Good Pendulum Question

[MedPR]

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I thought it was tricky. Maybe I'm dumb though.

http://www.mcatquestion.com/findquestion.php?arg1=186

 

[milski]

Yes, that's a nice one. You really have to work through it, no shortcuts.👍

 

[MedPR]

elastic collision = mv & KE are conserved?
inelastic collision = only mv is conserved?

What about total energy? PE?

 

[milski]

elastic collision = mv & KE are conserved?
inelastic collision = only mv is conserved?

What about total energy? PE?

TE is not preserved, at least not in a way that you can use. The lost KE is transformed to heat or some form of internal potential energy (deformations) which are hard to measure and you won't be able to use them in calculations.

Potential energy (from gravity) is whatever it is - mgh. The collision itself will not change it, any changes of h due to the collision obviously will.

 

[MedPR]

So why couldn't you use conservation of PE in that problem?

 

[milski]

So why couldn't you use conservation of PE in that problem?

What is conservation of PE? I'm not trying to be argumentative but I don't understand exactly what you mean by that. Conservation of two quantities allows us to write Q1=Q2 at two different moments, express the two in some way and solve the equation. For example, the preservation of momentum can give you m1v1+m2v2=m1v1'+m2v2' for the momentum before and after the collision. What are the two PE energies that you want to say are equal?

If you want to calculate how much PE the ball on the right has and say that it was all converted to PE when both balls go up, that would be incorrect. You can use that only when PE+KE=const, meaning there is additional energy added to/lost from the system. The inelastic collision will convert some of the KE to heat (or some other energy), for which you are not accounting.

They have the proper steps in the question:
From PE+KE=const, get the KE at the bottom.
From KE, get the speed and from that the momentum at the moment of collision.
Since momentum is preserved, it stays the same after the collision.
Get the new speed from the momentum - you already know it, since it did not change.
Get KE from the momentum.
Since KE+PE=const' (different one), you know what PE will be when KE=0.
Get the height from the new PE.

 

[MedPR]

If PE is conserved, then mgh=(m+M)gh. You say the collision will not change the PE, and from that I interpret that PE is conserved.

 

[milski]

If PE is conserved, then mgh=(m+M)gh. You say the collision will not change the PE, and from that I interpret that PE is conserved.

Ah, that. What you're really saying is that PE+KE=const and when KE=0 PE stays the same. The problem with that is that during the collision KE changes, so you cannot say that PE+KE=const anymore.

 

[MedPR]

Ah, that. What you're really saying is that PE+KE=const and when KE=0 PE stays the same. The problem with that is that during the collision KE changes, so you cannot say that PE+KE=const anymore.

Oh that's right. I understand what they are saying is the correct method, I just got confused when you said PE is conserved.

So, just so I'm clear can you tell me what is conserved in elastic collisions, and inelastic collisions?

 

[chiddler]

Oh that's right. I understand what they are saying is the correct method, I just got confused when you said PE is conserved.

So, just so I'm clear can you tell me what is conserved in elastic collisions, and inelastic collisions?

Inelastic - momentum conserved, kinetic energy is not.
Elastic - both conserved.

you had it right.

 

[chiddler]

I'm getting stuck on one of the steps in their solution.

Energy is conserved before the collision, use conservation of energy.
ngh = 1/2 nv1^2
v1 = sqrt(2gh)

momentum conserved during collision

nv1 = (n+m)v2
v2 = nv1 / (n+m)

plug it in plug it in
v2 = n*sqrt(2gh)/(n+m)

Ok so this is velocity after the inelastic collision. Now plug it into kinetic energy equation and find mgh's height.

1/2 mv2^2 = mgh and m's cancel out
1/2 v2^2 = gh
1/2 v2^2 / g = h
plug in the v2 i got earlier to find h
1/2 (n*sqrt(2gh)/(n+m))^2 / g = h

So where am I wrong?

 

[milski]

Oh that's right. I understand what they are saying is the correct method, I just got confused when you said PE is conserved.

So, just so I'm clear can you tell me what is conserved in elastic collisions, and inelastic collisions?

If I've said PE is conserved, kick me in the ankles - it's confusing way to explain things.

elastic collisions: momentum and kinetic energy of the objects before and after the collision are the same (aka preserved).
inelastic collisions: momentum of the objects before and after the collision is the same.

The other thing, not directly related to collisions:
ΔE=net work. If no work was done, ΔE=0 or PE+KE=const.

 

[milski]

I'm getting stuck on one of the steps in their solution.

Energy is conserved before the collision, use conservation of energy.
ngh = 1/2 nv1^2
v1 = sqrt(2gh)

momentum conserved during collision

nv1 = (n+m)v2
v2 = nv1 / (n+m)

plug it in plug it in
v2 = n*sqrt(2gh)/(n+m)

Ok so this is velocity after the inelastic collision. Now plug it into kinetic energy equation and find mgh's height.

1/2 mv2^2 = mgh and m's cancel out
1/2 v2^2 = gh
1/2 v2^2 / g = h
plug in the v2 i got earlier to find h
1/2 (n*sqrt(2gh)/(n+m))^2 / g = h

So where am I wrong?

I don't see any error - you got it right.

1/2 (n*sqrt(2gh)/(m+n))^2/g = n^2*2gh/(2 * g * (m+n)^2)=n^2h/(m+n)^2

 

[chiddler]

I don't see any error - you got it right.

1/2 (n*sqrt(2gh)/(m+n))^2/g = n^2*2gh/(2 * g * (m+n)^2)=n^2h/(m+n)^2

oh good. so my algebra sucks.

i was trying to simplify the squared terms 🙁 thanks very much

 

[milski]

oh good. so my algebra sucks.

i was trying to simplify the squared terms 🙁 thanks very much

If you're getting correct answers with it, I don't think it's a big deal that it sucks. 😎