# Conservation of Energy and change in the length of a spring

• jisbon
In summary, The spring has no momentum at all times. When the velocities of block a and block b have become equal, the spring will stop compressing.
jisbon
Homework Statement
Three objects lie on frictionless surface with the last one connected to a spring with constant k. Object on left travels at velocity to right v to the right and the other 2 remain stationery at first. Show that maximum change in length of spring is ##\sqrt\frac{4mv^2}{9k}## assuming collsion between the first 2 objects is elastic
Relevant Equations
COM: ##mv= mv_{a} + mv_{b} ##
COE: ##\frac{1}{2}mv^2 =\frac{1}{2}mv_{a}^2 + \frac{1}{2}mv_{b}^2##

Hola!

So my first approach to this is use both the conservation of energy and momentum equations since collision between the first two objects are elastic.
Let the 3 blocks be a,b and c (from left to right)
Does this means the following:
whereby
##v_{a} ##= speed of block a after collision
##v_{b}##= speed of block b after collision

COE:
##\frac{1}{2}mv^2 =\frac{1}{2}mv_{a}^2 + \frac{1}{2}2mv_{b}^2##
COM:
##mv= mv_{a} + 2mv_{b} ##

I'm not supposed to relate any of this equation to block C since it isn't directly involved in the collision with block a, is this correct?

Thanks

Correct. All you do is find out ##v_b##

Nitpicking: your homework equations should distinguish an ##m_a## and an ##m_b##

BvU said:
Correct. All you do is find out ##v_b##

Nitpicking: your homework equations should distinguish an ##m_a## and an ##m_b##
Thanks.
After finding out ##v_b## , will I use the gain in KE from B = lose in EPE in the spring?
Which means :
##\frac{1}{2}(2m)v_{b^2} = \frac{1}{2}(k)x^2##
where x is supposed to be the 'shown' answer?
Thanks

jisbon said:
Thanks.
After finding out ##v_b## , will I use the gain in KE from B = lose in EPE in the spring?
Which means :
##\frac{1}{2}(2m)v_{b^2} = \frac{1}{2}(k)x^2##
where x is supposed to be the 'shown' answer?
Thanks
No I am afraid this is not correct because the other end of the spring is not fixed but will also start moving as the spring starts to compress. The spring will stop compressing when block b and block c have equal velocity v' and at that time it will be the maximum compression x of the spring. Use conservation of momentum and conservation of mechanical energy for the system (block b +block c+spring ) to find v' and x.

By the way, equations provided are not just to annoy you. An important possible way to do elastic collisions is to first transform to the center of momentum (COM) frame. In that frame the momentum of block a and block b are equal-but-opposite. After the collision they are still equal-but-opposite, since that's the definition of COM. So after the collision they must simply have flipped the sign of their respective momentum. Then you transform back to the lab frame to get the momentum after the collision.

Then you look at the collision of block b and block c, again in the COM, but the COM of this new system with the new momentum of block b. Again, the things will be so much easier in the COM.

Delta2 said:
No I am afraid this is not correct because the other end of the spring is not fixed but will also start moving as the spring starts to compress. The spring will stop compressing when block b and block c have equal velocity v' and at that time it will be the maximum compression x of the spring. Use conservation of momentum and conservation of mechanical energy for the system (block b +block c+spring ) to find v' and x.
Hmm okay since block will also move as well, is it okay to assume the following equations?
COE:
##\frac{1}{2}2m(v_{b}^2)=\frac{1}{2}kx^2 + \frac{1}{2}2m(v_{c}^2)##
COM:
##2m(v_{b})= 2m(v_{c}) + ?## Does the spring have any momentum in this case?

Thanks

jisbon said:
Hmm okay since block will also move as well, is it okay to assume the following equations?
COE:
##\frac{1}{2}2m(v_{b}^2)=\frac{1}{2}kx^2 + \frac{1}{2}2m(v_{c}^2)##
This equation is missing a term. It is the kinetic energy of block b at the moment the velocity of block b and block c have become equal so its missing a term ##\frac{1}{2}2m(v_{c}^2)## in the right side of the equation.

COM:
##2m(v_{b})= 2m(v_{c}) + ?## Does the spring have any momentum in this case?

Thanks
I guess the spring is considered massless so it doesn't have any momentum or kinetic energy at all times. Again the momentum equation is missing the momentum of block b at the time the velocities of block b and c have become equal.

Delta2 said:
This equation is missing a term. It is the kinetic energy of block b at the moment the velocity of block b and block c have become equal so its missing a term ##\frac{1}{2}2m(v_{c}^2)## in the right side of the equation.I guess the spring is considered massless so it doesn't have any momentum or kinetic energy at all times. Again the momentum equation is missing the momentum of block b at the time the velocities of block b and c have become equal.

I only understood the part where you mentioned that the spring will stop compressing after block a and b reach the same velocity, but I don't really get how to put that in the COM/COE equation.

OK.

Lets look what happens at the start and end of the process.

At the start of the process block b has velocity ##v_b## so it has momentum and kinetic energy. Also at the start, block c has zero velocity so its momentum and kinetic energy is zero. Also the spring is uncompressed so the PE is zero.

At the end of the process (when the velocities of block b and block c become equal), block b and block c have equal velocity let's call it ##v_c##. So each block has momentum and kinetic energy (which happens to be equal since block b and block c have also equal mass). Also the spring is compressed at length x so it has PE.

So I think now you should be able to apply CoM and CoE for the start and end of the process.

Last edited:
Based on what I understood from your statements:

COE:
##\frac{1}{2}2m(v_{b}^2)=\frac{1}{2}kx^2 + \frac{1}{2}2m(v_{c}^2) + \frac{1}{2}2m(v_{c}^2)##
COM:
##2m(v_{b})= 2m(v_{c}) + 2m(v_{c})##

Will this be correct? Thanks

Delta2
jisbon said:
Based on what I understood from your statements:

COE:
##\frac{1}{2}2m(v_{b}^2)=\frac{1}{2}kx^2 + \frac{1}{2}2m(v_{c}^2) + \frac{1}{2}2m(v_{c}^2)##
COM:
##2m(v_{b})= 2m(v_{c}) + 2m(v_{c})##

Will this be correct? Thanks
Yes I think these two equations are correct.

Delta2 said:
Yes I think these two equations are correct.
Hi Delta. Just a enquiry while I continue with this question.
Since I need to prove that the maximum change in terms of m, v and k, should I be substituting such that my ##V_b## is in terms of v?

If it is, as such I have done the following workings, but am kind of stuck at this step:##COE: \frac{1}{2}mv^2 = \frac{1}{2}mv_a^2 + \frac{1}{2}2mv_b^2##
##COM: mv = mv_a +mv_b##
##mv^2 = mv_a^2 + 2mv_b^2##
##mv_a^2+ 2mv_b^2 = mvv_a^2 + 2mvv_b^2##
##v_a^2+ 2v_b^2 = vv_a^2 + 2vv_b^2##
##2v_b^2 - 2vv_b^2 = vv_a^2 - v_a^2##
##v_a = v - 2v_b##
##2v_b^2 - 2vv_b^2 = v(v - 2v_b)^2 - (v - 2v_b)^2##
##2v_b^2 - 2vv_b^2 = (v-1)(v - 2v_b)^2##
##2v_b^2(1-v) = (v-1)(v - 2v_b)^2##
##-2v_b^2 = (v - 2v_b)^2##
##-2v_b^2 = (v-2v_b)^2##
##-2v_b^2 = v^2-4vv_b+4v_b^2##
##6v_b^2-4vv_b+v^2 =0##
Am I suppose to solve for the root v_b, or have I done something wrong along these lines?
Thank you.

yes you should find ##v_b## in terms of ##v##.

At the moment I can't understand how you infer this line
##mv_a^2+2mv_b^2=mvv_a^2+2mvv_b^2##
Better to proceed as follows:
We know that ##v=v_a+2v_b## (this follows from CoM) and ##v^2=v_a^2+2v_b^2## (this follows from CoE)
From these two you can conclude (by squaring both sides of the first and then substituting the second) that ##4v_av_b+4v_b^2=2v_b^2## or ##4v_av_b+2v_b^2=0##
From this substitute ##v_a=v-2v_b## and you can solve easily for ##v_b##.

Delta2 said:
yes you should find ##v_b## in terms of ##v##.

At the moment I can't understand how you infer this line
##mv_a^2+2mv_b^2=mvv_a^2+2mvv_b^2##
Better to proceed as follows:
We know that ##v=v_a+2v_b## (this follows from CoM) and ##v^2=v_a^2+2v_b^2## (this follows from CoE)
From these two you can conclude (by squaring both sides of the first and then substituting the second) that ##4v_av_b+4v_b^2=2v_b^2## or ##4v_av_b+2v_b^2=0##
From this substitute ##v_a=v-2v_b## and you can solve easily for ##v_b##.
So the end result will be a quadratic equation,
##4V - 8v_b + 2 v_b^2 = 0##
In which I will need to use the formula to solve for v_b I assume?
Which will be
##\frac {8 \pm \sqrt{64-32v}}{4}##?
Thanks

jisbon said:
So the end result will be a quadratic equation,
##4V - 8v_b + 2 v_b^2 = 0##
In which I will need to use the formula to solve for v_b I assume?
Which will be
##\frac {8 \pm \sqrt{64-32v}}{4}##?
Thanks
That is not the quadratic equation I get. Please be a bit more carefull when you substitute ##v_a=v-2v_b## in ##4v_av_b+2v_b^2=0##

Sorry, corrected my mistake, but can't seem to proceed from here.
I got ##v_b = \frac{2}{3}v ##
There after, using both COE and Com equations as derived from eariler posts:
jisbon said:
COE:
##\frac{1}{2}2m(v_{b}^2)=\frac{1}{2}kx^2 + \frac{1}{2}2m(v_{c}^2) + \frac{1}{2}2m(v_{c}^2)##
COM:
##2m(v_{b})= 2m(v_{c}) + 2m(v_{c})##
##(v_{b})^2= 2(v_{c})##
##(v_{c})= \frac{(v_{b})^2}{2} ##
Subbing in this to the COE equation,
##m(v_{b}^2)=\frac{1}{2}kx^2 + 2m(\frac{(v_{b})^2}{2}) ^2 ##
##m(v_{b}^2)=\frac{1}{2}kx^2 + (\frac{1}{2}m (v_{b})^4)##
##\frac{1}{2}kx^2 = m(v_{b}^2)- (\frac{1}{2}m (v_{b})^4)##
Is everything correct so far? After I subbed in the appropriate v the answer doesn't seem to tally :/

jisbon said:
##(v_{b})^2= 2(v_{c})##
Where did that come from? You have a velocity on one side and a velocity squared on the other.

Delta2
jisbon said:
Sorry, corrected my mistake, but can't seem to proceed from here.
I got ##v_b = \frac{2}{3}v ##
Correct so far.
There after, using both COE and Com equations as derived from eariler posts:

##(v_{b})^2= 2(v_{c})##
As @haruspex also noted, this equation is wrong and we know that this equation cannot be derived by any other equations that describe physical laws, let it be CoE or CoM, by just looking at it and doing a so-called quick dimensional analysis. The units of ##v_b^2## is ##m^2/sec^2## while the units of ##v_c## are ##m/sec## and because the units don't match we know that this equation must be wrong. The correct equation as it follows from the CoM equation
$$2mv_b=2mv_c+2mv_c$$ is ##v_b=2v_c##.

haruspex said:
Where did that come from? You have a velocity on one side and a velocity squared on the other.
Delta2 said:
Correct so far.

As @haruspex also noted, this equation is wrong and we know that this equation cannot be derived by any other equations that describe physical laws, let it be CoE or CoM, by just looking at it and doing a so-called quick dimensional analysis. The units of ##v_b^2## is ##m^2/sec^2## while the units of ##v_c## are ##m/sec## and because the units don't match we know that this equation must be wrong. The correct equation as it follows from the CoM equation
$$2mv_b=2mv_c+2mv_c$$ is ##v_b=2v_c##.
Oh gosh.
Its ##\sqrt \frac{4mv^2}{9k}## :)
Thanks for all the help

BvU and Delta2

## What is the concept of conservation of energy?

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or converted from one form to another.

## How does the length of a spring change when energy is applied?

When energy is applied to a spring, it can stretch or compress, causing a change in its length. This is due to the potential energy stored in the spring being converted into kinetic energy.

## What factors affect the change in length of a spring?

The change in length of a spring is affected by the amount of energy applied, the stiffness of the spring, and the mass of the object attached to the spring.

## How can the conservation of energy be applied to the length of a spring?

The conservation of energy can be applied to the length of a spring by using the equation for potential energy, which is equal to the force applied multiplied by the distance the spring is stretched or compressed. This potential energy can then be converted into kinetic energy.

## Why is the conservation of energy important in understanding the behavior of springs?

The conservation of energy is important in understanding the behavior of springs because it allows us to predict how they will respond to different amounts of energy applied. It also helps us understand how springs store and release energy, making them useful in many applications such as in mechanical devices and engineering systems.

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