Work-energy v.s. impulse-momentum bar charts

[otownsend]

My question unfortunately does not fit in the Homework template, so I hope this is okay.

I attached a question below which involves multiple parts using the concepts related to energy, work, impulse, and momentum. Under the "Simplify and Diagram" section (you will notice it is in bold text), I want to better understand why Part II cannot be understood with a work-energy bar chart. The texts reasoning for why it is not a good idea to use a work-energy bar charts is because "we don't have a way to independently determine internal energy change". Why is this the reason? Isn't the reason more that we want to find the velocity of Jane and Spiderman after the collision occurred and using a work-energy expression (in this case the work energy expression would be something like kinetic energy = change in internal energy) wouldn't allow us to determine this velocity?

 

[haruspex]

kinetic energy = change in internal energy

Did you mean that? Which kinetic energy, exactly?

 

[otownsend]

Did you mean that? Which kinetic energy, exactly?

Well there's the kinetic energy coming from the collision, which then gets changed into internal energy (since its an inelastic collision)

 

[haruspex]

Well there's the kinetic energy coming from the collision, which then gets changed into internal energy (since its an inelastic collision)

That's the lost KE. But in that case you seem to be saying the same thing as the text, that we don't know what the lost KE is because we don't know the gained internal energy.

 

[neilparker62]

The texts reasoning for why it is not a good idea to use a work-energy bar charts is because "we don't have a way to independently determine internal energy change".

In a perfectly inelastic collision the "internal energy change" is (in practise) crash energy lost. For example in a car accident it would be all the energy dissipated as the cars crumple up and deform along with various other forms of irreversible energy dissipation. So - yes - it can't be independently determined but one can use conservation of momentum to determine energy before and after collision and hence the difference. It turns out that the energy loss (internal energy change in your textbook's lexicon) in a perfectly inelastic collision (collide and coalesce collision) is given by the following relatively simple equation: $$ΔE=½μΔv^2$$ where μ is the reduced mass of the colliding objects: $$μ=\frac{m_1m_2}{m_1+m_2}$$ and Δv is their relative velocity along the line of collision. In this case $$Δv=\sqrt{2gh_1}$$ as shown in your textbook.

Using this equation you could have just one energy bar chart showing U_i and U_f with the W column showing the inelastic collision loss per formula as Spiderman swoops to the rescue!

Edit: not quite true - I guess you would still need the first work-energy bar chart to determine $Δv=\sqrt{2gh_1}$.

 

[gneill]

Just to be annoyingly pedantic and almost entirely unhelpful , note that according to the supplied diagram Spiderman will not encounter Mary Jane at the bottom of his arc:

 

[haruspex]

Spiderman will not encounter Mary Jane at the bottom of his arc:

Which, at that speed, will be a relief for all concerned.

 

[gneill]

Which, at that speed, will be a relief for all concerned.

Very much !

 

[neilparker62]

according to the supplied diagram Spiderman will not encounter Mary Jane at the bottom of his arc:

In the interests of appropriate drama (and Science!), I think we have to assume that Mary Jane - with precious seconds ticking away - rushes to the right hand side of the platform and launches herself into the air just as Spiderman swoops in ...

 

[gneill]

In the interests of appropriate drama (and Science!), I think we have to assume that Mary Jane - with precious seconds ticking away - rushes to the right hand side of the platform and launches herself into the air just as Spiderman swoops in ...

...and smacks into her at 50 kph.