To analyze the situation with the three identical balls suspended on springs, we need to consider the forces acting on each ball and how they respond when the thread or the spring is cut. Let's break this down step by step.
Understanding the System
We have three identical balls, labeled 1, 2, and 3, suspended vertically on springs. Ball 1 is at the top, followed by ball 2 in the middle, and ball 3 at the bottom. The thread OA is holding the entire system in place. When we cut the thread or the spring, the dynamics of the system change significantly.
Part (a): Cutting the Thread OA
When the thread OA is cut, the entire system of balls is released from rest. At this moment, we need to determine the acceleration of each ball. Since they are in free fall, we can apply Newton's second law.
- **Force on each ball:** The only force acting on the balls after the thread is cut is gravity. Each ball experiences a gravitational force equal to its weight, which is given by \( F = mg \), where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- **Acceleration of the balls:** Since there are no other forces acting on the balls (the springs are not exerting any force because they are not stretched), all three balls will accelerate downwards at the same rate as gravity.
Thus, the acceleration of all three balls (1, 2, and 3) immediately after cutting the thread is:
a = g = 9.81 m/s²
Part (b): Cutting the Spring BC
Now, let's consider the scenario where we cut the spring BC, which is supporting ball 2. In this case, we need to analyze the forces acting on each ball again.
- **Ball 1:** Ball 1 is still supported by the spring above it, so it will remain in equilibrium until the spring is cut. However, once the spring BC is cut, ball 2 will fall, and ball 1 will start to experience a downward force due to the sudden release of tension in the spring.
- **Ball 2:** When spring BC is cut, ball 2 will fall freely under the influence of gravity, just like in part (a). Therefore, its acceleration will also be \( g \).
- **Ball 3:** Ball 3 is still supported by the spring below it. However, once ball 2 falls, ball 3 will also start to experience a downward force due to the lack of support from ball 2. Initially, it will not fall immediately but will eventually accelerate downwards as the tension in the spring is released.
At the initial instant after cutting spring BC:
- **Ball 1:** It will experience a downward acceleration due to the sudden loss of support from ball 2, but it will not fall immediately. Its acceleration will be less than \( g \) until the spring tension is fully released.
- **Ball 2:** The acceleration of ball 2 will be \( g \) as it falls freely.
- **Ball 3:** Initially, ball 3 will remain at rest, but it will eventually start to accelerate downwards as the tension in the spring is released. Its initial acceleration is \( 0 \) until the spring tension is fully released.
In summary, immediately after cutting the spring BC:
Ball 1: a < g (decelerating), Ball 2: a = g, Ball 3: a = 0
Final Thoughts
This analysis shows how the dynamics of the system change based on which component is cut. Understanding the forces at play helps us predict the motion of each ball accurately. If you have any further questions or need clarification on any part, feel free to ask!
