Ball being thrown against a wall

  • Thread starter turnip
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In summary: yes. distance = average velocity * time. d = \frac{(v1+v2)}{2}*tthanks that was very clearly explained :)
  • #1
turnip
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Homework Statement


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

Homework Equations


f= delta p/t
= mv-mu/t

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

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 don't have any formulas needed please tell me!
 
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  • #2
Try work energy theorem here.
Fd = change in Ek
 
  • #3
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
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
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:
  • #6
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
learningphysics said:
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.

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
turnip said:
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 :)

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
right. just one more thing i don't get:
d = (v1 + v2)/2 * t
i understand the formula, but what's with the /2 part of it?
is that to get the average velocity?
 
  • #10
turnip said:
right. just one more thing i don't get:
d = (v1 + v2)/2 * t
i understand the formula, but what's with the /2 part of it?
is that to get the average velocity?

yes. distance = average velocity * time. [tex]d = \frac{(v1+v2)}{2}*t[/tex]
 
  • #11
thanks that was very clearly explained :)
 

1. How does the speed of the ball affect its rebound against the wall?

The speed of the ball has a direct impact on its rebound against the wall. The faster the ball is thrown, the harder it will hit the wall and the higher it will bounce back. This is due to the conservation of energy and the elastic properties of the ball and wall.

2. What factors affect the distance the ball travels after being thrown against the wall?

The distance the ball travels depends on several factors, including the initial velocity of the ball, the angle at which it was thrown, and the height of the wall. Other factors such as air resistance and the surface of the wall can also play a role in the distance the ball travels.

3. Why does the ball bounce back when thrown against a wall?

The ball bounces back because of the law of conservation of energy. When the ball hits the wall, it compresses and stores potential energy. As it bounces back, the potential energy is converted back into kinetic energy, causing the ball to bounce back in the opposite direction.

4. How does the material of the ball and wall affect the rebound?

The material of the ball and wall can greatly affect the rebound. A more elastic ball and wall, such as a rubber ball and a concrete wall, will result in a higher rebound compared to a less elastic ball and wall, such as a foam ball and a wooden wall.

5. What happens to the energy of the ball after it hits the wall?

The energy of the ball is not lost after it hits the wall. Some of the energy is transferred to the wall upon impact, while the rest is converted into potential energy as the ball compresses. As the ball bounces back, the potential energy is converted back into kinetic energy, resulting in the ball's movement.

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