Consider the system of three identical balls A, B, and C, each of mass $m$, forming an equilateral triangle. The system is suspended by a thread from ball A.

Let $T_{thread}$ be the tension in the thread above A. Let $F_{s}$ be the spring force in the springs connecting A to B and A to C. The angle each spring makes with the vertical is $30^\circ$ (since the triangle is equilateral).

For the entire system (A+B+C):
The total downward force is the weight of three balls, $3mg$. For equilibrium, the upward tension must balance this: $$T_{thread} = 3mg$$

For the lower balls (B and C):
They are in equilibrium under gravity ($mg$) and the spring forces. The vertical component of the spring force from A balances the weight of B (and similarly for C). $$F_s \cos(30^\circ) = mg$$

When the thread is cut, the tension $T_{thread}$ instantaneously becomes zero. However, the spring forces do not change instantaneously because the positions of the balls (and thus the extension of the springs) cannot change instantly.

Acceleration of Ball A:
The forces acting on A are:

• Gravity ($mg$) acting downwards.
• Spring force from B ($F_s$) acting downwards and left.
• Spring force from C ($F_s$) acting downwards and right.

The net force on A ($F_{net, A}$) is the vector sum of these forces. By symmetry, the horizontal components of the spring forces cancel out. The vertical components add to gravity.

$$F_{net, A} = mg + 2 F_s \cos(30^\circ)$$

From our initial equilibrium analysis, we know that $2 F_s \cos(30^\circ)$ represents the upward force that was holding B and C, which equals $2mg$. Alternatively, looking at the free body diagram of A before the cut: $T_{thread} = mg + 2 F_s \cos(30^\circ)$. Since $T_{thread} = 3mg$, the downward spring contribution is $2mg$.

Substituting this back: $$F_{net, A} = mg + 2mg = 3mg \quad (\text{downwards})$$ $$a_A = \frac{F_{net, A}}{m} = \frac{3mg}{m} = 3g \quad (\text{downwards})$$

For balls B and C, the forces acting are gravity ($mg$ downwards) and the spring forces. Since the spring forces have not changed instantaneously, the net force on B and C remains exactly what it was before the thread was cut.

Before the cut, B and C were in equilibrium ($F_{net} = 0$). Immediately after the cut, the forces are identical. $$F_{net, B} = 0 \quad \text{and} \quad F_{net, C} = 0$$ Therefore, their initial accelerations are vanishingly small (zero).

Correct Answer: Acceleration of the upper ball is $3g$ downwards and the accelerations of the lower balls are vanishingly small.