# Energy conservation in superball collision

• Alettix
In summary: It should be##mu + Mu = m(u + 2v_{ZMF}) + M\color{red}{v}##You have some signs wrong in the second equation. It should be##mu + Mu = m(u + 2v_{ZMF}) + M\color{red}{v}##In summary, the conversation discusses the problem of two elastic "superballs" of mass M and m, placed on top of each other with a small distance and released from a height h with velocity u. The question is to prove that the top ball will have velocity ## v = \frac{3M-m}{M+m} \cdot u ## when it leaves the bigger ball. The discussion includes using
Alettix

## Homework Statement

There are two elastic "superballs" of mass M and m placed on top of each other with a smal distance. The lighter ball of mass m is on top of the bigger ball of mass M. The balls are released from a height h and have velocity u when they hit the ground. Prove that the top ball will have velocity ## v = \frac{3M-m}{M+m} \cdot u ## when it leaves the bigger ball.

## Homework Equations

momentum conservation ## p_{before} = p_{after} ##
energy conservation ##E_{before} = E_{after} ##
zero momentum frame velocity: ##v_{ZMF} = \frac{p_{tot}}{m_tot} ##

## The Attempt at a Solution

I didn't find it problematic to prove the requried result (see attached solution). What I find problematic is the result itself. I must have made some very silly misstake in my reasoning, but the end result I get is that the speed of the top ball after all collisions is ## u + 2v_{ZMF}## and of the bottom ball ##u##. However, before the two balls collide, they both had velocity ##u##. Clearly, energy conservation cannot be satisfied!

Can somebody please point out my probably very stupid misstake to me? Thank you! :)

#### Attachments

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Perhaps the big ball loses some speed when colliding with the smaller ball ? How did you find this ##u## for the bottom ball ?

BvU said:
Perhaps the big ball loses some speed when colliding with the smaller ball ? How did you find this ##u## for the bottom ball ?
Oh wait, I really messed it up! The speed of the big ball should be ## u - 2v_{ZMF}##, shouldn't it?

Alettix said:
Oh wait, I really messed it up! The speed of the big ball should be ## u - 2v_{ZMF}##, shouldn't it?
##E_{ki} = 1/2mu^2 + 1/2Mu^2##
##E_{kf} = 1/2m(u^2 + 2(2v_{ZMF})u + (2v_{ZMF})^2) \\+ 1/2M(u^2 + 2(2v_{ZMF})u - (2v_{ZMF})^2)##
##E_{kf} = (1/2M+1/2m)u^2 + m(2v_{ZMF})u + 2m(v_{ZMF})^2 + M(2v_{ZMF})u - 2M(v_{ZMF})^2##
##E_{kf} = (1/2M+1/2m)u^2 + 2(v_{ZMF})^2(m - M) + (M + m)(2v_{ZMF})u##
##E_{ki} \ne E_{kf}##
I think I did this correctly.

Buffu said:
##E_{ki} = 1/2mu^2 + 1/2Mu^2##
##E_{kf} = 1/2m(u^2 + 2(2v_{ZMF})u + (2v_{ZMF})^2) \\+ 1/2M(u^2 + 2(2v_{ZMF})u - (2v_{ZMF})^2)##
##E_{kf} = (1/2M+1/2m)u^2 + m(2v_{ZMF})u + 2m(v_{ZMF})^2 + M(2v_{ZMF})u - 2M(v_{ZMF})^2##
##E_{kf} = (1/2M+1/2m)u^2 + 2(v_{ZMF})^2(m - M) + (M + m)(2v_{ZMF})u##
##E_{ki} \ne E_{kf}##
I think I did this correctly.
Sooo, ##u - 2v_{ZMF}## is not right either?

I think using momentum conservation will give right answer.
Alettix said:
Sooo, ##u - 2v_{ZMF}## is not right either?

Buffu said:
I think using momentum conservation will give right answer.
As far as I am concerned ZMF is meant to make it quicker and easier, so I would like to use that method. But yes, momentum conservation should always give the right answer.

Alettix said:
As far as I am concerned ZMF is meant to make it quicker and easier, so I would like to use that method. But yes, momentum conservation should always give the right answer.
No that's not what I meant.
I mean that we can get right velocity for big ball after collision using momentum conservation.

##mu + Mu = m(u + 2v_{ZMF}) + M\color{red}{V}##

Last edited:
Alettix said:
Oh wait, I really messed it up! The speed of the big ball should be ## u - 2v_{ZMF}##, shouldn't it?
Yes, and with that I think you will find the energy is conserved in the lab frame.

Alettix
haruspex said:
Yes, and with that I think you will find the energy is conserved in the lab frame.
Yes. Thank you for helping me get it right! :)

haruspex said:
Yes, and with that I think you will find the energy is conserved in the lab frame.
That looks correct but I did not get it.

Initial velocity in lab frame :-
Big mass - u
Small mass - u

Final velocity in lab frame :-
Big mass - ##u - 2v_{ZMF}##
Small mass - ##u + 2v_{ZMF}##
As per OP,

That is not going to get conserved. Either I am getting final velocity of small mass wrong or the big mass.
Can you help me please ?

Buffu said:
That is not going to get conserved
Are you sure, or are you just guessing?

haruspex said:
Are you sure, or are you just guessing?
I checked if it is going to conserve in post number 4.
Can you please check my calculations ?

Buffu said:
I checked if it is going to conserve in post number 4.
Can you please check my calculations ?
You have some signs wrong in the second equation.

## 1. How does energy conservation apply to superball collisions?

Energy conservation is a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the case of superball collisions, the total kinetic energy of the system (the superballs) before and after the collision must be equal, demonstrating the conservation of energy.

## 2. What factors affect energy conservation in superball collisions?

The factors that affect energy conservation in superball collisions include the mass and velocity of the superballs, the elasticity of the balls, and the angle of impact. These factors determine how much kinetic energy is transferred between the balls during the collision.

## 3. Why is energy conservation important in superball collisions?

Energy conservation is important in superball collisions because it helps us understand and predict the behavior of the system. By knowing the initial conditions of the collision and applying the principle of energy conservation, we can calculate the final velocities of the superballs and determine how much energy was transferred during the collision.

## 4. Is energy always conserved in superball collisions?

In an ideal scenario where there is no external force or energy dissipation, energy is always conserved in superball collisions. However, in real-world scenarios, there may be some energy lost due to factors such as air resistance and friction. This means that energy conservation may not be completely accurate in practical applications, but it is still a useful concept to understand the behavior of the system.

## 5. How can we apply energy conservation in real-life situations involving superball collisions?

Energy conservation can be applied in real-life situations involving superball collisions to predict the behavior of the system and determine the outcome of the collision. For example, it can be used to design and engineer better superballs for sports or entertainment purposes, or to understand the impact of collisions in car accidents and develop safety measures to reduce energy transfer and potential injuries.

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