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  1. This may look like a stupid question, but it is really getting to me. Imagine a train moving with an acceleration $a$, and a person drops a stone from the window. To an observer on the ground, the stone follows a parabolic path, as it is a projectile with initial velocity the same as the velocity of train when dropped. However, the person who dropped it sees that it falls down in a straight line. Why? Can someone explain the reason to me?

  2. What will the acceleration of the stone measured by an observer on the ground. I think it should be $a_{net}=\sqrt{g^2 + a^2}$, but I have a book which says it should be $g$. No explanation is provided. Help me out please.

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    $\begingroup$ For both questions, because the stone isn't accelerating - it's moving horizontally at the same speed as that of the train when the stone was dropped, but acceleration would require a constant force being applied on it. $\endgroup$ – Abhimanyu Pallavi Sudhir Sep 9 '15 at 11:40
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  1. You're not correct that the stone will appear to fall straight down if the train's acceleration is $\neq 0$. It would appear to fall in a (straight) slanted line (with angle $\arctan\left(\frac{a}{g}\right)$) because the stone is accelerating in the $y$ axis (due to gravity) and the observer in the train is accelerating in the $x$ axis. From outside the train, it would appear to fall parabolically because it has an initial velocity in the $x$ axis (due to the fact it was moving with the train when it was dropped) and it accelerates in the $y$ axis.

  2. The answer is $g$ because the only force acting on the stone is gravity, and the acceleration on all free falling bodies on earth is $g$.

It might help if you draw a free body diagram of the problem to see why this is the case. It looks like you're a bit confused about why things accelerate. Acceleration is always due to a force, and if you can't point to a force that causes the acceleration (like: gravity, friction on the rails moving the train forward, etc) then there can't be acceleration in that direction. What forces are acting on the stone that make you think it'll keep accelerating with the train once it's separated from the train?

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  • $\begingroup$ I agree that there is no apparent force that would make it accelerate,and since it is no longer in contact with the train, it loses the acceleration. But then compare it to an arrow released from a bow. The string pushes the arrow, accelerates it, and even when the arrow is no longer in contact with the string, shouldn't it retain the acceleration? $\endgroup$ – GRrocks Sep 9 '15 at 12:08
  • $\begingroup$ No, arrows don't accelerate when they're not experiencing a force. By definition a force is what causes acceleration (recall $F = ma$, that is the definition of a force). So if you can't point to an agent that provides a force to accelerate the object, the object cannot be accelerating. $\endgroup$ – cyphar Sep 9 '15 at 12:09
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    $\begingroup$ It's got nothing to do with reference frames. Take a step back from relativity and go back to the fundamentals of Newtonian motion. This behaviour is entirely defined by Newton's laws of motion. The reason the stone "retains" (a bad word to use, more accurately it isn't accelerated by a force) its velocity is because of Newton's 1st law (there's no force acting to affect the $x$ velocity, so it is unchanged). The reason the acceleration is not "retained" is for the same reason -- once it's no longer part of the train there is no force in the $x$ direction and thus no acceleration in $x$. $\endgroup$ – cyphar Sep 9 '15 at 12:32
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    $\begingroup$ @Aniket uses the term "jumps between reference frames" to describe this, but I take issue with that terminology because it's not caused by the change in reference frames it's caused by the lack of a force on the object. The fact it isn't in an accelerating frame of reference (in terms of an $x$ acceleration) any more is the effect of there not being a force on the object in that direction anymore. $\endgroup$ – cyphar Sep 9 '15 at 12:33
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    $\begingroup$ thanks a bunch @cyphar....sometimes we tend to get so invloved in the complexities that we forget the basics....guess it was a stupid question after all :)....thanks $\endgroup$ – GRrocks Sep 9 '15 at 12:35
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To an observer on the ground, the stone follows a parabolic path, as it is a projectile with initial velocity the same as the velocity of train when dropped. However, the person who dropped it sees that it falls down in a straight line.

The stationary observer himself being at rest observes that the stone has a horizontal velocity and a vertical downward acceleration. So to him the stone has a parabolic path.But the man in the train will actually see the stone fall in a straight slanted line. Because the man in the train has an additional acceleration of 'a' along the horizontal direction, which the stone does not have. So he sees the stone with a relative horizontal deceleration and a downward acceleration. Hence he sees the stone fall in a 'straight slanted line' and not in a straight line as the train is accelerating**.

What will the acceleration of the stone measured by an observer on the ground. I think it should be $a_{net}=(g^2+a^2)^{1/2}$, but I have a book which says it should be g.

Here, I must say that when a stone is dropped from an accelerating train, the moment the stone loses contact with the man on the train, it no longer experiences the horizontal acceleration. It has a free fall ,but it retains the horizontal velocity it had at the moment the stone loses contact with the man on the train. Hence the net acceleration is 'g'.

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  • $\begingroup$ I'm fairly sure it'll appear to fall in a straight diagonal line, because from the frame of reference of the train observer, there's acceleration in two axes. $\endgroup$ – cyphar Sep 9 '15 at 11:44
  • $\begingroup$ It moves backwards but not in a parabola. From the PoV of the train, the stone moves in a straight line down and to the rear. We have to add two orthogonal acceleration vectors; g downwards and -a backwards. We get a diagonal straight line whose gradient is -a/g. $\endgroup$ – Oscar Bravo Sep 9 '15 at 11:50
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    $\begingroup$ @Aniket - Don't like to be picky, but diagonal is exactly accurate. There's nothing about a diagonal having to be at 45 deg.. If that were so, you couldn't talk about the diagonal of a rectangle; and you can [mathsisfun.com/definitions/diagonal.html] $\endgroup$ – Oscar Bravo Sep 9 '15 at 12:03
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    $\begingroup$ @GRocks in order for an object to accelerate, there must be a force acting upon it. Can you point to the force which would cause it to keep accelerating in the same direction as the train? $\endgroup$ – cyphar Sep 9 '15 at 12:06
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    $\begingroup$ @cyphar is right. Force is the cause and acceleration is its effect. When the stone jumps from the accelerating frame to a non-accelerating frame, it loses the acceleration because the force that was acting on it in the train no longer acts on it now. $\endgroup$ – SchrodingersCat Sep 9 '15 at 12:17
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The path of the stone for the observer on the train should be a straight line as, when releasing the stone, it had the velocity same as train. However, air resistance may affect the stationary path. For the observer on ground it should be parabolic due to the action of gravity on the moving stone. The net acceleration of stone should be $g$ since after dropping it from the train it is only accelerated by gravity.

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  • $\begingroup$ While this is correct, it does not talk about how the train in the question has a constant acceleration and not a constant velocity. $\endgroup$ – Martin Dec 7 '15 at 17:02
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First, shall I mention that you have not mentioned air resistance in your problem. Because in this example, air resistance would mean that the observer on the train will actually not see the stone fall in a straight line (including the diagonal mentioned by @Aniket) and there is a horizontal deceleration experienced.

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    $\begingroup$ This is more a comment than an answer. Please for this type of writing use the comment section below the question. $\endgroup$ – rmhleo Sep 9 '15 at 12:38
  • $\begingroup$ I would if it would let me. $\endgroup$ – nitarshs Sep 21 '15 at 15:46

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