To solve this problem, we need to analyze the motion of the two masses connected by the string over the pulley. Let's break it down step by step.
Understanding the System
We have two masses: one is 1 kg (let's call it Mass A) and the other is 2 kg (Mass B). When the system is released, the heavier mass (2 kg) will accelerate downward due to gravity, while the lighter mass (1 kg) will accelerate upward. The acceleration of the system can be calculated using Newton's second law.
Calculating the Acceleration
The net force acting on the system can be expressed as:
- Weight of Mass B (2 kg): F_B = m_B \cdot g = 2 \cdot 9.8 = 19.6 N
- Weight of Mass A (1 kg): F_A = m_A \cdot g = 1 \cdot 9.8 = 9.8 N
The net force acting on the system is:
F_net = F_B - F_A = 19.6 N - 9.8 N = 9.8 N
Using Newton's second law (F = ma), we can find the acceleration (a) of the system:
a = F_net / (m_A + m_B) = 9.8 N / (1 kg + 2 kg) = 9.8 N / 3 kg = 3.27 m/s²
Time Until the String is Tight Again
When the system is released, both masses start moving. After 1 second, Mass B (2 kg) is stopped momentarily. We need to find out how far Mass A (1 kg) has moved in that time.
Distance Traveled by Mass A
The distance traveled by Mass A in the first second can be calculated using the equation of motion:
s = ut + (1/2)at²
Here, u (initial velocity) is 0, a is 3.27 m/s², and t is 1 second:
s = 0 + (1/2) * 3.27 * (1)² = 1.635 m
Analyzing the Situation After 1 Second
At this point, Mass A has moved up 1.635 m, while Mass B has moved down the same distance. When Mass B is stopped, Mass A will continue to rise until the string becomes taut again. We need to find out how long it takes for Mass A to reach the point where the string is tight.
Time for Mass A to Move Back Down
After 1 second, Mass A is at a height of 1.635 m. To find the time it takes for Mass A to move back down to the original position (where the string becomes tight), we can use the same equation of motion:
Now, we need to find the time it takes for Mass A to fall back down 1.635 m under the acceleration of 3.27 m/s²:
Using the equation:
s = ut + (1/2)at²
Here, s is 1.635 m, u is 0 (since it starts from rest), and a is -3.27 m/s² (negative because it’s moving down):
1.635 = 0 + (1/2)(-3.27)t²
Solving for t² gives:
t² = (2 * 1.635) / 3.27 = 1.000 s²
Thus, t = 1 s.
Final Calculation
Since Mass A takes 1 second to fall back down after Mass B is stopped, the total time elapsed before the string is tight again is:
Total time = 1 second (initial) + 1 second (falling back) = 2 seconds
In summary, the time elapsed before the string becomes tight again is 2 seconds. This analysis shows how the forces and motions interact in a system of connected masses.
