Impulse and momentum in one dimension - Two carts are moving

  • #1
Specter
120
8

Homework Statement


[/B]
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k=1200 N/m).

a) Find the velocity of each cart after the collision
b) Find the maximum compression of the spring.

Homework Equations


Conservation of energy
1/2 mv^2

The Attempt at a Solution


[/B]
I'm not sure if this question is correct so any help would be great.

a) Find the velocity of each cart after the collision.

Let east be positive.

Drawing the carts in different frames of reference

https://i.imgur.com/56uRFk8.png

Cart 1 (.80kg) is the moving mass and cart 2 is stationary.

v1f=m1-m2/m1+m2 v1o

= 0.80-0.60 / 0.80+0.60 (7.0)

= 1.0

Switch back to Earth's frame of reference by adding (-5.0).
1.0 + (-5.0)= -4.0

Cart 1 is moving at 4.0 m/s [W] after the collision

Cart #2:

v2f=2m1/m1+m2v1o

=2(0.80)/0.80+0.60 (7.0)

= 8.0

Switch back to Earth's frame of reference by adding (-5.0).

8.0 + (-5.0) = 3.0

Cart 2 is moving at 3.0 m/s [E] after the collision.

b) Find the maximum compression of the spring.

This is where I am having the most trouble. I have seen different answers online than what I arrive at.

To find the maximum compression of the spring I need to know fast the carts are moving when the spring is at max compression.

PTO=PTF

m1v1o+m2v2o=(m1+m2)vf

(0.80)(2.0)+(0.60)(0)=(0.80+0.60)vf

1.6=1.4

=1.14 m/s

At max compression the two carts are moving at 1.14 m/s.

Now use conservation of energy to find the maximum compression.

1/2 m2v1o2+1/2 m2v2o2=1/2(m1+m2)v2f2+1/2 kx2

1/2 (0.8)(2.0)2+1/2 (0.6)(-5)2=1/2 (0.8+0.6)(3.0)2+1/2 (1200)x2

9.1 = 6.3+600x2
9.1 = √606.3
9.1 = 24.62315
2.17 m

I don't know what I did wrong but the answers I have seen online are around 0.12m. Where did I go wrong? I've looked it over many times and I can't figure it out.


 
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  • #2
Specter said:
b) Find the maximum compression of the spring.

This is where I am having the most trouble. I have seen different answers online than what I arrive at.

To find the maximum compression of the spring I need to know fast the carts are moving when the spring is at max compression.

PTO=PTF

m1v1o+m2v2o=(m1+m2)vf

(0.80)(2.0)+(0.60)(0)=(0.80+0.60)vf

You seem to have gone wrong here. You have speeds on ##2.0 m/s## and ##0 m/s##.
 
  • #3
PeroK said:
You seem to have gone wrong here. You have speeds on ##2.0 m/s## and ##0 m/s##.
Okay so this is what I have done.

m1v1o+m2v2o=(m1+m2)vf

(0.8)(2.0)+(0.6)(-5)=(0.8+0.6)Vf
-1.4 =1.4
=-1.0
=1.0 m/s [W]

So is this answer v2f? Is this the number that will get substituted into the conservation of energy for v2f? Would I use -1.0 or 1.0 [W] when substituting?

Thanks!
 
  • #4
Specter said:
Okay so this is what I have done.

m1v1o+m2v2o=(m1+m2)vf

(0.8)(2.0)+(0.6)(-5)=(0.8+0.6)Vf
-1.4 =1.4
=-1.0
=1.0 m/s [W]

So is this answer v2f? Is this the number that will get substituted into the conservation of energy for v2f? Would I use -1.0 or 1.0 [W] when substituting?

Thanks!

Okay, although maybe you could have stayed in the rest frame of one of the masses, as you did for part a).

KE involves ##v^2## so the direction doesn't matter.
 
  • #5
PeroK said:
Okay, although maybe you could have stayed in the rest frame of one of the masses, as you did for part a).

KE involves ##v^2## so the direction doesn't matter.

What do you mean by stayed in the rest frame? Do you mean use the speeds 7.0 and 0.0 instead of 2.0 and -5.0? I'm a bit confused.
 
  • #6
The calculations look easier, as in part a), with only one mass initially moving.
 
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  • #7
PeroK said:
The calculations look easier, as in part a), with only one mass initially moving.
So if I use the speeds of the reference frame where only one mass is moving I would have to add (-5) to my end answer to switch it back to Earth's reference frame just like in part a, right?m1v1o+m2v2o=(m1+m2)vf
(0.8)(7.0)+(0.6)(0.0)=(0.8+0.6)Vf
5.6=1.4
=4

Switch back to Earth's reference frame by adding -5.

4+(-5)=-1 m/s
 
  • #8
Specter said:
So if I use the speeds of the reference frame where only one mass is moving I would have to add (-5) to my end answer to switch it back to Earth's reference frame just like in part a, right?m1v1o+m2v2o=(m1+m2)vf
(0.8)(7.0)+(0.6)(0.0)=(0.8+0.6)Vf
5.6=1.4
=4

Why not keep going in that frame?
 
  • #9
PeroK said:
Why not keep going in that frame?
I didn't know I could do that! So would I substitute 4 into the conservation of energy? When do I switch back to Earth's frame of reference?
 
  • #10
Specter said:
I didn't know I could do that! So would I substitute 4 into the conservation of energy? When do I switch back to Earth's frame of reference?

The compression of the spring is the same in all frames of reference, so why are you so anxious to switch back?
 
  • #11
PeroK said:
The compression of the spring is the same in all frames of reference, so why are you so anxious to switch back?
I thought that I would get a different answer if the velocity was different. Wouldn't the maximum compression be different if the velocity were 1 instead of 4?
 
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  • #12
Specter said:
I thought that I would get a different answer if the velocity was different. Wouldn't the maximum compression be different if the velocity were 1 instead of 4?

You mean the maximum compression is frame dependent?

PS why not do the calculations in both frames - just to see?
 
  • #13
PeroK said:
You mean the maximum compression is frame dependent?

PS why not do the calculations in both frames - just to see?
I guess it wouldn't be frame dependant. I tried this but I guess I'm still doing something wrong.

Here's my work for the frame when only one mass is moving:

1/2 (0.8)(7.0)2+1/2(0.6)(0)2=1/2(0.8+0.6)(4)2+1/2(1200)x2

19.6=11.2+600x2
8.4=600x2
x=√0.014
x=0.12Here's my work for the frame when both masses are moving:

1/2 (0.8)(2.0)2+1/2(0.6)(5.0)2=1/2(0.8+0.6)(1)2+1/2(1200)x2
9.1 = 0.7 +600x2
8.4=600x2
x=√0.014
x=0.12Edit: Forgot to add my answers
 
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  • #14
Specter said:
I guess it wouldn't be frame dependant. I tried this but I guess I'm still doing something wrong.

Here's my work for the frame when only one mass is moving:

1/2 (0.8)(7.0)2+1/2(0.6)(0)2=1/2(0.8+0.6)(4)2+1/2(1200)x2

19.6=611.2x2

That's just an error. I've no idea where you got ##611.2## from. ##\frac12(1200) = 600##, surely?

Specter said:
Here's my work for the frame when both masses are moving:

1/2 (0.8)(2.0)2+1/2(0.6)(5.0)2=1/2(0.8+0.6)(1)2+1/2(1200)x2
9.1 = 0.7 +600x2

I'm still pretty confused.

That looks right. Just solve for ##x##.
 
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  • #15
PeroK said:
That's just an error. I've no idea where you got ##611.2## from. ##\frac12(1200) = 600##, surely?
That looks right. Just solve for ##x##.
From my post above. I edited it to include my answers.

Here's my work for the frame when only one mass is moving:

1/2 (0.8)(7.0)2+1/2(0.6)(0)2=1/2(0.8+0.6)(4)2+1/2(1200)x2

19.6=11.2+600x2
8.4=600x2
x=√0.014
x=0.12Here's my work for the frame when both masses are moving:

1/2 (0.8)(2.0)2+1/2(0.6)(5.0)2=1/2(0.8+0.6)(1)2+1/2(1200)x2
9.1 = 0.7 +600x2
8.4=600x2
x=√0.014
x=0.12

Does this look correct? I assume it is because both answers are the same.

Thanks for your help!
 
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  • #16
Specter said:
From my post above. I edited it to include my answers.

Here's my work for the frame when only one mass is moving:

1/2 (0.8)(7.0)2+1/2(0.6)(0)2=1/2(0.8+0.6)(4)2+1/2(1200)x2

19.6=11.2+600x2
8.4=600x2
x=√0.014
x=0.12Here's my work for the frame when both masses are moving:

1/2 (0.8)(2.0)2+1/2(0.6)(5.0)2=1/2(0.8+0.6)(1)2+1/2(1200)x2
9.1 = 0.7 +600x2
8.4=600x2
x=√0.014
x=0.12

Does this look correct? I assume it is because both answers are the same.

Thanks for your help!
Looks right.

Are you familiar with Newton's Experimental Law? It's mainly for partially elastic cases, but in the special case of fully elastic it can be derived from the momentum and energy conservation laws and has the advantage of avoiding quadratics. It says the relative velocity (regardless of mass) merely swaps sign. So if the first mass bounces back at speed u West and the second bounces back at speed v East then u+v=5+2=7 m/s.
 
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  • #17
haruspex said:
Looks right.

Are you familiar with Newton's Experimental Law? It's mainly for partially elastic cases, but in the special case of fully elastic it can be derived from the momentum and energy conservation laws and has the advantage of avoiding quadratics. It says the relative velocity (regardless of mass) merely swaps sign. So if the first mass bounces back at speed u West and the second bounces back at speed v East then u+v=5+2=7 m/s.
I've never heard of it but I will keep that in mind. Thanks!
 

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