically at a velocity of 3. Velocity down the hypotenuse is 5. So air resistance is 25.

The vertical component of air resistance is 25 sin theta -- which is 25 * 3/5 = 15.
The horizontal component of air resistance is 25 cos theta -- which is 25 * 4/5 = 20.
[We can sanity check with Pythagoras and see that total air resistance is 25 as expected]

Now we let a second bullet drop straight down at a speed of 3. Air resistance is 9.

The two bullets have identical vertical velocity components. Three in both cases. But they have different vertical components of air resistance. Fifteen versus nine.

The bullet moving faster horizontally has higher air resistance vertically.

One can contrast this with linear drag. The differential equations uncouple and one can solve the vertical and horizontal components of motion independently.

Or one can go with constant drag (kinetic friction). The differential equations couple again, but this time high speed motion in one direction translates to low drag in the orthogonal direction.

 

[sysprog]

I think that I should have corrected "vice" to "vise" -- the former being a moral inrectitude, the latter being a clamp . . .