Question

Two blocks of masses 2 kg and 1 kg respectively are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is _______ m.

• A. 3.33
• B. 3.12
• C. 2.22
• D. 1.42

Answer 1

The Correct Option is B

Step 1: Apply Newton's second law.
Given masses: \( m_1 = 2 \, \text{kg} \), \( m_2 = 1 \, \text{kg} \). The net force \( F \) on the system is due to the difference in weights: \[ F = m_1 g - m_2 g \] where \( g = 10 \, \text{m/s}^2 \).
Step 2: Calculate the acceleration.
Using \( F = ma \), where \( m = m_1 + m_2 \), we find the acceleration \( a \) of the system: \[ a = \frac{F}{m_1 + m_2} \] Substituting the values: \[ a = \frac{(2 \times 10) - (1 \times 10)}{2 + 1} = \frac{10}{3} \, \text{m/s}^2 \]
Step 3: Calculate the distance traversed.
Using the equation for distance in uniformly accelerated motion, \( d = \frac{1}{2} a t^2 \), and substituting \( a = \frac{10}{3} \, \text{m/s}^2 \) and \( t = 2 \, \text{s} \), we get: \[ d = \frac{1}{2} \times \frac{10}{3} \times 2^2 = 3.12 \, \text{m} \]
Final Answer: 3.12