# Not sure of a homework solution involving average acceleration

A 50.0 g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval?

I am using the equation Aavg = (Vf-Vi)/(Tf-Ti) where Vf = 22, Vi=25, Tf = .0035 and Ti = 0

So, (25 - 22)/.00035ms = -857.14m/s? But that doesn't seem right.

I think I am getting confused because it is in milliseconds and am forgetting a conversion or something. Could anyone lend a hand? thanks

• 3/0.0035=857.14 (positive), but this not my answer... Sep 20 '11 at 17:59

First of all, velocity has a sign. After rebounding the velocity is in the opposite direction so $\delta V = (25 - (-22))= 47 m/s$

$47 m/s / 0.0035 s = 1,342.9 m/s^2$ [correction $13,429 m/s^2$]

• 47/0.0035 = 13428.57 and not 1342.9. Sep 20 '11 at 16:33
• Touche. Nice catch. Sep 20 '11 at 17:08

I will use the conservation of kinetic energy , $E_{in} - E_{out}=E_{loss}$.
Since $E_{loss}=m\cdot \delta v^2/2=3.525, \delta v=11.87$
The magnitude of the average acceleration is $\delta a=\delta v/\delta t=$ 3392.67 $ms^{-2}$
The result is independent of the mass of the ball.

using free sage math package (online server availabe at sagenb.org)

m=0.050 #mass in Kg
dt=0.0035 # in seconds, vin e vout in  m/s
Ein=m*25^2/2
Eout=m*22^2/2
E_loss=Ein-Eout  # you can factor out m/2 and v_loss=sqrt(25.0^2-22.0^2)
v_loss=sqrt(2*E_loss/m)
da=v_loss/dt
Ein, Eout, E_loss  , v_loss, da


results
(15.6250000000000, 12.1000000000000, 3.52500000000000, 11.8743420870379, 3392.66916772512)

EDIT add: to detail the reasoning

The question is

magnitude of the average acceleration of the ball during this time interval

To answer about the "magnitude of the average acc or speed" only the magnitude of the vectors are relevant.
$acc_{in}=0,acc_{out}=0$. (make a simple sketch if you like)
$v_{in}=25$, from 0 to dt/2: $v$ drop to 0, from dt/2 to dt: $v$ grows to 22
$a_{in}=0$, from 0 to dt/2: $a$ step to -max1, from dt/2 to dt: $a$ step to max2, near dt goes down to 0=$a_{out}$. The answer asks the integral from 0 to dt of $acc$ divided by dt (the average), i.e. the subtractions of the area of the negative step minus the area of the positive step, divided by dt.

But I dont know the factual detailed form of the speed/acceleration of the center of mass of the ball and I have to try a different reasoning: I wil use the conservation of energy to describe the problem:

I will substitute the ball/energy at the entrance by two objects: one that traverses the region of contact without changing energy and other that models what happens during the contact time.

my novel configuration is:
at entrance : $E_{in}$ is substituted by two objects: $E_{out} + E_{loss}$.
at exit : $E_{out}$.
The Question asks: What happen during the dt time interval? Answer: $E_{loss}$! To this energy correspond an average speed of 11.87, that divided by dt gives the acceleration during dt of the object (part of it, i.e. momentum) that describe the contact time.

In this way I dont have to do find the detailed speed and acceleration and the integral is avoided.