Question

For the following question, enter the correct numerical value up to TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numerical value 5 will be written as 5.00 and 2.346 will be written as 2.35) A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light string passing over a pulley. At the end of 5 s the string breaks. How much high from now the 2 kg mass will go? _____ m

Hint:

After the string breaks, use kinematics with initial upward velocity and acceleration due to gravity to find the maximum height.

Answer 1

Correct Answer: 4.9

Step 1: The system is an Atwood machine with masses \[ m_1 = 3\,\text{kg}, \quad m_2 = 2\,\text{kg}. \] Acceleration of the system: \[ a = \frac{m_1 - m_2}{m_1 + m_2}g = \frac{1}{5}g = 1.96\,\text{m s}^{-2}. \]
Step 2: Velocity of the 2 kg mass after 5 s: \[ v = at = 1.96 \times 5 = 9.8\,\text{m s}^{-1}. \]
Step 3: After the string breaks, the 2 kg mass moves upward with initial velocity \(9.8\,\text{m s}^{-1}\) against gravity. Maximum additional height reached: \[ h = \frac{v^2}{2g} = \frac{(9.8)^2}{2 \times 9.8} = 4.9\,\text{m}. \]
Step 4: Hence, from the moment the string breaks, the 2 kg mass rises by \[ \boxed{4.90\,\text{m}}. \]