Problem 2: Spring Extension at Collision
Concept: The collision occurs when the system loses stable equilibrium. This happens when the attractive electrostatic force gradient overcomes the spring’s restoring stiffness.
Step 1: Equilibrium Condition
Let $x$ be the spring extension and $d$ be the position of the approaching charge $-q$ (measured from the spring’s relaxed position). The separation between charges is $r = d – x$.
Forces balance when the spring force equals the electrostatic attraction:
$$ kx = \frac{1}{4\pi\epsilon_0} \frac{qQ}{(d-x)^2} \quad \dots(1) $$Step 2: Condition for Instability (Collision)
The equilibrium becomes unstable when the rate of change of the electric force exceeds the spring constant (tangency condition):
$$ \frac{d}{dx}(kx) = \frac{d}{dx}\left( \frac{C}{(d-x)^2} \right) $$ $$ k = \frac{2C}{(d-x)^3} \quad \dots(2) $$Where constant $C = \frac{qQ}{4\pi\epsilon_0}$.
Step 3: Solving for Critical Points
Substituting $C = kx(d-x)^2$ from (1) into (2):
$$ k = \frac{2[kx(d-x)^2]}{(d-x)^3} = \frac{2kx}{d-x} $$ $$ 1 = \frac{2x}{d-x} \implies d = 3x $$This reveals that instability occurs when the charge $-q$ is at a distance $d = 3x$ from the origin. The block $Q$ is at $x$.
Now, find the critical extension $x_c$. From Eq (2) with $d-x = 2x$:
$$ k = \frac{2C}{(2x)^3} = \frac{2C}{8x^3} = \frac{C}{4x^3} $$ $$ x^3 = \frac{C}{4k} = \frac{qQ}{16\pi\epsilon_0 k} $$ $$ x_c = \left( \frac{qQ}{16\pi\epsilon_0 k} \right)^{1/3} $$Step 4: Extension at Impact
The problem states $-q$ is moved “very slowly”. Once the block reaches $x_c$, equilibrium breaks, and it snaps towards $-q$. Since $-q$ is effectively held stationary during this fast snap, the block collides with it at position $d_c$.
The total extension of the spring at the moment of collision is the position of $-q$, which is $d_c$.
$$ \text{Extension} = d_c = 3x_c $$