How Does a Ping Pong Ball Behave in Zero Gravity?

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SUMMARY

The discussion revolves around the behavior of a ping pong ball in zero gravity, specifically how it interacts with an astronaut and a wall. The astronaut, moving at 4 m/s, reaches the wall in 3 seconds, while the ball bounces back and forth at 11 m/s. The total distance the ball travels during this time is calculated to be 21 meters, considering the relative velocities of both the astronaut and the ball. The average acceleration of the ball during a bounce lasting 25 ms is also discussed, emphasizing the complexities of momentum transfer and the implications of bounce duration on the ball's motion.

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  • Familiarity with momentum transfer principles
  • Basic grasp of physics in microgravity environments
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Students and enthusiasts of physics, particularly those interested in mechanics, kinematics, and the effects of microgravity on motion. This discussion is beneficial for anyone studying space physics or engaging in practical applications of physics in zero-gravity environments.

robax25
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1. Problem Statement
An astronaut is playing with a ping pong paddle and ball in zero gravity.
The astronaut is 12m from a flat wall and floating toward it at 4m/s .
A ping pong ball bounces back and forth between his paddle and the wall at 11m/s.
Eventually, the astronaut reaches the wall and traps the ball against it.

1)how long does it take the Astronaut to reach the wall?
Ans: 3s.
2) Through what distance does the ball travel during this process in m?

3) If the bounce lasts 25ms. what is the average acceleration of the ball m/s²?

Homework Equations


Height h=ar^(n)
a= first amplitude ,
r=second amplitude
n=number of bounce...1,2,3,4,...

The Attempt at a Solution


t=X/V so time is 3s.

relative velocity=11m/s-4m/s=7m/s
 
Last edited:
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Instead of trying to calculate and sum the distances of each individual bounce, what do you know about the average speed of the ball and the total time it is in motion?
 
Average speed of the ball is 11m/s and total time in motion is 3s
 
robax25 said:
Average speed of the ball is 11m/s and total time in motion is 3s
Yes, so what distance does it cover in that time?
 
12m
 
robax25 said:
12m
How did you calculate that?
 
x=vt so 33m
 
robax25 said:
x=vt so 33m
Yes, much better :smile:
 
Yes but it is not the right answer. We need to consider relative velocity because Astronaut and ball both are moving.
relative velocity= 11-4=7m/s

x=7m*3s=21m.
 
  • #10
robax25 said:
Yes but it is not the right answer. We need to consider relative velocity because Astronaut and ball both are moving.
relative velocity= 11-4=7m/s

x=7m*3s=21m.
Then I submit that the "right answer" isn't correct. The ball is moving at 11 m/s for three seconds.

Besides, sometimes the ball is moving away from the astronaut, sometimes towards him. There would be different relative velocities for the two cases.
 
  • #11
Yes you are right. I am wrong. However, why there is two different relative velocities. yes, first case, the ball is moving away and second case ball is moving towards him but the velocity is constant for Astronaut(4m/s) and ball (11m/s). For my point of view, there can be two relative velocities because of displacement.
 
  • #12
robax25 said:
Yes you are right. I am wrong. However, why there is two different relative velocities. yes, first case, the ball is moving away and second case ball is moving towards him but the velocity is constant for Astronaut(4m/s) and ball (11m/s). For my point of view, there can be two relative velocities because of displacement.
What precisely do you mean by relative velocity? The relative velocity between what two things?

An observer in the rest frame of the spacecraft would judge the astronaut to have a constant velocity, but would see the ball's velocity changing between +11 m/s and -11 m/s.
 
  • #13
For fun, be aware that if the bounces were instantaneous, then the ball would bounce an infinite number ot times (in this mathematical model), but if we decide that each bounce requires a non-zero finite amount of time (it would be a different model), then the number of bounces would be finite, and you can calculate that exact finite number of bounces (just in case you want to have some fun).
 
  • #14
mattt said:
if we decide that each bounce requires a non-zero finite amount of time (it would be a different model), then the number of bounces would be finite
If the sum of the series of bounce times converges then there could be infinitely many.
 
  • #15
haruspex said:
If the sum of the series of bounce times converges then there could be infinitely many.

Yeah, I mean, if each bounce has the same non-zero duration, then there can only be a finite number of bounces, but yes, if we allow the duration of the bounces to be smaller and smaller in an appropriate way, then yes, there could still be an infinite number of bounces.
 
  • #16
robax25 said:
Yes you are right. I am wrong. However, why there is two different relative velocities. yes, first case, the ball is moving away and second case ball is moving towards him but the velocity is constant for Astronaut(4m/s) and ball (11m/s). For my point of view, there can be two relative velocities because of displacement.
(Somewhat belatedly) The question implies the ball's speed is constant. In practice, it slows a little each time it bounces from the wall because some momentum is transferred to the spacecraft .
Likewise, the astronaut has to hold the paddle nearly still in the reference frame of the spacecraft during each contact in order to get the ball back up to 11m/s in the spacecraft 's frame, and no more.
And then there is the slowing of the astronaut. Is it clear that the astronaut can reach the wall?
 
  • #17
haruspex said:
(Somewhat belatedly) The question implies the ball's speed is constant. In practice, it slows a little each time it bounces from the wall because some momentum is transferred to the spacecraft .
Likewise, the astronaut has to hold the paddle nearly still in the reference frame of the spacecraft during each contact in order to get the ball back up to 11m/s in the spacecraft 's frame, and no more.
And then there is the slowing of the astronaut. Is it clear that the astronaut can reach the wall?
This is a variant of the well-known "Bird and train problem" (bird flying back and forth between 2 approaching trains. With a bird there is no problem keeping the speed constant.

I think the astronaut can reach the wall. There is a minimum time given for a bounce, so there can only be a finite number of bounces. and only a finite amount of momentum can be transferred.
If there was no minimum bounce time, there's no limit to the number of bounces, and since a constant amount of momentum gets transferred every time, the astronaut will stop and reverse direction.

BTW 25 ms is much too long for the bounce time. The ball travels 275 mm in that time. the paddle/walls are apparently very elastic. This makes the total distanced traveled by the ball unclear, since we don't really know how much it travels during a bounce.
 
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  • #18
I think for questions 1) and 2) the problem wants us to assume that
  • the bounce time is zero
  • the momentum transfer from the ball to the spaceship is negligible
  • the momentum transfer from the ball to the astronaut is also negligible
 
  • #19
If the bounce lasts 25ms. what is the average acceleration of the ball m/s²?
Can anybody solve this?
which equation to use?
What will be the vf, time, displacement and vi?
 
  • #20
G9raz said:
If the bounce lasts 25ms. what is the average acceleration of the ball m/s²?
Can anybody solve this?
which equation to use?
What will be the vf, time, displacement and vi?
I think we should use $$a=\frac{\Delta v}{\Delta t}$$, with ##\Delta t=0.025## , ##\Delta v=11-(-11)##.
 
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Likes G9raz
  • #21
Delta2 said:
I think we should use $$a=\frac{\Delta v}{\Delta t}$$, with ##\Delta t=0.025## , ##\Delta v=11-(-11)##.
Thank u very much
I'm very grateful
 
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  • #22
10 The engines of a rocket sitting on the ground fire to accelerate the rocket at 3.5\,\mathrm{m/s^2}3.5m/s2 until it reaches an altitude of 1.2\,\mathrm{km}1.2km. At that point, the engines fail and the rocket is in free fall.

What height does the rocket reach in \mathrm{m}m? Assume a constant acceleration of gravity of 9.8\,\mathrm{m/s^2}9.8m/s2 even at these high altitudes.

Can anybody solve this too?
 
  • #23
G9raz said:
10 The engines of a rocket sitting on the ground fire to accelerate the rocket at 3.5\,\mathrm{m/s^2}3.5m/s2 until it reaches an altitude of 1.2\,\mathrm{km}1.2km. At that point, the engines fail and the rocket is in free fall.

What height does the rocket reach in \mathrm{m}m? Assume a constant acceleration of gravity of 9.8\,\mathrm{m/s^2}9.8m/s2 even at these high altitudes.

Can anybody solve this too?
Per forum rules, please post this as a new thread AND show some attempt.
 
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  • #24
G9raz said:
10 The engines of a rocket sitting on the ground fire to accelerate the rocket at 3.5\,\mathrm{m/s^2}3.5m/s2 until it reaches an altitude of 1.2\,\mathrm{km}1.2km. At that point, the engines fail and the rocket is in free fall.

What height does the rocket reach in \mathrm{m}m? Assume a constant acceleration of gravity of 9.8\,\mathrm{m/s^2}9.8m/s2 even at these high altitudes.

Can anybody solve this too?
Welcome to physics forums. Here in PF we don't solve your problems for you. You have to write your attempt at the solution and we can guide you by telling you where you go wrong, giving you hints on how is the correct solution e.t.c.. As a first step post this problem as a new thread and post also your solution...
 

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