# Verify this concept?

1. Mar 12, 2006

### GregA

EPE...Verify this concept?

I think I am on the verge of solving a question that has been bugging me for an hour or so but before I give myself a pat on the back can someone just have a look at a certain facet of the problem (that I will explain) and tell me if what I think is happening *is* actually happening...(ie: I'm gonna hit the solution but I want to be able to use the correct methodology in the future)

imagine a piece of elastic string of length l stretched to a length 5l with both ends attached to particles of mass m and 3m...firstly I know that the tension on both sides of the string is equal...but would I be right in saying that although the EPE of the system is $$\frac {\lambda(4l)^2)}{2l}$$ the work done in moving the lighter particle is a quarter of this value? (the combined mass is 4m)

(the book has left it for me to figure out what happens between 2 objects free to move attached to a spring or string...all the examples and questions I had solved in that topic where such that one end of the string was always attached to a fixed object like a wall etc...)

Last edited: Mar 12, 2006
2. Mar 12, 2006

### lightgrav

PE_elastic should be "(k(4l)^2)/2" , not "/2l" .

or the Work done BY the elastic after it is released?

Why would the lo-mass object moves less than the hi-mass object?

3. Mar 12, 2006

### GregA

The topic I'm currently working on is collisions and I need the express the speed of those particles *as they are just about to collide* symbolically...Algrebraically I have reached the solution to the part of the question that has been bugging me (which for clarity I shall post below)...but I don't want to proceed with flaws in my understanding that will become major weaknesses further down the line.

The question:

Two particles P and Q, of masses m and 3m respectively, are connected by a light elastic string of lenght l and modulus of elasticity 3mg. They are held at rest on a smooth horizontal plane, with the string stretched toa length 5l, and released from rest.

a) By considering momentum, show that if $$V_p$$ and $$V_q$$ are their speeds toward each other at any time between being released and collision then $$V_p = kV_q$$, where k is a constant and state the value of k.

b) By considering energy find in terms of g and l, their speeds just before collision

Given that they adhere to each other after collision

c) find their common velocity just after impact

d) find in terms of m, g, l the impulse each exerts on the other at impact

Part (a) relies on knowing that the tension at both ends of the string is equal such that the total momentum of the system is zero (ie: they both cancel)...the masses of each object determine their speeds...k = 3

Parts (c) and (d) I haven't started on yet but i see them as nothing more than flinging them all the info I have at them and simplifying

Its part (b) where I have been stuck and though I'm 90% sure I have the right idea I only have myself to back me up.

I have to disagree with you though about the expression I used for EPE... the denominator should be 2l as it refers to the unstretched length of the string

Last edited: Mar 12, 2006
4. Mar 12, 2006

### lightgrav

Elastic PE is ks^2/2 , where k is the "spring stiffness constant" [N/m].
Nobody uses "k" for an Elastic Modulus ... it would be too ambiguous!
I presume that your modulus is the Force needed to elongate by 100 %
... so your PE is okay, since you know what YOUR "k" is.

You haven't answered the key question:
The Work done by the Tension on each object is W = F dx ...
Why would the lo-mass object move less than the hi-mass object?

You also seem to be thinking that somehow, the two objects have
the same speed... otherwise their KE can't be proportional to their mass.

Last edited: Mar 12, 2006
5. Mar 12, 2006

### GregA

sorry..I shall look up the text code for the correct symbol and edit my post (only just started using Tex)...but yes k in my second post is meant to denote the modulus of elasticity and i should (and will) substitute either for the correct symbol or '3mg' (I overlooked that one trying to post too quickly )

my reasoning and method for part a):
The tension in the string is equal at both ends..ie Particle P is feeling the same force as particle Q...upon release those two particles will reach a point where there is no tension but the force at Q's end is pulling an object that is three times heavier than that at P's end...Since the tension at both ends are equal then the momentum of both particles is also equal (apart from their directions of motion being opposite) therefore...
m$$V_p - 3mV_q = 0...mV_p = 3mV_q$$ where 3 = k

my reasoning for part (b)
Firstly I have to hit answers 2(sqrt(gl)) and 6(sqrt(gl))

If I am correct in thinking that the EPE of P is 3 times that of Q (I probably could have worded my initial post better) then...
$$\lambda \frac {x^2}{2a} = \frac {3mg(4l)^2} {2l}$$

for the velocity of P

$$\frac {3}{4} \frac {(3mg(4l)^2)}{2l} = \frac {1}{2}mv^2 ... = \frac {9(16)gl}{8}$$
$$v = 6\sqrt{gl}$$
for the velocity of Q

$$\frac {1}{4} \frac {(3mg(4l)^2)}{2l} = \frac {1}{2}3mv^2 ... \frac {3(16)gl}{3(8)}$$
$$v = 2\sqrt{gl}$$

I am not so good with tex [frown]...(bits are missing)

Last edited: Mar 12, 2006
6. Mar 12, 2006

### lightgrav

in part (a), you said that the VELOCITY of the lo-mass one is 3 times the velocity of the hi-mass one.
(CORRECT, except for the negative sign which indicates opposite direction)

KE = 1/2 * mometum * velocity ... momentums are opposite ...
so the (3x faster) lo-mass object has 3x the KE ...
that is 3/4 of the original PE, if Energy is conserved.

What does this imply about the Work done by the Tension?
(the lo-mass object moves 3x farther)

7. Mar 12, 2006

### GregA

I expect I'm being really thick here by answering your question with: the work done by tension is 3 times as much for moving P as it is for Q such that the total work done is equal to the total EPE of the system?

Last edited: Mar 12, 2006
8. Mar 12, 2006

### lightgrav

Yeah, since the lo-mass object moves 3x as far.
so that's the other way of seeing that its KE is 3x as much.
(besides the "1/2 p v" argument)

9. Mar 12, 2006

### GregA

Lightgrav I have to say that I am very grateful to you for helping me to crystallize and better define my thoughts...thankyou very much
looking back the whole thing seems so damned obvious now!!!

Last edited: Mar 12, 2006