Acceleration right after ball bounces

  • Context: Undergrad 
  • Thread starter Thread starter SweatingBear
  • Start date Start date
  • Tags Tags
    Acceleration Ball
Click For Summary

Discussion Overview

The discussion revolves around the acceleration of a ball immediately after it bounces from the ground following a fall from a high altitude. Participants explore concepts related to elastic collisions, gravitational forces, and the effects of air resistance, while questioning the conditions under which the acceleration might be considered as 2g or simply g.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the acceleration immediately after the bounce could be 2g, questioning how to derive this value without a change in time.
  • Others argue that the acceleration should remain g after the bounce, suggesting that the only force acting on the ball is gravity if drag is not considered.
  • One participant highlights the change in velocity before and after the collision, noting that if the velocity before the collision is +v, then after it is -v, leading to a change of 2v.
  • Several participants discuss the role of air resistance and terminal velocity, with some asserting that the upward force during the bounce must be considered to understand the acceleration dynamics.
  • There is a suggestion that if air resistance is ignored, the average rate of change of velocity between collisions could be 2g, but this is contested by others who emphasize the necessity of including drag forces in the analysis.
  • One participant points out that during the contact with the ground, the upward acceleration is significantly greater than g, while after contact, the acceleration is downward as the ball's upward speed decreases.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the acceleration immediately after the bounce is 2g or g. Multiple competing views exist regarding the influence of air resistance and the conditions under which different accelerations apply.

Contextual Notes

Some participants note that the problem may be framed as a challenge, and the discussion reflects varying assumptions about the presence of air resistance and the nature of forces acting on the ball during and after the bounce.

SweatingBear
Messages
119
Reaction score
0
Suppose a ball falls from a very high altitude to the ground and bounces in form of a completely elastic collision. What then is the acceleration on the ball immediately after the bounce?

It is supposed to be 2g, but I have no idea how you are supposed to obtain that. We are not given the change in time, so how on Earth would one be able to calculate the acceleration?
 
Physics news on Phys.org
I guess I'm not understanding the problem. (Which is your problem as well, I think.)

After the bounce, why would the acceleration be anything other than g?
 
Are you sure you are not confusing velocity with acceleration?
If the velocity before the collision is +v then the velocity after is -v so the change in velocity is 2v
 
There are a few different steps to this. First off, what will be ##\Delta v##, the change in velocity instantaneously before the collision to instantaneously after? You will have to find the velocity immediately before hitting the ground - this should be straight forward for this system.

EDIT: Actually I just reread the way you phrased your question - are you sure it wasn't asking for the average rate of change of velocity between collisions?
 
Last edited:
I have posted the question just as it is formulated.
 
I think I get it.

If the ball has fallen from a great height, it must strike the ground at terminal velocity. That is the velocity where the upward force of air resistance is equal to the downward force of gravity for a combined net force of zero.

Immediately after the bounce, what is the new net force on the ball?
 
jbriggs444 said:
I think I get it.

If the ball has fallen from a great height, it must strike the ground at terminal velocity.
You got it! :approve:
 
jbriggs444 said:
Immediately after the bounce, what is the new net force on the ball?

I have no idea, care to tell me?
 
SweatingBear said:
I have no idea, care to tell me?

The net force on the ball will be the sum of all the forces acting it:
1) Gravity.
2) Air resistance. This is a function of the speed of the ball, whether it's moving down or up.
3) Anything touching the ball and pushing on it. After the bounce, is it in contact with ground?

Add these to get the force, then use F=ma to get the acceleration.
 
  • #10
Nugatory said:
The net force on the ball will be the sum of all the forces acting it:
1) Gravity.
2) Air resistance. This is a function of the speed of the ball, whether it's moving down or up.
3) Anything touching the ball and pushing on it. After the bounce, is it in contact with ground?

Add these to get the force, then use F=ma to get the acceleration.

At this stage, we are not considering the drag force. Other than that, I am not getting anywhere. Care to show me?
 
  • #11
If there is no drag force at all (i.e. the particle is in free fall) then the average rate of change of the velocity between collisions is in fact ##2g## and this is easily shown. This is just the change in velocity per collision divided by the time interval between collisions. Otherwise, I have to fall back on Doc Al's original post (post #2).
 
  • #12
WannabeNewton said:
If there is no drag force at all (i.e. the particle is in free fall) then the average rate of change of the velocity between collisions is in fact ##2g## and this is easily shown. This is just the change in velocity per collision divided by the time interval between collisions. Otherwise, I have to fall back on Doc Al's original post (post #2).

The average rate of change of velocity over time under the influence of gravity alone is equal to the average acceleration of gravity over time. If one is modelling gravity as a constant then this average is trivially one g. 2g is wrong.

But that's irrelevant to the problem as posed.

SweatingBear:

1. Suppose that the ball has mass m. What is the force of gravity on the ball?
2. Suppose that the ball is at terminal velocity. What is the force of drag on the ball?
3. Suppose that the ball hits the ground at velocity v. What is its velocity after a completely elastic collision with the ground?
4. Air resistance is often modeled as a force whose magnitude is purely a function of speed and whose direction is opposite to an object's movement. What is the force of drag on the ball just after it bounces off the ground?
 
Last edited:
  • #13
WannabeNewton said:
If there is no drag force at all (i.e. the particle is in free fall) then the average rate of change of the velocity between collisions is in fact ##2g## and this is easily shown.
How in the world are you deducing this?

The key to the problem, as pointed out by jbriggs444, is to consider air resistance and terminal velocity.
 
  • #14
Doc Al, the OP said, in post #10, "At this stage, we are not considering the drag force".

Sweating Bear, you have been told repeatedly that if there is no drag force and no one is touching the ball, then the only force acting on the ball is gravity and the acceleration of the ball is g, downward.
 
  • #15
HallsofIvy said:
Doc Al, the OP said, in post #10, "At this stage, we are not considering the drag force".
I know what he said. And he's obviously wrong! The only way the problem makes sense is if air resistance is considered.

I suspect that this might have been just tossed out as a challenge problem. You don't really need to know details about drag force, just the concept of terminal velocity.
 
  • #16
SweatingBear said:
At this stage, we are not considering the drag force.
Then the only force acting on the ball is gravity -- and you know what the answer must be in that case.

If the answer is not g, then some force other than gravity must be considered. Agreed?
 
  • #17
Doc Al said:
How in the world are you deducing this?
I forgot to divide by 2 from doubling the height :smile: Yeah the average rate of change still comes out to g.
 
  • #18
The problem is not that I do not understand that the downward acceleration is g, I understand that already.

Fact is that something must accelerate the balls upwards since it rebounds and begins to rise with the same velocity it had just before rebounding. How come that particular acceleration which it gains immediately after the rebound is 2g?
 
  • #19
SweatingBear said:
The problem is not that I do not understand that the downward acceleration is g, I understand that already.
That would be the acceleration if the ball were in free fall with only gravity acting. But as explained above, that's not the case here.

Fact is that something must accelerate the balls upwards since it rebounds and begins to rise with the same velocity it had just before rebounding.
Sure. The ground exerts an upward force on the ball during the bounce, but that's irrelevant here.

How come that particular acceleration which it gains immediately after the rebound is 2g?
See post #12.
 
  • #20
SweatingBear said:
Fact is that something must accelerate the balls upwards since it rebounds and begins to rise with the same velocity it had just before rebounding. How come that particular acceleration which it gains immediately after the rebound is 2g?
But those are two different accelerations you are talking about. During the contact with the ground, acceleration is upward and a lot larger than g (not merely 2g). After contact with the ground, the acceleration will be downward, not upward, as the ball's upward speed is decreasing from that point onward.

You really do need to consider air resistance here. If you were told not to, then you were told wrong.
 

Similar threads

Graduate Simulating an elastic bouncy ball
  • · Replies 10 ·
Replies
10
Views
3K
High School Understanding Physics for Coding Collision of 2 Balls
  • · Replies 14 ·
Replies
14
Views
4K
Undergrad Why don't we "fly up" in an accelerating elevator?
  • · Replies 11 ·
Replies
11
Views
4K
Undergrad Terminal velocity of two balls with different masses
  • · Replies 22 ·
Replies
22
Views
23K
Undergrad Calculate the *Acceleration* after a (mostly) elastic collision?
  • · Replies 3 ·
Replies
3
Views
2K
High School Help me understand jerk of a falling object hitting an "ideal foam"
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Momentum is not conserved with elastic bouncing ball?
  • · Replies 2 ·
Replies
2
Views
2K
High School The physics of flywheel launchers (like tennis ball shooters)
  • · Replies 60 ·
3
Replies
60
Views
7K
Undergrad How would you include gravity in a momentum problem?
  • · Replies 4 ·
Replies
4
Views
1K
High School Forces inside a moving elevator
  • · Replies 10 ·
Replies
10
Views
2K