10
$\begingroup$

Suppose a ball is flying towards me at a speed of 10m/s and that, on impact, I feel "x" amount of pain.

If, instead, it was me flying towards the ball at the same speed, with all other conditions being the same, would I feel the same amount of pain?

$\endgroup$
0

7 Answers 7

34
$\begingroup$

Look at it this way:

Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball at 10 m/s (= your second experiment). Since it is the same experiment, you will feel exactly the same pain.

$\endgroup$
0
14
$\begingroup$

Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames.

However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if their function is symmetric or anti-symmetric.

$\endgroup$
0
10
$\begingroup$

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation)

Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.

$\endgroup$
0
4
$\begingroup$

No mater the inertial referential: The pain is linked to the energy dissipated by the change of speed of the 2 objects in any inertial referential.

Remarque 1: a part of the energy can be absorbed by the ball by deformation or heating. So a soft ball would make less pain.

Remarque 2: The pain depends on what is static behind you and what is static behind the ball. Example in a squash court: you at 50km/h or ball at 50km/h will not produce the same pain further after the ball contact.

More over if the thing behind you or behind the ball is really heavy, it will produce gravity attraction. As attraction is proportional to the square of the distance with the attractive object, you and the ball will not undergo the same acceleration. So speed will be altered.

$\endgroup$
1
$\begingroup$

Pavel Petrman gave a different perspective, but since impact remains same pain should remain same, given that properties of the situation remain same. To try take a ball in your hand and run towards a weighing scale and hit it and in next case exchange ball and weighing scale and repeat the procedure. Take a note of readings on weighing scale.

$\endgroup$
1
$\begingroup$

A lot of the answers get distracted by the "feeling pain" part of the question. So let me start by focusing on that - simplifying a very complex psycho-physiological issue to a simple physical statement (this is a physics forum - we leave the other stuff for other sites):

During the impact of two bodies, there will be an exchange of momentum. The force required for this exchange of momentum is responsible for the pain.

The momentum exchange is mediated by a force $F(t)$ by one body on the other (and, by Newton's law, this force is reciprocal); the exchanged momentum is given by $$\Delta p = \int F(t) dt$$ with the integral taken over the duration of the collision (or any longer time than that, given that $F(t)=0$ for other times by my definition of it). "Pain" is a result of both the magnitude and duration of $F$ - so the question can be restated as

For a collision between two bodies, does the force-time curve describing the impact depend on which of the bodies is moving?

The answer to that question is (within the bounds of classical mechanics) a categorical "no". The force depends only on the initial difference in velocity and the shape / materials properties of the two objects. Several answers already addressed that issue in detail - I quite like the example by Rob who describes the experiment taking place on a train, but here's an even better one:

You are on the baseball field, and the ball is traveling towards you. There are two observers: one is standing right next to you, while the other is traveling on a train at the same speed as the ball. For both observers, it is impossible to see any of the ball park - so they don't have an "external" frame of reference, only their own.

They observe the same event - same look of surprise, eyes closing in anticipation, face distorting during impact, cry of pain. One of them thinks you hit the ball; the other thinks the ball hit you. Same experiment, different observer, same outcome.

IT DOES NOT MAKE A DIFFERENCE.

$\endgroup$
1
-5
$\begingroup$

Yes you will.. what really matters here is the relative momentum !

$\endgroup$
3
  • 12
    $\begingroup$ very short answer. Could be improved. $\endgroup$
    – tom
    Commented Dec 9, 2014 at 12:32
  • $\begingroup$ relative momentum is its answer. nothing is there to explain further. $\endgroup$
    – Vinayak
    Commented Dec 10, 2014 at 2:44
  • $\begingroup$ Maybe an equation that shows that force/impulse is a result of "relative momentum" (however you define that) would improve the answer. $\endgroup$ Commented Jul 30, 2015 at 13:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.