Loss of kinetic energy due to heat

[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.

 

[BvU]

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 !

 

[Cocoleia]

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 !

Ok, I added it. I didn't think about this concept playing a role in the question

 

[BvU]

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

 

[Cocoleia]

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 ?

 

[BvU]

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 !

Ok, I added it. I didn't think about this concept playing a role in the question

Yeah, you added it in words. Now work it out in terms of $\ v_0\$ and $\ \gamma\$,

 

[Cocoleia]

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

 

[BvU]

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

 

[Cocoleia]

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

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?

 

[BvU]

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:

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.

 

[Cocoleia]

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 ?

 

[BvU]

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 ?

 

[haruspex]

I don't think it will be used for the kinetic energy ?

For the KE of what, exactly? [hint]

 

[Cocoleia]

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:
m_ballv
after
m_ballv/2 + M_blockV_block

 

[Cocoleia]

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.

 

[haruspex]

you have the kinetic energy of the ball with the reduced speed

Yes, and of what else?

 

[Cocoleia]

Yes, and of what else?

Of the block ?

 

[haruspex]

Of the block ?

Yes

 

[Cocoleia]

Yes

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

 

[BvU]

Will the block be moving at the same speed as the ball

If the ball comes out the other end, then: no.

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

 

[Cocoleia]

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

 

[BvU]

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

 

[Cocoleia]

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

I am able to get for loss of kinetic energy
3/4m_ballv^2-1/4M_blockγv
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

 

[haruspex]

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?

 

[haruspex]

$\frac 34m_{ball}v^2-\frac 14M_{block}γv$

That is not quite right. You omitted the power of 2 on the second velocity (γv) and have dropped a factor.

 

[Cocoleia]

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