Why don't we account for energy lost in collision here? SHM

In summary: So only the very topmost mass of the spring will collide with the block, and the rest of the spring is irrelevant.
  • #1
Tea_Aficionado
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Homework Statement
A block of mass m = 2.0kg is dropped from height h = 40cm onto a spring of k = 1960N/m. Find the maximum distance the spring is compressed.
Relevant Equations
mgh = 1/2 * kx[SUB]max[/SUB][SUP]2[/SUP]

delta x = mg/k

h = x_max
We know that the Ug is converted to KE and Us. I thought that since the system loses energy after the collision that we shouldn't use the equation hnew= delta x + h.

I thought instead that maybe the h we should use is xmax, because that's when there is maximum Ug and there is no other energy present (having been converted all to Ug).

Therefore, mg * xmax= 1/2 * kxmax2

Solving for that with quad. eq, we get x = 0.02m


The solution, however, was 0.10m.

I looked at the worked-through examples online, and I'm confused as to why we don't consider the energy before and after the collision to be partially lost: their examples said to use hnew = h + Δx rather than just Δx.

If anyone could please take a bit of time to explain my thinking errors, I'd greatly appreciate it!


Post Notes:

  • I don't know if I'm allowed to link the referenced examples here, but they seem for the most part agree with each other and can be found by copying and pasting the problem into the google search bar.
  • I'm new, so if I made any formatting errors I apologize!
 
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  • #2
Tea_Aficionado said:
the system loses energy after the collision
Why do you say that the system loses energy after the collision. Why would it not be elastic?
 
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  • #3
jbriggs444 said:
Why do you say that the system loses energy after the collision. Why would it not be elastic?
Oh, that's true. I guessed that it would be inelastic because I did another problem before that had the collision be inelastic. That would explain why we include h in the total height... but at the same time aren't we assuming that the mass sticks onto the spring because the motion is sustained as SHM, and therefore assume that it's inelastic?
 
  • #4
Tea_Aficionado said:
aren't we assuming that the mass sticks onto the spring because the motion is sustained as SHM, and therefore assume that it's inelastic?
If the mass is stuck to the spring and keeps oscillating forever, the energy is still there. If it bounces away, the energy is still there.
 
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  • #5
jbriggs444 said:
If the mass is stuck to the spring and keeps oscillating forever, the energy is still there. If it bounces away, the energy is still there.
But the energy transferred is different for the mass stuck on the spring (inelastic) vs. the mass bouncing away (elastic) right? Or is it that whether it's inelastic or elastic doesn't depend on whether or not it sticks or bounces?

How do we know that the problem is an elastic collision?
 
  • #6
If you assume no energy is lost what do you get for the distance the spring is compressed?

What assumption do you have to make to come up with the maximum distance the spring can compress?
 
  • #7
We have done a lot of problems like this here in physics forums and I think the problem wants us to assume that the kinetic energy is conserved before and after the collision of the block with the spring.

After all we consider the spring to have negligible mass hence the formula for inelastic collision ##v_f=\frac{mv_i}{m+M}## will give ##v_f=v_i## when ##M=0##.
 
  • #8
Tea_Aficionado said:
But the energy transferred is different for the mass stuck on the spring (inelastic) vs. the mass bouncing away (elastic) right? Or is it that whether it's inelastic or elastic doesn't depend on whether or not it sticks or bounces?

How do we know that the problem is an elastic collision?
The problem does not specify a mass for the spring, so its mass should be ignored; and even if it did give a mass, the inelastic collision is only at first contact, involving only a tiny amount of spring mass. The mechanical energy lost as a result is negligible.
 
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  • #9
BTW I did the problem and I get an irrational number as solution that is very close to 0.1 though.
Your quadratic equation ##mgx=\frac{1}{2}kx^2## is not correct, in this equation you ignore the initial potential energy of the block because it is at height 0.4m above from the spring. The correct equation is $$mgx+mgh=\frac{1}{2}kx^2$$ where ##h=0.4##.
 
  • #10
haruspex said:
the inelastic collision is only at first contact, involving only a tiny amount of spring mass
This is an interesting thought, can you elaborate a bit more why only the mass on the very top of the spring participates in the collision?
 
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  • #11
Delta2 said:
This is an interesting thought, can you elaborate a bit more why only the mass on the very top of the spring participates in the collision?
In a standard coalescence problem, one mass impacts another and thereafter they move together. But looking at it in detail, there will be some compression on impact. We can think of it as a mass hitting a spring that is very inefficient; the spring compresses but does not rebound. (Or it may rebound a little, but the two bodies are now stuck together so the rebound energy becomes lost in decaying oscillations.)
The difference between that and the present problem is that the spring is presumed efficient, losing negligible mechanical energy.
 
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  • #12
haruspex said:
In a standard coalescence problem, one mass impacts another and thereafter they move together. But looking at it in detail, there will be some compression on impact. We can think of it as a mass hitting a spring that is very inefficient; the spring compresses but does not rebound. (Or it may rebound a little, but the two bodies are now stuck together so the rebound energy becomes lost in decaying oscillations.)
The difference between that and the present problem is that the spring is presumed efficient, losing negligible mechanical energy.
Would you say that this is presumed true for all ideal springs? Or is it just a presumption we're supposed to make in this problem only?

On a side note, thanks to everyone for chiming in! I woke up today and I was pleasantly surprised. Thanks y'all! (not end of thread though lol)
 
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  • #13
Mister T said:
If you assume no energy is lost what do you get for the distance the spring is compressed?

What assumption do you have to make to come up with the maximum distance the spring can compress?
1. It'd just be h + delta x right?

2. The assumption that gravitational potential energy before the collision is completely converted into potential spring energy after the collision?



Oh I also just realized. Maybe by "maximum" it means max. in general considering favorable conditions (assuming that energy is conserved before and after the collision)? Or am I going through some mental gymnastics? I feel like I'm especially confused because I did another very similar problem but with one of the conditions being that the collision was inelastic. The energy was similarly assumed to be conserved before and after the collision, which is why I asked haru if that assumption for these problems came from the fact that they include ideal springs... if someone can clarify that would be great!

Edit: Question summarized in below post
 
Last edited:
  • #14
Delta2 said:
BTW I did the problem and I get an irrational number as solution that is very close to 0.1 though.
Your quadratic equation ##mgx=\frac{1}{2}kx^2## is not correct, in this equation you ignore the initial potential energy of the block because it is at height 0.4m above from the spring. The correct equation is $$mgx+mgh=\frac{1}{2}kx^2$$ where ##h=0.4##.
Wouldn't it be $$mgdeltax+mgh=\frac{1}{2}kxmax^2 $$ where ##h=0.4##, and ##deltax = 0.098... oh, no, we're saying that xmax + h = hnew

we don't use deltax at all

so then, $$mgxmax+mgh=\frac{1}{2}kxmax^2 $$

and solving for xmax we get exactly 0.1m


Should we always assume for these SHM problems that energy is conserved before and after collision? How come in some problems we consider energy lost to not be negligible but here to be considered negligible? How would we make the distinction?

Edit: Okay, so reading through more carefully from what haru said, we presume the spring to be "efficient." Why?
 
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  • #15
Tea_Aficionado said:
Edit: Okay, so reading through more carefully from what haru said, we presume the spring to be "efficient." Why?
Primarily because ideal springs are simple while non-ideal springs are complicated. Also, because we try to make real springs as ideal as possible and we do fairly well at it.

Problems with balls dropping on springs are first year physics. And in first year physics, all springs are ideal.

There are several ways in which a spring can fail to be ideal. It could fail to obey Hooke's law. Then you could not use ##PE=\frac 1 2 kx^2##. It could fail to exert the same force when extending as it does when compressing (also known as hysteresis). It could (as alluded to in multiple posts above) have non-zero mass and absorb energy into internal vibrations.

We design springs so that they operate with small deflections where Hooke's law holds pretty much automatically. We design springs from materials that deform elastically without significant loss to hysteresis. And we design springs from materials with a high strength to weight ratio. So real springs tend to be close to ideal.
 
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  • #16
jbriggs444 said:
Primarily because ideal springs are simple while non-ideal springs are complicated. Also, because we try to make real springs as ideal as possible and we do fairly well at it.

Problems with balls dropping on springs are first year physics. And in first year physics, all springs are ideal.

There are several ways in which a spring can fail to be ideal. It could fail to obey Hooke's law. Then you could not use ##PE=\frac 1 2 kx^2##. It could fail to exert the same force when extending as it does when compressing (also known as hysteresis). It could (as alluded to in multiple posts above) have non-zero mass and absorb energy into internal vibrations.

We design springs so that they operate with small deflections where Hooke's law holds pretty much automatically. We design springs from materials that deform elastically without significant loss to hysteresis. And we design springs from materials with a high strength to weight ratio. So real springs tend to be close to ideal.
Huh, okay. So we assume collisions onto ideal springs conserve energy before and after collision?
 
  • #17
Tea_Aficionado said:
Huh, okay. So we assume collisions onto ideal springs conserve energy before and after collision?
In classical physics, energy is *always* conserved - even for non-ideal springs.

But for the non-ideal case, some of the energy gets turned to ‘heat’ and sound. The temperatures of the spring and block slightly increase and you can hear the collision.

We sometimes use the term 'mechanical energy' to mean the sum of the kinetic and potential energies. (Generally ‘mechanical energy’ excludes heat, light and sound.)

For the ideal block/spring system, mechanical and total energy are the same thing. They are conserved.

For the non-ideal system, mechanical energy is not conserved (because of heat and sound production). But total energy is conserved.
 
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  • #18
Steve4Physics said:
In classical physics, energy is *always* conserved - even for non-ideal springs.

But for the non-ideal case, some of the energy gets turned to ‘heat’ and sound. The temperatures of the spring and block slightly increase and you can hear the collision.

We sometimes use the term 'mechanical energy' to mean the sum of the kinetic and potential energies. (Generally ‘mechanical energy’ excludes heat, light and sound.)

For the ideal block/spring system, mechanical and total energy are the same thing. They are conserved.

For the non-ideal system, mechanical energy is not conserved (because of heat and sound production). But total energy is conserved.
I understand that generally energy is always conserved based on the law of conservation of energy, but if we define the system to be just that of the block and the spring, wouldn't the block/spring system lose energy due to deformation? Wiki states that in "such a collision, kinetic energy is lost by bonding the two bodies together," (wiki link) and even if it was just a partially inelastic collision I was taught the system loses energy too. So wouldn't the current problem solving of considering the energy of the system to be the same before and after the collision be incorrect?

Is this just a moot point? I'm sorry -- I just feel really confused, and a lot of questions come up. Do we consider collisions of ideal pendulums to also follow this pattern of energy conservation before and after the collision? Is the energy conservation of the system intrinsic to just a collision with an "ideal" object? Or is it that the problem supposes elastic collision? How would I then differentiate between elastic and inelastic collision without any keywords relating to either? How would the collision be elastic if we assume the object sticks onto the spring (isn't that inherently inelastic)?


I think my biggest confusion is: even if it's an ideal system, isn't there still energy lost from deformation? I think the rest stem from there
 
  • #19
Tea_Aficionado said:
if we define the system to be just that of the block and the spring, wouldn't the block/spring system lose energy due to deformation?
It loses mechanical energy to, say, heat. But that heat is initially still in the spring-block system. Only over time does it dissipate further afield.
 
  • #20
haruspex said:
It loses mechanical energy to, say, heat. But that heat is initially still in the spring-block system. Only over time does it dissipate further afield.
Right. I think I'll make another thread about the similar problem that I kept on referring to -- maybe it'll help illustrate my confusion.

But once again, thank you all so much for helping me throughout this thread! I understand that it's pretty mentally taxing to try to fill the gaps in my knowledge :oldbiggrin:
 
  • #21
Tea_Aficionado said:
isn't there still energy lost from deformation?
For an elastic deformation such as with a spring or a well insulated and lubricated gas cylinder, mechanical energy is not lost when the device is deformed. It is still available as potential energy in the spring or as energy in the compressed gas. If the spring is released or the gas cylinder allowed to expand, the "lost" mechanical energy is restored.

For a plastic deformation such as an impact into a blob of putty or a dent in a mild steel door panel, most of the mechanical energy may be dissipated as heat and will result in permanent deformation.
 
  • #22
Tea_Aficionado said:
but if we define the system to be just that of the block and the spring, wouldn't the block/spring system lose energy due to deformation?
Deformation simply means changing shape.

An ideal spring deforms elastically – you can get all the input energy back from an ideal compressed spring when it expands. At an atomic level, we have simply temporarily stored energy by pushing atoms together against a repulsive force between them (like a load of nano-ideal-springs!). We can recover all this stored energy.

But in a non-ideal situation, some of the energy supplied during deformation ends up as additional random thermal motion of the atoms (temperature increases). This energy cannot be recovered when the spring expands because of the randomness of the atomic motion. The thermal energy spreads through the spring and block and is lost to the environment. And some energy is radiated away as sound (You are now into the realms of thermodynamics and have increased the entropy of the universe!)

Tea_Aficionado said:
Wiki states that in "such a collision, kinetic energy is lost by bonding the two bodies together," (wiki link) and even if it was just a partially inelastic collision I was taught the system loses energy too. So wouldn't the current problem solving of considering the energy of the system to be the same before and after the collision be incorrect?
The current problem is simply an introductory level exercise. To simplify the maths we are allowed (expected!) to assume all the mechnical energy is conserved. In real life it wouldn't be.

Tea_Aficionado said:
Is this just a moot point? I'm sorry -- I just feel really confused, and a lot of questions come up. Do we consider collisions of ideal pendulums to also follow this pattern of energy conservation before and after the collision?
If we are told (or can approximate) that the pendulums behave ideally, yes - we can assume mechanical (hence total) energy is conserved.

Tea_Aficionado said:
Is the energy conservation of the system intrinsic to just a collision with an "ideal" object? Or is it that the problem supposes elastic collision?
Energy is always conserved. You mean 'Is the mechnical energy conservation of the system intrinsic...'.
We are assuming ideality which leads to conservation of mechanical energy in this particular problem.

Tea_Aficionado said:
How would I then differentiate between elastic and inelastic collision without any keywords relating to either? How would the collision be elastic if we assume the object sticks onto the spring (isn't that inherently inelastic)?
Have you met inelastic collisions which can be solved by conservation of momentum to find velocity just after impact? In the current problem we are assuming the spring is massless. Read what @haruspex said in Post #8.

Tea_Aficionado said:
think my biggest confusion is: even if it's an ideal system, isn't there still energy lost from deformation? I think the rest stem from there
You have to decide from the context and wording. Experience helps. If a question is genuinely ambiguous you can make and state your assumed interpretation and solve that problem. Or if you are very keen you can solve both elastic and inelastic versions!

Edit - typo's
 
  • #23
Steve4Physics said:
Deformation simply means changing shape.

An ideal spring deforms elastically – you can get all the input energy back from an ideal compressed spring when it expands. At an atomic level, we have simply temporarily stored energy by pushing atoms together against a repulsive force between them (like a load of nano-ideal-springs!). We can recover all this stored energy.

But in a non-ideal situation, some of the energy supplied during deformation ends up as additional random thermal motion of the atoms (temperature increases). This energy cannot be recovered when the spring expands because of the randomness of the atomic motion. The thermal energy spreads through the spring and block an is lot to the environment. And some energy is radiated away as sound (You are now into the realms of thermodynamics and have increased the entropy of the universe!)The current problem is simply an introductory level exercise. To simplify the maths we are allowed (expected!) to assume all the mechnical energy is conserved. In real life it wouldn't be.If we are told (or can approximate) that the pendulums behave ideally, yes - we can assume mechnical (hence total) energy is conserved.Energy is always conserved. You mean 'Is the mechnical energy conservation of the system intrinsic...'.
We are assuming ideality which leads to conservation of mechanical energy in this particular problemHave you met inelastic collisions which can be solved by conservation of momentum to find velocity just after impact? In the current problem we are assuming the spring is massless. Read what @haruspex said in Post #8.You have to decide from the context and wording. Experience helps. If a question is genuinely ambiguous you can make and state your assumed interpretation and solve that problem. Or if you are very keen you can solve both elastic and inelastic versions!
OMG, thank you so much! Read through it all. That makes sense -- that we are assuming that mechanical energy is conserved due to ideality. I'll still post my second thread -- just for reference sake ig.

But again, thank you all so much! :bow::oldbiggrin::oldbiggrin::oldbiggrin:
 
  • #24
haruspex said:
The problem does not specify a mass for the spring, so its mass should be ignored; and even if it did give a mass, the inelastic collision is only at first contact, involving only a tiny amount of spring mass. The mechanical energy lost as a result is negligible.
ah sorry haru, I didn't fully process that until I carefully read through it. Thank you!
 

Related to Why don't we account for energy lost in collision here? SHM

1. Why is energy lost in collisions not accounted for in simple harmonic motion (SHM) equations?

In SHM, the system is assumed to be ideal and without any external forces acting on it. This means that there is no energy lost due to friction or other forms of resistance. Therefore, the equations do not account for energy loss in collisions.

2. How does energy loss in collisions affect the accuracy of SHM predictions?

If there is significant energy loss in collisions, the actual motion of the system may deviate from the predicted SHM motion. This can result in inaccurate predictions and may require more complex equations to accurately model the system.

3. Can energy loss in collisions be ignored in SHM calculations?

In most cases, yes. The amount of energy lost in collisions is usually very small compared to the total energy of the system. Therefore, it can be ignored without significantly affecting the accuracy of SHM predictions.

4. Are there any real-world systems that exhibit perfect SHM without any energy loss in collisions?

No, there are no real-world systems that can exhibit perfect SHM without any energy loss in collisions. In reality, all systems experience some form of resistance or friction, which results in energy loss. However, some systems may exhibit SHM behavior that is close enough to ideal to be accurately modeled using SHM equations.

5. How can energy loss in collisions be accounted for in SHM calculations?

If the energy loss in collisions is significant and cannot be ignored, more complex equations such as the damped harmonic oscillator equation can be used to account for it. These equations take into account the energy lost in each collision and can provide more accurate predictions for the system's motion.

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