Motion of dropped object relative to an accelerating observer

Question

I had a question from my physics test today, as we think the supplied answer is wrong (even my teacher does).

So the question goes like this: A mass (ball, let's say) is hanging from the top of an accelerating train. (on Earth) The train is also travelling on a horizontal track. The string holding the mass is shown to be at an angle due to inertia.

If cut, what will be the path traced by the mass relative to the train.

We hypothesised in class that it would follow a "cut off $\frac{1}{x}$ relationship" since the ball is falling at the same rate however the train is getting further and further away from its release point faster.

However, we thought this would only work for when the train is accelerating at >$9.8\frac{m}{s²}$ because the vectors would cancel out when both objects are accelerating at $9.8\frac{m}{s²}$ down, thus leaving a straight (diagonal) line for the ball's path.

If these hypotheses are correct so far (which we hope so) then the question begs to be asked.. What is the path when the train is accelerating slower than $9.8\frac{m}{s²}$ but greater than zero? This has us stumped.

So to recap:

What path does the mass take (when cut) when the train is accelerating at >$9.8\frac{m}{s²}$.

What path does the mass take (when cut) when the train is accelerating at $9.8\frac{m}{s²}$

What path does the mass take (when cut) when the train is accelerating at <$9.8\frac{m}{s²}$

These are all relative to the train.

A visualisation would be great, but a good explanation would suffice.

Thanks

Answer 1

From the point of view of an observer on the train the ball starts from rest.
It undergoes a downward acceleration of $g$ and an acceleration in the opposite direction to that of the train $a$.

This will result in a straight line trajectory for the ball along a line whose angle below the horizontal is $\tan^{-1}\left (\frac {g}{a} \right)$ which is exactly the same angle that the string made with the horizontal before it was cut.

Since the downward displacement $z=\frac 12 gt^2$ and the horizontal displacement is $x=\frac 12 at^2$, where $t$ is the time from which the string was cut, $z= \frac g a \,x$ which is of the form $z = mx +c$, the general equation of a straight line.

Answer 2

If not relativity mechanics then i can provide you a simple answer.

I''m not exactly familiar with "cut off $\frac{1}{x}$ relationship" yet, What do I think is that it's trajectory each and every time will be a simple parabola following the given particulars.

$v_o=0\vec{i} + 0\vec {j} $ $ms^{-1}$ and $a=-a_t\vec{i}-g\vec{j}$ $ms^{-2}$

I assume that train is accelerating along positive $x$ axis at $a_t$ $ms^{-2}$.

So, with respect to a frame inside the train, the ball will be experiencing a psuedo force opposite to the motion of train.

Rest, I guess , is simple mathematics