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Find extension in the spring when the charged blocks collide

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A small block having charge +Q placed on a frictionless horizontal table is attached to one end of a spring of force constant k. The other end of the spring is attached to a fixed support and the spring stays horizontally in relaxed state.
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Another small block carrying a negative charge -q is brought very slowly from a great distance towards the former block along the line coinciding the axis of the spring.
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Find extension in the spring when the blocks collide.

But the answer is given as :
$l=\frac{3}{2} (\frac{Qq}{2\pi \epsilon k})^{\frac{1}{3}}$

asked May 28 in Physics Problems by koolman (4,116 points)
edited May 28 by sammy gerbil

1 Answer

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Observations

Start by trying to understand what is happening here.

Block #2 is moved to the left very slowly from infinity, held in place by some external force. At each step block #1 is in quasi-static equilibrium : the electrostatic (ES) attractive force to the right is matched by the elastic force to the left.

The natural response of block #1 to a sudden change in the position of block #2 is to oscillate : the ES force is suddenly bigger while the elastic force remains the same. Block #2 oscillates until it loses its KE somehow and finds a new equilibrium position. Regarding the spring and 2 blocks as the system, the external force does -ve work on this system, taking away the kinetic energy as it develops.

So energy is not conserved, and since it is not obvious how the unknown external force varies, the work-energy theorem is not useful here. What we can say is that the KE removed equals the total PE stored at equilibrium (which will be -ve), but that is not useful either.

Your observation that $r=0$ when the blocks collide is correct and warns us that block #1 cannot be in equilibrium at all times. At some point the equilibrium becomes unstable. We have to find out when that happens.

Suppose at this point the separation between the blocks is $r_0$ and the extension of the spring is $x_0$. At this point of unstable equilibrium a small displacement of block #1 causes it to accelerate the rest of the distance $r_0$ towards block #2 while block #2 is held stationary by the external force. Block #1 collides with block #2 and the KE it gained is again dissipated somehow, but we don't need to know what this KE is. The final extension of the spring is $x_0+r_0$.

Solution

The resultant force on block #1 is $$F=\frac{KQq}{r^2}-kx$$ where $K$ is the Coulomb constant. This is zero right up to the unstable position so $$x_0=\frac{KQq}{kr_0^2}$$

If block #2 is held fixed, an increase of $dx$ in $x$ is equal to an increase of $-dr$ in separation $r$. That is, $dr=-dx$. Therefore equilibrium becomes unstable when $$\frac{dF}{dx}|_{r_0}=\frac{2KQq}{r_0^3}-k=0$$ $$r_0=(\frac{2KQq}{k})^{\frac13}$$ $$x_0=\frac12 (\frac{2KQq}{k})\frac{1}{r_0^2}=\frac12 r_0^3 \frac{1}{r_0^2}=\frac12 r_0$$

The final extension of the spring is therefore $$x_0+r_0=\frac32 r_0=\frac32 (\frac{2KQq}{k})^{\frac13}=\frac32 (\frac{Qq}{2\pi \epsilon_0 k})^{\frac13}$$

answered May 28 by sammy gerbil (26,096 points)
selected May 28 by koolman
If you are bringing a small charge -q slowly then why would there be a sudden increment in force ?
I used the idea of sudden small steps to illustrate the need to remove energy from the system. Even if block #2 is moved smoothly but relatively fast block #1 will oscillate. With slow small smooth steps there are still oscillations, but in practice these die out without being detected because of friction or hysteresis in the spring.
If block #1 has reached a position of unstable equilibrium then a small force towards block #2 is all that is needed to make block #1 move.
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