Bullet fired from rear of vehicle of identical velocity?

Question

Here's a question I have trouble wrapping my mind around:

A person riding some futuristic speedy train pulls a Colt 1911 and fires a round from the rear of the caboose (exact opposite direction the train is traveling). If the train was traveling at a velocity of 750 feet per second and the projectile from the handgun was also traveling 750 fps, to the stationary observer standing by the tracks, would the bullet appear to simply emerge from the barrel and fall directly to the ground?

I try to simplify this by applying the same idea to lower velocities. I try to figure what a baseball would do if a person threw one from the rear of a modern Amtrak at 40 MPH when the locomotive was ALSO traveling at 40 MPH at the time. Still, I cannot figure what would appear to happen to the ball to the stationary track-side observer.... it hurts my brain!

Any ideas?

Answer 1

Its easy to imagine these few cases for the observer on the road:

1. a person jumping down from a stationary bus
observer finds everything stationary

2. a person simply jumping down from a moving bus
observer finds bus moving with speed and person moves with an initial velocity of the bus and deccelerating to zero due to gravity and air friction. His head will still continue to move forward with some speed. so he would run a little to prevent falling down. In this case, considering the size of the bullet, we can ignore this.

3. a person jumping from a moving bus in the opposite direction with the same velocity of the bus
Observer finds the person's jumping straight down if no forces are acting on him.
But due to gravity and air friction, he is deccelerating and hence he require more speed to make the observer feel as if he is dropped from the bus.
As he jumps with the same speed of the train, he falls a bit backward in the direction of the train, his path will be slightly curved towards the train's direction.

Headache questions...

Answer 2

‘The train’s velocity is 750 ft per second’

This is the measurement of speed by the one who is on the station and not by the person who is riding in the train.

In fact it is wrong to say, ‘by the person who is riding in the train” Because it is also the statement that can be made only by the one who is at the station.

The person in the train is always at rest with respect to the train.

The person in the train fires a bullet which travels with a speed of 750 ft/s.

This speed is the speed of the bullet with respect to the person in the train.

For the one who is on the station, the bullet (before firing), the hand gun, the person and the train are all are moving in one direction with a speed of 750 ft/s.

After firing, the one in the station observes that except the bullet all the other is moving with a speed of 750ft/s.

If v is the speed of the bullet ( as per the person inside the train) in the direction of the train, the one in the station will see that the bullet is overtaking the train, that is, it is moving with a speed 750 + v.

If the bullet is fired in the opposite direction, then its speed will be 750 - v.

Since in the present case, v = 750 opposite to train’s motion, 750 - 750 = 0

Thus the one in the station will see as if the bullet is dropped from one point to the ground.

However, with respect to the one in the station, for the bullet ( inside the hand gun) which was initially moving with a speed of 750 ft/s in the direction of motion has to change its speed to zero and then change its speed to the opposite direction.

Of course, it takes a very small time. The observer in the station will see this also.

That is slowing of the bullet from 750m/ s in the direction of motion to zero during this small time and at the same time falling to the ground.

Hence the motion of the bullet will not be in perfect vertical straight line but along a steep parabola.

Answer 3

I'm so glad that you asked. The bullet would travel 750 feet per second away from the back of the train. Normally a bullet quickly loses velocity from air resistance, but in this case it would not have any because the net velocity to the air and bystander would be zero feet per second. The bystander would see the bullet fall straight down to the ground as if it where dropped.

Answer 4

the stationary observer will find that the bullet leaves the gun and just falls down as the net speed on the ball relative to the ground in zero. however to the person in the train the bullet would appear to travel at 750 feet per second as their relative velocity is 750 feet per second. it is all about the frame of reference used.

Answer 5

Yes, it's all a matter of point of reference.
(Provided the speeds are low enough for relativity effects to be ignored. It does not work for someone in a space ship at 75% of lightspeed shining a torch over the stern. Sorry, this might hurt your brain even more. (URL 1) )

And were the gangster on the train unaware he was moving (OK, it's a night, the trains' quiet, and he's not feeling acceleration: how's he to know?) the behaviour of the bullet would come as a surprise, and reveal the situation.

Now try the Coriolis effect. You are on a rotating roundabout, and you throw a baseball to someone directly opposite. But your throw does not appear straight, and the ball curves to one side.
It will look straight to an outside observer, but because you and your target are moving in a circle, the ball will seem to be drawn to one side.(URL 2)

Answer 6

Yes, the bullet will fall to the ground, as described. To the shooter on the train (or the pitcher of the ball), it appears that it's gone out on a trajectory slightly arcing down to the ground, but that's not what the ground observer sees.