Physical Significance of Numbers

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Discussion Overview

The discussion revolves around the physical significance of the two possible solutions for the final velocity of a ball falling freely under constant acceleration. Participants explore the implications of these solutions in the context of physics, particularly focusing on the conditions under which each solution might be considered valid.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a scenario involving a ball falling freely with a constant acceleration of 10 m/s² and calculates the final velocity after traveling 10 m, arriving at two possible solutions: +10 m/s and -10 m/s.
  • Another participant questions the validity of dividing by acceleration in the context of vector quantities, suggesting that the signs of the vectors must be consistent.
  • A different perspective is introduced, considering a stationary ball below the falling ball and calculating its velocity in relation to the free-falling ball, as well as relating it to a rocket's vertical takeoff.
  • A participant reiterates the existence of two solutions and seeks clarification on the physical reality of the positive velocity solution, suggesting that it may relate to the ball's trajectory prior to reaching the initial condition of 0 m/s.
  • A later reply expresses understanding of the previous explanations, indicating a resolution of confusion for that participant.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the physical reality of the positive velocity solution, and multiple interpretations of the problem remain. The discussion includes both technical clarifications and speculative reasoning.

Contextual Notes

There are unresolved assumptions regarding the interpretation of velocity signs and the implications of vector analysis in this context. The discussion does not fully address the mathematical steps leading to the two solutions.

gamemania1986
Consider this case: A ball is allowed to fall freely with a constant acceleration of 10 m/s^2. What will its speed be after traveling a distance of 10 m?

We get that:
v[initial] = 0 m/s
d = -5 m (downward direction is taken as negative)
a = -10 m/s^2

To find v[final], we will use the formula d = (v[final]^2-v[initial]^2) / (2a)

Rearranging, we will get:

v[final]^2 = 2ad + v[initial]^2

By plugging the numbers, we will get:

v[final]^2 = 100 m^2/s^2

By mathematics, there are 2 solutions to this problem, that is v[initial][1] = +10 m/s and v[initial][2] = -10 m/s. I can only think of the situation where the ball has a downward velocity. Is the other answer (the one with +10 m/s velocity) possible/does it has a physical reality? If so, can anyone explain the situation? Thanks a lot!
 
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I think to obtain your first formula, the identity
vf - vi = a*t
is used.
Next, you divide by a to get t, and plug that into d = a/2 * t^2.

But strictly speaking, d, v, and a are vectors.
So you are in fact not allowed to divide by a.

You could work around this by saying you just talk about one component (the z-component) of each vector. But this implies that (vf-vi) has the same sign as a (i.e., minus).

Of course, this is still true in the answer.
 
Hi,
u consider a stationary ball 'b' directly below the falling ball and calculate the velocity of 'b' in + with respect to free falling ball.

Else, if free fall is not crucial, vertical take off of a rocket (containing the ball) with acceleration 10 m/s^2. v = u + at = final velocity of both rocket & ball.
 
Last edited:
By mathematics, there are 2 solutions to this problem, that is v[initial][1] = +10 m/s and v[initial][2] = -10 m/s. I can only think of the situation where the ball has a downward velocity. Is the other answer (the one with +10 m/s velocity) possible/does it has a physical reality? If so, can anyone explain the situation? Thanks a lot!


The -10 m/s velocity occurs when you trace the evolution of the system going forward in time. Consider the trajectory of the ball before it had the 0 m/s velocity.

Hurkyl
 
Aaaah... I got it! Thanks a lot Hurkyl!
 

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