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Physics, Astronomy & Cosmology / Re: A reported object travelling faster than the speed of light?

« on: 28/11/2019 22:04:14 »

Think of it this way,  If You were at rest with respect to the start point, after 1 sec, A and B would be 1.2 light sec apart. ( assume A move to the left and B to the right)
A sends a signal.  That signal travels at c relative to you.  After other second the front of that signal is 1 light sec from where it
was emitted.  A has moved 0.6 ls to the left and B another 0.6 ls to the right. Distance between signal and B is 1.8-1 = .8 light sec.   In one more sec the signal has moved another 1 ls to to right from and B 0.6 ls to the right.  New distance = 1.4-1 = 0.4 ls   In one more sec, the light travels 1 ls, B .06 ls, and the distance is 1-1 =0. the light has caught up with B.  3 sec after emission, the light reached B. 

Perfectly true, but irrelevant. The light reaching B will be Doppler shifted so that
f_B = f_A√(1-β)/(1+β) where β = v/c = 1.2
So if A transmits, say, a 1 GHz pulse of radio waves, B will be able to calculate their relative speed as the received pulse will be at  0.4 GHz.

And indeed we have observed astronomical objects with extreme redshifts.

The conundrum is resolved by remembering that you only require infinite energy to accelerate a massive object from rest to c with respect to its starting point. There is no reason why two objects shouldn't travel in opposite directions at > 0.5c relative to their origin.

The formula you give is for Doppler shift do to ordinary motion ( rather than Cosmological Redshift)  In this case, then v/c has to be determined using the Relativistic addition of velocities as noted by Halc. This gives a value of 0.882...  for v/c and ~0.7288 for the Doppler shift.  B would see a frequency of 0.7288 hz.  If you try to use 1.2 for v/c,  you get (1-1.2)/(1+1.2) = -0.909 under the radical, which leaves you with trying to get the square root of a negative number.

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Physics, Astronomy & Cosmology / Re: A reported object travelling faster than the speed of light?

« on: 28/11/2019 06:07:18 »

If A and B were simultaneously ejected from a point in opposite directions at 0.6 relative to the start point, each would be travelling at 1.2c relative to the other, but neither would know it because any signal from A could only travel towards B at c.

Think of it this way,  If You were at rest with respect to the start point, after 1 sec, A and B would be 1.2 light sec apart. ( assume A move to the left and B to the right)
A sends a signal.  That signal travels at c relative to you.  After other second the front of that signal is 1 light sec from where it
was emitted.  A has moved 0.6 ls to the left and B another 0.6 ls to the right. Distance between signal and B is 1.8-1 = .8 light sec.   In one more sec the signal has moved another 1 ls to to right from and B 0.6 ls to the right.  New distance = 1.4-1 = 0.4 ls   In one more sec, the light travels 1 ls, B .06 ls, and the distance is 1-1 =0. the light has caught up with B.  3 sec after emission, the light reached B. 

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New Theories / Re: Is the designation "positive" and "negative" in electricity arbitrary?

« on: 27/11/2019 18:01:16 »

The reason behind which pole we called positive and which is negative was that it was a random choice made by Benjamin Franklin.  He assumed the electricity involved the movement of something from an excess(+) to a deficit(-), but had no way of knowing which direction this was.  So he just took a guess.   It wasn't until this convention had been long established that the current carrier was learned to be the electron and they flowed from - to +.     "conventional current" still treats current as flowing from positive to negative, even though this is opposite to the actual electron flow.   The schematic symbols of components like transistors and diodes are based on conventional current. ( the arrows  are supposed to point in the direction of current flow).

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Physics, Astronomy & Cosmology / Re: A reported object travelling faster than the speed of light?

« on: 26/11/2019 18:13:31 »

 A key is understanding the difference between "apparently" moving faster than light and actually moving faster than light.
Simple example:  And object starts 8 light mins away traveling towards you at 0.8c.
The light from the moment it starts at the distance of 8 light min, arrives 8 min after it starts.  The object itself arrives in 10 min, or only 2 min after the light of it being 8 light min away does.  Thus it would appear to you that the object crossed the 8 light min distance in just 2 min and thus traveled at 4c.   But this is just an visual effect caused by the fact that the object is following closely behind its own light. The object was always traveling at 0.8c You merely saw its 10 min trip compressed into 2 min.

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Physics, Astronomy & Cosmology / Re: How do protons and electrons react?

« on: 22/11/2019 18:56:08 »

When you have a reaction that produces electrons, you can grab them with an electrode.

When you have a reaction that produces protons, how can you grab them?

So that then you join them together to produce hydrogen?

The only type of "reaction" that produces protons or electrons are nuclear reactions.  For example, A neutron decaying into a proton and electron, either as a free neutron, in the nucleus of a radioisotope.

I may be wrong, but from this and your other thread, I get the impression that you think that electricity involves the creation of new electrons.  It doesn't, it generally involves just moving existing electrons around.

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Physics, Astronomy & Cosmology / Re: How can Mars' sunset be blue?

« on: 22/11/2019 18:40:41 »

Well, quite obviously, the dust doesn't absorb all the blue light. Mars atmosphere is very thin compared to our own, and there just isn't enough dust in it to absorb all the blue in the sunlight. 

During the day on Mars, the dust scatters the red light and lets the blue through. The scattered red makes the sky look pink.  On the Earth the thicker air scatters the blue, making the sky look blue.
On Mars during sunset, the light has to pass through more atmosphere, which scatters out more red light, filtering it out from the light coming at you from the direction of the Sun.  Thus looking in the direction of the sunset, you see the blue light that isn't scattered.

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Physics, Astronomy & Cosmology / Re: How much fuel would we need to get to Mars?

« on: 22/11/2019 18:22:15 »

Daniel wrote to us to ask us this:
If the same amount of fuel that is used to get a rocket to the moon from the earth, is used on a rocket that is launched from the moon, would the rocket make the journey to Mars?
What do you think?

The Apollo Saturn V was able to get the CM/LM pair up to about 10.9 km/sec.  we'll use this as the approximate delta v budget.
escape velocity from the Moon is ~2.4 km/sec
escape velocity from the Earth, at Moon orbit distance is ~ 0.414 km/sec
Minimum delta v to leave Earth's Solar orbit and intersect Mars' solar orbit is ~ 3 km/sec
Delta v needed to match velocity with Mars upon arrival is ~ 2.6 km/ sec.
Total required for trip is ~ 8.4 km/sec, which is less than the Saturn V is capable of, so, yes, it could make the trip.

However, this assumes that the fuel and rocket are already on the Moon.
The payload the Saturn V could get to trans-lunar injection was 48,600 kg.  The total mass of the Saturn V was 2,957,700 kg, or ~61 times that.   IOW, it would take about 61 Saturn V rockets to deliver a Saturn V equivalent mass to the Moon. (and this is not counting soft landing it on the Moon. But you probably wouldn't want to do that, as why would you want to land all that stuff on the Moon just to have burn fuel to lift it back off again.)

What this means is that to make a Moon-Mars shot really feasible, you would need to be able to manufacture the rocket and create fuel for it on the Moon itself.  This is quite an endeavor, and would likely require quite an expansive colony on the Moon. And by the time we have established such a colony on the Moon, we will have likely come up with better methods of space propulsion than the Saturn V's late 1960's rocket technology.   

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Physics, Astronomy & Cosmology / Re: Can we detect acceleration in free fall?

« on: 19/11/2019 19:56:53 »

Well this is not the question you might think it is. If a spacecraft is falling freely towards a gravitating body, and then starts accelerating towards the body, would an accelerometer show the acceleration?

I have not seen anything about this online. I would say it would. Is there any observational evidence anyone can point to? Have you tried this personally?

If the ship fires it engines such that the ship accelerates towards the gravitating body at a rate greater than that due to the gravity of the body, then yes, the occupants will sense that extra acceleration.  It doesn't matter what direction the ship is pointed, any thrust by provided by the engines will be felt the same way by the occupants.  In other words, if the occupants were unable to look outside and had no prior knowledge of their situation in regards to the gravitating body, firing their engines, no matter what their orientation, would not give them any information regarding their situation with respect to that body.

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New Theories / Re: Does Einstein’s Equivalence Principle have any physical standing?

« on: 19/11/2019 17:08:40 »

My point is that astronauts in my example (i.e., in a spaceship far from any gravitational influences), would also be weightless and in free fall and hence the force of acceleration exerted on the ship would not affect them. Ergo, Einstein’s equivalence principle is wrong.

Your conclusion is demonstratively wrong.  Take the Dawn space probe to Ceres for example.  It was put on a course to Ceres under a gentle constant acceleration provided by its ION engine.  This required quite precise maneuvering.   While you can design the Engine to produce a given thrust, there will always be variations.   While at Ceres, the craft made a number of maneuvers to change its orbit from time to time.  Due to the great distance of the Ceres and the time lag, they could not be directly done from the Earth. Instead a signal was sent telling the craft to change its velocity by a certain amount.  This means the craft needed a means of telling when it had made the proper velocity change.   This was done by an accelerometer, a device which measures how much acceleration the probe is undergoing by measuring the very effect you claim doesn't exist.
If you were right, the Dawn probe's accelerometers would have not functioned as designed, and the craft wouldn't have been able to perform the maneuver  as instructed.   As it happened, the probe performed exactly as expected.
The Apollo mission craft also made use of accelerometers in their computer guidance systems so that when they made mid-course corrections, the computer would know just how long to burn the engines to effect the exact right amount of correction.  Again, according to your belief, this shouldn't have worked, but it did.

When the facts run contrary to your position, it is time to re-evaluate your position.

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New Theories / Re: Does Einstein’s Equivalence Principle have any physical standing?

« on: 18/11/2019 04:54:09 »

Kryptid: The point I was trying to make is that an acceleration of 9.8m/s², 50 billion kilometres from earth or other gravitational influences,is in no way representative of the gravitational force on earth. Instead as Janus points out the force that would be experienced by the spaceship would be the same force that it would experience if it were on an earth that possessed a radius of 50 billion kilometres ! This reinforces my original conjecture that in the absence of a large mass close by, Einstein's thought experiment makes no sense. He is using gravity as an explanation of itself.

This has no bearing on the equivalence principle. 
An example illustrating the principle is this:
Take your ship, put it just above the surface of the Earth firing its engines with just enough thrust to hold it hovering above the Ground at a constant height.  An occupant in the rocket standing on a scale would weight the same as he would standing of the surface, if he drops an object, it falls to the floor, etc.
Now take that rocket and move out to 50 billion miles from the Earth.  But still firing its engines at the sane thrust as before.  Now this of course will be more thrust than the pull of gravity from the Earth and the rocket will accelerate away from the Earth.  An occupant inside the ship however will still weigh the same on a scale in the ship and objects will still fall towards the floor of the ship just like they did when the ship was hovering just at the surface of the Earth.  Relative to the rocket, objects in this accelerating rocket behave just like they would in a gravitational field.
This is the gist of the equivalence principle.  The importance of this is that you can imagine how something like light would behave in the rocket in the second scenario For example, light emitted from the tail would appear red-shifted to an observer in the nose, or if fired sideways to the acceleration would trace a curved path relative to the ship, and then use this to conclude conclude that it would exhibit the same behavior in a gravity field (be red shifted climbing against it or have it path curved by it.
You would also conclude that an observer in the nose of the accelerating rocket would measure a clock at the tail as running slow, and by extension that someone higher in a gravitational field would measure a clock lower than them as running slow.

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New Theories / Re: Does Einstein’s Equivalence Principle have any physical standing?

« on: 17/11/2019 16:54:02 »

Let’s just suppose (as a special case) that in interstellar space there is a planet exactly like earth about 50 billion miles away from the space ship, which weighs 2 tons or 2000 Kg. Then the acceleration due to that  planet would be 9.8m/s2 and the gravitational  attraction between the planet and the space ship would be  G x (5.972 x 1024 x 2x 103)/((5X1010)2) = 0.000318860846 m3 kg-1 s-2 .

For one, you screwed up your equation.  You put the (2 x 10^3), which I assume was a rough attempt to convert miles to meters*,  in the wrong half of the equation   It needed to be multiplied by the 5 x 10 so that the you had
 G x (5.972 x 1024 )/((5X1010x 2x 103)2) . In other words, the numerical value of your number was off by a factor of at ~ 8e10. in addition to being expressed in the wrong units.  Correctly used, the equation gives an answer in m/s^2 or acceleration.  For example, the equation used for the surface of the Earth (a distance of ~6378000 meters from the Earth center gives
G(5.972e24/6378000^2 = 9.8 m/s^2 or the acceleration due to gravity at the surface.)
A more accurate value for your scenario is 6.23e-14 m/s^2.    Which is the acceleration due to gravity at this distance from the Earth. 

It is now possible to see that the gravity exerted on the space ship is very small, therefore if the mass of the spaceship were to be multiplied by the force of gravity a total force of 0.00031 N which is an extremely small force would be exerted by gravity. If the ship were being accelerated at 9.8m/s2 then the force exerted on the space ship would be equal to F = 2000 Kg x 0.00003 m/s1 = 0.62 N which is a ridiculously small force and certainly would not have the effect of pushing anyone in spaceship back against their seats.

The above arrived at value 6.23e-14 m/s^2 multiplied by the 2000 kg of the rocket gives the amount of thrust or force the rocket would have to generate in order to "hold station" at this distance and not slowly begin to drift towards the planet (1.25e-10 N).
A 80 kg man standing on the floor of said rocket would feel a force of ~5e-12 N Small, yes, but no more or less than what he would feel if standing on the surface of a planet with a mass equal to that of the Earth and 50 billion miles in radius.  It is also exactly the same force he would feel pressing him against the floor if the rocket were accelerating at 6.23e10-14 m/s^2 in empty space with no gravitational forces present.  The point of the equivalence principle is that our passenger, would not be able to distinguish between "hovering" 50 billion miles from an Earth massed planet and the rocket accelerating at 6.23e10-14 m/s^2 through free space with there being no gravity.  The fact that the force he feels is in all practical effect too small to be measured doesn't mean that it is the same in both cases.

Where am I going with this? I am trying to show that the equivalence principle of Einstein’s has absolutely no real meaning, it is just using gravitation as an explanation of itself. Here we have an object in space being accelerated to an equivalent acceleration of the acceleration due to the gravity on earth (i.e., 9.8 m/s2 ) yet it has no effect, some gravitational mass is needed close by to make it work.

No. What you have shown here is that you don't actually grasp the meaning behind the equivalence principle. The flaw is in your understanding, not the principle itself.

*more accurate would have been 8/5 x 10^3. And if you are going to rough out this by this degree (2 instead of 1.6, or only to to 1 significant digit), it really doesn't make sense to carry out the Earth's mass to 5 significant digits.

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That CAN'T be true! / Re: Do you weigh less when the moon is directly overhead?

« on: 09/11/2019 16:59:33 »

As for the shielding effect I know of course when they are roughly in line there is enhanced gravity but is there any effect for the few minutes of the total eclipse, now that high resolution gravity meters are available this should be easy to check.
As matter is almost transparent to Neutrinos I would expect it to be even more so to the vastly less massive Gravitons but I would like to see experimental verification.

It is pretty obvious that matter doesn't block gravity.  If it did, then the gravity produced by the matter at the center of our own Earth would be partially blocked by the matter between the surface and the it.  The gravity on the surface would be less than the what it should be for the mass of the Earth.  This would mean that The effective gravitational mass of any planet, moon, etc. would be less than its inertial mass, and this would play havoc with orbital mechanics.

In addition, do not confuse  gravitons, which are quanta of gravitational radiation (gravitational waves), with the "virtual gravitons" that would be expected to mediate the gravitational field in quantum gravity theory. ( in the same way, it is virtual photons, and not actual photons that mediate the magnetic and electric fields.)

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Physics, Astronomy & Cosmology / Re: What would happen to a bullet fired from the International Space Station?

« on: 03/11/2019 01:33:01 »

Assuming a muzzle velocity of 1200 m/sec ( about mid range for small arms), The new bullet orbit would climb to a new apogee of ~400 km higher than ISS orbit ( perigee would remain at ISS orbit distance) and would take ~ 5 min longer to complete its orbit.

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Physics, Astronomy & Cosmology / Re: Pulsars, why is it we can't see them?

« on: 25/10/2019 17:18:49 »

Is there a phase change with time?  I have doubts about a rotating disk structure, it would cause a phase change of the signal.  If the spectrum is much higher than light, then it's direct particle emission, which means fast dynamic power.

Perhaps some interaction between the poles.   Cutting a large amount of M flux could certainly do it.

Very mysterious.  Could life survive in a galaxy with one?  Are the ones we know of in galaxies?

The disk evan_au is referring to a a device for viewing a pulsar, not anything doing with the structure of the pulsar.
The pulsar itself is a rotating neutron star with a strong magnetic field.  The magnetic poles and rotational axis are not aligned ( just like for the Earth).  The high spin rate and magnetic field combine to funnel particles into beams emitted from the magnetic poles.  The beams are constant but sweep like a search light.  If you are in the path of the beam your will see it "flash" as it sweeps by.  (This means that there could be more pulsars out there that we no nothing about as their beams never sweep past the Earth.
There are a few classes of Pulsar. 
Rotational pulsars tap the rotational energy of the star to drive the beam and slowly decrease their spin and flash rate over time*
Accretion pulsars have a binary companion that has reached the stage where it has begun to expand. The pulsar starts to gather in material from the companion, and it is the gravitational potential of this material that provides the energy for the beam.
Magnetars have extremely strong magnetic fields and it is the collapse of the field that provides the energy.

*The fact the pulsars are only noticeable from point through which their beams sweep and the rate decays over time is the reason The Voyager probes included "pulsar maps"   Which showed what pulsars we had detected from the Earth, their relative distances and pulse rates.  The idea being that if some alien civilization discovered them in in the far, far future, they would be able to work out not only from where the probes were launched, but also when, as there would have only been one place at one time where that particular pattern would have been seen.

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Physics, Astronomy & Cosmology / Re: Pulsars, why is it we can't see them?

« on: 24/10/2019 16:07:58 »

The "brightest" pulsar is some 1000 ly away, has an apparent magnitude of 23.6. under ideal viewing conditions, the best the unaided human eye could see would be about a magnitude 6.  It also pulses at a rate of ~ 11 pulses per sec, which is close to the limit of what we could even detect as a being a flicker.

The closest known pulsar is some 500 ly away, and pulses approximately every 6 milliseconds, much faster than we could see.
One factor also in play is that the peak frequency of the electromagnetic radiation emitted by these pulsars lies outside of the visible spectrum.

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Physics, Astronomy & Cosmology / Re: What motion will weights released from a spinning spacecraft describe?

« on: 22/10/2019 01:41:30 »

Most take one v as side a and the other v as side b, with the hypotenuse as resultant v.

After leaving the muzzle, we can not increase the speed of the bullet.

And because of the cross displacement, a longer flight path and time.

If we use the magnitude trolley v for side b, and the magnitude of bullet v for the hypotenuse.......the resulting side a........will be the resulting magnitude of velocity of incident impact.

The side a, will always be shorter than the hypotenuse(muzzle v)........less velocity.

Does the impact hole in wall, go straight in?  Does that hole go deeper with trolley velocity?  The bullet's mass remains constant.

How does this really work?

If the gun is fired while pointing in the A direction, then  the A side is the muzzle velocity and the the resultant velocity( along the hypotenuse) is the vector addition of the trolley velocity and  muzzle velocity.  It will be greatwer than either trolley or muzzle velocity.   If muzzle velocity is 100 m/sec and trolley velocity is 15 m/sec, then the velocity along the hypotenuse will be  ~101.119 m/sec.   The bullet will strike the target at this speed and at an angle of ~8.627 degrees from straight in.
What makes you think that the hypotenuse velocity would be equal to the muzzle velocity?

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Physics, Astronomy & Cosmology / Re: What motion will weights released from a spinning spacecraft describe?

« on: 21/10/2019 18:55:05 »

Ok, so my target moves parallel to the trolley, the further I move away from trolley(staying parallel), the faster the bullet travels?

Let me ask you this.  And thanks to all for participating.

Does the flight time change?

The distance between target and trolley path has no effect on the bullet speed.  It still travels at the same speed, it just travels a longer distance and take more time to hit the target.  It is no different than if the trolley and target were not moving with respect to each other.  if the bullet leaves the gun at 100 m/s and the target is 100 m away, it takes 1 sec for the bullet to hit the target, if the target is 200 m away, it takes 2 sec.   Having either the target or trolley moving at some velocity perpendicular to the direction the gun is aimed has no effect on this time.

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Physics, Astronomy & Cosmology / Re: What motion will weights released from a spinning spacecraft describe?

« on: 21/10/2019 18:46:49 »

ok, what happened to the bullet's velocity?

Bullet velocity( along black arrow) would be equal to  sqrt((trolley velocity)^2+(muzzle velocity)^2)*

* to a good approximation when trolley velocity and bullet velocity are low compared to c.
If they were significant fractions of c, you'd need to use
sqrt(u^2+v^2- (uv)^2)
where u is the speed of the bullet as measured from the trolley, and v is the velocity of the Trolley, expressed in units of c.
(Actually, this is the correct equation for all cases, it is just that for low values of v and u the answer comes out to being almost exactly what you get using the first equation above)
The bullet still remains abreast of the trolley during its entire flight.

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Physics, Astronomy & Cosmology / Re: What motion will weights released from a spinning spacecraft describe?

« on: 21/10/2019 18:17:39 »

ok, what happened to the bullet's velocity?

Bullet velocity is the vector sum of trolley velocity and muzzle velocity relative to the trolley. This is true in any frame.

Janus must be an Aussie because he drew the North wall at the bottom. 

Actually, I didn't pay any attention to the cardinal directions (they really don't matter in the final analysis).  I originally started drawing the diagram with the trolley going from bottom to top, changed my mind, and rotated it.  I had already added some of the labels, and quite frankly, was too lazy to start over.

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Physics, Astronomy & Cosmology / Re: What motion will weights released from a spinning spacecraft describe?

« on: 21/10/2019 17:02:04 »

Ok, I will try to use metric.  Evidently, no one is concerned with period of acceleration.

That diagram shows when the cut release is at the same time as muzzle exit, not firing time.

Let's try something different.  Imagine an American football field.  On the close side of the field we have a trolley, and on the other side a wall with the field yard lines extended.  The field lays east-west.  Trolley at south, wall at north.  With trolley stationary at 50 yard line, we shoot wall along the line....hit wall.   No gravity, no air.  The bullet will strike wall 50 yard line at muzzle velocity.

Are we all together?

Now, the trolley is moving from right to left, and we fire incident at 50 yard line point.  OOPS....I shouldn't say fire.....the bullet leaves muzzle at 50 line incidence.

Does the bullet hit the line?....or veer to the left?....or veer to right? 

If we keep the muzzle v constant, and vary the trolley v, will the bullet always land on the same side of the 50 wall line.  Even if trolley v, exceeds bullet v?

bullet.png (25.1 kB . 480x418 - viewed 368 times)
If the bullet leaves the gun perpendicular to the path of the trolley ( as measured from the trolley), it will follow the path indicated in the diagram, and strike the wall/target at a point exactly opposite of where the trolley is when the bullet hits ( as shown by the cyan line).  The relative speed difference between trolley and bullet makes no difference.