Bouncing Steel Ball (Elastic Collision)

  • Thread starter Awwnutz
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  • #1
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http://img395.imageshack.us/img395/6158/steelballxu7.gif [Broken]

A 35 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 20.8 m.

a) With what speed does the ball leave the plate?

b) What is the magnitude of the total change in momentum of the ball with each bounce?

c) What is the magnitude of the total momentum the ball transfers to the plate at each bounce?

d) What is the time interval between bounces?

e) Averaged over a long time interval, what is the magnitude of the rate at which momentum is transferred to the plate?

f) What is the magnitude of the average force exerted by the ball on the plate?

g) What is the weight of the ball?



Conservation of energy and momentum i believe will have to be used here.


I figured out part a.)
I set the Kinetic energy equal to the work done by gravity and found the ball moves at 20.199 m/s.

It's part b.) that i'm having the problem with. I initially thought the magnitude would have to be zero since the ball would have to moving at the same speeds to continue reaching the same height. But now i feel that it will have to be moving faster coming back up because it will counteract gravity whereas when its moving down it's moving with gravity. I just can't figure out a way to put that into an equation.
 
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Answers and Replies

  • #2
LowlyPion
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http://img395.imageshack.us/img395/6158/steelballxu7.gif [Broken]

A 35 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 20.8 m.

a) With what speed does the ball leave the plate?

b) What is the magnitude of the total change in momentum of the ball with each bounce?

c) What is the magnitude of the total momentum the ball transfers to the plate at each bounce?

d) What is the time interval between bounces?

e) Averaged over a long time interval, what is the magnitude of the rate at which momentum is transferred to the plate?

f) What is the magnitude of the average force exerted by the ball on the plate?

g) What is the weight of the ball?



Conservation of energy and momentum i believe will have to be used here.


I figured out part a.)
I set the Kinetic energy equal to the work done by gravity and found the ball moves at 20.199 m/s.

It's part b.) that i'm having the problem with. I initially thought the magnitude would have to be zero since the ball would have to moving at the same speeds to continue reaching the same height. But now i feel that it will have to be moving faster coming back up because it will counteract gravity whereas when its moving down it's moving with gravity. I just can't figure out a way to put that into an equation.

In b) they are asking for ΔM. What is that if it reverses upon striking the plate? Remember V is a vector.
 
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  • #3
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The change in momentum?

The vector as the ball moves downward would be negative, but as it bounces and begins to move upward it would then be positive.
 
  • #4
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the momentum of the ball moving back up would be (.035kg * 20.199m/sec) = .706965J

-.707 - .707 = 1.41J total. Ok got it.
 
  • #5
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Now i'm on to the time interval between bounces. Would i just use 2-D Kinematics?
 
  • #6
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1-D dynamics will do...
 
  • #7
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Alright i figured that one out.

I'm stuck on finding the weight of the ball...what equation would i want to do to find that?

Should i set the change in Kinetic energy = the work done by gravity to find the mass?
 
  • #8
LowlyPion
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Alright i figured that one out.

I'm stuck on finding the weight of the ball...what equation would i want to do to find that?

Should i set the change in Kinetic energy = the work done by gravity to find the mass?

You might do better to work it out as the variables.

For instance the Δp (change in momentum) is as you have already found 2*m*v. where:

v = (2*g*h)1/2

Now the average Force then is Total force over ΔT which is the period you found.

And T you can find from h = 1/2*g*t2. But you have to double it for the time up and then down again

T = 2*t = 2*(2h/g)1/2

Hence ΔT = 2*(2h/g)1/2

So ... Favg = Δp/ΔT = 2*m*v/ 2*(2h/g)1/2

But what did we find the Velocity could be expressed as? Try substituting that into this equation.
 
  • #9
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Alright i figured that one out.

I'm stuck on finding the weight of the ball...what equation would i want to do to find that?

Should i set the change in Kinetic energy = the work done by gravity to find the mass?

Huh? The mass is given.
 
  • #10
LowlyPion
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Huh? The mass is given.

I think that the idea is to show a derivation of weight.
 
  • #11
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I think that the idea is to show a derivation of weight.

I know, but the poster thinks he needs to find the mass.
 
  • #12
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Yeah i definitely overlooked what i was given. They give me the mass and all i needed to do was find the weight. I start doing all these equations and formula's that i figure everything needs to be some type of formula so i overlook something so simple. I feel like an idiot :)

Thanks for all the help everyone!
 

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