# Ball being thrown against a wall

1. Sep 29, 2007

### turnip

1. The problem statement, all variables and given/known data
a 120kg ball moving at 18m/s strikes a wall perpendicularly and rebounds straight back at 12m/s. after the initial contact, the centre of the ball moves 0.27cm closer to the wall. Assuming uniform deceleration, show that the time of the contact is 0.00075s. How large an average forced does the ball exert on the wall?

i figure that:
m=.12
u= 18
v=12
s1 - s2 = 0.0027
s1 = s2 + 0.0027

2. Relevant equations
f= delta p/t
= mv-mu/t

s=ut+1/2at(squared)
p=mv

3. The attempt at a solution
what threw me off is the the time, i can get the average force no problem after i have the time.

my question is do i put the time (0.00075) into the equation s=ut+1/2at(squared), because i have done that which worked out wrong.
or
do i rearrange and substitute some other formulas? if i dont have any formulas needed please tell me!

2. Sep 29, 2007

### rootX

Try work energy theorem here.
Fd = change in Ek

3. Sep 29, 2007

### rootX

Trying one way I get this ridiculously large # for force: 7 x 10^6 N
I assumed that it took 0.27 cm to stop the ball from speed of 18 m/s

4. Sep 30, 2007

### turnip

thank you for trying, but the way i read it, the actual displacement is unknown
you see the ball hits the wall starting at displacement = x
after it rebounds off the wall it is .27 cm closer from its original starting point of x
therefore it is: displacement after rebound = x - .27 cm

anymore ideas anyone?

5. Sep 30, 2007

### learningphysics

To get the time, you need to divide the problem into two parts... first find the time it takes to be brought to 0m/s from 18m/s... then find the time it takes to go from 0m/s up to 12m/s.

I get the given time of 0.00075s.

For the second part, Faverage*time = change in momentum. (oops nevermind, I just saw that you got this part).

Last edited: Sep 30, 2007
6. Sep 30, 2007

### learningphysics

The trick to this problem is that the accelerations for the two parts are not the same... ie: the deceleration from 18m/s to 0m/s is different from the acceleration from 0m/s to 12m/s.

7. Sep 30, 2007

### turnip

okay i get that, but could you please explain how you algebraically got the acceleration becuase i can only think of f=ma and a = v-u/t but both have two variables!

if you could please explain the acceleration i will be right from there thanks :)

8. Sep 30, 2007

### learningphysics

well, you don't need acceleration... you need time:

so for the first part:

d = (v1 + v2)/2 * t

0.0027 = (18 + 0)/2 * t

solve for t

get t = 0.0003s

do the same thing for the second part... v1 = 0 v2 = 12.

if you want to get acceleration (you don't need to get it)

use v2^2 = v1^2 + 2ad.

9. Oct 1, 2007

### turnip

right. just one more thing i dont get:
d = (v1 + v2)/2 * t
i understand the formula, but whats with the /2 part of it?
is that to get the average velocity?

10. Oct 1, 2007

### learningphysics

yes. distance = average velocity * time. $$d = \frac{(v1+v2)}{2}*t$$

11. Oct 1, 2007

### turnip

thanks that was very clearly explained :)