Impulse and momentum in one dimension - Two carts are moving

[Specter]

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.

v_1f=m₁-m₂/m₁+m₂ v_1o

= 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:

v_2f=2m₁/m₁+m₂v_1o

=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.

P_TO=P_TF

m₁v_1o+m₂v_2o=(m₁+m₂)v_f

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

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 m₂v_1o²+1/2 m₂v_2o²=1/2(m₁+m₂)v_2f²+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.

 

[PeroK]

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.

P_TO=P_TF

m₁v_1o+m₂v_2o=(m₁+m₂)v_f

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

You seem to have gone wrong here. You have speeds on $2.0 m/s$ and $0 m/s$.

 

[Specter]

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.

m₁v_1o+m₂v_2o=(m₁+m₂)v_f

(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 v_2f? Is this the number that will get substituted into the conservation of energy for v_2f? Would I use -1.0 or 1.0 [W] when substituting?

Thanks!

 

[PeroK]

Okay so this is what I have done.

m₁v_1o+m₂v_2o=(m₁+m₂)v_f

(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 v_2f? Is this the number that will get substituted into the conservation of energy for v_2f? 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.

 

[Specter]

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.

 

[PeroK]

The calculations look easier, as in part a), with only one mass initially moving.

 

[Specter]

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?m₁v_1o+m₂v_2o=(m₁+m₂)v_f
(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

 

[PeroK]

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?m₁v_1o+m₂v_2o=(m₁+m₂)v_f
(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?

 

[Specter]

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?

 

[PeroK]

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?

 

[Specter]

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?

 

[PeroK]

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?

 

[Specter]

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 dependent. 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

 

[PeroK]

I guess it wouldn't be frame dependent. 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?

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 $x$.

 

[Specter]

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!

 

[haruspex]

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.

 

[Specter]

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!