Bouncing Steel Ball (Elastic Collision)

In summary, the conversation covers a 35 g steel ball bouncing elastically on a steel plate, with a maximum height of 20.8 m. The speed at which the ball leaves the plate is 20.199 m/s, and the magnitude of the total change in momentum of the ball with each bounce is 1.41 J. The time interval between bounces can be found using 1-D dynamics, and the weight of the ball can be calculated using the given mass and the formula for weight (W = mg). The conversation also discusses the use of conservation of energy and momentum, as well as the average force exerted by the ball on the plate.
  • #1
Awwnutz
43
0
http://img395.imageshack.us/img395/6158/steelballxu7.gif

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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  • #2
Awwnutz said:
http://img395.imageshack.us/img395/6158/steelballxu7.gif

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
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
the momentum of the ball moving back up would be (.035kg * 20.199m/sec) = .706965J

-.707 - .707 = 1.41J total. Ok got it.
 
  • #5
Now I'm on to the time interval between bounces. Would i just use 2-D Kinematics?
 
  • #6
1-D dynamics will do...
 
  • #7
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
Awwnutz said:
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
Awwnutz said:
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
borgwal said:
Huh? The mass is given.

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

Related to Bouncing Steel Ball (Elastic Collision)

1. How does a bouncing steel ball demonstrate an elastic collision?

When a steel ball bounces off a surface, it experiences a sudden change in velocity due to the impact. An elastic collision is one in which the total kinetic energy of the objects involved is conserved. In the case of the steel ball, the impact with the surface causes it to compress slightly, storing potential energy. As it bounces back, this potential energy is released as kinetic energy, resulting in the ball bouncing back to its original height.

2. What factors affect the amount of bounce in a steel ball?

The amount of bounce in a steel ball can be affected by various factors, including the material and surface properties of the ball and the surface it is bouncing on. The elasticity of the ball and the surface, as well as the angle and speed of impact, can all impact the amount of bounce. Additionally, air resistance and any imperfections on the surface can also affect the bounce.

3. How does the elasticity of the steel ball impact its bounce?

The elasticity of a steel ball refers to its ability to deform under stress and return to its original shape. A highly elastic ball will experience a greater change in velocity during an impact, resulting in a higher bounce. On the other hand, a less elastic ball will experience a smaller change in velocity and therefore have a lower bounce.

4. Can a steel ball experience a perfectly elastic collision?

In theory, a perfectly elastic collision is one in which there is no loss of kinetic energy. However, in reality, all collisions involve some loss of energy due to factors such as friction and air resistance. Therefore, it is not possible for a steel ball to experience a perfectly elastic collision, but it can come close under ideal conditions.

5. How is the coefficient of restitution related to the bounce of a steel ball?

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach. In the case of a steel ball bouncing off a surface, a higher coefficient of restitution indicates a more elastic collision, resulting in a higher bounce. The coefficient of restitution can be calculated by measuring the heights of the ball's bounces and using the formula e = √(h2/h1), where h1 is the initial height and h2 is the height after the bounce.

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