Conservation of energy with kinetic energy and spring

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Homework Help Overview

The discussion revolves around an elastic collision between two dynamics carts, focusing on the conservation of energy and momentum. The problem involves calculating the compression of a spring during the collision, given the masses and velocities of the carts, as well as the spring constant.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the kinetic energy of each cart before and after the collision, questioning how to account for energy transferred to the spring.
  • Some participants attempt to apply conservation of momentum to find the velocities after the collision, while others express uncertainty about the relationship between kinetic energy and potential energy in the spring.
  • There are discussions about the total kinetic energy before and after the collision and how it relates to the energy stored in the spring.
  • Questions arise regarding the assumptions made about energy conservation and the need to find the velocity of cart 1 during the collision.

Discussion Status

The discussion is ongoing, with participants providing various calculations and questioning the validity of their approaches. Some guidance has been offered regarding the use of conservation of momentum and energy, but there is no explicit consensus on the correct method or final outcome.

Contextual Notes

Participants note the complexity of the problem due to the interplay between kinetic and potential energy, and the need to consider both conservation laws in their calculations. There is also mention of imposed homework rules that may limit the methods used.

Lolagoeslala
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Homework Statement


A dynamics cart 1 ahs a mass of 1.8 kg and is moving with a velocity of 4.0 m/s
along a frictionless track. Dynamics cart 2 has a mass of 2.2 kg and is moving at 6.0 m/s
. The carts collide in a head-on elastic collision cushioned by a spring with spring constant k = 8.0 x 104 N/m.

Determine the compression of the spring in cm, during the collision when cart 2 is moving at 4.0m/s


and

Calculate the maximum compression of the spring in cm

The Attempt at a Solution



so the givens are these:
m1 = 1.8 kg
m2 = 2.2 kg
v1=4 m/s

v2 = 6 m/s

K = 80000N/ma)

Ek = 1/2m2v2`
Ek = 1/2(2.2kg)(4m/s
)^2
Ek = 17.6 J

17.6 J = 1/2kx^2
17.6 J = 1/2(80000N/m)x^2
0.020976176 m = x

b)

I am not sure what to do for this particular part?

v1`=(m1-m2)/(m1+m2)v1 + (m1xm2)/(m1+m2)v2
v1`=0.4 m/s
+5.94 m/s

v1`= 6.34 m/s


v2` = (2m1)/(m1+m2)v1 + (m2-m1)/(m1+m2)v2
v2` = 3.6 m/s
+ 0.6 m/s

v2` = 3 m/s


Ektot = 1/2m1v1^2 + 1/2m2v2^2
Ektot = 14.4 J +39.6 J
Ektot = 54 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`^2
Ektot` = 36.17604 J + 9.9 J
Ektot` = 46.07604 - but they are not the same since its elastic they are supposed to be the same​
 
Last edited:
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Lolagoeslala said:

Homework Statement


A dynamics cart 1 ahs a mass of 1.8 kg and is moving with a velocity of 4.0 m/s
along a frictionless track. Dynamics cart 2 has a mass of 2.2 kg and is moving at 6.0 m/s
. The carts collide in a head-on elastic collision cushioned by a spring with spring constant k = 8.0 x 104 N/m.

Determine the compression of the spring in cm, during the collision when cart 2 is moving at 4.0m/s


and

Calculate the maximum compression of the spring in cm

The Attempt at a Solution



so the givens are these:
m1 = 1.8 kg
m2 = 2.2 kg
v1=4 m/s

v2 = 6 m/s

K = 80000N/m

a)

Ek = 1/2m2v2`
Ek = 1/2(2.2kg)(4m/s
)^2
Ek = 17.6 J

17.6 J = 1/2kx^2
17.6 J = 1/2(80000N/m)x^2
0.020976176 m = x

b)

I am not sure what to do for this particular part?​

For part a), which is incorrect:

What is the KE of cart 1, when cart 2 is moving at 4.0m/s
? You need to include this in your total KE.​
 
SammyS said:
For part a), which is incorrect:

What is the KE of cart 1, when cart 2 is moving at 4.0m/s
? You need to include this in your total KE.​


would it be Ek = 1/2m1v1^2
Ek = 1/2(1.8kg)(4m\s)^2
Ek= 14.4 J

so then itll be

Ektot = 14.4 J + 17.6 J
Ektot = 32 J

32 J = 1/2(80000N/m)x^2
0.02828 m = x

how about this?​
 
Lolagoeslala said:
would it be Ek = 1/2m1v1^2
Ek = 1/2(1.8kg)(4m\s)^2
Ek= 14.4 J

so then itll be

Ektot = 14.4 J + 17.6 J
Ektot = 32 J

32 J = 1/2(80000N/m)x^2
0.02828 m = x

how about this?
Incorrect.

Use conservation of momentum to find the velocity of Cart 1, when Cart 2 is moving at 4.0m/s
​
 
SammyS said:
Incorrect.

Use conservation of momentum to find the velocity of Cart 1, when Cart 2 is moving at 4.0m/s
​


Okay .. but my other tutor said you can't use momentum because the energy is being conserved into the spring not the balls itself...? But if you suggest it i can try it that way...

It would be m1v1 + m2v2 = m1v1` + m2v2`

(1.8 kg)(4 m/s
) + (2.2 kg)(6 m/s
) = (1.8 kg)v1` + (2.2 kg)(4m/s
)
7.2 kgm/s
+ 13.2kgm\s
= (1.8kg)v1` + 8.8kgm/s

6 kgm/s
- 8.8kgm\s
= (1.8kg)v1`
2.8 kgm\s
/ 1.8 kg = v1`
1.556 m/s
= v1`

Ektot= 1/2m1v1`^2 + 1/2m2v2`^2
Ektot = 2.1790 J + 17.6 J
Ektot = 19.779 J

Ektot = 1/2kx^2
19.779 J / 40000 N/m = x^2
0.022236 m = x​
 
SammyS said:
Incorrect.

Use conservation of momentum to find the velocity of Cart 1, when Cart 2 is moving at 4.0m/s
​


but it doesn't say that the m1 would also be changing in the velocity... then why do we need to find the v1` ?​
 
Lolagoeslala said:
but it doesn't say that the m1 would also be changing in the velocity... then why do we need to find the v1` ?
... because the velocity of m1 does change! Conservation of momentum tells you that, without it being stated in the problem .
 
SammyS said:
... because the velocity of m1 does change! Conservation of momentum tells you that, without it being stated in the problem .

Oh i see what you are talking about ... so how is my part a calculation i just did?
 
ANY BODY?!:confused:
 
  • #10
Lolagoeslala said:
ANY BODY?!:confused:
Patience !

You need to quit prematurely bumping your posts !
 
  • #11
Lolagoeslala said:
Okay .. but my other tutor said you can't use momentum because the energy is being conserved into the spring not the balls itself...? But if you suggest it i can try it that way...

It would be m1v1 + m2v2 = m1v1` + m2v2`

(1.8 kg)(4 m/s
) + (2.2 kg)(6 m/s
) = (1.8 kg)v1` + (2.2 kg)(4m/s
)
7.2 kgm/s
+ 13.2kgm\s
= (1.8kg)v1` + 8.8kgm/s

6 kgm/s
- 8.8kgm\s
= (1.8kg)v1`
2.8 kgm\s
/ 1.8 kg = v1`
1.556 m/s
= v1`

Ektot= 1/2m1v1`^2 + 1/2m2v2`^2
Ektot = 2.1790 J + 17.6 J
Ektot = 19.779 J

Ektot = 1/2kx^2   This is the Potential Energy of the spring.
19.779 J / 40000 N/m = x^2
0.022236 m = x​

Why are you equating the Potential Energy of the spring to the total KE of the two carts?​
 
  • #12
SammyS said:
Why are you equating the Potential Energy of the spring to the total KE of the two carts?

Well the amount of energy that the carts have would be transferring into the spring wouldn't it?
 
  • #13
SammyS said:
Why are you equating the Potential Energy of the spring to the total KE of the two carts?

Well when i come to think of it...
Ektot = 19.779 J is the kinetic energy when they have come together compressing the spring..
but how much did they have the initial energy...
To do this i did this:

Ektot(initial) = 1/2m1v1^2 + 1/2m2v2^2
14.4 J + 39.6 J = 54 J

And if i subtract the 54 J - 19.779 J = 34.221 that's the energy lost in the spring correct?

so then i can use that ...

34.221 J = 1/2(80000N/m)x^2
0.029249 m = x
 
  • #14
Lolagoeslala said:
Well the amount of energy that the carts have would be transferring into the spring wouldn't it?
The amount of (kinetic) energy the carts lose will be transferred to PE of the spring. The Total energy is conserved.
 
  • #15
SammyS said:
The amount of (kinetic) energy the carts lose will be transferred to PE of the spring. The Total energy is conserved.

Yes that is correct similarely what i have done like this..

Ektot(initial) = 1/2m1v1^2 + 1/2m2v2^2
14.4 J + 39.6 J = 54 J

And if i subtract the 54 J - 19.779 J = 34.221 that's the energy lost in the spring correct?

so then i can use that ...

34.221 J = 1/2(80000N/m)x^2
0.029249 m = x

im i correct on this calculation?
 
  • #16
Lolagoeslala said:
Yes that is correct similarely what i have done like this..

Ektot(initial) = 1/2m1v1^2 + 1/2m2v2^2
14.4 J + 39.6 J = 54 J

And if i subtract the 54 J - 19.779 J = 34.221 that's the energy [STRIKE]lost[/STRIKE] in the spring correct?

so then i can use that ...

34.221 J = 1/2(80000N/m)x^2
0.029249 m = x

I'm i correct on this calculation?

Yes. now convert that to cm.
 
  • #17
SammyS said:
Yes. now convert that to cm.

Okay but now what about the the sexond part where you need to find the maximum spring compression ? i showed my work above... but the energy before and after is different.. is that the way u should do it?
 
  • #18
Lolagoeslala said:
...

Calculate the maximum compression of the spring in cm

The Attempt at a Solution



so the givens are these:

m1 = 1.8 kg  m2 = 2.2 kg  v1=4 m/s
  v2 = 6 m/s
  K = 80000N/m

b)

I am not sure what to do for this particular part?

v1`=(m1-m2)/(m1+m2)v1 + (m1xm2)/(m1+m2)v2
v1`=0.4 m/s
+5.94 m/s

v1`= 6.34 m/s


v2` = (2m1)/(m1+m2)v1 + (m2-m1)/(m1+m2)v2
v2` = 3.6 m/s
+ 0.6 m/s

v2` = 3 m/s


Ektot = 1/2m1v1^2 + 1/2m2v2^2
Ektot = 14.4 J +39.6 J
Ektot = 54 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`^2
Ektot` = 36.17604 J + 9.9 J
Ektot` = 46.07604 - but they are not the same since its elastic they are supposed to be the same​


Where did you get those equations for v1' & v2'? For what conditions are those equations true? I suspect that they're for the velocities resulting from an elastic collision. As you point out, there's something wrong because there is kinetic energy lost.

While you have an error somewhere when you used these. They have nothing to do with solving part b).

At the instant the spring is at maximum compression, the two carts have the same velocity, v1 = v2 . Why is that the case?​
 
Last edited:
  • #19
SammyS said:
Where did you get those equations for v1' & v2'? For what conditions are those equations true? I suspect that they're for the velocities resulting from an elastic collision. As you point out, there's something wrong because there is kinetic energy lost.

While you have an error somewhere when you used these. They have nothing to do with solving part b).

At the instant the spring is at maximum compression, the two carts have the same velocity, v1 = v2 . Why is that the case?

well those equations are true for two moving object.. and i found the mistake where i was wrong... for the v1` . the second mart is (2m1/m1+m2)v2

i used these to find the kinetic energy i just wanted to see if they equal the kinetic energy before and after.. and its true... so now to find the maximum compression.. i need to find the velocity of them in the middle... right?
 
  • #20
Lolagoeslala said:
well those equations are true for two moving object.. and i found the mistake where i was wrong... for the v1` . the second mart is (2m1/(m1+m2))v2

i used these to find the kinetic energy i just wanted to see if they equal the kinetic energy before and after.. and its true... so now to find the maximum compression.. i need to find the velocity of them in the middle... right?
In the middle of what ?

Use conservation of momentum to find v2 & v2, when they're equal. Of course that's when v1 = v2 = vcenter of mass. And, I asked earlier, if you know why that gives the distance of closest approach.
 
Last edited:
  • #21
SammyS said:
In the middle of what ?

Use conservation of momentum to find v2 & v2, when they're equal. Of course that's when v2 v2 = vcenter of mass. And, I asked earlier, if you know why that gives the distance of closest approach.

Alright this is what i did...

m1v1 + m2v2 = (m1+m2) Vmin
(1.8kg)(4m/s
) + (2.2kg)(6m/s
) = ( 4kg ) Vmin
7.2 kgm/s
+13.2kgm/s
/4kg = vmin
1.5 m\s
= Vmin

Ekmin = 1/2m1Vmin^2 + 1/2m1Vmin^2
Ekmin = 1/2(1.8kg)(1.5m\s)^2 + 1/2(2.2kg)(1.5m\s)^2
Ekmin=2.025 J + 2.475 J
Ekmin = 4.5 J

Ektot - Wdef = Ekmin
54 J - Wdef = 4.5 J
Wdef = 49.5 J

Ekspring = 1/2kx^2
Ekspring = 1/2(80000N/m)x^2
49.5 J/40000N/m = x^2
0.035178 m =x
3.178 cm = x​
 
  • #22
Why is it that when v1 = v2, then the spring is at maximum compression. In other words, How is it that you know that this is the condition for which the carts are at a minimum distance from each other?
 
  • #23
SammyS said:
Why is it that when v1 = v2, then the spring is at maximum compression. In other words, How is it that you know that this is the condition for which the carts are at a minimum distance from each other?

Well we did a long prove question in class. and that i found out.. but like in general just think about it .. if the spring is connected to both of them.. then the velocity have to be the same inorder and the distance have to be short inorder to cause the maximum compression of the spring but are my calculation correct u would say?
 
  • #24
Lolagoeslala said:
Well we did a long prove question in class. and that i found out.. but like in general just think about it .. if the spring is connected to both of them.. then the velocity have to be the same inorder and the distance have to be short inorder to cause the maximum compression of the spring but are my calculation correct u would say?
They look fine to me.
 

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