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PhysicsElectrostaticsPotential & potential energyMedium2 minQB

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Passage / Comprehension

Three identical charged particles each of mass m, carrying charge q each, are placed on the vertices of an equilateral triangle ABC of side $ℓ$ and are tied with the help of massless strings AB, BC and CA of length $ℓ$ each. If string BC is burnt and system is released, answer the following.

Find the velocity of particle B , when the three particles become collinear. ( $k = \frac{1}{4 π ε _{0}}$ )

Options:

Answer:

Solution:

As B and C are symmetrically placed, momentum conservation along the line gives $m v_{A} = 2 m v$ , so $v_{A} = 2 v$ when B and C each move with speed v.

Initial potential energy is $3 \frac{k q ^{2}}{ℓ}$ . When the particles become collinear, AB and AC are at distance $ℓ$ and BC at $2 ℓ$ , giving $2 \frac{k q ^{2}}{ℓ} + \frac{k q ^{2}}{2 ℓ} = \frac{5 k q ^{2}}{2 ℓ}$ .

Energy conservation: $\frac{1}{2} m (v_{A}^{2} + 2 v^{2}) = \frac{k q ^{2}}{2 ℓ} \Rightarrow 3 m v^{2} = \frac{k q ^{2}}{2 ℓ}$ .

Hence $v = q \frac{k}{6 m ℓ}$ .

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: QB

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