# Loss of kinetic energy due to heat

• Cocoleia
In summary, the fraction of the initial energy that is lost due to heat is 3/4 - γ/4, where γ is mass of ball/mass of block (γ<1).
Cocoleia

## Homework Statement

I have a bloc sitting on a horizontal table, and we shoot a ball through it. The speed right before entering the block is v and the speed when it exits the block is v/2. I need to prove that the fraction of the initial energy that is lost due to heat is 3/4 - γ/4, where γ is mass of ball/mass of block (γ<1)

## Homework Equations

Conservation of momentum

## The Attempt at a Solution

I know that usually kinetic energy is given by 1/2 mv^2, so then the kinetic energy of the ball once it exits the block will be 1/8 mv^2. I am not sure how to begin with this question.

Hi Coco,

There is always the conservation of momentum to consider. Should have been listed as a relevant equation under item 2 of the template !

BvU said:
Hi Coco,

There is always the conservation of momentum to consider. Should have been listed as a relevant equation under item 2 of the template !

Good. Now check your math. the 1/4 is not correct.

BvU said:
Good. Now check your math. the 1/4 is not correct.
Ok, I fixed that too. What I don't understand is how to find the heatloss from the energy or the conservation of momentum. What is the relationship between them ?

if you keep fixing like that, my comments will make me look dumb ... never mind.

Who mentioned heat loss ? I don't see it in the problem statement !
Cocoleia said:
Yeah, you added it in words. Now work it out in terms of ##\ v_0\ ## and ##\ \gamma\ ##,

BvU said:
if you keep fixing like that, my comments will make me look dumb ... never mind.

Who mentioned heat loss ? I don't see it in the problem statement !
Yeah, you added it in words. Now work it out in terms of ##\ v_0\ ## and ##\ \gamma\ ##,
I don't quite understand what γ is

##\gamma = m_{\rm ball} / M_{\rm block}##

BvU said:
##\gamma = m_{\rm ball} / M_{\rm block}##
Would I have to separate this into three parts, before the block, inside the block and after the block? Do I start by working with momentum or kinetic energy?

Questions, questions, questions...

They tell you the ball speed before the block (##v_0##) and after (##{1\over 2}v_0##).
And they ask for something after the block. So no need to worry about the violent path through the block
And you have the corresponding kinetic energies of the ball already

I have to correct myself:
BvU said:
Who mentioned heat loss ? I don't see it in the problem statement !
It's there bright and clear, sorry .

Note that you are already very close to where you want to be: the ball goes from ##{1\over 2} mv_0^2## to ##{1\over 8} mv_0^2## so 3/4 is the maximum loss due to conversion to heat. Apparently that's overestimated and needs a correction.

I don't get what the mass of the block has to do with the problem, how do i incorporate it in the forumlas? I don't think it will be used for the kinetic energy ?

Can you express the conservation of momentum in ## \ m_{\rm ball} , \ M_{\rm block},\ v_{\rm ball} , \ V_{\rm block}\ ## before the block is hit and after the ball has come out ?

Cocoleia said:
I don't think it will be used for the kinetic energy ?
For the KE of what, exactly? [hint]

BvU said:
Can you express the conservation of momentum in ## \ m_{\rm ball} , \ M_{\rm block},\ v_{\rm ball} , \ V_{\rm block}\ ## before the block is hit and after the ball has come out ?
Before:
mballv
after
mballv/2 + MblockVblock

haruspex said:
For the KE of what, exactly? [hint]
You have the kinetic energy of the ball before it hits the block, and the conservation of momentum before
after it hits you have the kinetic energy of the ball with the reduced speed and for the conservation of momentum I assume now the block will come into play, but I don't know to tie all of this together.

Cocoleia said:
you have the kinetic energy of the ball with the reduced speed
Yes, and of what else?

haruspex said:
Yes, and of what else?
Of the block ?

Cocoleia said:
Of the block ?
Yes

haruspex said:
Yes
Will the block be moving at the same speed as the ball? I don't know how to set up the equations

Cocoleia said:
Will the block be moving at the same speed as the ball
If the ball comes out the other end, then: no.
Cocoleia said:
I don't know how to set up the equations
Not true: you do know. You did it in your post #14. How many unknowns are there ? Remember ##\gamma = m_{\rm ball} / M_{\rm block}##

BvU said:
If the ball comes out the other end, then: no.
Not true: you do know. You did it in your post #14. How many unknowns are there ? Remember ##\gamma = m_{\rm ball} / M_{\rm block}##
Yes, but how do I relate this to the conservation of momentum and how will this lead me to finding my heat loss

Post #14 is conservation of momentum if you equate 'before' and 'after'. It helps you express Vblock in terms of v and ##\gamma##

BvU said:
Post #14 is conservation of momentum if you equate 'before' and 'after'. It helps you express Vblock in terms of v and ##\gamma##
I am able to get for loss of kinetic energy
3/4mballv^2-1/4Mblockγv
How can I refine this to get rid of the m, M and v (like in the initial question)
If I divide this by the initial kinetic energy I end up with
1/4-(γ^2/2)
which is not what I want, I need to get 3/4-γ/4

Cocoleia said:
If I divide this by the initial kinetic energy I end up with
1/4-(γ^2/2)
I do not get that from so dividing. In particular, how does the 3 disappear?

Cocoleia said:
##\frac 34m_{ball}v^2-\frac 14M_{block}γv##
That is not quite right. You omitted the power of 2 on the second velocity (γv) and have dropped a factor.

haruspex said:
That is not quite right. You omitted the power of 2 on the second velocity (γv) and have dropped a factor.
Ok. I fixed the problem. Thanks

## 1. What is loss of kinetic energy due to heat?

Loss of kinetic energy due to heat is the process in which the kinetic energy of an object decreases as it transfers heat energy to its surroundings. This can occur through various mechanisms such as conduction, convection, and radiation.

## 2. How does loss of kinetic energy due to heat affect objects?

The loss of kinetic energy due to heat can have various effects on objects depending on the amount of heat transferred and the properties of the object. It can cause a change in temperature, phase change, or even structural changes in the material.

## 3. Can loss of kinetic energy due to heat be prevented?

In most cases, it is not possible to completely prevent the loss of kinetic energy due to heat. However, it can be minimized by using materials with lower thermal conductivity, insulating the object, or reducing the temperature difference between the object and its surroundings.

## 4. How is loss of kinetic energy due to heat measured?

The loss of kinetic energy due to heat is measured using a unit called joules (J). This unit represents the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.

## 5. What are some real-life examples of loss of kinetic energy due to heat?

Some common examples of loss of kinetic energy due to heat include the cooling of a hot cup of coffee, the melting of an ice cube in a warm room, and the transfer of heat from a stove to a pot while cooking. It is a natural process that occurs in many everyday situations.

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