# Cue Ball Moving on a Train Problem

Imagine a pool table on a train, and ball on the pool table. The train's forward direction is to the left. Imagine a completely horizontal force applied on the cue ball to the right. The path of the cue ball however, is as shown by the orange line below. Is train, therefore, moving to the left or the right (As in from the viewpoint of the front of the train looking in the direction of its motion, is it to its left or its right)?

I am almost 100% certain it is to the left, which can be best shown by taking a sheet of paper, putting a pen to it, and moving the paper like the train and the pen backwards like a cue ball (Which confirms my answer). Why does this work specifically, because presumably if not forces act on the ball, it can never vary from the straight line, and the simulation should work?

Consider NO centripetal or centrifugal force acting on the ball, or any other forces for that matter. The ball itself in all reality is only moving in a straight line - however, when compared to the table, the ball's path seems curved due to the motion and changing of the orientation of the table, which will turn in the exact same way as the train.

EDIT: I now know the answer is that it turns right (It was an HSC Question, 2015 Q14 on the Multiple Choice). I understand the issue now, and the problem has to do with context. When the question was in an HSC paper (Being a year 12 Physics paper), they expected you to take into account inertia, centripetal and centrifugal force into the problem. The thing was, the question was transposed then into my situation - that is a class who just started high school physics, and as such should have no real understanding of any of those forces. For all practical purposes, the answer should be right. I never disputed that. However, in the context of where it was asked, I thought the best interpretation would be if it was on a positionally stationary, but rotating table, not on a train that was moving. This also explains why the pen experiment didn't work - because the pen didn't take into account the fact that forces would act on the ball itself.

Thanks so much for the explanations!

• The train is moving to the left. The blue arrow says so. I think you want to know whether the train is turning left or right. Feb 7, 2020 at 14:58
• Can you edit the screenshot to include the actual problem statement? Something doesn't add up here. Feb 7, 2020 at 15:13
• "the inertia acted like an actual force." That's exactly the right idea, and you've even anticipated the name of this phenomenon. Such apparent forces that result from being in a non-inertial reference frame are often called "inertial forces." Feb 8, 2020 at 0:54
• @CharlesHudgins Inertia is momentum, not force. Feb 8, 2020 at 1:06
• @Acccumulation See here. "A fictitious force (also called a pseudo force,[1] d'Alembert force,[2][3] or inertial force[4][5]) is a force that appears to act on a mass whose motion is described using a non-inertial frame of reference, such as an accelerating or rotating reference frame." Here the acceleration is due to the train turning. Also, to be clear, inertia is not momentum. It's the tendency of matter to maintain its state of motion. In GR, inertia is the tendency of matter to follow geodesic trajectories. Feb 8, 2020 at 2:16

If the train is moving in a straight line at an unchanging speed, say 30 kilometers per hour, then everything inside the train is moving with it, all of the physics inside the train will work the same as if the train were not moving. The table, the ball, and you, if you are standing still in the train, are all going the same speed and direction as the train. So if you sit the ball on the table, it will not move. Now if the train accelerates to 40 kilometers per hour straight forwards, then the ball will seem to move straight backwards, as the train has accelerated. This is the same effect you feel pushing you back into a car seat when it accelerates forward quickly. If the train decelerates to 20 kilometers per hour, the ball will seem to roll forwards, just as you would feel a forward pull when slowing a car quickly. If the train turns left, the ball will seem to roll right, if the train turns right, the ball seems to roll left. Just as you feel pushed right or left in a turning car. This is due to the ball's inertia and to Newton's first law of motion. See; https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion

• I understand where you are coming from (And how inertia works), but the ball is moving in the opposite direction. A lot of people are just saying this doesn't make a difference to the question, but in actual fact, it makes a considerable difference. I edited the question with a series of diagrams hopefully outlining what I see differently. Feb 7, 2020 at 12:58
• @Pedro Hablespanyos this answer is correct. In the non inertial frame of the train the ball is accelerating to the left. Therefore the train is accelerating to the right in an inertial frame. The velocity of the ball is not relevant in this case, just the acceleration.
– Dale
Feb 7, 2020 at 13:15
• I should have clarified better, but - no forces are acting on anything in this scene. This is not a practical problem in real life, where centripetal force, centrifugal force and all other factors would obviously mean that the ball moves that way because it is turning right. This is a problem where everything moves at a constant velocity, the direction of velocity changes but not the magnitude, and it does so seamlessly. The ball is not accelerating - the curved path comes from a viewer outside the train. This isn't a question about how the ball acts - it's how the environment changes. Feb 7, 2020 at 13:45
• @PedroHablespanyos So there is no friction between the ball and the train? Feb 7, 2020 at 14:58
• @PedroHablespanyos the path of the ball is curved in the train's frame therefore there is a force in that frame. If there is no real force then that force is an inertial force which indicates that the train is non-inertial and is accelerating in the opposite direction with respect to an inertial frame.
– Dale
Feb 7, 2020 at 16:14

# This answer is incorrect because it does not account for the balls moving backwards with respect to the train.

In my pictures the train will be moving upwards (positive Y direction).

If the train retained its velocity in the positive Y direction when it turned, we can get the ball to hit either wall, depending on where the ball is, in relation to the centre of rotation. You can see how the balls have been moved by hitting the walls. The green ball hits the right wall (or, we should say the wall hits the ball!), the yellow ball hits neither wall, and the purple ball hits the left wall.

However, trains do not move sideways! Trains always move forward or backward. If the train turns 90 degrees to the left, it won't continue moving along the positive Y direction. It will now be moving along the negative X direction. If the train stays the same speed, once it starts turning left, all the balls will hit the right wall (actually the right wall will hit the balls) because the train is now moving left (negative X direction) and the balls are not:

They will also roll towards what used to be the front of the train (not shown), because the train now has less velocity in the positive Y direction than the balls do.

Conclusion: The balls will hit the right wall if the train turns left; by mirror symmetry, they will hit the left wall if the train turns right.

• But does this not only work if the balls are moving forward WRT the train? Looking at your second diagram, the ball is moving in a forward direction. But if the ball was hit back with velocity while on the train, it will always be moving slower than the train, meaning the ball would move backwards and hit the left wall. If you were to draw the same diagram but turning to the right, the balls would hit the left wall in a different path, not the one in the q. I don't mean to be stubborn here, and I know the answer is supposed to be right (Australian HSC Q14 2015), but I just don't understand why Feb 7, 2020 at 23:35
• Also, presumably, if they were moving forward WRT the train, then if the train turns right, they would make a different curve to that of the one asked in the question - that curve would hit the left wall in the top left corner. That isn't what the question asked, and since neither going left or right works if the ball's velocity is greater than the train's while turning, we must assume that the ball's velocity is less than that of the train's. In other words - even with this description, the balls make the wrong path.The only possible scenario is if they moved w less velocity than the train. Feb 7, 2020 at 23:51
• @PedroHablaspanyos again, the velocity is irrelevant here. Only the acceleration matters.
– Dale
Feb 8, 2020 at 0:51
• @PedroHablespanyos The balls are moving at the same speed as the train (so if you're inside the train you won't see them move). The train is going up, and when the train turns, the train is going up and left, but the balls are still only going up - they aren't going left - so once the train goes far enough left, the balls have to hit the wall. Feb 9, 2020 at 12:49
• @Dale My answer does not use acceleration, only velocity in an inertial frame. Feb 9, 2020 at 12:50

Is train, therefore, moving to the left or the right ?

With respect to what ? There is no absolute movement and as such every movement must be analyzed from some reference point. So the answer is - depends. Therefore train has speed $$v_{train/ground}$$ with respect to ground, but at the same time it has also speed $$v_{train/ball} = -v_{ball}$$ with respect to cue ball. Train speed with respect to pool table is $$v_{train/table} = 0$$ and if there is some other train B at the moment going by near first train A, but in opposite direction, then train's A speed with respect to train B is: $$v_{trainA/trainB} = v_{trainA/ground}+v_{trainB/ground}$$ If train is moving away from some level crossing and at same moment some car is also receding from that crossing in a road, then train speed with respect to that car is : $$v_{train/car} = \sqrt{v^2_{train/crossing}+v^2_{car/crossing}}$$, according to Pythagorean theorem. Hope that helps !

You have to distinguish between velocity and acceleration. If the train is moving with constant speed you will notice nothing different, the ball will move in a straight line. You can't tell in which direction the train is moving without looking outside the window. You can only tell if the train is being accelerated. Only in this case you will see the path of the ball deviate from a straight line as if there was a force applied to it, a so called pseudo force.

• The reason that the path of the ball deviates is not because a force is applied, but rather it seems as if the path has deviated because the pool table itself has moved. So, thats why the path changes from being perfectly straight. By moving left or right, I do not mean that it moves to the left or right of the diagram - I mean it would move either up the diagram or down the diagram (Basically rotate the diagram 90 degrees so the train motion faces upwards, then consider if it moves to the left or the right. Feb 7, 2020 at 5:35
• When you say "the pool table itself has moved" you mean that it has changed from being stand still to having a velocity > 0, in other words, it has accelerated. It is the acceleration that causes the path of the ball to deviate. If, on the other hand, the table was originally moving and it continued moving, the path will be a straight line. Velocity does not cause any observable effect, acceleration does.
– Leo
Feb 7, 2020 at 8:53

If we ignore all forces and only thinking about the relative motion of the train and the ball (which I think is what you're asking) then we can simplify the scenario.

Assuming that initially the train isn't turning there is no relative motion so we can ignore the fact that the train is moving forward (since the ball is also moving forward). For now I'll also ignore the table since it has a limited effect on the outcome.

As the train begins turning, regardless of whether the ball is pushed back or not there is now relative motion.

If we take the ball as our reference point and imagine it inside a rectangle, the train turning will manifest as the rectangle rotating. As other answers have mentioned, where the ball is inside the rectangle relative to the centre of rotation will decide which wall it hits.

This still applies when the ball is moving backwards. In fact the backwards motion only affects which side is hit if it causes the ball to cross the centre of rotation.

edit: because of how train tracks work I think that the center of rotation begins at the front of the carriage so the ball will always be behind the centre of rotation. This isn't a very scientific answer but maybe it helps you visualise what's happening

You are 100% right. The cue ball moved(not actually moved but appeared to be moving for you) because the table along with the train moved towards left . The pen and paper experiment you performed will also give a similar result if you use a ball instead of a pen(do not apply a force to the ball, it will make it complicated and difficult to perform. Just move the paper in forward+leftward direction and you will get the result).

• Why that downvote Feb 12, 2020 at 5:08