Question

Three blocks A, B and C are arranged as shown in the figure such that the distance between two successive blocks is 10 m. Block A is displaced towards block B by 2 m and block C is displaced towards block B by 3 m. The distance through which the block B should be moved so that the centre of mass of the system does not change is

• A. 1.4 m, towards block C
• B. 1.5 m, towards block A
• C. 2 m, towards block A
• D. 1 m, towards block C

Hint:

The centre of mass of a system of particles is given by \( X_{CM} = \frac{\sum m_i x_i}{\sum m_i} \). If the centre of mass of the system does not change, \( \Delta X_{CM} = 0 \). This implies \( \sum m_i \Delta x_i = 0 \), where \( \Delta x_i \) is the displacement of the \(i\)-th particle. \( m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C = 0 \). \( \Delta x_A = +2 \) m (towards B is positive x). \( \Delta x_C = -3 \) m (towards B is negative x relative to C's original frame, or if origin is A, C moves from 20 to 17). Let's use actual displacements: Displacement of A, \( \delta x_A = +2 \) m. Displacement of C, \( \delta x_C = -3 \) m. Let displacement of B be \( \delta x_B \). \( 10(+2) + 25(\delta x_B) + 15(-3) = 0 \) \( 20 + 25\delta x_B - 45 = 0 \) \( 25\delta x_B - 25 = 0 \) \( 25\delta x_B = 25 \implies \delta x_B = 1 \) m. Positive \( \delta x_B \) means in the direction from A to C. So, 1m towards block C.

Answer 1

The Correct Option is D

Let the initial positions of the blocks be: \( x_A = 0 \) m, \( m_A = 10 \) kg \( x_B = 10 \) m, \( m_B = 25 \) kg \( x_C = 20 \) m, \( m_C = 15 \) kg Initial centre of mass \( X_{CM,i} = \frac{m_A x_A + m_B x_B + m_C x_C}{m_A+m_B+m_C} \).
Total mass \( M = 10+25+15 = 50 \) kg.
\( X_{CM,i} = \frac{10(0) + 25(10) + 15(20)}{50} = \frac{0 + 250 + 300}{50} = \frac{550}{50} = 11 \) m.
New positions after displacement (before B is moved): Block A is displaced towards B by 2 m: \( x_A' = x_A + 2 = 0+2 = 2 \) m.
Block C is displaced towards B by 3 m: \( x_C' = x_C - 3 = 20-3 = 17 \) m.
Let block B be moved by a distance \( \Delta x_B \).
Its new position is \( x_B' = x_B + \Delta x_B = 10 + \Delta x_B \).
The new centre of mass \( X_{CM,f} = \frac{m_A x_A' + m_B x_B' + m_C x_C'}{M} \).
We want the centre of mass to not change, so \( X_{CM,f} = X_{CM,i} = 11 \) m.
\[ 11 = \frac{10(2) + 25(10+\Delta x_B) + 15(17)}{50} \] \[ 11 \times 50 = 20 + 250 + 25\Delta x_B + 255 \] \[ 550 = 20 + 250 + 255 + 25\Delta x_B \] \[ 550 = 525 + 25\Delta x_B \] \[ 550 - 525 = 25\Delta x_B \] \[ 25 = 25\Delta x_B \] \[ \Delta x_B = 1 \] m.
Since \( \Delta x_B = 1 \) is positive, block B should be moved 1 m in the positive x-direction.
In the setup, positive x-direction is from A towards C.
So, block B should be moved 1 m towards block C.
This matches option (4).