Finding equilibrium position after inelastic collision of charged block with spring

[erisedk]

Homework Statement

In the figure mA = mB = 1kg . Block A is neutral while qB = -1C, sizes of A and B are negligible. B is released from rest at a distance 1.8m from A . Initially springs is neither compressed nor elongated.
https://brilliant.org/discussions/thread/amplitude-of-oscillation/
View the figure above.
Equilibrium position of the combined mass is at x = ... m

Note: Collision between A and B is perfectly inelastic.

Homework Equations

The Attempt at a Solution

a= 10m/s (of block B using qE/m)
v = root(2as) = 6m/s
mv = 2mv'
v' = 3m/s (combined velocity of the system)

1/2 mv^2 = 1/2 kx^2 - qEx
9 = 9x^2 - 10x
x = -0.588
The answer is -5/9, which is quite close to my answer but not exactly.
Can someone just tell me if what I've done is right?

 

[rude man]

1/2 mv^2 = 1/2 kx^2 - qEx
9 = 9x^2 - 10x
x = -0.588
The answer is -5/9, which is quite close to my answer but not exactly.
Can someone just tell me if what I've done is right?

Your equation makes no sense. It is not even dimensionally correct. qE is a force, the other terms are energy.
At equilibrium the electrostatic force must be equal and opposite to the spring force. That gives x = -5/9 exactly.

 

[Brian T]

Your equation makes no sense. It is not even dimensionally correct. qE is a force, the other terms are energy.
At equilibrium the electrostatic force must be equal and opposite to the spring force. That gives x = -5/9 exactly.

He put qEx, not qE, which is an attempt to add in a work contribution i. E. Fd.

Although your method seems correct, just didn't want to give him the wrong idea

 

[rude man]

He put qEx, not qE, which is an attempt to add in a work contribution i. E. Fd.

OK, I thought he meant E-sub-x. Still doesn't make sense, the equation I mean.

Although your method seems correct, just didn't want to give him the wrong idea

Well, my method has to be correct. May be some other way is too.

 

[erisedk]

Okay thanks, got it!