Let the mass of each block $A$, $B$, and $C$ be $m$. The angle of inclination is $\theta$. Let $T$ be the tension in the thread. The thread is continuous, so the tension is uniform throughout initially.

Analyzing Block B:

Block B is supported by a movable pulley. The string passes under this pulley, exerting an upward force of $2T$. For equilibrium: $$2T = mg \implies T = \frac{mg}{2}$$

Analyzing Block A:

Block A is on a frictionless incline. It is held by a spring connected to the string. The force exerted by the spring is equal to the tension $T$ (since the spring is light). For equilibrium along the incline: $$T = mg \sin\theta$$ Substituting $T = \frac{mg}{2}$: $$\frac{mg}{2} = mg \sin\theta \implies \sin\theta = \frac{1}{2} \implies \theta = 30^\circ$$

Analyzing Block C:

Block C is attached to a movable pulley which is suspended by Spring S. The string passes over this pulley, exerting a downward force of $T$ on both sides ($2T$ total downward). Let the force in Spring S be $F_S$. For equilibrium of the pulley-block C system: $$F_S = 2T + mg$$ Substituting $T = \frac{mg}{2}$: $$F_S = 2\left(\frac{mg}{2}\right) + mg = 2mg$$

Concept: A string is an object that cannot sustain tension if one end is free. Immediately after cutting the thread at P, the thread becomes slack.

• The tension in the thread drops to zero instantly: $T’ = 0$.
• The spring forces depend on extension ($F = kx$). Since position does not change in an infinitesimal time interval, the force in Spring S ($F_S$) and the spring attached to A remains unchanged initially. However, for the spring attached to A, it is in series with the string. A massless spring connected to a slack string cannot exert force (net force on massless element must be zero). Thus, the force exerted by the spring on A also vanishes instantly.

Acceleration of Block A ($a_A$):

Forces along the incline: Gravity ($mg \sin\theta$) down, Spring force ($0$) up. $$m a_A = mg \sin\theta$$ $$a_A = g \sin\theta = \frac{g}{2}$$ Direction: Down the incline.

Acceleration of Block B ($a_B$):

Forces: Gravity ($mg$) down, Tension ($2T’ = 0$) up. $$m a_B = mg$$ $$a_B = g$$ Direction: Downwards ($\downarrow$).

Acceleration of Block C ($a_C$):

Forces: Gravity ($mg$) down, Rope force ($2T’ = 0$) down, Spring S force ($F_S = 2mg$) up. $$m a_C = F_{up} – F_{down}$$ $$m a_C = 2mg – mg = mg$$ $$a_C = g$$ Direction: Upwards ($\uparrow$).

Concept: When Spring S is cut, its force vanishes instantly ($F_S’ = 0$). However, the string is not cut; it remains intact.

Key Insight: The tension in the string is determined by the forces acting on its ends. One end is attached to the ground (fixed), and the other is attached to Block A via a spring. The spring attached to A has a specific extension $x_0$ corresponding to the initial equilibrium force $mg/2$. Since the block A has not moved yet, the extension $x_0$ is unchanged. Therefore, this spring continues to pull the string with force $mg/2$. Consequently, the tension in the string remains $T = \frac{mg}{2}$ immediately after the cut.

Acceleration of Block A ($a_A$):

Forces: Tension/Spring force ($T = mg/2$) up the incline, Gravity ($mg \sin\theta = mg/2$) down the incline. $$F_{net} = \frac{mg}{2} – \frac{mg}{2} = 0$$ $$a_A = 0$$

Acceleration of Block B ($a_B$):

Forces: Tension ($2T = mg$) up, Gravity ($mg$) down. $$F_{net} = mg – mg = 0$$ $$a_B = 0$$

Acceleration of Block C ($a_C$):

Forces: Spring S force ($0$) up, Gravity ($mg$) down, Rope tension on pulley ($2T = mg$) down. $$F_{net} = mg \text{ (gravity)} + 2T \text{ (rope)}$$ $$F_{net} = mg + mg = 2mg$$ $$m a_C = 2mg \implies a_C = 2g$$ Direction: Downwards ($\downarrow$).