To solve this problem, we need to analyze the motion of the two masses connected by a string over a pulley. We have two unequal masses: 1 kg and 2 kg. When the system is released, the heavier mass (2 kg) will accelerate downward due to gravity, while the lighter mass (1 kg) will move upward. Let's break down the steps to find the time elapsed before the string becomes tight again after the larger mass is stopped for 1 second.
Understanding the Initial Motion
Initially, when the system is released, both masses will start to accelerate. The net force acting on the system can be calculated using Newton's second law. The acceleration of the system can be determined by the difference in weights of the two masses:
- Weight of 2 kg mass: W1 = 2 kg × 9.81 m/s² = 19.62 N
- Weight of 1 kg mass: W2 = 1 kg × 9.81 m/s² = 9.81 N
- Net force (F_net) = W1 - W2 = 19.62 N - 9.81 N = 9.81 N
- Total mass (m_total) = 1 kg + 2 kg = 3 kg
Using Newton's second law (F = ma), we can find the acceleration (a) of the system:
a = F_net / m_total = 9.81 N / 3 kg = 3.27 m/s²
Calculating the Distance Traveled
Now, let's determine how far the 1 kg mass moves upward in the first second before the larger mass is stopped. We can use the equation of motion:
s = ut + (1/2)at²
Since the system starts from rest (u = 0), the distance (s) traveled by the 1 kg mass in the first second is:
s = 0 + (1/2) × 3.27 m/s² × (1 s)² = 1.635 m
Analyzing the Situation After 1 Second
After 1 second, the 2 kg mass is stopped, and the 1 kg mass has moved up by 1.635 m. At this point, the 2 kg mass is still at its original position, and the string is slack. The 1 kg mass will continue to move upward for a brief moment due to its inertia.
Finding the Time Until the String is Tight Again
Once the 2 kg mass is stopped, the 1 kg mass will continue to rise until it comes to a stop. To find the time it takes for the 1 kg mass to stop, we can use the same acceleration (but in the opposite direction) to find the time taken to come to rest:
v² = u² + 2as
Here, the final velocity (v) will be 0 when the 1 kg mass stops, and we can rearrange the equation to find the distance it travels upward:
0 = (3.27 m/s)² + 2(-3.27 m/s²)(s)
Solving for s gives:
s = (3.27 m/s)² / (2 × 3.27 m/s²) = 0.5 m
Now, the total distance the 1 kg mass has moved upward is:
1.635 m + 0.5 m = 2.135 m
Final Calculation of Time Elapsed
To find the time taken for the 1 kg mass to move this distance, we can use the equation of motion again:
s = ut + (1/2)at²
Here, we know s = 2.135 m, u = 0, and a = -3.27 m/s² (deceleration). Rearranging gives:
2.135 m = 0 + (1/2)(-3.27 m/s²)t²
Solving for t² gives:
t² = (2 × 2.135 m) / 3.27 m/s² = 1.304 s²
Thus, t = √1.304 s² ≈ 1.14 s.
Finally, the total time elapsed before the string becomes tight again is:
Total time = 1 s (initial) + 1.14 s ≈ 2.14 s.
In summary, the time elapsed before the string is tight again after stopping the larger mass for 1 second is approximately 2.14 seconds. This problem illustrates the principles of motion, forces, and the effects of inertia in a connected mass system.
