Two balls, A and B, are suspended from the ceiling by a spring and a string, respectively. Both balls have the same mass. The string holding Ball B is cut. What happens to Ball A and Ball B, and which

Ball B will hit the ground first.

Here's why:

1. Ball B (String): When the string holding Ball B is cut, the tension in the string immediately drops to zero. The only force acting on Ball B is gravity, which pulls it downwards. Therefore, Ball B begins to accelerate downwards instantly with an acceleration equal to the acceleration due to gravity ($g$).

2. Ball A (Spring): Before the cut, the spring is stretched by the weight of Ball A, exerting an upward force equal to $mg$ (where $m$ is the mass of the ball and $g$ is the acceleration due to gravity) to keep it in equilibrium. When the suspension point of the spring is cut, the spring is still attached to Ball A and is still stretched by the same amount at that instant. This means the spring continues to exert an upward force of $mg$ on Ball A. The forces acting on Ball A immediately after the cut are gravity ($mg$ downwards) and the spring force ($mg$ upwards). The net force on Ball A is $mg - mg = 0$. Therefore, Ball A's initial acceleration is zero.

Since Ball B starts accelerating downwards immediately at $g$, while Ball A has an initial acceleration of 0, Ball B will begin to fall first and reach the ground before Ball A.

$$\text{Ball B}$$