# Acceleration right after ball bounces

1. Mar 28, 2013

### SweatingBear

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?

2. Mar 28, 2013

### Staff: Mentor

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?

3. Mar 28, 2013

### technician

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

4. Mar 28, 2013

### WannabeNewton

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: Mar 28, 2013
5. Mar 29, 2013

### SweatingBear

I have posted the question just as it is formulated.

6. Mar 29, 2013

### jbriggs444

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?

7. Mar 29, 2013

### Staff: Mentor

You got it!

8. Mar 29, 2013

### SweatingBear

I have no idea, care to tell me?

9. Mar 29, 2013

### Staff: Mentor

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. Mar 30, 2013

### SweatingBear

At this stage, we are not considering the drag force. Other than that, I am not getting anywhere. Care to show me?

11. Mar 30, 2013

### WannabeNewton

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. Mar 30, 2013

### jbriggs444

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: Mar 30, 2013
13. Mar 30, 2013

### Staff: Mentor

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. Mar 30, 2013

### HallsofIvy

Staff Emeritus
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. Mar 30, 2013

### Staff: Mentor

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. Mar 30, 2013

### Redbelly98

Staff Emeritus
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. Mar 30, 2013

### WannabeNewton

I forgot to divide by 2 from doubling the height :rofl: Yeah the average rate of change still comes out to g.

18. Mar 30, 2013

### SweatingBear

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. Mar 30, 2013

### Staff: Mentor

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.

Sure. The ground exerts an upward force on the ball during the bounce, but that's irrelevant here.

See post #12.

20. Mar 30, 2013

### Redbelly98

Staff Emeritus
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.