When does the ball come to rest?

In summary: The time between bounces is?It's a bit easier to see what is going on if you look at the time between bounces.Between first and second bounce, how long is it in the air?Between second and third bounces, how long is it in the air?It is 3 seconds between first and second bounce, and 5 seconds between second and third bounce.The time between bounces is 3 seconds.It is 5 seconds between second and third bounce.
  • #1
Gopal Mailpalli
28
2

Homework Statement


A ball is dropped vertically from a height H on to a plane surface and permitted to bounce repeatedly along a vertical line. After every bounce, its kinetic energy becomes a quarter of its kinetic energy before the bounce. The ball will come to rest after time?

Homework Equations


Kinetic Energy = ##\frac{1}{2}mv^{2}##
Velocity = ##\sqrt{2gRh}## , where R is the fraction of hight ball reaches after every bounce

The Attempt at a Solution


I could recognize the geometrical series and attempted the solution which gave me a wrong time.
 
Physics news on Phys.org
  • #2
Gopal Mailpalli said:
I could recognize the geometrical series and attempted the solution which gave me a wrong time
Did you ? And how did you interpret the sum of this series to get the answer ?
 
  • #3
BvU said:
Did you ? And how did you interpret the sum of this series to get the answer ?

Excuse me! Some where I did a mistake, i assumed the the initial Kinetic energy be K and next consecutive terms are (1/4)K, (1/4)^2 K ...
This represents a Geometric Progression, this is how I approached the problem.
 
  • #4
Yes, that's what it says in the problem statement. So what series did you get and how did you work the sum of that series around to the answer you gave ?

There is a simple mistake that's easy to make here, but if I ask specifically I give away the answer, which is a bit against PF culture.
 
  • #5
BvU said:
Yes, that's what it says in the problem statement. So what series did you get and how did you work the sum of that series around to the answer you gave ?

There is a simple mistake that's easy to make here, but if I ask specifically I give away the answer, which is a bit against PF culture.

I found the coefficient of restitution which is 0.5, found out the height using the respective COR - Height relation and formed a geometric progression. I calculated the total height and went further to solve the time it takes using basic laws of motion.
 
  • #6
Gopal Mailpalli said:
I found the coefficient of restitution which is 0.5, found out the height using the respective COR - Height relation and formed a geometric progression. I calculated the total height and went further to solve the time it takes using basic laws of motion.

You have to post your working, equations and all. We can't spot a mistake from a description of what you did.
 
  • #7
I don't understand what you did. Can you post your working ?
Why bother to find a total height ? The exercise asks for a time.
What total height did you find ? Something like ##4h\over 3## ?
 
  • #8
BvU said:
I don't understand what you did. Can you post your working ?
Why bother to find a total height ? The exercise asks for a time.
What total height did you find ? Something like ##4h\over 3## ?

I mean sum of the geometric progression, which i got 4h/3.
 
  • #9
Nice, but useless: the total time has nothing to do with the time needed to fall 4h/3 (why not ?).
 
  • #10
BvU said:
Nice, but useless: the total time has nothing to do with the time needed to fall 4h/3 (why not ?).

Then how do I approach the problem? I don't have an idea.
 
  • #11
You will also get a geometric series if you calculate the sum of the times the ball needs to get back to the ground after each bounce.
 
  • #12
BvU said:
You will also get a geometric series if you calculate the sum of the times the ball needs to get back to the ground after each bounce.

Still, i couldn't solve the problem.
 
  • #13
Gopal Mailpalli said:
Still, i couldn't solve the problem.
It's a bit easier to see what is going on if you look at the time between bounces.
Between first and second bounce, how long is it in the air?
What about between second and third bounces?
What is the pattern?
 

Related to When does the ball come to rest?

1. When does the ball come to rest?

The ball comes to rest when all external forces acting on it have been balanced out and it is no longer in motion. This can occur when it reaches a stationary position or when it comes to a complete stop.

2. What factors affect when the ball comes to rest?

The factors that affect when the ball comes to rest include the initial velocity of the ball, the surface it is rolling on, the presence of any external forces such as friction or air resistance, and the mass and shape of the ball.

3. How can you calculate when the ball will come to rest?

The time it takes for the ball to come to rest can be calculated using the equations of motion, taking into account the initial velocity, acceleration, and final velocity of the ball. However, this calculation may be affected by other factors such as external forces and the shape of the ball.

4. Can the ball come to rest in different ways?

Yes, the ball can come to rest in different ways depending on the surface it is rolling on and the external forces acting on it. For example, a ball rolling on a flat surface may come to rest by gradually slowing down until it reaches a stationary position, while a ball rolling on a slope may come to rest by reaching the bottom of the slope.

5. Why is it important to know when the ball will come to rest?

Knowing when the ball will come to rest is important in many situations, such as in sports where the movement of a ball affects the outcome of the game. It is also important in physics experiments and calculations where the motion of a ball needs to be accurately measured and analyzed.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
13
Views
6K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
2K
Back
Top