Let's say we have a car moving on the horizontal ground with a velocity of $10 \hat{i} \frac{m}{s}$. A person inside throws a ball vertically upward with a velocity of $3 \hat{j} \frac{m}{s}$. Now I wanted to know a few things, first of all the initial velocity of the ball as seen by a ground observer would $(10 \hat{i} + 3 \hat{j}) \frac{m}{s}$. Now if the car maintains that uniform velocity, at every moment of the motion of the car, the velocity of the ball in the horizontal direction for the ground obsever would always be $ 10 \hat{i} \frac{m}{s}$ right? And secondly, if the car were to accelerate, then does the velocity of the ball in the horizontal direction also change with that for the ground observer? I personally think it does because the Car and the Ball are a system as a whole and hence the ground person cannot differentiate it, but I am not completely sure about it either. I would like to basically know if acceleration in the horizontal direction causes the horizontal velocity of the ball thrown in the car (upwards or with an angle, doesn't really matter) to change for a ground observer?
3 Answers
The car, the person throwing the ball, and the ball --- these three constitute a whole system as long as they are moving together. And when is that? When the person sits on the car and keeps the ball in his hand, the three move together with the same velocity. Now as soon as the person throws the ball, he no longer has contact with it. Hence, that person and the car can no longer affect the motion of the ball. Your confusion arises because you think that the ball is a part of the system of the car (and the person). Why? Because it is "inside" the car? Why would being "inside" the boundaries of the structure of the car make you a part of "system"? Once the person has thrown the ball, the car's acceleration cannot possibly alter the velocity of the ball because there is no force on it. If you're still confused about the ball being part of the system, think about throwing a ball. Go in an open area, throw a ball high in the air and start running at high speed in some random direction. A friend of yours standing nearby would not observe any change in the motion of the ball except in the vertical direction. To conclude, ---- to be able to change the velocity of an object, you must apply force on it. One way of applying a force is to keep it in contact with you. Since the nature of the force is complicated, it is simply referred to as being a "system" of all the objects touching each other. As soon as the "touching" is over, it is no longer a part of the system, hence no force, hence no changes in velocity for a ground-based observer.
-
$\begingroup$ Thank you for your answer, it makes sense to me. Although the last part where you tell about self experimentation, I think you meant that I throw the ball while running vertically upwards and then run with a higher velocity which in no way does interfere with the ground observer as they would still see that the ball is having the same horizontal velocity I had before I threw it. $\endgroup$– FloatossCommented Nov 18, 2021 at 8:14
-
$\begingroup$ Are the person and the ball not in contact via the air inside the car, though? If the air inside the car accelerates with the car, would that not accelerate the ball as well, albeit to a lesser extent? $\endgroup$ Commented Nov 18, 2021 at 16:25
-
$\begingroup$ @JoãoMendes That is an extremely negligible aspect, and I guessed that the question does not ask for such minor factors. $\endgroup$ Commented Nov 19, 2021 at 10:45
-
If the car is constantly accelerating after throwing the ball up, then the ball will retain the horizontal velocity which it gained at the moment it was thrown up(due to N2L) and if the car accelerates,
- the ball will move backwards with respect to the guy in the car,
- and for the guy on the ground, the velocity will be the same as it did when the ball was thrown up, and acceleration of the car won't affect his observations regarding the ball
Correct me if I am wrong.
-
$\begingroup$ Yes, it does make sense to me now. The "contact" is what is needed to transfer all such information. Thank you. $\endgroup$– FloatossCommented Nov 18, 2021 at 8:16
Yes your predictions are absolutely right. The ground observer seeing the ball thrown upwards in a car would see the velocity of the ball to be moving with $(10 \hat{i} + 3 \hat{j}) \frac{m}{s}$ and a person in the car will be seeing it as $3 \hat{j} \frac{m}{s}$. Yes if the car was moving at uniform velocity, the observer would be still seeing it as $(10 \hat{i} + 3 \hat{j}) \frac{m}{s}$ in the positive x direction. Also the person in viewing it in the ground frame sees the ball as a parabola(projectile) this is because it has a horizontal velocity and vertical velocity and the person in the car frame sees as if the ball is thrown vertically upward ( due to relative motion) If the car had accelerated in the horizontal x direction, it does affect horizontal motion but never affects vertical motion, the horiontal and vertical motion are completely independent. If the ball was thrown horizontally say in the same direction as Car's motion with $5\frac{m}{s}$ and person in ground would see it moving with $(10 \hat{i} + 5 \hat{i}) \frac{m}{s}$, i.e $15\frac{m}{s}$. If thrown at an angle find the respective components in the horizontal and vertical motion. So while solving problems use the kinematic equations independently in the horizontal and vertical motions and solve for what is being asked for. Aceeleration of the car does not affect velocity of the ball in the horizontal direction. The ball acquires the velocity of the car at the instant it was thrown and after that no matter what was the motion of car, it doesn't affect motion of the ball
-
$\begingroup$ You miss the point of my question, the previous two answers have cleared my doubt. "If the car had accelerated in the horizontal "x" direction, it does affect horizontal motion but never affects vertical motion" I never really asked about this. $\endgroup$– FloatossCommented Nov 18, 2021 at 8:19
-
1$\begingroup$ I hope now it's clear. Please let me know if you have any more doubts $\endgroup$– user318937Commented Nov 18, 2021 at 9:03