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: Understanding the Concept:
We have an Atwood machine. The two blocks will accelerate with the same magnitude but in opposite directions.
After finding the acceleration of the individual blocks, we can calculate the net acceleration of the center of mass of the two-block system.
Using kinematics, we can then find the displacement of the center of mass over the given time interval.
Step 2: Key Formula or Approach:
Acceleration of blocks in an Atwood machine: \(a = \frac{m_1 - m_2}{m_1 + m_2} g\).
Acceleration of the center of mass: \(\vec{a}_{cm} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}\).
Displacement under constant acceleration from rest: \(S = \frac{1}{2} a_{cm} t^2\).
Step 3: Detailed Explanation:
Let \(m_1 = 2 \text{ kg}\) and \(m_2 = 1 \text{ kg}\).
Calculate the common acceleration \(a\) of the blocks:
\[ a = \frac{2 - 1}{2 + 1} (10) = \frac{1}{3} \times 10 = \frac{10}{3} \text{ m/s}^2 \] The heavier block \(m_1\) moves downwards (let's define this as the positive direction), so \(\vec{a}_1 = \frac{10}{3}\).
The lighter block \(m_2\) moves upwards, so \(\vec{a}_2 = -\frac{10}{3}\).
Now, find the acceleration of the center of mass:
\[ a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \] \[ a_{cm} = \frac{(2)\left(\frac{10}{3}\right) + (1)\left(-\frac{10}{3}\right)}{2 + 1} \] \[ a_{cm} = \frac{\frac{20}{3} - \frac{10}{3}}{3} = \frac{\frac{10}{3}}{3} = \frac{10}{9} \text{ m/s}^2 \] Now, calculate the distance traversed by the center of mass in \(t = 2 \text{ s}\):
\[ S_{cm} = \frac{1}{2} a_{cm} t^2 = \frac{1}{2} \left(\frac{10}{9}\right) (2)^2 \] \[ S_{cm} = \frac{1}{2} \left(\frac{10}{9}\right) (4) = \frac{20}{9} \text{ m} \] Converting the fraction to a decimal:
\[ \frac{20}{9} \approx 2.22 \text{ m} \] Step 4: Final Answer:
The distance traversed by the center of mass is \(2.22 \text{ m}\).