# Avg. Acceleration of ball against a wall with two velocities

1. Sep 15, 2009

### shizupple

I searched the forum before posting and couldnt seem to find anything on the acceleration off a wall that wasnt dealing with force as well.

1. The problem statement, all variables and given/known data
A tennis ball with a velocity of +9.6 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of -5.2 m/s (to the left). If the ball is in contact with the wall for 0.008 s, what is the average acceleration of the ball while it is in contact with the wall?

Vi=9.6m/s
Vf=-5.2m/s
t=.008s

2. Relevant equations
Aavg = Vf-Vi / Tf-Ti
Vf = Vi+a(Tf-Ti)

3. The attempt at a solution
9.6-5.2 / .008 = 550 m/s^2 left ???
I realize that answer is nowhere close to correct. I must be using an incorrect formula because its dealing with the wall. If anyone could guide me to a correct formula that would be awesome!

2. Sep 15, 2009

### Gear300

The answer is actually correct (unless the book says it isn't). Collisions produce a large amount of force that varies through a small interval of time (this, I think, could be modeled under the Dirac delta). However, the force varies throughout the time interval, so the acceleration upon leaving the wall is probably not 550 m/s^2; this value, however, looks to be the average acceleration across the time interval (which is what you're asked for).

If you want to look this sort of thing up under the forums or anywhere else, it would usually fall under impulse or momentum.

3. Sep 15, 2009

### shizupple

Hmm well i tried entering 550 (m/s^2) left (on webassign) and it said the 550 was wrong but the direction was correct? I just dont see how it can be accelerating at 550 m/s^2 when it is pretty much in contact with the wall?

4. Sep 15, 2009

### Gear300

Sorry about that - I didn't do the math so I overlooked it...your method is correct, but remember that the final velocity is in the opposite direction of the initial, so the two values should have opposite signs. It looks like the actual answer is even larger than 550m/s^2.

Try to think of the amount of force needed to send a ball coming at you at 10m/s in the opposite direction with a speed of 6m/s all in the range of .008s.

Last edited: Sep 15, 2009
5. Sep 15, 2009

### shizupple

we havent talked about force yet. i'm in a high school honors physics class. there shouldnt need to be anything dealing with force in this problem though should there? its just acceleration? and are you saying i need to change the sign on the 550 or i need to do (9.6+5.2)/.008 = 1850? And the time interval is not .008s, that is the time it is in contact with the wall.

6. Sep 15, 2009

### Gear300

(9.6+5.2)/.008 = 1850 - this is right (and the direction is to the left). Just know that the ball accelerated from 9.6m/s in one direction to 5.2m/s in the opposite direction in a time interval of .008s.

7. Sep 15, 2009

### shizupple

Wow! This is amazing. I would have never thought that was right. Thats crazy to think that its accelerating at 1850 m/s^2 in .008s. Thanks for your help!

8. Sep 15, 2009

### pr0blumz

Why wouldn't it be (-5.2m/s - 9.6m/s)/(.008s)= -1850m/s/s. The negative tells you it's to the left. Am I wrong for thinking this way?

9. Sep 16, 2009

### Gear300

You're doing it the more precise way...though it is relatively the same as saying 1850m/s/s to the left (the negative indicates to the left).

10. Sep 16, 2009

### semc

hello shizupple!! i read through the post and in the 1st post you said acceleration off the wall. personally i think its the wrong concept if you are using projectile motion to solve this problem. I think the question is trying to ask something like F=I/t where I is impluse. Although you get the same answer but i believe that is the correct concept just my 20c ::

11. Sep 17, 2009

### shizupple

Oh ok. Well we havent talked about any of that yet. I'm just in a high school honors physics class.